3.759 \(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt{c-i c \tan (e+f x)} \, dx\)

Optimal. Leaf size=142 \[ \frac{2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac{8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac{8 a^3 (B+i A) \sqrt{c-i c \tan (e+f x)}}{f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{7/2}}{7 c^3 f} \]

[Out]

(8*a^3*(I*A + B)*Sqrt[c - I*c*Tan[e + f*x]])/f - (8*a^3*(I*A + 2*B)*(c - I*c*Tan[e + f*x])^(3/2))/(3*c*f) + (2
*a^3*(I*A + 5*B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(7/2))/(7*c^3*f)

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Rubi [A]  time = 0.181702, antiderivative size = 142, normalized size of antiderivative = 1., 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} \[ \frac{2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac{8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac{8 a^3 (B+i A) \sqrt{c-i c \tan (e+f x)}}{f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{7/2}}{7 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^3*(I*A + B)*Sqrt[c - I*c*Tan[e + f*x]])/f - (8*a^3*(I*A + 2*B)*(c - I*c*Tan[e + f*x])^(3/2))/(3*c*f) + (2
*a^3*(I*A + 5*B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(7/2))/(7*c^3*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]

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]))))

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \sqrt{c-i c \tan (e+f x)} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^2 (A+B x)}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{4 a^2 (A-i B)}{\sqrt{c-i c x}}-\frac{4 a^2 (A-2 i B) \sqrt{c-i c x}}{c}+\frac{a^2 (A-5 i B) (c-i c x)^{3/2}}{c^2}+\frac{i a^2 B (c-i c x)^{5/2}}{c^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{8 a^3 (i A+B) \sqrt{c-i c \tan (e+f x)}}{f}-\frac{8 a^3 (i A+2 B) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac{2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{5/2}}{5 c^2 f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{7/2}}{7 c^3 f}\\ \end{align*}

Mathematica [A]  time = 6.78404, size = 124, normalized size = 0.87 \[ \frac{a^3 \sec ^2(e+f x) (\cos (3 f x)+i \sin (3 f x)) \sqrt{c-i c \tan (e+f x)} ((-98 A+100 i B) \tan (e+f x)+\cos (2 (e+f x)) ((-98 A+130 i B) \tan (e+f x)+322 i A+290 B)+280 i A+170 B)}{105 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[e + f*x]^2*(Cos[3*f*x] + I*Sin[3*f*x])*Sqrt[c - I*c*Tan[e + f*x]]*((280*I)*A + 170*B + (-98*A + (100*
I)*B)*Tan[e + f*x] + Cos[2*(e + f*x)]*((322*I)*A + 290*B + (-98*A + (130*I)*B)*Tan[e + f*x])))/(105*f*(Cos[f*x
] + I*Sin[f*x])^3)

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Maple [A]  time = 0.085, size = 121, normalized size = 0.9 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{3}} \left ({\frac{i}{7}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}+{\frac{-5\,iBc+Ac}{5} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{-4\, \left ( -iBc+Ac \right ) c+4\,iB{c}^{2}}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+4\, \left ( -iBc+Ac \right ){c}^{2}\sqrt{c-ic\tan \left ( fx+e \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.13744, size = 146, normalized size = 1.03 \begin{align*} \frac{2 i \,{\left (15 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} B a^{3} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}{\left (21 \, A - 105 i \, B\right )} a^{3} c -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (140 \, A - 280 i \, B\right )} a^{3} c^{2} + \sqrt{-i \, c \tan \left (f x + e\right ) + c}{\left (420 \, A - 420 i \, B\right )} a^{3} c^{3}\right )}}{105 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

2/105*I*(15*I*(-I*c*tan(f*x + e) + c)^(7/2)*B*a^3 + (-I*c*tan(f*x + e) + c)^(5/2)*(21*A - 105*I*B)*a^3*c - (-I
*c*tan(f*x + e) + c)^(3/2)*(140*A - 280*I*B)*a^3*c^2 + sqrt(-I*c*tan(f*x + e) + c)*(420*A - 420*I*B)*a^3*c^3)/
(c^3*f)

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Fricas [A]  time = 1.20107, size = 390, normalized size = 2.75 \begin{align*} \frac{\sqrt{2}{\left ({\left (840 i \, A + 840 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (1960 i \, A + 1400 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (1568 i \, A + 1120 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (448 i \, A + 320 \, B\right )} a^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{105 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/105*sqrt(2)*((840*I*A + 840*B)*a^3*e^(6*I*f*x + 6*I*e) + (1960*I*A + 1400*B)*a^3*e^(4*I*f*x + 4*I*e) + (1568
*I*A + 1120*B)*a^3*e^(2*I*f*x + 2*I*e) + (448*I*A + 320*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(6*I*f*
x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e)),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \sqrt{-i \, c \tan \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

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