3.2.75 \(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\) [175]

Optimal. Leaf size=231 \[ \frac {(4-4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (10 i A+7 B) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (130 i A+133 B) \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]

[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+4/105*a^2*(130*I*A+
133*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/35*a^2*(10*I*A+7*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)
^(5/2)+2/105*a^2*(80*A-77*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)-2/7*a*A*(a+I*a*tan(d*x+c))^(3/2)/d/
tan(d*x+c)^(7/2)

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Rubi [A]
time = 0.51, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 (7 B+10 i A) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (133 B+130 i A) \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{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]^(9/2),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*((10*I)*A + 7*B)*Sqrt[a + I*a*Tan[c + d*x]])/(35*d*Tan[c + d*x]^(5/2)) + (2*a^2*(80*A - (77*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2)) + (4*a^2*((130*I)*A + 133*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d
*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(7*d*Tan[c + d*x]^(7/2))

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 {9}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (10 i A+7 B)-\frac {1}{2} a (4 A-7 i B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (10 i A+7 B) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4}{35} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (80 A-77 i B)-\frac {3}{4} a^2 (20 i A+21 B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (10 i A+7 B) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^3 (130 i A+133 B)+\frac {1}{4} a^3 (80 A-77 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{105 a}\\ &=-\frac {2 a^2 (10 i A+7 B) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (130 i A+133 B) \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {16 \int \frac {105 a^4 (A-i B) \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{105 a^2}\\ &=-\frac {2 a^2 (10 i A+7 B) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (130 i A+133 B) \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\left (4 a^2 (A-i B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (10 i A+7 B) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (130 i A+133 B) \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {\left (8 a^4 (i A+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} (i A+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 (10 i A+7 B) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (130 i A+133 B) \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 9.47, size = 363, normalized size = 1.57 \begin {gather*} \frac {4 \sqrt {2} e^{-2 i c} \sqrt {e^{i d x}} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \left (7 i B \left (-15+50 e^{2 i (c+d x)}-61 e^{4 i (c+d x)}+26 e^{6 i (c+d x)}\right )-5 A \left (-21+70 e^{2 i (c+d x)}-77 e^{4 i (c+d x)}+40 e^{6 i (c+d x)}\right )\right )+105 (A-i B) \left (-1+e^{2 i (c+d x)}\right )^4 \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{105 d \left (-1+e^{2 i (c+d x)}\right )^{9/2} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \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]^(9/2),x]

[Out]

(4*Sqrt[2]*Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(E^(I*(c + d*x))*
Sqrt[-1 + E^((2*I)*(c + d*x))]*((7*I)*B*(-15 + 50*E^((2*I)*(c + d*x)) - 61*E^((4*I)*(c + d*x)) + 26*E^((6*I)*(
c + d*x))) - 5*A*(-21 + 70*E^((2*I)*(c + d*x)) - 77*E^((4*I)*(c + d*x)) + 40*E^((6*I)*(c + d*x)))) + 105*(A -
I*B)*(-1 + E^((2*I)*(c + d*x)))^4*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]])*(a + I*a*Tan[c + d*x]
)^(5/2)*(A + B*Tan[c + d*x]))/(105*d*E^((2*I)*c)*(-1 + E^((2*I)*(c + d*x)))^(9/2)*Sqrt[E^(I*(c + d*x))/(1 + E^
((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 797 vs. \(2 (190 ) = 380\).
time = 0.12, size = 798, normalized size = 3.45 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)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/105/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(7/2)*(532*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)+520*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)-420*I*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)^4+160*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)-154*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)+420*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)^4+210*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)^
4-105*(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)^4+105*I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^4+210*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)^4-90*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)-42*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)-30*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)

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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)^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 657 vs. \(2 (177) = 354\).
time = 0.65, size = 657, normalized size = 2.84 \begin {gather*} -\frac {2 \, {\left (105 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (i \, \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 ) - 105 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (-i \, \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 (100 \, A - 91 i \, B\right )} a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} - 5 \, {\left (37 \, A - 49 i \, B\right )} a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 7 \, {\left (5 \, A - 11 i \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 245 \, {\left (A - i \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 105 \, {\left (A - i \, 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 )}}{105 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, 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)^(9/2),x, algorithm="fricas")

[Out]

-2/105*(105*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) +
6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*log((I*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)) - 105*sq
rt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x
+ 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*log((-I*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)) + 2*sqrt(2)*(2*(100*A
- 91*I*B)*a^2*e^(9*I*d*x + 9*I*c) - 5*(37*A - 49*I*B)*a^2*e^(7*I*d*x + 7*I*c) - 7*(5*A - 11*I*B)*a^2*e^(5*I*d
*x + 5*I*c) + 245*(A - I*B)*a^2*e^(3*I*d*x + 3*I*c) - 105*(A - I*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^(8*I*d*x + 8*I*c) - 4*d*e^(6*I
*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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)**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3879 deep

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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)^(9/2),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 [

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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 )}^{9/2}} \,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)^(9/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(9/2), x)

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