3.409 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=493 \[ \frac{b^{5/2} \left (9 a^2 A b-7 a^3 B-3 a b^2 B+5 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{7/2} d \left (a^2+b^2\right )^2}+\frac{b (A b-a B)}{a d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}+\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}-\frac{2 a^2 A-3 a b B+5 A b^2}{3 a^2 d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b-2 a^3 B-3 a b^2 B+5 A b^3}{a^3 d \left (a^2+b^2\right ) \sqrt{\tan (c+d x)}}+\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2}-\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2} \]

[Out]

-(((2*a*b*(A - B) - a^2*(A + B) + b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*
d)) + ((2*a*b*(A - B) - a^2*(A + B) + b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2
)^2*d) + (b^(5/2)*(9*a^2*A*b + 5*A*b^3 - 7*a^3*B - 3*a*b^2*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(a
^(7/2)*(a^2 + b^2)^2*d) + ((a^2*(A - B) - b^2*(A - B) + 2*a*b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Ta
n[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - ((a^2*(A - B) - b^2*(A - B) + 2*a*b*(A + B))*Log[1 + Sqrt[2]*Sqrt[T
an[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - (2*a^2*A + 5*A*b^2 - 3*a*b*B)/(3*a^2*(a^2 + b^2)*d
*Tan[c + d*x]^(3/2)) + (4*a^2*A*b + 5*A*b^3 - 2*a^3*B - 3*a*b^2*B)/(a^3*(a^2 + b^2)*d*Sqrt[Tan[c + d*x]]) + (b
*(A*b - a*B))/(a*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 1.53286, antiderivative size = 493, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.394, Rules used = {3609, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac{b^{5/2} \left (9 a^2 A b-7 a^3 B-3 a b^2 B+5 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{7/2} d \left (a^2+b^2\right )^2}+\frac{b (A b-a B)}{a d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}+\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}-\frac{2 a^2 A-3 a b B+5 A b^2}{3 a^2 d \left (a^2+b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b-2 a^3 B-3 a b^2 B+5 A b^3}{a^3 d \left (a^2+b^2\right ) \sqrt{\tan (c+d x)}}+\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2}-\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2),x]

[Out]

-(((2*a*b*(A - B) - a^2*(A + B) + b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*
d)) + ((2*a*b*(A - B) - a^2*(A + B) + b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2
)^2*d) + (b^(5/2)*(9*a^2*A*b + 5*A*b^3 - 7*a^3*B - 3*a*b^2*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(a
^(7/2)*(a^2 + b^2)^2*d) + ((a^2*(A - B) - b^2*(A - B) + 2*a*b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Ta
n[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - ((a^2*(A - B) - b^2*(A - B) + 2*a*b*(A + B))*Log[1 + Sqrt[2]*Sqrt[T
an[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - (2*a^2*A + 5*A*b^2 - 3*a*b*B)/(3*a^2*(a^2 + b^2)*d
*Tan[c + d*x]^(3/2)) + (4*a^2*A*b + 5*A*b^3 - 2*a^3*B - 3*a*b^2*B)/(a^3*(a^2 + b^2)*d*Sqrt[Tan[c + d*x]]) + (b
*(A*b - a*B))/(a*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*(a + b*Tan[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[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(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 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

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

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 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{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2} \, dx &=\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (2 a^2 A+5 A b^2-3 a b B\right )-a (A b-a B) \tan (c+d x)+\frac{5}{2} b (A b-a B) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac{2 \int \frac{\frac{3}{4} \left (4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B\right )+\frac{3}{2} a^2 (a A+b B) \tan (c+d x)+\frac{3}{4} b \left (2 a^2 A+5 A b^2-3 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx}{3 a^2 \left (a^2+b^2\right )}\\ &=-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B}{a^3 \left (a^2+b^2\right ) d \sqrt{\tan (c+d x)}}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{4 \int \frac{-\frac{3}{8} \left (2 a^4 A-4 a^2 A b^2-5 A b^4+4 a^3 b B+3 a b^3 B\right )+\frac{3}{4} a^3 (A b-a B) \tan (c+d x)+\frac{3}{8} b \left (4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{3 a^3 \left (a^2+b^2\right )}\\ &=-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B}{a^3 \left (a^2+b^2\right ) d \sqrt{\tan (c+d x)}}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{4 \int \frac{-\frac{3}{4} a^3 \left (a^2 A-A b^2+2 a b B\right )+\frac{3}{4} a^3 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{3 a^3 \left (a^2+b^2\right )^2}+\frac{\left (b^3 \left (9 a^2 A b+5 A b^3-7 a^3 B-3 a b^2 B\right )\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a^3 \left (a^2+b^2\right )^2}\\ &=-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B}{a^3 \left (a^2+b^2\right ) d \sqrt{\tan (c+d x)}}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{8 \operatorname{Subst}\left (\int \frac{-\frac{3}{4} a^3 \left (a^2 A-A b^2+2 a b B\right )+\frac{3}{4} a^3 \left (2 a A b-a^2 B+b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{3 a^3 \left (a^2+b^2\right )^2 d}+\frac{\left (b^3 \left (9 a^2 A b+5 A b^3-7 a^3 B-3 a b^2 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{2 a^3 \left (a^2+b^2\right )^2 d}\\ &=-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B}{a^3 \left (a^2+b^2\right ) d \sqrt{\tan (c+d x)}}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{\left (b^3 \left (9 a^2 A b+5 A b^3-7 a^3 B-3 a b^2 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^3 \left (a^2+b^2\right )^2 d}-\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac{b^{5/2} \left (9 a^2 A b+5 A b^3-7 a^3 B-3 a b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{7/2} \left (a^2+b^2\right )^2 d}-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B}{a^3 \left (a^2+b^2\right ) d \sqrt{\tan (c+d x)}}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=\frac{b^{5/2} \left (9 a^2 A b+5 A b^3-7 a^3 B-3 a b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{7/2} \left (a^2+b^2\right )^2 d}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B}{a^3 \left (a^2+b^2\right ) d \sqrt{\tan (c+d x)}}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}\\ &=-\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{b^{5/2} \left (9 a^2 A b+5 A b^3-7 a^3 B-3 a b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{a^{7/2} \left (a^2+b^2\right )^2 d}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{2 a^2 A+5 A b^2-3 a b B}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 A b+5 A b^3-2 a^3 B-3 a b^2 B}{a^3 \left (a^2+b^2\right ) d \sqrt{\tan (c+d x)}}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 3.81302, size = 287, normalized size = 0.58 \[ \frac{\frac{-2 a^2 A+3 a b B-5 A b^2}{a \tan ^{\frac{3}{2}}(c+d x)}+\frac{3 \left (4 a^2 A b-2 a^3 B-3 a b^2 B+5 A b^3\right )}{a^2 \sqrt{\tan (c+d x)}}+\frac{3 \left (b^{5/2} \left (9 a^2 A b-7 a^3 B-3 a b^2 B+5 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )+\sqrt [4]{-1} a^{7/2} (a+i b)^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+\sqrt [4]{-1} a^{7/2} (a-i b)^2 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )}{a^{5/2} \left (a^2+b^2\right )}+\frac{3 b (A b-a B)}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}}{3 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2),x]

[Out]

((3*((-1)^(1/4)*a^(7/2)*(a + I*b)^2*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + b^(5/2)*(9*a^2*A*b + 5*A
*b^3 - 7*a^3*B - 3*a*b^2*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] + (-1)^(1/4)*a^(7/2)*(a - I*b)^2*(A +
I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]))/(a^(5/2)*(a^2 + b^2)) + (-2*a^2*A - 5*A*b^2 + 3*a*b*B)/(a*Tan[c
+ d*x]^(3/2)) + (3*(4*a^2*A*b + 5*A*b^3 - 2*a^3*B - 3*a*b^2*B))/(a^2*Sqrt[Tan[c + d*x]]) + (3*b*(A*b - a*B))/
(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])))/(3*a*(a^2 + b^2)*d)

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Maple [B]  time = 0.061, size = 1198, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x)

[Out]

1/2/d/(a^2+b^2)^2*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^2-1/2/d/(a^2+b^2)^2*B*2^(1/2)*arctan(1+2^(1/
2)*tan(d*x+c)^(1/2))*a^2+1/2/d/(a^2+b^2)^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^2-1/2/d/(a^2+b^2)^2*
B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^2-1/2/d/(a^2+b^2)^2*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+t
an(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a*b+1/2/d/(a^2+b^2)^2*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c
)^(1/2))*b^2-1/2/d/(a^2+b^2)^2*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^2+1/d/(a^2+b^2)^2*A*2^(1/2)*arc
tan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b+1/4/d/(a^2+b^2)^2*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1
-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*b^2+1/d*b^4/(a^2+b^2)^2/a*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))*A+9/d*b^4/(
a^2+b^2)^2/a/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*A+1/2/d/(a^2+b^2)^2*A*2^(1/2)*ln((1-2^(1/2)*ta
n(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a*b-2/3/d/a^2*A/tan(d*x+c)^(3/2)-2/d/a^2/t
an(d*x+c)^(1/2)*B-3/d*b^5/a^2/(a^2+b^2)^2/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B+1/d*b^6/a^3/(a^
2+b^2)^2*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))*A+5/d*b^6/a^3/(a^2+b^2)^2/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*
b)^(1/2))*A-1/d*b^5/a^2/(a^2+b^2)^2*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))*B+1/d/(a^2+b^2)^2*A*2^(1/2)*arctan(1+2^(
1/2)*tan(d*x+c)^(1/2))*a*b-1/d/(a^2+b^2)^2*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b-1/d/(a^2+b^2)^2*B*
2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b-1/4/d/(a^2+b^2)^2*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan
(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^2+1/2/d/(a^2+b^2)^2*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)
^(1/2))*b^2+4/d/a^3/tan(d*x+c)^(1/2)*A*b-1/d*b^3/(a^2+b^2)^2*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))*B-7/d*b^3/(a^2+
b^2)^2/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))*B-1/2/d/(a^2+b^2)^2*A*2^(1/2)*arctan(1+2^(1/2)*tan(d
*x+c)^(1/2))*a^2-1/4/d/(a^2+b^2)^2*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^
(1/2)+tan(d*x+c)))*a^2+1/4/d/(a^2+b^2)^2*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d
*x+c)^(1/2)+tan(d*x+c)))*b^2

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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((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((A+B*tan(d*x+c))/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.48322, size = 760, normalized size = 1.54 \begin{align*} -\frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b + 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b + 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b + 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} + \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b + 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac{{\left (7 \, B a^{3} b^{3} - 9 \, A a^{2} b^{4} + 3 \, B a b^{5} - 5 \, A b^{6}\right )} \arctan \left (\frac{b \sqrt{\tan \left (d x + c\right )}}{\sqrt{a b}}\right )}{{\left (a^{7} d + 2 \, a^{5} b^{2} d + a^{3} b^{4} d\right )} \sqrt{a b}} - \frac{B a b^{3} \sqrt{\tan \left (d x + c\right )} - A b^{4} \sqrt{\tan \left (d x + c\right )}}{{\left (a^{5} d + a^{3} b^{2} d\right )}{\left (b \tan \left (d x + c\right ) + a\right )}} - \frac{2 \,{\left (3 \, B a \tan \left (d x + c\right ) - 6 \, A b \tan \left (d x + c\right ) + A a\right )}}{3 \, a^{3} d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(sqrt(2)*A*a^2 + sqrt(2)*B*a^2 - 2*sqrt(2)*A*a*b + 2*sqrt(2)*B*a*b - sqrt(2)*A*b^2 - sqrt(2)*B*b^2)*arcta
n(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))))/(a^4*d + 2*a^2*b^2*d + b^4*d) - 1/2*(sqrt(2)*A*a^2 + sqrt(2)*B
*a^2 - 2*sqrt(2)*A*a*b + 2*sqrt(2)*B*a*b - sqrt(2)*A*b^2 - sqrt(2)*B*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqr
t(tan(d*x + c))))/(a^4*d + 2*a^2*b^2*d + b^4*d) - 1/4*(sqrt(2)*A*a^2 - sqrt(2)*B*a^2 + 2*sqrt(2)*A*a*b + 2*sqr
t(2)*B*a*b - sqrt(2)*A*b^2 + sqrt(2)*B*b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/(a^4*d + 2*a^2*
b^2*d + b^4*d) + 1/4*(sqrt(2)*A*a^2 - sqrt(2)*B*a^2 + 2*sqrt(2)*A*a*b + 2*sqrt(2)*B*a*b - sqrt(2)*A*b^2 + sqrt
(2)*B*b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/(a^4*d + 2*a^2*b^2*d + b^4*d) - (7*B*a^3*b^3 -
9*A*a^2*b^4 + 3*B*a*b^5 - 5*A*b^6)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^7*d + 2*a^5*b^2*d + a^3*b^4*d)*s
qrt(a*b)) - (B*a*b^3*sqrt(tan(d*x + c)) - A*b^4*sqrt(tan(d*x + c)))/((a^5*d + a^3*b^2*d)*(b*tan(d*x + c) + a))
- 2/3*(3*B*a*tan(d*x + c) - 6*A*b*tan(d*x + c) + A*a)/(a^3*d*tan(d*x + c)^(3/2))