3.132 \(\int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=144 \[ \frac {8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {16 a^3 (6 A-5 i B)}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

8*(-1)^(1/4)*a^3*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+16/15*a^3*(6*A-5*I*B)/d/tan(d*x+c)^(1/2)-2/5*a*
A*(a+I*a*tan(d*x+c))^2/d/tan(d*x+c)^(5/2)-2/15*(9*I*A+5*B)*(a^3+I*a^3*tan(d*x+c))/d/tan(d*x+c)^(3/2)

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Rubi [A]  time = 0.38, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3593, 3591, 3533, 205} \[ \frac {8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {16 a^3 (6 A-5 i B)}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (16*a^3*(6*A - (5*I)*B))/(15*d*Sqrt[Tan
[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^2)/(5*d*Tan[c + d*x]^(5/2)) - (2*((9*I)*A + 5*B)*(a^3 + I*a^3*Tan[
c + d*x]))/(15*d*Tan[c + d*x]^(3/2))

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 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 {7}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+i a \tan (c+d x))^2 \left (\frac {1}{2} a (9 i A+5 B)-\frac {1}{2} a (A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {(a+i a \tan (c+d x)) \left (-2 a^2 (6 A-5 i B)-a^2 (3 i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {16 a^3 (6 A-5 i B)}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {-15 a^3 (i A+B)+15 a^3 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {16 a^3 (6 A-5 i B)}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (120 a^6 (i A+B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-15 a^3 (i A+B)-15 a^3 (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {8 \sqrt [4]{-1} a^3 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {16 a^3 (6 A-5 i B)}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 8.94, size = 273, normalized size = 1.90 \[ \frac {\cos ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (-\frac {8 i e^{-3 i c} (A-i B) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac {(\cos (3 c)-i \sin (3 c)) \csc ^2(c+d x) (5 (B+3 i A) \sin (2 (c+d x))+9 (7 A-5 i B) \cos (2 (c+d x))-57 A+45 i B)}{15 \sqrt {\tan (c+d x)}}\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(Cos[c + d*x]^4*(((-8*I)*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[S
qrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/(E^((3*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 +
E^((2*I)*(c + d*x)))]) - (Csc[c + d*x]^2*(Cos[3*c] - I*Sin[3*c])*(-57*A + (45*I)*B + 9*(7*A - (5*I)*B)*Cos[2*(
c + d*x)] + 5*((3*I)*A + B)*Sin[2*(c + d*x)]))/(15*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 = 0.85, size = 498, normalized size = 3.46 \[ -\frac {15 \, \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) - 15 \, \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) - {\left ({\left (624 i \, A + 400 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-288 i \, A - 320 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-528 i \, A - 400 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (384 i \, A + 320 \, 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}}}{60 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^(7/2),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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)*(d*e^(2*I*d*x + 2*I*c) + 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)) - 15*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(
4*I*d*x + 4*I*c) + 3*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 + 12
8*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) + 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)) - ((624*I*A + 400*B)*a^3*e^(6*I*d*x + 6*I*c) + (-288*I*A - 32
0*B)*a^3*e^(4*I*d*x + 4*I*c) + (-528*I*A - 400*B)*a^3*e^(2*I*d*x + 2*I*c) + (384*I*A + 320*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^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

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)^(7/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3/tan(d*x + c)^(7/2), x)

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maple [B]  time = 0.11, size = 538, normalized size = 3.74 \[ -\frac {2 a^{3} A}{5 d \tan \left (d x +c \right )^{\frac {5}{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 {2 a^{3} B}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 i a^{3} A}{d \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {8 a^{3} A}{d \sqrt {\tan \left (d x +c \right )}}-\frac {6 i a^{3} B}{d \sqrt {\tan \left (d x +c \right )}}-\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}-\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} 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 {2 a^{3} A \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)^(7/2),x)

[Out]

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

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maxima [A]  time = 0.87, size = 195, normalized size = 1.35 \[ \frac {15 \, {\left (\sqrt {2} {\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3} + \frac {2 \, {\left (15 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} - {\left (15 i \, A + 5 \, B\right )} a^{3} \tan \left (d x + c\right ) - 3 \, A a^{3}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{15 \, d} \]

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)^(7/2),x, algorithm="maxima")

[Out]

1/15*(15*(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)*sqr
t(tan(d*x + c)) + tan(d*x + c) + 1))*a^3 + 2*(15*(4*A - 3*I*B)*a^3*tan(d*x + c)^2 - (15*I*A + 5*B)*a^3*tan(d*x
+ c) - 3*A*a^3)/tan(d*x + c)^(5/2))/d

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mupad [B]  time = 7.39, size = 258, normalized size = 1.79 \[ -\frac {\frac {2\,A\,a^3}{5\,d}-\frac {8\,A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {A\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}-\frac {\frac {2\,B\,a^3}{3\,d}+\frac {B\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,6{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

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)^(7/2),x)

[Out]

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

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

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)**(7/2),x)

[Out]

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

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