3.316 \(\int \frac{1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=863 \[ \frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \Pi \left (-\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)} b^3}{a (a-b)^{3/2} (a+b)^{3/2} d e^2 \sqrt{\sin (c+d x)}}-\frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \Pi \left (\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)} b^3}{a (a-b)^{3/2} (a+b)^{3/2} d e^2 \sqrt{\sin (c+d x)}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right ) b^2}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right ) b^2}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}-\sqrt{2} \sqrt{e \tan (c+d x)}\right ) b^2}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}+\sqrt{2} \sqrt{e \tan (c+d x)}\right ) b^2}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{2 \cos (c+d x) (e \tan (c+d x))^{3/2} b}{\left (a^2-b^2\right ) d e^3}+\frac{2 \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)} b}{\left (a^2-b^2\right ) d e^2 \sqrt{\sin (2 c+2 d x)}}+\frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}+\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}} \]

[Out]

(a*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*(a^2 - b^2)*d*e^(3/2)) - (b^2*ArcTan[1 - (Sqrt
[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*(a^2 - b^2)*d*e^(3/2)) - (a*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c +
d*x]])/Sqrt[e]])/(Sqrt[2]*(a^2 - b^2)*d*e^(3/2)) + (b^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(S
qrt[2]*a*(a^2 - b^2)*d*e^(3/2)) - (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sq
rt[2]*(a^2 - b^2)*d*e^(3/2)) + (b^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqr
t[2]*a*(a^2 - b^2)*d*e^(3/2)) + (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt
[2]*(a^2 - b^2)*d*e^(3/2)) - (b^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[
2]*a*(a^2 - b^2)*d*e^(3/2)) - (2*(a - b*Sec[c + d*x]))/((a^2 - b^2)*d*e*Sqrt[e*Tan[c + d*x]]) + (2*Sqrt[2]*b^3
*Sqrt[Cos[c + d*x]]*EllipticPi[-(Sqrt[a - b]/Sqrt[a + b]), ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x]]],
-1]*Sqrt[e*Tan[c + d*x]])/(a*(a - b)^(3/2)*(a + b)^(3/2)*d*e^2*Sqrt[Sin[c + d*x]]) - (2*Sqrt[2]*b^3*Sqrt[Cos[c
 + d*x]]*EllipticPi[Sqrt[a - b]/Sqrt[a + b], ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x]]], -1]*Sqrt[e*Tan
[c + d*x]])/(a*(a - b)^(3/2)*(a + b)^(3/2)*d*e^2*Sqrt[Sin[c + d*x]]) + (2*b*Cos[c + d*x]*EllipticE[c - Pi/4 +
d*x, 2]*Sqrt[e*Tan[c + d*x]])/((a^2 - b^2)*d*e^2*Sqrt[Sin[2*c + 2*d*x]]) - (2*b*Cos[c + d*x]*(e*Tan[c + d*x])^
(3/2))/((a^2 - b^2)*d*e^3)

________________________________________________________________________________________

Rubi [A]  time = 1.24995, antiderivative size = 863, normalized size of antiderivative = 1., number of steps used = 39, number of rules used = 23, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.92, Rules used = {3893, 3882, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639, 3890, 2733, 2730, 2906, 2905, 490, 1213, 537} \[ \frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \Pi \left (-\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)} b^3}{a (a-b)^{3/2} (a+b)^{3/2} d e^2 \sqrt{\sin (c+d x)}}-\frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \Pi \left (\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)} b^3}{a (a-b)^{3/2} (a+b)^{3/2} d e^2 \sqrt{\sin (c+d x)}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right ) b^2}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right ) b^2}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}-\sqrt{2} \sqrt{e \tan (c+d x)}\right ) b^2}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}+\sqrt{2} \sqrt{e \tan (c+d x)}\right ) b^2}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{2 \cos (c+d x) (e \tan (c+d x))^{3/2} b}{\left (a^2-b^2\right ) d e^3}+\frac{2 \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)} b}{\left (a^2-b^2\right ) d e^2 \sqrt{\sin (2 c+2 d x)}}+\frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}+\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{e}+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2)),x]

[Out]

(a*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*(a^2 - b^2)*d*e^(3/2)) - (b^2*ArcTan[1 - (Sqrt
[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*(a^2 - b^2)*d*e^(3/2)) - (a*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c +
d*x]])/Sqrt[e]])/(Sqrt[2]*(a^2 - b^2)*d*e^(3/2)) + (b^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(S
qrt[2]*a*(a^2 - b^2)*d*e^(3/2)) - (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sq
rt[2]*(a^2 - b^2)*d*e^(3/2)) + (b^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqr
t[2]*a*(a^2 - b^2)*d*e^(3/2)) + (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt
[2]*(a^2 - b^2)*d*e^(3/2)) - (b^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[
2]*a*(a^2 - b^2)*d*e^(3/2)) - (2*(a - b*Sec[c + d*x]))/((a^2 - b^2)*d*e*Sqrt[e*Tan[c + d*x]]) + (2*Sqrt[2]*b^3
*Sqrt[Cos[c + d*x]]*EllipticPi[-(Sqrt[a - b]/Sqrt[a + b]), ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x]]],
-1]*Sqrt[e*Tan[c + d*x]])/(a*(a - b)^(3/2)*(a + b)^(3/2)*d*e^2*Sqrt[Sin[c + d*x]]) - (2*Sqrt[2]*b^3*Sqrt[Cos[c
 + d*x]]*EllipticPi[Sqrt[a - b]/Sqrt[a + b], ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x]]], -1]*Sqrt[e*Tan
[c + d*x]])/(a*(a - b)^(3/2)*(a + b)^(3/2)*d*e^2*Sqrt[Sin[c + d*x]]) + (2*b*Cos[c + d*x]*EllipticE[c - Pi/4 +
d*x, 2]*Sqrt[e*Tan[c + d*x]])/((a^2 - b^2)*d*e^2*Sqrt[Sin[2*c + 2*d*x]]) - (2*b*Cos[c + d*x]*(e*Tan[c + d*x])^
(3/2))/((a^2 - b^2)*d*e^3)

Rule 3893

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)/(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/(a^2 - b^
2), Int[(e*Cot[c + d*x])^m*(a - b*Csc[c + d*x]), x], x] + Dist[b^2/(e^2*(a^2 - b^2)), Int[(e*Cot[c + d*x])^(m
+ 2)/(a + b*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m + 1/2, 0]

Rule 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
 b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
 x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
 f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
 x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3890

Int[Sqrt[cot[(c_.) + (d_.)*(x_)]*(e_.)]/(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/a, Int[Sqr
t[e*Cot[c + d*x]], x], x] - Dist[b/a, Int[Sqrt[e*Cot[c + d*x]]/(b + a*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c,
 d, e}, x] && NeQ[a^2 - b^2, 0]

Rule 2733

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[g^(2*
IntPart[p])*(g*Cot[e + f*x])^FracPart[p]*(g*Tan[e + f*x])^FracPart[p], Int[(a + b*Sin[e + f*x])^m/(g*Tan[e + f
*x])^p, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rule 2730

Int[1/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(g_)*tan[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[
e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[g*Tan[e + f*x]]), Int[Sqrt[Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]*(a + b*Sin[e +
 f*x])), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2906

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x
_)])), x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]], Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]
*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2905

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> Dist[(-4*Sqrt[2]*g)/f, Subst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sq
rt[g*Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
 s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
 Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
 c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rubi steps

\begin{align*} \int \frac{1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx &=\frac{\int \frac{a-b \sec (c+d x)}{(e \tan (c+d x))^{3/2}} \, dx}{a^2-b^2}+\frac{b^2 \int \frac{\sqrt{e \tan (c+d x)}}{a+b \sec (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}\\ &=-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}+\frac{2 \int \left (-\frac{a}{2}-\frac{1}{2} b \sec (c+d x)\right ) \sqrt{e \tan (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}+\frac{b^2 \int \sqrt{e \tan (c+d x)} \, dx}{a \left (a^2-b^2\right ) e^2}-\frac{b^3 \int \frac{\sqrt{e \tan (c+d x)}}{b+a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right ) e^2}\\ &=-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}-\frac{a \int \sqrt{e \tan (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}-\frac{b \int \sec (c+d x) \sqrt{e \tan (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\sqrt{x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a \left (a^2-b^2\right ) d e}-\frac{\left (b^3 \sqrt{e \cot (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \frac{1}{(b+a \cos (c+d x)) \sqrt{e \cot (c+d x)}} \, dx}{a \left (a^2-b^2\right ) e^2}\\ &=-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}-\frac{2 b \cos (c+d x) (e \tan (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e^3}+\frac{(2 b) \int \cos (c+d x) \sqrt{e \tan (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}-\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{\left (a^2-b^2\right ) d e}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a \left (a^2-b^2\right ) d e}-\frac{\left (b^3 \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \frac{\sqrt{\sin (c+d x)}}{\sqrt{-\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a \left (a^2-b^2\right ) e^2 \sqrt{\sin (c+d x)}}\\ &=-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}-\frac{2 b \cos (c+d x) (e \tan (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e^3}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{\left (a^2-b^2\right ) d e}-\frac{b^2 \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a \left (a^2-b^2\right ) d e}+\frac{b^2 \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a \left (a^2-b^2\right ) d e}+\frac{\left (2 b \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2 \sqrt{\sin (c+d x)}}-\frac{\left (b^3 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a \left (a^2-b^2\right ) e^2 \sqrt{\sin (c+d x)}}\\ &=-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}-\frac{2 b \cos (c+d x) (e \tan (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e^3}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}+2 x}{-e-\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}-2 x}{-e+\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{a \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{\left (a^2-b^2\right ) d e}-\frac{a \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{\left (a^2-b^2\right ) d e}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 a \left (a^2-b^2\right ) d e}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 a \left (a^2-b^2\right ) d e}-\frac{\left (4 \sqrt{2} b^3 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^4} \left (a+b+(-a+b) x^4\right )} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{a \left (a^2-b^2\right ) d e^2 \sqrt{\sin (c+d x)}}+\frac{\left (2 b \cos (c+d x) \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\sin (2 c+2 d x)} \, dx}{\left (a^2-b^2\right ) e^2 \sqrt{\sin (2 c+2 d x)}}\\ &=\frac{b^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{b^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}+\frac{2 b \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{\left (a^2-b^2\right ) d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{2 b \cos (c+d x) (e \tan (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e^3}-\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}+2 x}{-e-\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}-2 x}{-e+\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \left (a^2-b^2\right ) d e}-\frac{a \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \left (a^2-b^2\right ) d e}-\frac{\left (2 \sqrt{2} b^3 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a+b}-\sqrt{a-b} x^2\right ) \sqrt{1-x^4}} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{a \sqrt{a-b} \left (a^2-b^2\right ) d e^2 \sqrt{\sin (c+d x)}}+\frac{\left (2 \sqrt{2} b^3 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a+b}+\sqrt{a-b} x^2\right ) \sqrt{1-x^4}} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{a \sqrt{a-b} \left (a^2-b^2\right ) d e^2 \sqrt{\sin (c+d x)}}\\ &=-\frac{b^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{b^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}+\frac{b^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{b^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}+\frac{2 b \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{\left (a^2-b^2\right ) d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{2 b \cos (c+d x) (e \tan (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e^3}-\frac{a \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}+\frac{a \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{\left (2 \sqrt{2} b^3 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (\sqrt{a+b}-\sqrt{a-b} x^2\right )} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{a \sqrt{a-b} \left (a^2-b^2\right ) d e^2 \sqrt{\sin (c+d x)}}+\frac{\left (2 \sqrt{2} b^3 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (\sqrt{a+b}+\sqrt{a-b} x^2\right )} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{a \sqrt{a-b} \left (a^2-b^2\right ) d e^2 \sqrt{\sin (c+d x)}}\\ &=\frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{b^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}+\frac{b^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}+\frac{b^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}+\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2-b^2\right ) d e^{3/2}}-\frac{b^2 \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a \left (a^2-b^2\right ) d e^{3/2}}-\frac{2 (a-b \sec (c+d x))}{\left (a^2-b^2\right ) d e \sqrt{e \tan (c+d x)}}+\frac{2 \sqrt{2} b^3 \sqrt{\cos (c+d x)} \Pi \left (-\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)}}{a (a-b)^{3/2} (a+b)^{3/2} d e^2 \sqrt{\sin (c+d x)}}-\frac{2 \sqrt{2} b^3 \sqrt{\cos (c+d x)} \Pi \left (\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)}}{a (a-b)^{3/2} (a+b)^{3/2} d e^2 \sqrt{\sin (c+d x)}}+\frac{2 b \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{\left (a^2-b^2\right ) d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{2 b \cos (c+d x) (e \tan (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e^3}\\ \end{align*}

Mathematica [C]  time = 27.1128, size = 1571, normalized size = 1.82 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2)),x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*((-2*(b - a*Cos[c + d*x])*Csc[c + d*x])/(-a^2 + b^2) + (2*b*Sin[c + d*x])/(
-a^2 + b^2))*Tan[c + d*x]^2)/(d*(a + b*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2)) + ((b + a*Cos[c + d*x])*Sec[c + d
*x]*Tan[c + d*x]^(3/2)*(-((-a^2 + 3*b^2)*Sec[c + d*x]*(6*Sqrt[2]*(a^2 - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d
*x]]] - 6*Sqrt[2]*a^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + 6*Sqrt[2]*b^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]]] - (6 + 6*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)]
 + (6 + 6*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] - 3*
Sqrt[2]*a^2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 3*Sqrt[2]*b^2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*
x]] + Tan[c + d*x]] + 3*Sqrt[2]*a^2*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - 3*Sqrt[2]*b^2*Log[1 +
 Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + (3 + 3*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*Log[Sqrt[a^2 - b^2] - (1 + I
)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]] - (3 + 3*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*Log[S
qrt[a^2 - b^2] + (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]] + 8*a*b*AppellF1[3/4
, 1/2, 1, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(3/2))*(a + b*Sqrt[1 + Tan[c +
d*x]^2]))/(12*(a^3 - a*b^2)*(b + a*Cos[c + d*x])*(1 + Tan[c + d*x]^2)) + (b*Cos[2*(c + d*x)]*Sec[c + d*x]^2*(-
84*Sqrt[2]*b*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] + 84*Sqrt[2]*b*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + ((
42 + 42*I)*(-a^2 + 2*b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(Sqrt[b]*(a^2 -
b^2)^(1/4)) + ((42 + 42*I)*(a^2 - 2*b^2)*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(
Sqrt[b]*(a^2 - b^2)^(1/4)) + 42*Sqrt[2]*b*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - 42*Sqrt[2]*b*Lo
g[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + ((21 + 21*I)*(a^2 - 2*b^2)*Log[Sqrt[a^2 - b^2] - (1 + I)*Sq
rt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]])/(Sqrt[b]*(a^2 - b^2)^(1/4)) + ((21 + 21*I)*(-a
^2 + 2*b^2)*Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]])/(S
qrt[b]*(a^2 - b^2)^(1/4)) + (112*a^3*AppellF1[3/4, 1/2, 1, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b
^2)]*Tan[c + d*x]^(3/2))/(a^2 - b^2) - (168*a*b^2*AppellF1[3/4, 1/2, 1, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x
]^2)/(a^2 - b^2)]*Tan[c + d*x]^(3/2))/(a^2 - b^2) - (24*a*b^2*AppellF1[7/4, 1/2, 1, 11/4, -Tan[c + d*x]^2, (b^
2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(7/2))/(a^2 - b^2) - (168*a*Tan[c + d*x]^(3/2))/Sqrt[1 + Tan[c + d
*x]^2])*(a + b*Sqrt[1 + Tan[c + d*x]^2]))/(84*a*(b + a*Cos[c + d*x])*(-1 + Tan[c + d*x]^2)*Sqrt[1 + Tan[c + d*
x]^2])))/((a - b)*(a + b)*d*(a + b*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2))

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Maple [B]  time = 0.251, size = 6426, normalized size = 7.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(d*x + c) + a)*(e*tan(d*x + c))^(3/2)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \tan{\left (c + d x \right )}\right )^{\frac{3}{2}} \left (a + b \sec{\left (c + d x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))**(3/2),x)

[Out]

Integral(1/((e*tan(c + d*x))**(3/2)*(a + b*sec(c + d*x))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c) + a)*(e*tan(d*x + c))^(3/2)), x)