3.8.58 \(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx\) [758]

Optimal. Leaf size=144 \[ \frac {8 a^3 (i A+B) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {8 a^3 (i A+2 B) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{7/2}}{7 c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{9/2}}{9 c^3 f} \]

[Out]

8/3*a^3*(I*A+B)*(c-I*c*tan(f*x+e))^(3/2)/f-8/5*a^3*(I*A+2*B)*(c-I*c*tan(f*x+e))^(5/2)/c/f+2/7*a^3*(I*A+5*B)*(c
-I*c*tan(f*x+e))^(7/2)/c^2/f-2/9*a^3*B*(c-I*c*tan(f*x+e))^(9/2)/c^3/f

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Rubi [A]
time = 0.14, antiderivative size = 144, 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) (c-i c \tan (e+f x))^{7/2}}{7 c^2 f}-\frac {8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {8 a^3 (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{9/2}}{9 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)*(c - I*c*Tan[e + f*x])^(3/2))/(3*f) - (8*a^3*(I*A + 2*B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*c*f
) + (2*a^3*(I*A + 5*B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(9/2))/(9*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 (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 (a+i a x)^2 (A+B x) \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (4 a^2 (A-i B) \sqrt {c-i c x}-\frac {4 a^2 (A-2 i B) (c-i c x)^{3/2}}{c}+\frac {a^2 (A-5 i B) (c-i c x)^{5/2}}{c^2}+\frac {i a^2 B (c-i c x)^{7/2}}{c^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {8 a^3 (i A+B) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {8 a^3 (i A+2 B) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{7/2}}{7 c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{9/2}}{9 c^3 f}\\ \end {align*}

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Mathematica [A]
time = 3.09, size = 130, normalized size = 0.90 \begin {gather*} -\frac {2 a^3 c \sec ^3(e+f x) (\cos (e-2 f x)-i \sin (e-2 f x)) \sqrt {c-i c \tan (e+f x)} (7 (-12 i A+B)+(81 A-62 i B) \tan (e+f x)+\cos (2 (e+f x)) (-129 i A-113 B+(81 A-97 i B) \tan (e+f x)))}{315 f (\cos (f x)+i \sin (f x))^3} \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]

(-2*a^3*c*Sec[e + f*x]^3*(Cos[e - 2*f*x] - I*Sin[e - 2*f*x])*Sqrt[c - I*c*Tan[e + f*x]]*(7*((-12*I)*A + B) + (
81*A - (62*I)*B)*Tan[e + f*x] + Cos[2*(e + f*x)]*((-129*I)*A - 113*B + (81*A - (97*I)*B)*Tan[e + f*x])))/(315*
f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]
time = 0.46, size = 121, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}+\frac {\left (-5 i B c +A c \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (-4 \left (-i B c +A c \right ) c +4 i B \,c^{2}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 \left (-i B c +A c \right ) c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f \,c^{3}}\)	\(121\)
default	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}+\frac {\left (-5 i B c +A c \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (-4 \left (-i B c +A c \right ) c +4 i B \,c^{2}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {4 \left (-i B c +A c \right ) c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f \,c^{3}}\)	\(121\)

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*(1/9*I*B*(c-I*c*tan(f*x+e))^(9/2)+1/7*(-5*I*B*c+A*c)*(c-I*c*tan(f*x+e))^(7/2)+1/5*(-4*(-I*B*c+A*
c)*c+4*I*B*c^2)*(c-I*c*tan(f*x+e))^(5/2)+4/3*(-I*B*c+A*c)*c^2*(c-I*c*tan(f*x+e))^(3/2))

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Maxima [A]
time = 0.32, size = 108, normalized size = 0.75 \begin {gather*} \frac {2 i \, {\left (35 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} B a^{3} + 45 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} {\left (A - 5 i \, B\right )} a^{3} c - 252 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A - 2 i \, B\right )} a^{3} c^{2} + 420 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a^{3} c^{3}\right )}}{315 \, c^{3} 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]

2/315*I*(35*I*(-I*c*tan(f*x + e) + c)^(9/2)*B*a^3 + 45*(-I*c*tan(f*x + e) + c)^(7/2)*(A - 5*I*B)*a^3*c - 252*(
-I*c*tan(f*x + e) + c)^(5/2)*(A - 2*I*B)*a^3*c^2 + 420*(-I*c*tan(f*x + e) + c)^(3/2)*(A - I*B)*a^3*c^3)/(c^3*f
)

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Fricas [A]
time = 4.96, size = 161, normalized size = 1.12 \begin {gather*} -\frac {16 \, \sqrt {2} {\left (105 \, {\left (-i \, A - B\right )} a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} + 63 \, {\left (-3 i \, A - B\right )} a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} + 36 \, {\left (-3 i \, A - B\right )} a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-3 i \, A - B\right )} a^{3} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{315 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 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]

-16/315*sqrt(2)*(105*(-I*A - B)*a^3*c*e^(6*I*f*x + 6*I*e) + 63*(-3*I*A - B)*a^3*c*e^(4*I*f*x + 4*I*e) + 36*(-3
*I*A - B)*a^3*c*e^(2*I*f*x + 2*I*e) + 8*(-3*I*A - B)*a^3*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(8*I*f*x +
8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int i A c \sqrt {- i c \tan {\left (e + f x \right )} + c}\, dx + \int \left (- 2 A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- 2 A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\right )\, dx + \int \left (- i A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\right )\, dx + \int i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\, dx + \int \left (- i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )}\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*c*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-2*A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e +
f*x), x) + Integral(-2*A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3, x) + Integral(-2*B*c*sqrt(-I*c*tan(e +
f*x) + c)*tan(e + f*x)**2, x) + Integral(-2*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x) + Integral(-I
*A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x) + Integral(I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)
, x) + Integral(-I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**5, 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 = 14.08, size = 335, normalized size = 2.33 \begin {gather*} -\frac {\left (\frac {a^3\,c\,\left (A-B\,1{}\mathrm {i}\right )\,16{}\mathrm {i}}{5\,f}+\frac {a^3\,c\,\left (A-B\,3{}\mathrm {i}\right )\,16{}\mathrm {i}}{5\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\left (\frac {a^3\,c\,\left (A-B\,1{}\mathrm {i}\right )\,16{}\mathrm {i}}{9\,f}-\frac {a^3\,c\,\left (A+B\,1{}\mathrm {i}\right )\,16{}\mathrm {i}}{9\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {\left (\frac {64\,B\,a^3\,c}{7\,f}+\frac {a^3\,c\,\left (A-B\,1{}\mathrm {i}\right )\,16{}\mathrm {i}}{7\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {a^3\,c\,\left (A-B\,1{}\mathrm {i}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{3\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+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*(A - B*1i)*16i)/(7*f) + (64*B*a^3*c)/(7*f))*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*
2i) + 1))^(1/2))/(exp(e*2i + f*x*2i) + 1)^3 - (((a^3*c*(A - B*1i)*16i)/(9*f) - (a^3*c*(A + B*1i)*16i)/(9*f))*(
c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2))/(exp(e*2i + f*x*2i) + 1)^4 - (((a^3*c
*(A - B*1i)*16i)/(5*f) + (a^3*c*(A - B*3i)*16i)/(5*f))*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*
x*2i) + 1))^(1/2))/(exp(e*2i + f*x*2i) + 1)^2 + (a^3*c*(A - B*1i)*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(ex
p(e*2i + f*x*2i) + 1))^(1/2)*16i)/(3*f*(exp(e*2i + f*x*2i) + 1))

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