3.749 \(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=105 \[ -\frac {2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac {4 a^2 (B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{9/2}}{9 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 105, 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 = {3588, 77} \[ -\frac {2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac {4 a^2 (B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{9/2}}{9 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(I*A + B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*f) - (2*a^2*(I*A + 3*B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*c*f
) + (2*a^2*B*(c - I*c*Tan[e + f*x])^(9/2))/(9*c^2*f)

Rule 77

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 3588

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))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int (a+i a x) (A+B x) (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (2 a (A-i B) (c-i c x)^{3/2}-\frac {a (A-3 i B) (c-i c x)^{5/2}}{c}-\frac {i a B (c-i c x)^{7/2}}{c^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {4 a^2 (i A+B) (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (i A+3 B) (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{9/2}}{9 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 7.48, size = 112, normalized size = 1.07 \[ \frac {a^2 c^2 (\sin (2 e)+i \cos (2 e)) \sec ^4(e+f x) \sqrt {c-i c \tan (e+f x)} (5 (13 B+9 i A) \sin (2 (e+f x))+(81 A-61 i B) \cos (2 (e+f x))+81 A+9 i B)}{315 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^2*Sec[e + f*x]^4*(I*Cos[2*e] + Sin[2*e])*(81*A + (9*I)*B + (81*A - (61*I)*B)*Cos[2*(e + f*x)] + 5*((9*I
)*A + 13*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(315*f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [A]  time = 2.69, size = 134, normalized size = 1.28 \[ \frac {\sqrt {2} {\left ({\left (1008 i \, A + 1008 \, B\right )} a^{2} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (1296 i \, A - 144 \, B\right )} a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (288 i \, A - 32 \, B\right )} a^{2} c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{315 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*((1008*I*A + 1008*B)*a^2*c^2*e^(4*I*f*x + 4*I*e) + (1296*I*A - 144*B)*a^2*c^2*e^(2*I*f*x + 2*I*e
) + (288*I*A - 32*B)*a^2*c^2)*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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.30, size = 83, normalized size = 0.79 \[ -\frac {2 i a^{2} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}+\frac {\left (-3 i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}-\frac {2 \left (-i B c +c A \right ) c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

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

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maxima [A]  time = 0.45, size = 78, normalized size = 0.74 \[ -\frac {2 i \, {\left (35 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} B a^{2} + 45 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} {\left (A - 3 i \, B\right )} a^{2} c - 126 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A - i \, B\right )} a^{2} c^{2}\right )}}{315 \, c^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2/315*I*(35*I*(-I*c*tan(f*x + e) + c)^(9/2)*B*a^2 + 45*(-I*c*tan(f*x + e) + c)^(7/2)*(A - 3*I*B)*a^2*c - 126*
(-I*c*tan(f*x + e) + c)^(5/2)*(A - I*B)*a^2*c^2)/(c^2*f)

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mupad [B]  time = 15.34, size = 132, normalized size = 1.26 \[ \frac {16\,a^2\,c^2\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,18{}\mathrm {i}-2\,B+A\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,81{}\mathrm {i}+A\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,63{}\mathrm {i}-9\,B\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+63\,B\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}{315\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4} \]

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)^2*(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

(16*a^2*c^2*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*(A*18i - 2*B + A*exp(e*2i
+ f*x*2i)*81i + A*exp(e*4i + f*x*4i)*63i - 9*B*exp(e*2i + f*x*2i) + 63*B*exp(e*4i + f*x*4i)))/(315*f*(exp(e*2
i + f*x*2i) + 1)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- A c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- 2 A c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- A c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\right )\, dx + \int \left (- B c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- 2 B c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- B c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

-a**2*(Integral(-A*c**2*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-2*A*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e
+ f*x)**2, x) + Integral(-A*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x) + Integral(-B*c**2*sqrt(-I*c
*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(-2*B*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3, x) + Int
egral(-B*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**5, x))

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