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

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

[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 = 0.76, 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 = {3593, 3598, 12, 3544, 205} \[ \frac {2 a^2 (80 A-77 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 (7 B+10 i A) \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (133 B+130 i A) \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}+\frac {(4-4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]

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 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]

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 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]

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 ) \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 {(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 = 14.62, size = 363, normalized size = 1.57 \[ \frac {4 \sqrt {2} e^{-2 i c} \sqrt {e^{i d x}} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \left (7 i B \left (50 e^{2 i (c+d x)}-61 e^{4 i (c+d x)}+26 e^{6 i (c+d x)}-15\right )-5 A \left (70 e^{2 i (c+d x)}-77 e^{4 i (c+d x)}+40 e^{6 i (c+d x)}-21\right )\right )+105 (A-i B) \left (-1+e^{2 i (c+d x)}\right )^4 \log \left (\sqrt {-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right ) (A+B \tan (c+d x))}{105 d \left (-1+e^{2 i (c+d x)}\right )^{9/2} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[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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fricas [B]  time = 0.56, size = 641, normalized size = 2.77 \[ -\frac {8 \, \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}} - 105 \, \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 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 (\sqrt {2} {\left ({\left (4 i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 i \, A + 4 \, 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}} + i \, \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right ) + 105 \, \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 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 (\sqrt {2} {\left ({\left (4 i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 i \, A + 4 \, 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}} - i \, \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right )}{210 \, {\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 )}} \]

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]

-1/210*(8*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)) - 105*sqr
t((-32*I*A^2 - 64*A*B + 32*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((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*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)) + I*sqrt((-32
*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((4*I*A + 4*B)*a^2)) + 105*sqrt((-32*
I*A^2 - 64*A*B + 32*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((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*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)) - I*sqrt((-32*I*A^2
- 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((4*I*A + 4*B)*a^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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

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]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(9/2), x)

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maple [B]  time = 0.33, size = 798, normalized size = 3.45 \[ \frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (532 B \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-154 i B \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+210 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right ) a +160 A \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-420 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right ) a +420 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right ) a -90 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-105 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a +520 i A \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+210 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right ) a +105 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a -42 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-30 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{105 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}} \]

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)

[Out]

1/105/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(7/2)*(532*B*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*(I*a)^(1/2)*(-I*a)^(1/2)-154*I*B*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/
2)+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)*tan(d*x+c)^4*a+160*A*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(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)*tan(
d*x+c)^4*a+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)*tan(d*x+c)^4*a-90*I*A*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-1
05*(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)^4*a+520*I*A*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(
1/2)+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)*tan(d*x+c)^4*a+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*tan(d*x+c)))^(1/
2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a-42*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(
-I*a)^(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))/(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \]

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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