3.2.76 \(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx\) [176]

Optimal. Leaf size=277 \[ \frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)} \]

[Out]

(4+4*I)*a^(5/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-8/315*a^2*(197*A-19
5*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/21*a^2*(4*I*A+3*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^
(7/2)+2/105*a^2*(46*A-45*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)+8/315*a^2*(59*I*A+60*B)*(a+I*a*tan(d
*x+c))^(1/2)/d/tan(d*x+c)^(3/2)-2/9*a*A*(a+I*a*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(9/2)

________________________________________________________________________________________

Rubi [A]
time = 0.64, antiderivative size = 277, normalized size of antiderivative = 1.00, 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 = {3674, 3679, 12, 3625, 211} \begin {gather*} \frac {(4+4 i) a^{5/2} (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 {8 a^2 (60 B+59 i A) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 (3 B+4 i A) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(11/2),x]

[Out]

((4 + 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a^2*((4*I)*A + 3*B)*Sqrt[a + I*a*Tan[c + d*x]])/(21*d*Tan[c + d*x]^(7/2)) + (2*a^2*(46*A - (45*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (8*a^2*((59*I)*A + 60*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Ta
n[c + d*x]^(3/2)) - (8*a^2*(197*A - (195*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Sqrt[Tan[c + d*x]]) - (2*a*A
*(a + I*a*Tan[c + d*x])^(3/2))/(9*d*Tan[c + d*x]^(9/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 3625

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 3674

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] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3679

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]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2}{9} \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} a (4 i A+3 B)-\frac {3}{2} a (2 A-3 i B) \tan (c+d x)\right )}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {4}{63} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (46 A-45 i B)-\frac {3}{4} a^2 (38 i A+39 B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{2} a^3 (59 i A+60 B)+\frac {3}{2} a^3 (46 A-45 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{315 a}\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^4 (197 A-195 i B)+\frac {3}{2} a^4 (59 i A+60 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{945 a^2}\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {32 \int \frac {945 a^5 (i A+B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\left (4 a^2 (i A+B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {\left (8 a^4 (A-i B)\right ) \text {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 {(4+4 i) a^{5/2} (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^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 9.25, size = 246, normalized size = 0.89 \begin {gather*} \frac {a^2 \left (\frac {1260 (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 ^2(2 (c+d x)) (-2331 A+2205 i B+12 (251 A-260 i B) \cos (2 (c+d x))+(-961 A+915 i B) \cos (4 (c+d x))+282 i A \sin (2 (c+d x))+390 B \sin (2 (c+d x))-331 i A \sin (4 (c+d x))-285 B \sin (4 (c+d x)))}{\tan ^{\frac {5}{2}}(c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{315 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(11/2),x]

[Out]

(a^2*((1260*(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[2*(c + d*x)]^2*(-23
31*A + (2205*I)*B + 12*(251*A - (260*I)*B)*Cos[2*(c + d*x)] + (-961*A + (915*I)*B)*Cos[4*(c + d*x)] + (282*I)*
A*Sin[2*(c + d*x)] + 390*B*Sin[2*(c + d*x)] - (331*I)*A*Sin[4*(c + d*x)] - 285*B*Sin[4*(c + d*x)]))/Tan[c + d*
x]^(5/2))*Sqrt[a + I*a*Tan[c + d*x]])/(315*d)

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 886 vs. \(2 (228 ) = 456\).
time = 0.12, size = 887, normalized size = 3.20 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/315/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(9/2)*(-1576*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)-315*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))*a*tan(d*x+c)^5-190*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)+480*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)+472*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)+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)*a*tan(d*x+c)^5
+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)*a*tan(d*x+c)^5-315*(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))*a*tan(d*x+c)^5+276*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)-270*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)-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)*a*ta
n(d*x+c)^5+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)*a*tan(d*x+c)^5+1560*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)-90*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)-70*A*(I*a)^(1/2)*(-I*a)^(
1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 722 vs. \(2 (213) = 426\).
time = 0.58, size = 722, normalized size = 2.61 \begin {gather*} -\frac {2 \, {\left (315 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 315 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) + 2 \, \sqrt {2} {\left (2 \, {\left (323 i \, A + 300 \, B\right )} a^{2} e^{\left (11 i \, d x + 11 i \, c\right )} + 77 \, {\left (-13 i \, A - 15 \, B\right )} a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} + 18 \, {\left (38 i \, A + 25 \, B\right )} a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 42 \, {\left (23 i \, A + 20 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 1050 \, {\left (-i \, A - B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 315 \, {\left (i \, A + B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{315 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

-2/315*(315*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/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)*sqrt(-(-I*A
^2 - 2*A*B + I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^2)
*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)/((-I*A - B)*a^2)) - 315*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/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(-(sq
rt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) - sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c)
+ (-I*A - B)*a^2)*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)/((-I*A - B)*a^2)) + 2*sqrt(2)*(2*(323*I*A + 300*B)*a^2*e^(11*I*d*x + 11*I*c) + 77*(-13*I*
A - 15*B)*a^2*e^(9*I*d*x + 9*I*c) + 18*(38*I*A + 25*B)*a^2*e^(7*I*d*x + 7*I*c) + 42*(23*I*A + 20*B)*a^2*e^(5*I
*d*x + 5*I*c) + 1050*(-I*A - B)*a^2*e^(3*I*d*x + 3*I*c) + 315*(I*A + B)*a^2*e^(I*d*x + I*c))*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)))/(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)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(11/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu
lar value [

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(11/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(11/2), x)

________________________________________________________________________________________