3.368 \(\int \frac{1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx\)
Optimal. Leaf size=306 \[ -\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 d^{3/2} f}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 d^{3/2} f}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{2 \sqrt{2} a^2 d^{3/2} f}+\frac{\log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{\log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt{d \tan (e+f x)}} \]
[Out]
(-5*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*d^(3/2)*f) + ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d
]]/(2*Sqrt[2]*a^2*d^(3/2)*f) - ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(2*Sqrt[2]*a^2*d^(3/2)*f) +
Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(3/2)*f) - Log[Sqrt[d] + S
qrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(3/2)*f) - 5/(2*a^2*d*f*Sqrt[d*Tan[e + f*
x]]) + 1/(2*d*f*Sqrt[d*Tan[e + f*x]]*(a^2 + a^2*Tan[e + f*x]))
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Rubi [A] time = 0.73018, antiderivative size = 306, normalized size of antiderivative = 1.,
number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used
= {3569, 3649, 3653, 12, 3476, 329, 211, 1165, 628, 1162, 617, 204, 3634, 63, 205}
\[ -\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 d^{3/2} f}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 d^{3/2} f}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{2 \sqrt{2} a^2 d^{3/2} f}+\frac{\log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{\log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2),x]
[Out]
(-5*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*a^2*d^(3/2)*f) + ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d
]]/(2*Sqrt[2]*a^2*d^(3/2)*f) - ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(2*Sqrt[2]*a^2*d^(3/2)*f) +
Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(3/2)*f) - Log[Sqrt[d] + S
qrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]]/(4*Sqrt[2]*a^2*d^(3/2)*f) - 5/(2*a^2*d*f*Sqrt[d*Tan[e + f*
x]]) + 1/(2*d*f*Sqrt[d*Tan[e + f*x]]*(a^2 + a^2*Tan[e + f*x]))
Rule 3569
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))
Rule 3649
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Rule 3653
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !GtQ[n, 0] && !LeQ[n, -
1]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 211
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))
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 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 3634
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx &=\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}+\frac{\int \frac{\frac{5 a^2 d}{2}-a^2 d \tan (e+f x)+\frac{3}{2} a^2 d \tan ^2(e+f x)}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx}{2 a^3 d}\\ &=-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\int \frac{\frac{7 a^3 d^3}{4}+\frac{1}{2} a^3 d^3 \tan (e+f x)+\frac{5}{4} a^3 d^3 \tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{a^4 d^4}\\ &=-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\int \frac{a^4 d^3}{\sqrt{d \tan (e+f x)}} \, dx}{2 a^6 d^4}-\frac{5 \int \frac{1+\tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a d}\\ &=-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\int \frac{1}{\sqrt{d \tan (e+f x)}} \, dx}{2 a^2 d}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{4 a d f}\\ &=-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (d^2+x^2\right )} \, dx,x,d \tan (e+f x)\right )}{2 a^2 f}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{a+\frac{a x^2}{d}} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 a d^2 f}\\ &=-\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 d^{3/2} f}-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 f}\\ &=-\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 d^{3/2} f}-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 a^2 d f}-\frac{\operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 a^2 d f}\\ &=-\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 d^{3/2} f}-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}+2 x}{-d-\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 d^{3/2} f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}-2 x}{-d+\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{\operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^2 d f}-\frac{\operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^2 d f}\\ &=-\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 d^{3/2} f}+\frac{\log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{\log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 d^{3/2} f}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 d^{3/2} f}\\ &=-\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 a^2 d^{3/2} f}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 d^{3/2} f}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{2 \sqrt{2} a^2 d^{3/2} f}+\frac{\log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{\log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{4 \sqrt{2} a^2 d^{3/2} f}-\frac{5}{2 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.23758, size = 203, normalized size = 0.66 \[ \frac{\tan ^{\frac{3}{2}}(e+f x) \left (\sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )-\sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-10 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )+\frac{\log \left (-\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}-1\right )-\log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )}{\sqrt{2}}-\frac{2 (5 \sin (e+f x)+4 \cos (e+f x))}{\sqrt{\tan (e+f x)} (\sin (e+f x)+\cos (e+f x))}\right )}{4 a^2 f (d \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2),x]
[Out]
((Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] - 10*ArcTan[
Sqrt[Tan[e + f*x]]] + (Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]
] + Tan[e + f*x]])/Sqrt[2] - (2*(4*Cos[e + f*x] + 5*Sin[e + f*x]))/((Cos[e + f*x] + Sin[e + f*x])*Sqrt[Tan[e +
f*x]]))*Tan[e + f*x]^(3/2))/(4*a^2*f*(d*Tan[e + f*x])^(3/2))
________________________________________________________________________________________
Maple [A] time = 0.037, size = 255, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{2}}{8\,f{a}^{2}{d}^{2}}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}}{4\,f{a}^{2}{d}^{2}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{\sqrt{2}}{4\,f{a}^{2}{d}^{2}}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-2\,{\frac{1}{{a}^{2}df\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{1}{2\,{a}^{2}df \left ( d\tan \left ( fx+e \right ) +d \right ) }\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{5}{2\,f{a}^{2}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ){d}^{-{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x)
[Out]
-1/8/f/a^2/d^2*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*t
an(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/4/f/a^2/d^2*(d^2)^(1/4)*2^(1/2)*arctan(2^(1
/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/4/f/a^2/d^2*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f
*x+e))^(1/2)+1)-2/a^2/d/f/(d*tan(f*x+e))^(1/2)-1/2/f/a^2/d*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)-5/2*arctan((d
*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(3/2)/f
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 5.81514, size = 5619, normalized size = 18.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="fricas")
[Out]
[-1/8*(5*(cos(f*x + e)^2 + 2*(cos(f*x + e)^3 - cos(f*x + e))*sin(f*x + e) - 1)*sqrt(-d)*log(-(6*d^2*cos(f*x +
e)*sin(f*x + e) - d^2 + 4*(d*cos(f*x + e)^2 - d*cos(f*x + e)*sin(f*x + e))*sqrt(-d)*sqrt(d*sin(f*x + e)/cos(f*
x + e)))/(2*cos(f*x + e)*sin(f*x + e) + 1)) - 4*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqr
t(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-
sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*
d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f
^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) - 1) - 4*(sqrt(2)*a^2*d^2*f*co
s(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f
*x + e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4
))^(3/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d
*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^6*f
^4))^(3/4) + 1) + (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3
- sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*log((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))
*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*s
in(f*x + e))/cos(f*x + e)) - (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(
f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*log((a^4*d^4*f^2*sqrt(1/(a^
8*d^6*f^4))*cos(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x
+ e) + d*sin(f*x + e))/cos(f*x + e)) + 4*(9*cos(f*x + e)^4 - 9*cos(f*x + e)^2 + (cos(f*x + e)^3 - 5*cos(f*x +
e))*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(a^2*d^2*f*cos(f*x + e)^2 - a^2*d^2*f + 2*(a^2*d^2*f*cos
(f*x + e)^3 - a^2*d^2*f*cos(f*x + e))*sin(f*x + e)), -1/8*(20*(cos(f*x + e)^2 + 2*(cos(f*x + e)^3 - cos(f*x +
e))*sin(f*x + e) - 1)*sqrt(d)*arctan(sqrt(d*sin(f*x + e)/cos(f*x + e))/sqrt(d)) - 4*(sqrt(2)*a^2*d^2*f*cos(f*x
+ e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x +
e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3
/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(
f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(1/(a^8*d^6*f^4))^
(3/4) - 1) - 4*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - s
qrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*arctan(-sqrt(2)*a^6*d^4*f^3*sqrt(d*sin(f*
x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) + sqrt(2)*a^6*d^4*f^3*sqrt((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos
(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f
*x + e))/cos(f*x + e))*(1/(a^8*d^6*f^4))^(3/4) + 1) + (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a^2*d^2*f +
2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4))^(1/4)*lo
g((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) + sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x + e))*(1/(a^
8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) - (sqrt(2)*a^2*d^2*f*cos(f*x + e)^2 - sqrt(2)*a
^2*d^2*f + 2*(sqrt(2)*a^2*d^2*f*cos(f*x + e)^3 - sqrt(2)*a^2*d^2*f*cos(f*x + e))*sin(f*x + e))*(1/(a^8*d^6*f^4
))^(1/4)*log((a^4*d^4*f^2*sqrt(1/(a^8*d^6*f^4))*cos(f*x + e) - sqrt(2)*a^2*d^2*f*sqrt(d*sin(f*x + e)/cos(f*x +
e))*(1/(a^8*d^6*f^4))^(1/4)*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e)) + 4*(9*cos(f*x + e)^4 - 9*cos(f*x +
e)^2 + (cos(f*x + e)^3 - 5*cos(f*x + e))*sin(f*x + e))*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(a^2*d^2*f*cos(f*x +
e)^2 - a^2*d^2*f + 2*(a^2*d^2*f*cos(f*x + e)^3 - a^2*d^2*f*cos(f*x + e))*sin(f*x + e))]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan ^{2}{\left (e + f x \right )} + 2 \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan{\left (e + f x \right )} + \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx}{a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d*tan(f*x+e))**(3/2)/(a+a*tan(f*x+e))**2,x)
[Out]
Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x)**2 + 2*(d*tan(e + f*x))**(3/2)*tan(e + f*x) + (d*tan(e + f*x)
)**(3/2)), x)/a**2
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Giac [A] time = 1.37079, size = 397, normalized size = 1.3 \begin{align*} -\frac{1}{8} \, d^{3}{\left (\frac{2 \, \sqrt{2} \sqrt{{\left | d \right |}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} + 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{a^{2} d^{5} f} + \frac{2 \, \sqrt{2} \sqrt{{\left | d \right |}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} - 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{a^{2} d^{5} f} + \frac{\sqrt{2} \sqrt{{\left | d \right |}} \log \left (d \tan \left (f x + e\right ) + \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{a^{2} d^{5} f} - \frac{\sqrt{2} \sqrt{{\left | d \right |}} \log \left (d \tan \left (f x + e\right ) - \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{a^{2} d^{5} f} + \frac{20 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{2} d^{\frac{9}{2}} f} + \frac{4 \,{\left (5 \, d \tan \left (f x + e\right ) + 4 \, d\right )}}{{\left (\sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + \sqrt{d \tan \left (f x + e\right )} d\right )} a^{2} d^{4} f}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e))^2,x, algorithm="giac")
[Out]
-1/8*d^3*(2*sqrt(2)*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d
)))/(a^2*d^5*f) + 2*sqrt(2)*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/s
qrt(abs(d)))/(a^2*d^5*f) + sqrt(2)*sqrt(abs(d))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d))
+ abs(d))/(a^2*d^5*f) - sqrt(2)*sqrt(abs(d))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) +
abs(d))/(a^2*d^5*f) + 20*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*d^(9/2)*f) + 4*(5*d*tan(f*x + e) + 4*d)/((
sqrt(d*tan(f*x + e))*d*tan(f*x + e) + sqrt(d*tan(f*x + e))*d)*a^2*d^4*f))