3.133 \(\int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=169 \[ -\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 B+11 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^3 (B+i A)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]

[Out]

-8*(-1)^(1/4)*a^3*(A-I*B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+8*a^3*(I*A+B)/d/tan(d*x+c)^(1/2)+8/105*a^3*(23
*A-21*I*B)/d/tan(d*x+c)^(3/2)-2/7*a*A*(a+I*a*tan(d*x+c))^2/d/tan(d*x+c)^(7/2)-2/35*(11*I*A+7*B)*(a^3+I*a^3*tan
(d*x+c))/d/tan(d*x+c)^(5/2)

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Rubi [A]  time = 0.42, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3593, 3591, 3529, 3533, 205} \[ -\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 B+11 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^3 (B+i A)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(23*A - (21*I)*B))/(105*d*Tan[c
+ d*x]^(3/2)) + (8*a^3*(I*A + B))/(d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^2)/(7*d*Tan[c + d*x]
^(7/2)) - (2*((11*I)*A + 7*B)*(a^3 + I*a^3*Tan[c + d*x]))/(35*d*Tan[c + d*x]^(5/2))

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 3529

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

Rule 3533

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

Rule 3591

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^3 (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))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+i a \tan (c+d x))^2 \left (\frac {1}{2} a (11 i A+7 B)-\frac {1}{2} a (3 A-7 i B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {(a+i a \tan (c+d x)) \left (-a^2 (23 A-21 i B)-2 a^2 (6 i A+7 B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {-35 a^3 (i A+B)+35 a^3 (A-i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (i A+B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {35 a^3 (A-i B)+35 a^3 (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (i A+B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {\left (280 a^6 (A-i B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{35 a^3 (A-i B)-35 a^3 (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (i A+B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [B]  time = 12.58, size = 495, normalized size = 2.93 \[ \frac {8 e^{-3 i c} (A-i B) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d \sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {\cos ^4(c+d x) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (\csc (c) \left (\frac {2}{5} \cos (3 c)-\frac {2}{5} i \sin (3 c)\right ) \csc ^3(c+d x) (B \sin (d x)+3 i A \sin (d x))+\csc (c) \left (\frac {2}{105} \cos (3 c)-\frac {2}{105} i \sin (3 c)\right ) \csc ^2(c+d x) (170 A \sin (c)-63 i A \cos (c)-105 i B \sin (c)-21 B \cos (c))+\csc (c) \left (\frac {2}{5} \cos (3 c)-\frac {2}{5} i \sin (3 c)\right ) \csc (c+d x) (-21 B \sin (d x)-23 i A \sin (d x))+\csc (c) \left (\frac {2}{105} \cos (3 c)-\frac {2}{105} i \sin (3 c)\right ) (-155 A \sin (c)+483 i A \cos (c)+105 i B \sin (c)+441 B \cos (c))+\left (-\frac {2}{7} A \cos (3 c)+\frac {2}{7} i A \sin (3 c)\right ) \csc ^4(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

(8*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c
+ d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*E^((3*I)
*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*S
in[c + d*x])) + (Cos[c + d*x]^4*(Csc[c]*Csc[c + d*x]^2*((-63*I)*A*Cos[c] - 21*B*Cos[c] + 170*A*Sin[c] - (105*I
)*B*Sin[c])*((2*Cos[3*c])/105 - ((2*I)/105)*Sin[3*c]) + Csc[c]*((483*I)*A*Cos[c] + 441*B*Cos[c] - 155*A*Sin[c]
+ (105*I)*B*Sin[c])*((2*Cos[3*c])/105 - ((2*I)/105)*Sin[3*c]) + Csc[c + d*x]^4*((-2*A*Cos[3*c])/7 + ((2*I)/7)
*A*Sin[3*c]) + Csc[c]*Csc[c + d*x]*((2*Cos[3*c])/5 - ((2*I)/5)*Sin[3*c])*((-23*I)*A*Sin[d*x] - 21*B*Sin[d*x])
+ Csc[c]*Csc[c + d*x]^3*((2*Cos[3*c])/5 - ((2*I)/5)*Sin[3*c])*((3*I)*A*Sin[d*x] + B*Sin[d*x]))*Sqrt[Tan[c + d*
x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*
x]))

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fricas [B]  time = 1.22, size = 557, normalized size = 3.30 \[ \frac {105 \, \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 105 \, \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 16 \, {\left ({\left (319 \, A - 273 i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, {\left (109 \, A - 133 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 5 \, {\left (19 \, A - 21 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (129 \, A - 133 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, {\left (41 \, A - 42 i \, B\right )} a^{3}\right )} \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}}}{420 \, {\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))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

1/420*(105*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/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((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A
^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3)) - 105*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/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)*l
og((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x + 2*
I*c) - I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*
a^3)) - 16*((319*A - 273*I*B)*a^3*e^(8*I*d*x + 8*I*c) - 3*(109*A - 133*I*B)*a^3*e^(6*I*d*x + 6*I*c) - 5*(19*A
- 21*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 3*(129*A - 133*I*B)*a^3*e^(2*I*d*x + 2*I*c) - 4*(41*A - 42*I*B)*a^3)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\tan \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(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)^3/tan(d*x + c)^(9/2), x)

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maple [B]  time = 0.11, size = 572, normalized size = 3.38 \[ -\frac {2 a^{3} A}{7 d \tan \left (d x +c \right )^{\frac {7}{2}}}+\frac {i a^{3} A \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}+\frac {8 a^{3} B}{d \sqrt {\tan \left (d x +c \right )}}-\frac {6 i a^{3} A}{5 d \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 a^{3} B}{5 d \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 i a^{3} B}{d \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {8 a^{3} A}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 i a^{3} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {2 i a^{3} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {8 i a^{3} A}{d \sqrt {\tan \left (d x +c \right )}}+\frac {2 a^{3} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {a^{3} A \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}+\frac {2 a^{3} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{3} B \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}+\frac {2 i a^{3} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {2 i a^{3} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {a^{3} B \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{d}+\frac {2 a^{3} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {2 a^{3} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x)

[Out]

-2/7/d*a^3*A/tan(d*x+c)^(7/2)+I/d*a^3*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)))+8/d*a^3/tan(d*x+c)^(1/2)*B-6/5*I/d*a^3/tan(d*x+c)^(5/2)*A-2/5/d*a^3/tan(d*x+c)^(5/2)*B-2
*I/d*a^3/tan(d*x+c)^(3/2)*B+8/3/d*a^3*A/tan(d*x+c)^(3/2)+2*I/d*a^3*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2
))-2*I/d*a^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+8*I/d*a^3/tan(d*x+c)^(1/2)*A+2/d*a^3*A*2^(1/2)*arct
an(-1+2^(1/2)*tan(d*x+c)^(1/2))+1/d*a^3*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)))+2/d*a^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-I/d*a^3*B*2^(1/2)*ln((1+2^(1/2)*t
an(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*I/d*a^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(
d*x+c)^(1/2))-2*I/d*a^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+1/d*a^3*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)))+2/d*a^3*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1
/2))+2/d*a^3*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))

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maxima [A]  time = 1.04, size = 213, normalized size = 1.26 \[ -\frac {105 \, {\left (\sqrt {2} {\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3} + \frac {2 \, {\left ({\left (-420 i \, A - 420 \, B\right )} a^{3} \tan \left (d x + c\right )^{3} - 35 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + {\left (63 i \, A + 21 \, B\right )} a^{3} \tan \left (d x + c\right ) + 15 \, A a^{3}\right )}}{\tan \left (d x + c\right )^{\frac {7}{2}}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

-1/105*(105*(sqrt(2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + sqrt(
2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*((I - 1)*A + (
I + 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A + (I + 1)*B)*log(-sqrt(2)*sq
rt(tan(d*x + c)) + tan(d*x + c) + 1))*a^3 + 2*((-420*I*A - 420*B)*a^3*tan(d*x + c)^3 - 35*(4*A - 3*I*B)*a^3*ta
n(d*x + c)^2 + (63*I*A + 21*B)*a^3*tan(d*x + c) + 15*A*a^3)/tan(d*x + c)^(7/2))/d

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mupad [B]  time = 9.52, size = 293, normalized size = 1.73 \[ -\frac {\frac {2\,A\,a^3}{7\,d}+\frac {A\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,6{}\mathrm {i}}{5\,d}-\frac {8\,A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3\,d}-\frac {A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}-\frac {\frac {2\,B\,a^3}{5\,d}-\frac {8\,B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {B\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (A\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,A\,a^3\,\ln \left (A\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (8\,B\,a^3\,d+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,B\,a^3\,\ln \left (8\,B\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

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)^3)/tan(c + d*x)^(9/2),x)

[Out]

(2^(1/2)*A*a^3*log(A*a^3*d*8i - 2^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)*(4 - 4i))*(2 - 2i))/d - ((2*B*a^3)/(5*d) +
(B*a^3*tan(c + d*x)*2i)/d - (8*B*a^3*tan(c + d*x)^2)/d)/tan(c + d*x)^(5/2) - ((2*A*a^3)/(7*d) + (A*a^3*tan(c +
d*x)*6i)/(5*d) - (8*A*a^3*tan(c + d*x)^2)/(3*d) - (A*a^3*tan(c + d*x)^3*8i)/d)/tan(c + d*x)^(7/2) - ((-16i)^(
1/2)*A*a^3*log(A*a^3*d*8i + 2*(-16i)^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)))/d + (2^(1/2)*B*a^3*log(8*B*a^3*d - 2^(
1/2)*B*a^3*d*tan(c + d*x)^(1/2)*(4 + 4i))*(2 + 2i))/d - (16i^(1/2)*B*a^3*log(8*B*a^3*d + 2*16i^(1/2)*B*a^3*d*t
an(c + d*x)^(1/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int \left (- \frac {3 A}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {A}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {3 B}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {B}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \frac {i A}{\tan ^{\frac {9}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {3 i A}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {i B}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {3 i B}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(9/2),x)

[Out]

-I*a**3*(Integral(-3*A/tan(c + d*x)**(7/2), x) + Integral(A/tan(c + d*x)**(3/2), x) + Integral(-3*B/tan(c + d*
x)**(5/2), x) + Integral(B/sqrt(tan(c + d*x)), x) + Integral(I*A/tan(c + d*x)**(9/2), x) + Integral(-3*I*A/tan
(c + d*x)**(5/2), x) + Integral(I*B/tan(c + d*x)**(7/2), x) + Integral(-3*I*B/tan(c + d*x)**(3/2), x))

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