3.8.61 \(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [761]

Optimal. Leaf size=140 \[ -\frac {8 a^3 (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {8 a^3 (i A+2 B)}{c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a^3 (i A+5 B) \sqrt {c-i c \tan (e+f x)}}{c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{3/2}}{3 c^3 f} \]

[Out]

8*a^3*(I*A+2*B)/c/f/(c-I*c*tan(f*x+e))^(1/2)+2*a^3*(I*A+5*B)*(c-I*c*tan(f*x+e))^(1/2)/c^2/f-8/3*a^3*(I*A+B)/f/
(c-I*c*tan(f*x+e))^(3/2)-2/3*a^3*B*(c-I*c*tan(f*x+e))^(3/2)/c^3/f

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Rubi [A]
time = 0.15, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {3669, 78} \begin {gather*} \frac {2 a^3 (5 B+i A) \sqrt {c-i c \tan (e+f x)}}{c^2 f}+\frac {8 a^3 (2 B+i A)}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {8 a^3 (B+i A)}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^3 B (c-i c \tan (e+f x))^{3/2}}{3 c^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(-8*a^3*(I*A + B))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (8*a^3*(I*A + 2*B))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) +
(2*a^3*(I*A + 5*B)*Sqrt[c - I*c*Tan[e + f*x]])/(c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(3/2))/(3*c^3*f)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 3669

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^2 (A+B x)}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {4 a^2 (A-i B)}{(c-i c x)^{5/2}}-\frac {4 a^2 (A-2 i B)}{c (c-i c x)^{3/2}}+\frac {a^2 (A-5 i B)}{c^2 \sqrt {c-i c x}}+\frac {i a^2 B \sqrt {c-i c x}}{c^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {8 a^3 (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {8 a^3 (i A+2 B)}{c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a^3 (i A+5 B) \sqrt {c-i c \tan (e+f x)}}{c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{3/2}}{3 c^3 f}\\ \end {align*}

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Mathematica [A]
time = 3.15, size = 168, normalized size = 1.20 \begin {gather*} \frac {a^3 (15 (i A+3 B) \cos (e+f x)+(7 i A+23 B) \cos (3 (e+f x))+2 (9 A-26 i B+(9 A-25 i B) \cos (2 (e+f x))) \sin (e+f x)) (\cos (2 e+5 f x)+i \sin (2 e+5 f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 c^2 f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(a^3*(15*(I*A + 3*B)*Cos[e + f*x] + ((7*I)*A + 23*B)*Cos[3*(e + f*x)] + 2*(9*A - (26*I)*B + (9*A - (25*I)*B)*C
os[2*(e + f*x)])*Sin[e + f*x])*(Cos[2*e + 5*f*x] + I*Sin[2*e + 5*f*x])*(A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e
+ f*x]])/(3*c^2*f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [A]
time = 0.26, size = 118, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 i B c \sqrt {c -i c \tan \left (f x +e \right )}+A c \sqrt {c -i c \tan \left (f x +e \right )}-\frac {4 c^{3} \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (-2 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(118\)
default	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 i B c \sqrt {c -i c \tan \left (f x +e \right )}+A c \sqrt {c -i c \tan \left (f x +e \right )}-\frac {4 c^{3} \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (-2 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 109, normalized size = 0.78 \begin {gather*} \frac {2 i \, {\left (\frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 2 i \, B\right )} a^{3} - {\left (A - i \, B\right )} a^{3} c\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} + \frac {i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B a^{3} + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 5 i \, B\right )} a^{3} c}{c^{2}}\right )}}{3 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*I*(4*(3*(-I*c*tan(f*x + e) + c)*(A - 2*I*B)*a^3 - (A - I*B)*a^3*c)/(-I*c*tan(f*x + e) + c)^(3/2) + (I*(-I*
c*tan(f*x + e) + c)^(3/2)*B*a^3 + 3*sqrt(-I*c*tan(f*x + e) + c)*(A - 5*I*B)*a^3*c)/c^2)/(c*f)

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Fricas [A]
time = 1.64, size = 122, normalized size = 0.87 \begin {gather*} -\frac {2 \, \sqrt {2} {\left ({\left (i \, A + B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, {\left (-i \, A - 3 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 \, {\left (-i \, A - 3 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, A - 3 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-2/3*sqrt(2)*((I*A + B)*a^3*e^(6*I*f*x + 6*I*e) + 3*(-I*A - 3*B)*a^3*e^(4*I*f*x + 4*I*e) + 12*(-I*A - 3*B)*a^3
*e^(2*I*f*x + 2*I*e) + 8*(-I*A - 3*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*
f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int \frac {i A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 A \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{3}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 B \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{4}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i A \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {i B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i B \tan ^{3}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

-I*a**3*(Integral(I*A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + In
tegral(-3*A*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) +
Integral(A*tan(e + f*x)**3/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x
) + Integral(-3*B*tan(e + f*x)**2/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) +
c)), x) + Integral(B*tan(e + f*x)**4/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x)
+ c)), x) + Integral(-3*I*A*tan(e + f*x)**2/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(
e + f*x) + c)), x) + Integral(I*B*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*ta
n(e + f*x) + c)), x) + Integral(-3*I*B*tan(e + f*x)**3/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt
(-I*c*tan(e + f*x) + c)), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^3/(-I*c*tan(f*x + e) + c)^(3/2), x)

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Mupad [B]
time = 10.71, size = 221, normalized size = 1.58 \begin {gather*} \frac {a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,20{}\mathrm {i}+60\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,23{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,2{}\mathrm {i}-A\,\cos \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}+69\,B\,\cos \left (2\,e+2\,f\,x\right )+8\,B\,\cos \left (4\,e+4\,f\,x\right )-B\,\cos \left (6\,e+6\,f\,x\right )-7\,A\,\sin \left (2\,e+2\,f\,x\right )-2\,A\,\sin \left (4\,e+4\,f\,x\right )+A\,\sin \left (6\,e+6\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,21{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,8{}\mathrm {i}-B\,\sin \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}\right )}{3\,c^2\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

(a^3*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*20i + 60*B + A*cos(2*e
+ 2*f*x)*23i + A*cos(4*e + 4*f*x)*2i - A*cos(6*e + 6*f*x)*1i + 69*B*cos(2*e + 2*f*x) + 8*B*cos(4*e + 4*f*x) -
B*cos(6*e + 6*f*x) - 7*A*sin(2*e + 2*f*x) - 2*A*sin(4*e + 4*f*x) + A*sin(6*e + 6*f*x) + B*sin(2*e + 2*f*x)*21
i + B*sin(4*e + 4*f*x)*8i - B*sin(6*e + 6*f*x)*1i))/(3*c^2*f*(cos(2*e + 2*f*x) + 1))

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