3.16 \(\int \frac {(c+d x)^3}{a+i a \cot (e+f x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3723, 3479, 8} \[ \frac {3 i d^2 (c+d x)}{4 f^3 (a+i a \cot (e+f x))}+\frac {3 d (c+d x)^2}{4 f^2 (a+i a \cot (e+f x))}-\frac {i (c+d x)^3}{2 f (a+i a \cot (e+f x))}-\frac {3 d (c+d x)^2}{8 a f^2}+\frac {i (c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+i a \cot (e+f x))}-\frac {3 i d^3 x}{8 a f^3} \]

Antiderivative was successfully verified.

[In]

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

Rule 8

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.63, size = 246, normalized size = 1.30 \[ \frac {i (\cos (2 e)+i \sin (2 e)) \cos (2 f x) \left (4 c^3 f^3+6 c^2 d f^2 (2 f x+i)+6 c d^2 f \left (2 f^2 x^2+2 i f x-1\right )+d^3 \left (4 f^3 x^3+6 i f^2 x^2-6 f x-3 i\right )\right )-(\cos (2 e)+i \sin (2 e)) \sin (2 f x) \left (4 c^3 f^3+6 c^2 d f^2 (2 f x+i)+6 c d^2 f \left (2 f^2 x^2+2 i f x-1\right )+d^3 \left (4 f^3 x^3+6 i f^2 x^2-6 f x-3 i\right )\right )+2 f^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )}{16 a f^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.76, size = 153, normalized size = 0.81 \[ \frac {2 \, d^{3} f^{4} x^{4} + 8 \, c d^{2} f^{4} x^{3} + 12 \, c^{2} d f^{4} x^{2} + 8 \, c^{3} f^{4} x + {\left (4 i \, d^{3} f^{3} x^{3} + 4 i \, c^{3} f^{3} - 6 \, c^{2} d f^{2} - 6 i \, c d^{2} f + 3 \, d^{3} + {\left (12 i \, c d^{2} f^{3} - 6 \, d^{3} f^{2}\right )} x^{2} + {\left (12 i \, c^{2} d f^{3} - 12 \, c d^{2} f^{2} - 6 i \, d^{3} f\right )} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{16 \, a f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 243, normalized size = 1.29 \[ \frac {2 \, d^{3} f^{4} x^{4} + 8 \, c d^{2} f^{4} x^{3} + 4 i \, d^{3} f^{3} x^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, c^{2} d f^{4} x^{2} + 12 i \, c d^{2} f^{3} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, c^{3} f^{4} x + 12 i \, c^{2} d f^{3} x e^{\left (2 i \, f x + 2 i \, e\right )} - 6 \, d^{3} f^{2} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c^{3} f^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 \, c d^{2} f^{2} x e^{\left (2 i \, f x + 2 i \, e\right )} - 6 \, c^{2} d f^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 6 i \, d^{3} f x e^{\left (2 i \, f x + 2 i \, e\right )} - 6 i \, c d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 3 \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )}}{16 \, a f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 2.07, size = 170, normalized size = 0.90 \[ \frac {d^{3} x^{4}}{8 a}+\frac {d^{2} c \,x^{3}}{2 a}+\frac {3 d \,c^{2} x^{2}}{4 a}+\frac {c^{3} x}{2 a}+\frac {c^{4}}{8 d a}+\frac {i \left (4 d^{3} x^{3} f^{3}+12 c \,d^{2} f^{3} x^{2}+6 i d^{3} f^{2} x^{2}+12 c^{2} d \,f^{3} x +12 i c \,d^{2} f^{2} x +4 c^{3} f^{3}+6 i c^{2} d \,f^{2}-6 d^{3} f x -6 c \,d^{2} f -3 i d^{3}\right ) {\mathrm e}^{2 i \left (f x +e \right )}}{16 a \,f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*d^3/a*x^4+1/2*d^2/a*c*x^3+3/4*d/a*c^2*x^2+1/2*c^3*x/a+1/8/d/a*c^4+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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 1.24, size = 423, normalized size = 2.24 \[ \frac {3\,d^3\,\cos \left (2\,e+2\,f\,x\right )+8\,c^3\,f^4\,x-4\,c^3\,f^3\,\sin \left (2\,e+2\,f\,x\right )+2\,d^3\,f^4\,x^4-6\,c^2\,d\,f^2\,\cos \left (2\,e+2\,f\,x\right )+12\,c^2\,d\,f^4\,x^2+8\,c\,d^2\,f^4\,x^3-6\,d^3\,f^2\,x^2\,\cos \left (2\,e+2\,f\,x\right )-4\,d^3\,f^3\,x^3\,\sin \left (2\,e+2\,f\,x\right )+6\,c\,d^2\,f\,\sin \left (2\,e+2\,f\,x\right )+6\,d^3\,f\,x\,\sin \left (2\,e+2\,f\,x\right )-12\,c\,d^2\,f^2\,x\,\cos \left (2\,e+2\,f\,x\right )-12\,c^2\,d\,f^3\,x\,\sin \left (2\,e+2\,f\,x\right )-12\,c\,d^2\,f^3\,x^2\,\sin \left (2\,e+2\,f\,x\right )+d^3\,\sin \left (2\,e+2\,f\,x\right )\,3{}\mathrm {i}+c^3\,f^3\,\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-c^2\,d\,f^2\,\sin \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+d^3\,f^3\,x^3\,\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-d^3\,f^2\,x^2\,\sin \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}-c\,d^2\,f\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}-d^3\,f\,x\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+c^2\,d\,f^3\,x\,\cos \left (2\,e+2\,f\,x\right )\,12{}\mathrm {i}-c\,d^2\,f^2\,x\,\sin \left (2\,e+2\,f\,x\right )\,12{}\mathrm {i}+c\,d^2\,f^3\,x^2\,\cos \left (2\,e+2\,f\,x\right )\,12{}\mathrm {i}}{16\,a\,f^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/(a + a*cot(e + f*x)*1i),x)

[Out]

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

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sympy [A]  time = 0.37, size = 318, normalized size = 1.68 \[ \begin {cases} - \frac {\left (- 4 i c^{3} f^{3} e^{2 i e} - 12 i c^{2} d f^{3} x e^{2 i e} + 6 c^{2} d f^{2} e^{2 i e} - 12 i c d^{2} f^{3} x^{2} e^{2 i e} + 12 c d^{2} f^{2} x e^{2 i e} + 6 i c d^{2} f e^{2 i e} - 4 i d^{3} f^{3} x^{3} e^{2 i e} + 6 d^{3} f^{2} x^{2} e^{2 i e} + 6 i d^{3} f x e^{2 i e} - 3 d^{3} e^{2 i e}\right ) e^{2 i f x}}{16 a f^{4}} & \text {for}\: 16 a f^{4} \neq 0 \\- \frac {c^{3} x e^{2 i e}}{2 a} - \frac {3 c^{2} d x^{2} e^{2 i e}}{4 a} - \frac {c d^{2} x^{3} e^{2 i e}}{2 a} - \frac {d^{3} x^{4} e^{2 i e}}{8 a} & \text {otherwise} \end {cases} + \frac {c^{3} x}{2 a} + \frac {3 c^{2} d x^{2}}{4 a} + \frac {c d^{2} x^{3}}{2 a} + \frac {d^{3} x^{4}}{8 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(-4*I*c**3*f**3*exp(2*I*e) - 12*I*c**2*d*f**3*x*exp(2*I*e) + 6*c**2*d*f**2*exp(2*I*e) - 12*I*c*d**
2*f**3*x**2*exp(2*I*e) + 12*c*d**2*f**2*x*exp(2*I*e) + 6*I*c*d**2*f*exp(2*I*e) - 4*I*d**3*f**3*x**3*exp(2*I*e)
+ 6*d**3*f**2*x**2*exp(2*I*e) + 6*I*d**3*f*x*exp(2*I*e) - 3*d**3*exp(2*I*e))*exp(2*I*f*x)/(16*a*f**4), Ne(16*
a*f**4, 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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