### 3.18 $$\int \frac{(c+d x)^3}{a+i a \tan (e+f x)} \, dx$$

Optimal. Leaf size=189 $-\frac{3 i d^2 (c+d x)}{4 f^3 (a+i a \tan (e+f x))}+\frac{3 d (c+d x)^2}{4 f^2 (a+i a \tan (e+f x))}+\frac{i (c+d x)^3}{2 f (a+i a \tan (e+f x))}-\frac{3 d (c+d x)^2}{8 a f^2}-\frac{i (c+d x)^3}{4 a f}+\frac{(c+d x)^4}{8 a d}-\frac{3 d^3}{8 f^4 (a+i a \tan (e+f x))}+\frac{3 i d^3 x}{8 a f^3}$

[Out]

(((3*I)/8)*d^3*x)/(a*f^3) - (3*d*(c + d*x)^2)/(8*a*f^2) - ((I/4)*(c + d*x)^3)/(a*f) + (c + d*x)^4/(8*a*d) - (3
*d^3)/(8*f^4*(a + I*a*Tan[e + f*x])) - (((3*I)/4)*d^2*(c + d*x))/(f^3*(a + I*a*Tan[e + f*x])) + (3*d*(c + d*x)
^2)/(4*f^2*(a + I*a*Tan[e + f*x])) + ((I/2)*(c + d*x)^3)/(f*(a + I*a*Tan[e + f*x]))

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Rubi [A]  time = 0.198783, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3723, 3479, 8} $-\frac{3 i d^2 (c+d x)}{4 f^3 (a+i a \tan (e+f x))}+\frac{3 d (c+d x)^2}{4 f^2 (a+i a \tan (e+f x))}+\frac{i (c+d x)^3}{2 f (a+i a \tan (e+f x))}-\frac{3 d (c+d x)^2}{8 a f^2}-\frac{i (c+d x)^3}{4 a f}+\frac{(c+d x)^4}{8 a d}-\frac{3 d^3}{8 f^4 (a+i a \tan (e+f x))}+\frac{3 i d^3 x}{8 a f^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c + d*x)^3/(a + I*a*Tan[e + f*x]),x]

[Out]

(((3*I)/8)*d^3*x)/(a*f^3) - (3*d*(c + d*x)^2)/(8*a*f^2) - ((I/4)*(c + d*x)^3)/(a*f) + (c + d*x)^4/(8*a*d) - (3
*d^3)/(8*f^4*(a + I*a*Tan[e + f*x])) - (((3*I)/4)*d^2*(c + d*x))/(f^3*(a + I*a*Tan[e + f*x])) + (3*d*(c + d*x)
^2)/(4*f^2*(a + I*a*Tan[e + f*x])) + ((I/2)*(c + d*x)^3)/(f*(a + I*a*Tan[e + f*x]))

Rule 3723

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

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{(c+d x)^3}{a+i a \tan (e+f x)} \, dx &=\frac{(c+d x)^4}{8 a d}+\frac{i (c+d x)^3}{2 f (a+i a \tan (e+f x))}-\frac{(3 i d) \int \frac{(c+d x)^2}{a+i a \tan (e+f x)} \, dx}{2 f}\\ &=-\frac{i (c+d x)^3}{4 a f}+\frac{(c+d x)^4}{8 a d}+\frac{3 d (c+d x)^2}{4 f^2 (a+i a \tan (e+f x))}+\frac{i (c+d x)^3}{2 f (a+i a \tan (e+f x))}-\frac{\left (3 d^2\right ) \int \frac{c+d x}{a+i a \tan (e+f x)} \, dx}{2 f^2}\\ &=-\frac{3 d (c+d x)^2}{8 a f^2}-\frac{i (c+d x)^3}{4 a f}+\frac{(c+d x)^4}{8 a d}-\frac{3 i d^2 (c+d x)}{4 f^3 (a+i a \tan (e+f x))}+\frac{3 d (c+d x)^2}{4 f^2 (a+i a \tan (e+f x))}+\frac{i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac{\left (3 i d^3\right ) \int \frac{1}{a+i a \tan (e+f x)} \, dx}{4 f^3}\\ &=-\frac{3 d (c+d x)^2}{8 a f^2}-\frac{i (c+d x)^3}{4 a f}+\frac{(c+d x)^4}{8 a d}-\frac{3 d^3}{8 f^4 (a+i a \tan (e+f x))}-\frac{3 i d^2 (c+d x)}{4 f^3 (a+i a \tan (e+f x))}+\frac{3 d (c+d x)^2}{4 f^2 (a+i a \tan (e+f x))}+\frac{i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac{\left (3 i d^3\right ) \int 1 \, dx}{8 a f^3}\\ &=\frac{3 i d^3 x}{8 a f^3}-\frac{3 d (c+d x)^2}{8 a f^2}-\frac{i (c+d x)^3}{4 a f}+\frac{(c+d x)^4}{8 a d}-\frac{3 d^3}{8 f^4 (a+i a \tan (e+f x))}-\frac{3 i d^2 (c+d x)}{4 f^3 (a+i a \tan (e+f x))}+\frac{3 d (c+d x)^2}{4 f^2 (a+i a \tan (e+f x))}+\frac{i (c+d x)^3}{2 f (a+i a \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.627074, size = 278, normalized size = 1.47 $\frac{\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (2 f^4 x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right ) (\cos (e)+i \sin (e))+(\cos (e)-i \sin (e)) \cos (2 f x) \left (6 c^2 d f^2 (1+2 i f x)+4 i c^3 f^3+6 c d^2 f \left (2 i f^2 x^2+2 f x-i\right )+d^3 \left (4 i f^3 x^3+6 f^2 x^2-6 i f x-3\right )\right )+(\cos (e)-i \sin (e)) \sin (2 f x) \left (6 c^2 d f^2 (2 f x-i)+4 c^3 f^3+6 c d^2 f \left (2 f^2 x^2-2 i f x-1\right )+d^3 \left (4 f^3 x^3-6 i f^2 x^2-6 f x+3 i\right )\right )\right )}{16 f^4 (a+i a \tan (e+f x))}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c + d*x)^3/(a + I*a*Tan[e + f*x]),x]

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*(((4*I)*c^3*f^3 + 6*c^2*d*f^2*(1 + (2*I)*f*x) + 6*c*d^2*f*(-I + 2*f*x +
(2*I)*f^2*x^2) + d^3*(-3 - (6*I)*f*x + 6*f^2*x^2 + (4*I)*f^3*x^3))*Cos[2*f*x]*(Cos[e] - I*Sin[e]) + 2*f^4*x*(4
*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*(Cos[e] + I*Sin[e]) + (4*c^3*f^3 + 6*c^2*d*f^2*(-I + 2*f*x) + 6*c*d^
2*f*(-1 - (2*I)*f*x + 2*f^2*x^2) + d^3*(3*I - 6*f*x - (6*I)*f^2*x^2 + 4*f^3*x^3))*(Cos[e] - I*Sin[e])*Sin[2*f*
x]))/(16*f^4*(a + I*a*Tan[e + f*x]))

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Maple [A]  time = 0.194, size = 159, normalized size = 0.8 \begin{align*}{\frac{{d}^{3}{x}^{4}}{8\,a}}+{\frac{c{d}^{2}{x}^{3}}{2\,a}}+{\frac{3\,{c}^{2}d{x}^{2}}{4\,a}}+{\frac{{c}^{3}x}{2\,a}}+{\frac{{\frac{i}{16}} \left ( 4\,{d}^{3}{x}^{3}{f}^{3}-6\,i{d}^{3}{f}^{2}{x}^{2}+12\,c{d}^{2}{f}^{3}{x}^{2}-12\,ic{d}^{2}{f}^{2}x+12\,{c}^{2}d{f}^{3}x-6\,i{c}^{2}d{f}^{2}+4\,{c}^{3}{f}^{3}-6\,{d}^{3}fx+3\,i{d}^{3}-6\,c{d}^{2}f \right ){{\rm e}^{-2\,i \left ( fx+e \right ) }}}{a{f}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3/(a+I*a*tan(f*x+e)),x)

[Out]

1/8/a*d^3*x^4+1/2/a*c*d^2*x^3+3/4/a*c^2*d*x^2+1/2/a*c^3*x+1/16*I*(4*d^3*x^3*f^3-6*I*d^3*f^2*x^2+12*c*d^2*f^3*x
^2-12*I*c*d^2*f^2*x+12*c^2*d*f^3*x-6*I*c^2*d*f^2+4*c^3*f^3-6*d^3*f*x+3*I*d^3-6*c*d^2*f)/a/f^4*exp(-2*I*(f*x+e)
)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.51438, size = 370, normalized size = 1.96 \begin{align*} \frac{{\left (4 i \, d^{3} f^{3} x^{3} + 4 i \, c^{3} f^{3} + 6 \, c^{2} d f^{2} - 6 i \, c d^{2} f - 3 \, d^{3} +{\left (12 i \, c d^{2} f^{3} + 6 \, d^{3} f^{2}\right )} x^{2} +{\left (12 i \, c^{2} d f^{3} + 12 \, c d^{2} f^{2} - 6 i \, d^{3} f\right )} x + 2 \,{\left (d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{16 \, a f^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*I*d^3*f^3*x^3 + 4*I*c^3*f^3 + 6*c^2*d*f^2 - 6*I*c*d^2*f - 3*d^3 + (12*I*c*d^2*f^3 + 6*d^3*f^2)*x^2 + (
12*I*c^2*d*f^3 + 12*c*d^2*f^2 - 6*I*d^3*f)*x + 2*(d^3*f^4*x^4 + 4*c*d^2*f^4*x^3 + 6*c^2*d*f^4*x^2 + 4*c^3*f^4*
x)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*f^4)

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Sympy [A]  time = 0.929476, size = 372, normalized size = 1.97 \begin{align*} \begin{cases} \frac{\left (4 i a^{3} c^{3} f^{9} e^{6 i e} + 12 i a^{3} c^{2} d f^{9} x e^{6 i e} + 6 a^{3} c^{2} d f^{8} e^{6 i e} + 12 i a^{3} c d^{2} f^{9} x^{2} e^{6 i e} + 12 a^{3} c d^{2} f^{8} x e^{6 i e} - 6 i a^{3} c d^{2} f^{7} e^{6 i e} + 4 i a^{3} d^{3} f^{9} x^{3} e^{6 i e} + 6 a^{3} d^{3} f^{8} x^{2} e^{6 i e} - 6 i a^{3} d^{3} f^{7} x e^{6 i e} - 3 a^{3} d^{3} f^{6} e^{6 i e}\right ) e^{- 8 i e} e^{- 2 i f x}}{16 a^{4} f^{10}} & \text{for}\: 16 a^{4} f^{10} e^{8 i e} \neq 0 \\\frac{c^{3} x e^{- 2 i e}}{2 a} + \frac{3 c^{2} d x^{2} e^{- 2 i e}}{4 a} + \frac{c d^{2} x^{3} e^{- 2 i e}}{2 a} + \frac{d^{3} x^{4} e^{- 2 i e}}{8 a} & \text{otherwise} \end{cases} + \frac{c^{3} x}{2 a} + \frac{3 c^{2} d x^{2}}{4 a} + \frac{c d^{2} x^{3}}{2 a} + \frac{d^{3} x^{4}}{8 a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3/(a+I*a*tan(f*x+e)),x)

[Out]

Piecewise(((4*I*a**3*c**3*f**9*exp(6*I*e) + 12*I*a**3*c**2*d*f**9*x*exp(6*I*e) + 6*a**3*c**2*d*f**8*exp(6*I*e)
+ 12*I*a**3*c*d**2*f**9*x**2*exp(6*I*e) + 12*a**3*c*d**2*f**8*x*exp(6*I*e) - 6*I*a**3*c*d**2*f**7*exp(6*I*e)
+ 4*I*a**3*d**3*f**9*x**3*exp(6*I*e) + 6*a**3*d**3*f**8*x**2*exp(6*I*e) - 6*I*a**3*d**3*f**7*x*exp(6*I*e) - 3*
a**3*d**3*f**6*exp(6*I*e))*exp(-8*I*e)*exp(-2*I*f*x)/(16*a**4*f**10), Ne(16*a**4*f**10*exp(8*I*e), 0)), (c**3*
x*exp(-2*I*e)/(2*a) + 3*c**2*d*x**2*exp(-2*I*e)/(4*a) + c*d**2*x**3*exp(-2*I*e)/(2*a) + d**3*x**4*exp(-2*I*e)/
(8*a), True)) + c**3*x/(2*a) + 3*c**2*d*x**2/(4*a) + c*d**2*x**3/(2*a) + d**3*x**4/(8*a)

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Giac [A]  time = 1.19524, size = 261, normalized size = 1.38 \begin{align*} \frac{{\left (2 \, d^{3} f^{4} x^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, c d^{2} f^{4} x^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, c^{2} d f^{4} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, d^{3} f^{3} x^{3} + 8 \, c^{3} f^{4} x e^{\left (2 i \, f x + 2 i \, e\right )} + 12 i \, c d^{2} f^{3} x^{2} + 12 i \, c^{2} d f^{3} x + 6 \, d^{3} f^{2} x^{2} + 4 i \, c^{3} f^{3} + 12 \, c d^{2} f^{2} x + 6 \, c^{2} d f^{2} - 6 i \, d^{3} f x - 6 i \, c d^{2} f - 3 \, d^{3}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{16 \, a f^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*d^3*f^4*x^4*e^(2*I*f*x + 2*I*e) + 8*c*d^2*f^4*x^3*e^(2*I*f*x + 2*I*e) + 12*c^2*d*f^4*x^2*e^(2*I*f*x +
2*I*e) + 4*I*d^3*f^3*x^3 + 8*c^3*f^4*x*e^(2*I*f*x + 2*I*e) + 12*I*c*d^2*f^3*x^2 + 12*I*c^2*d*f^3*x + 6*d^3*f^2
*x^2 + 4*I*c^3*f^3 + 12*c*d^2*f^2*x + 6*c^2*d*f^2 - 6*I*d^3*f*x - 6*I*c*d^2*f - 3*d^3)*e^(-2*I*f*x - 2*I*e)/(a
*f^4)