3.168 \(\int \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=269 \[ \frac{(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{4 a (57 B+61 i A) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (9 B+10 i A) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{4 a (193 A-201 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)} \]

[Out]

((2 + 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a*A*Sqrt[a + I*a*Tan[c + d*x]])/(9*d*Tan[c + d*x]^(9/2)) - (2*a*((10*I)*A + 9*B)*Sqrt[a + I*a*Tan[c + d*x]])/(
63*d*Tan[c + d*x]^(7/2)) + (4*a*(11*A - (12*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (4*
a*((61*I)*A + 57*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Tan[c + d*x]^(3/2)) - (4*a*(193*A - (201*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(315*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.936302, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3593, 3598, 12, 3544, 205} \[ \frac{(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{4 a (57 B+61 i A) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (9 B+10 i A) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{4 a (193 A-201 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(11/2),x]

[Out]

((2 + 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a*A*Sqrt[a + I*a*Tan[c + d*x]])/(9*d*Tan[c + d*x]^(9/2)) - (2*a*((10*I)*A + 9*B)*Sqrt[a + I*a*Tan[c + d*x]])/(
63*d*Tan[c + d*x]^(7/2)) + (4*a*(11*A - (12*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (4*
a*((61*I)*A + 57*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Tan[c + d*x]^(3/2)) - (4*a*(193*A - (201*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(315*d*Sqrt[Tan[c + d*x]])

Rule 3593

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

Rule 3598

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[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 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

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+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{11}{2}}(c+d x)} \, dx &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2}{9} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (10 i A+9 B)-\frac{1}{2} a (8 A-9 i B) \tan (c+d x)\right )}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a (10 i A+9 B) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{2} a^2 (11 A-12 i B)-\frac{3}{2} a^2 (10 i A+9 B) \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx}{63 a}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a (10 i A+9 B) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{4} a^3 (61 i A+57 B)+3 a^3 (11 A-12 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{315 a^2}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a (10 i A+9 B) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (61 i A+57 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{16 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{3}{8} a^4 (193 A-201 i B)+\frac{3}{4} a^4 (61 i A+57 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{945 a^3}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a (10 i A+9 B) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (61 i A+57 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{4 a (193 A-201 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}+\frac{32 \int \frac{945 a^5 (i A+B) \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{945 a^4}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a (10 i A+9 B) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (61 i A+57 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{4 a (193 A-201 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}+(2 a (i A+B)) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a (10 i A+9 B) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (61 i A+57 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{4 a (193 A-201 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}+\frac{\left (4 a^3 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac{(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a (10 i A+9 B) \sqrt{a+i a \tan (c+d x)}}{63 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a (11 A-12 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (61 i A+57 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{4 a (193 A-201 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 14.2096, size = 242, normalized size = 0.9 \[ \frac{a \sqrt{a+i a \tan (c+d x)} \left (\frac{2520 (A-i B) e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac{\csc ^4(c+d x) (12 (117 A-134 i B) \cos (2 (c+d x))+(-487 A+474 i B) \cos (4 (c+d x))+144 i A \sin (2 (c+d x))-172 i A \sin (4 (c+d x))-1197 A+138 B \sin (2 (c+d x))-159 B \sin (4 (c+d x))+1134 i B)}{\sqrt{\tan (c+d x)}}\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(11/2),x]

[Out]

(a*((2520*(A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(E
^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) + (Csc[c + d*x]^4*(-1197*A +
(1134*I)*B + 12*(117*A - (134*I)*B)*Cos[2*(c + d*x)] + (-487*A + (474*I)*B)*Cos[4*(c + d*x)] + (144*I)*A*Sin[
2*(c + d*x)] + 138*B*Sin[2*(c + d*x)] - (172*I)*A*Sin[4*(c + d*x)] - 159*B*Sin[4*(c + d*x)]))/Sqrt[Tan[c + d*x
]])*Sqrt[a + I*a*Tan[c + d*x]])/(1260*d)

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Maple [B]  time = 0.048, size = 885, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x)

[Out]

1/630/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(9/2)*(-1544*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4*(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)+1608*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)+488*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+456*B*(I*a)^(1/2)*(-I*
a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+1260*I*A*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^5*a+1260*B*ln(1/2*(2*I*a*tan(d*x+c
)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^5*a-288*I*B*(I*a
)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-315*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2
)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a+264*A*
(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+630*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(
a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^5*a-630*ln(1/2*(2*I*a
*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^5*a-31
5*I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))
/(tan(d*x+c)+I))*tan(d*x+c)^5*a-200*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)-180*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-140*A*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)

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Maxima [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+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.89244, size = 1971, normalized size = 7.33 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

1/630*(sqrt(2)*((-2636*I*A - 2532*B)*a*e^(10*I*d*x + 10*I*c) + (3556*I*A + 4452*B)*a*e^(8*I*d*x + 8*I*c) + (-3
384*I*A - 2088*B)*a*e^(6*I*d*x + 6*I*c) + (-4536*I*A - 3192*B)*a*e^(4*I*d*x + 4*I*c) + (3780*I*A + 4620*B)*a*e
^(2*I*d*x + 2*I*c) + (-1260*I*A - 1260*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) +
I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 315*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*(d*e^(10*I*d*x
+ 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2
*I*c) - d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + sqrt((8*I*A^2 + 16*A*B - 8*
I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) - 315*sqrt((8*I*A^2 + 16*A*B -
8*I*B^2)*a^3/d^2)*(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*
d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*
a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I
*c) - sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)
))/(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) +
5*d*e^(2*I*d*x + 2*I*c) - d)

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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+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.49962, size = 333, normalized size = 1.24 \begin{align*} -\frac{\left (i - 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{7} +{\left (\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{6} - \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{7}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{7} a - 8 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a^{2} + 27 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{3} - 50 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{4} + 55 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{5} - 36 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{6} + 13 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{7} - 2 \, a^{8}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="giac")

[Out]

-1/2*((I - 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^2*a^7 + ((2*I + 2)*(I*a*tan(d*x
+ c) + a)^2*a^6 - (2*I + 2)*(I*a*tan(d*x + c) + a)*a^7)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*ta
n(d*x + c) + a)*B)/(((I*a*tan(d*x + c) + a)^7*a - 8*(I*a*tan(d*x + c) + a)^6*a^2 + 27*(I*a*tan(d*x + c) + a)^5
*a^3 - 50*(I*a*tan(d*x + c) + a)^4*a^4 + 55*(I*a*tan(d*x + c) + a)^3*a^5 - 36*(I*a*tan(d*x + c) + a)^2*a^6 + 1
3*(I*a*tan(d*x + c) + a)*a^7 - 2*a^8)*d)