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

Optimal. Leaf size=103 \[ \frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3591, 3529, 3533, 205} \[ \frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 4.26, size = 265, normalized size = 2.57 \[ \frac {\cos ^2(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \left (-\frac {2 i e^{-i c} (A-i B) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac {(\cos (c)-i \sin (c)) \csc ^2(c+d x) (5 (B+i A) \sin (2 (c+d x))+3 (6 A-5 i B) \cos (2 (c+d x))-12 A+15 i B)}{15 \sqrt {\tan (c+d x)}}\right )}{d (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^2*(Cos[d*x] - I*Sin[d*x])*(((-2*I)*(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)))]])/(E^(I*c)*Sqrt[(-1 + E^((2*
I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]) - (Csc[c + d*x]^2*(Cos[c] - I*Sin[c])*(-12*A + (15*I)*B + 3*(6*A -
(5*I)*B)*Cos[2*(c + d*x)] + 5*(I*A + B)*Sin[2*(c + d*x)]))/(15*Sqrt[Tan[c + d*x]]))*(a + I*a*Tan[c + d*x])*(A
+ B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 0.79, size = 482, normalized size = 4.68 \[ -\frac {15 \, {\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 )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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 (i \, A + B\right )} a}\right ) - 15 \, {\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 )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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 (i \, A + B\right )} a}\right ) - {\left ({\left (184 i \, A + 160 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-8 i \, A - 80 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-88 i \, A - 160 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (104 i \, A + 80 \, B\right )} a\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))*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/60*(15*(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)*sqrt((4*I*A^2 + 8*A*
B - 4*I*B^2)*a^2/d^2)*log((2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4*I*A^2 + 8*A
*B - 4*I*B^2)*a^2/d^2)*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)/((I*
A + B)*a)) - 15*(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)*sqrt((4*I*A^2
+ 8*A*B - 4*I*B^2)*a^2/d^2)*log((2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4*I*A^2
+ 8*A*B - 4*I*B^2)*a^2/d^2)*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
)/((I*A + B)*a)) - ((184*I*A + 160*B)*a*e^(6*I*d*x + 6*I*c) + (-8*I*A - 80*B)*a*e^(4*I*d*x + 4*I*c) + (-88*I*A
- 160*B)*a*e^(2*I*d*x + 2*I*c) + (104*I*A + 80*B)*a)*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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.11, size = 505, normalized size = 4.90 \[ -\frac {2 a A}{5 d \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {i a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {2 a A}{d \sqrt {\tan \left (d x +c \right )}}-\frac {i a 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 )}{4 d}-\frac {2 a B}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {i a 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 )}{4 d}-\frac {i a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a 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 )}{4 d}-\frac {a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {2 i a B}{d \sqrt {\tan \left (d x +c \right )}}-\frac {2 i a A}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {i a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a 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 )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.87, size = 187, normalized size = 1.82 \[ -\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a - \frac {8 \, {\left (15 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )^{2} - {\left (5 i \, A + 5 \, B\right )} a \tan \left (d x + c\right ) - 3 \, A a\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

-1/60*(15*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*
((I - 1)*A + (I + 1)*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)*sqrt(tan
(d*x + c)) + tan(d*x + c) + 1))*a - 8*(15*(A - I*B)*a*tan(d*x + c)^2 - (5*I*A + 5*B)*a*tan(d*x + c) - 3*A*a)/t
an(d*x + c)^(5/2))/d

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mupad [B]  time = 8.74, size = 123, normalized size = 1.19 \[ -\frac {\frac {2\,B\,a}{3\,d}+\frac {B\,a\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,B\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d}-\frac {2\,A\,a\,\left (15\,{\left (-1\right )}^{1/4}\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )-15\,{\mathrm {tan}\left (c+d\,x\right )}^2+3+\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}\right )}{15\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \]

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))/tan(c + d*x)^(7/2),x)

[Out]

- ((2*B*a)/(3*d) + (B*a*tan(c + d*x)*2i)/d)/tan(c + d*x)^(3/2) - (2^(1/2)*B*a*atan(2^(1/2)*tan(c + d*x)^(1/2)*
(1/2 - 1i/2))*(1 + 1i))/d - (2*A*a*(tan(c + d*x)*5i - 15*tan(c + d*x)^2 + 15*(-1)^(1/4)*tan(c + d*x)^(5/2)*ata
nh((-1)^(1/4)*tan(c + d*x)^(1/2)) + 3))/(15*d*tan(c + d*x)^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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