\(\int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx\) [18]

Optimal result

Integrand size = 23, antiderivative size = 189 \[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\frac {3 i d^3 x}{8 a f^3}-\frac {3 d (c+d x)^2}{8 a f^2}-\frac {i (c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+i a \tan (e+f x))}-\frac {3 i d^2 (c+d x)}{4 f^3 (a+i a \tan (e+f x))}+\frac {3 d (c+d x)^2}{4 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.47 \[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\left (4 i c^3 f^3+6 c^2 d f^2 (1+2 i f x)+6 c d^2 f \left (-i+2 f x+2 i f^2 x^2\right )+d^3 \left (-3-6 i f x+6 f^2 x^2+4 i f^3 x^3\right )\right ) \cos (2 f x) (\cos (e)-i \sin (e))+2 f^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) (\cos (e)+i \sin (e))+\left (4 c^3 f^3+6 c^2 d f^2 (-i+2 f x)+6 c d^2 f \left (-1-2 i f x+2 f^2 x^2\right )+d^3 \left (3 i-6 f x-6 i f^2 x^2+4 f^3 x^3\right )\right ) (\cos (e)-i \sin (e)) \sin (2 f x)\right )}{16 f^4 (a+i a \tan (e+f x))} \] Input:

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

Output:

(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]))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4206, 3042, 4206, 3042, 4206, 3042, 3960, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d x)^3}{a+i a \tan (e+f x)}dx\)

\(\Big \downarrow \) 4206

\(\displaystyle -\frac {3 i d \int \frac {(c+d x)^2}{i \tan (e+f x) a+a}dx}{2 f}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 i d \int \frac {(c+d x)^2}{i \tan (e+f x) a+a}dx}{2 f}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 4206

\(\displaystyle -\frac {3 i d \left (-\frac {i d \int \frac {c+d x}{i \tan (e+f x) a+a}dx}{f}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 i d \left (-\frac {i d \int \frac {c+d x}{i \tan (e+f x) a+a}dx}{f}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 4206

\(\displaystyle -\frac {3 i d \left (-\frac {i d \left (-\frac {i d \int \frac {1}{i \tan (e+f x) a+a}dx}{2 f}+\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^2}{4 a d}\right )}{f}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 i d \left (-\frac {i d \left (-\frac {i d \int \frac {1}{i \tan (e+f x) a+a}dx}{2 f}+\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^2}{4 a d}\right )}{f}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3960

\(\displaystyle -\frac {3 i d \left (-\frac {i d \left (-\frac {i d \left (\frac {\int 1dx}{2 a}+\frac {i}{2 f (a+i a \tan (e+f x))}\right )}{2 f}+\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^2}{4 a d}\right )}{f}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}+\frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {i (c+d x)^3}{2 f (a+i a \tan (e+f x))}-\frac {3 i d \left (\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}-\frac {i d \left (\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^2}{4 a d}-\frac {i d \left (\frac {x}{2 a}+\frac {i}{2 f (a+i a \tan (e+f x))}\right )}{2 f}\right )}{f}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}+\frac {(c+d x)^4}{8 a d}\)

Input:

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

Output:

(c + d*x)^4/(8*a*d) + ((I/2)*(c + d*x)^3)/(f*(a + I*a*Tan[e + f*x])) - ((( 
3*I)/2)*d*((c + d*x)^3/(6*a*d) + ((I/2)*(c + d*x)^2)/(f*(a + I*a*Tan[e + f 
*x])) - (I*d*((c + d*x)^2/(4*a*d) + ((I/2)*(c + d*x))/(f*(a + I*a*Tan[e + 
f*x])) - ((I/2)*d*(x/(2*a) + (I/2)/(f*(a + I*a*Tan[e + f*x]))))/f))/f))/f
 

Defintions of rubi rules used

rule 24

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3960

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + 
b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[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 4206

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[(c + d*x)^(m + 1)/(2*a*d*(m + 1)), x] + (Simp[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]
 

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {d^{3} x^{4}}{8 a}+\frac {d^{2} c \,x^{3}}{2 a}+\frac {3 d \,c^{2} x^{2}}{4 a}+\frac {c^{3} x}{2 a}+\frac {c^{4}}{8 a d}+\frac {i \left (4 d^{3} x^{3} f^{3}+12 c \,d^{2} f^{3} x^{2}-6 i d^{3} f^{2} x^{2}+12 c^{2} d \,f^{3} x -12 i c \,d^{2} f^{2} x +4 c^{3} f^{3}-6 i c^{2} d \,f^{2}-6 d^{3} f x -6 c \,d^{2} f +3 i d^{3}\right ) {\mathrm e}^{-2 i \left (f x +e \right )}}{16 a \,f^{4}}\)	\(170\)

Input:

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

Output:

1/8/a*d^3*x^4+1/2/a*d^2*c*x^3+3/4/a*d*c^2*x^2+1/2/a*c^3*x+1/8/a/d*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))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\frac {{\left (4 i \, d^{3} f^{3} x^{3} + 4 i \, c^{3} f^{3} + 6 \, c^{2} d f^{2} - 6 i \, c d^{2} f - 3 \, d^{3} - 6 \, {\left (-2 i \, c d^{2} f^{3} - d^{3} f^{2}\right )} x^{2} - 6 \, {\left (-2 i \, c^{2} d f^{3} - 2 \, c d^{2} f^{2} + i \, d^{3} f\right )} x + 2 \, {\left (d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{16 \, a f^{4}} \] Input:

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

Output:

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 - 
6*(-2*I*c*d^2*f^3 - d^3*f^2)*x^2 - 6*(-2*I*c^2*d*f^3 - 2*c*d^2*f^2 + 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)
 

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.37 \[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\begin {cases} \frac {\left (4 i c^{3} f^{3} + 12 i c^{2} d f^{3} x + 6 c^{2} d f^{2} + 12 i c d^{2} f^{3} x^{2} + 12 c d^{2} f^{2} x - 6 i c d^{2} f + 4 i d^{3} f^{3} x^{3} + 6 d^{3} f^{2} x^{2} - 6 i d^{3} f x - 3 d^{3}\right ) e^{- 2 i e} e^{- 2 i f x}}{16 a f^{4}} & \text {for}\: a f^{4} e^{2 i e} \neq 0 \\\frac {c^{3} x e^{- 2 i e}}{2 a} + \frac {3 c^{2} d x^{2} e^{- 2 i e}}{4 a} + \frac {c d^{2} x^{3} e^{- 2 i e}}{2 a} + \frac {d^{3} x^{4} e^{- 2 i e}}{8 a} & \text {otherwise} \end {cases} + \frac {c^{3} x}{2 a} + \frac {3 c^{2} d x^{2}}{4 a} + \frac {c d^{2} x^{3}}{2 a} + \frac {d^{3} x^{4}}{8 a} \] Input:

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

Output:

Piecewise(((4*I*c**3*f**3 + 12*I*c**2*d*f**3*x + 6*c**2*d*f**2 + 12*I*c*d* 
*2*f**3*x**2 + 12*c*d**2*f**2*x - 6*I*c*d**2*f + 4*I*d**3*f**3*x**3 + 6*d* 
*3*f**2*x**2 - 6*I*d**3*f*x - 3*d**3)*exp(-2*I*e)*exp(-2*I*f*x)/(16*a*f**4 
), Ne(a*f**4*exp(2*I*e), 0)), (c**3*x*exp(-2*I*e)/(2*a) + 3*c**2*d*x**2*ex 
p(-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)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\frac {{\left (2 \, d^{3} f^{4} x^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, c d^{2} f^{4} x^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, c^{2} d f^{4} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, d^{3} f^{3} x^{3} + 8 \, c^{3} f^{4} x e^{\left (2 i \, f x + 2 i \, e\right )} + 12 i \, c d^{2} f^{3} x^{2} + 12 i \, c^{2} d f^{3} x + 6 \, d^{3} f^{2} x^{2} + 4 i \, c^{3} f^{3} + 12 \, c d^{2} f^{2} x + 6 \, c^{2} d f^{2} - 6 i \, d^{3} f x - 6 i \, c d^{2} f - 3 \, d^{3}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{16 \, a f^{4}} \] Input:

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

Output:

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)
 

Mupad [B] (verification not implemented)

Time = 8.92 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.24 \[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\frac {8\,c^3\,f^4\,x-3\,d^3\,\cos \left (2\,e+2\,f\,x\right )+4\,c^3\,f^3\,\sin \left (2\,e+2\,f\,x\right )+2\,d^3\,f^4\,x^4+6\,c^2\,d\,f^2\,\cos \left (2\,e+2\,f\,x\right )+12\,c^2\,d\,f^4\,x^2+8\,c\,d^2\,f^4\,x^3+6\,d^3\,f^2\,x^2\,\cos \left (2\,e+2\,f\,x\right )+4\,d^3\,f^3\,x^3\,\sin \left (2\,e+2\,f\,x\right )-6\,c\,d^2\,f\,\sin \left (2\,e+2\,f\,x\right )-6\,d^3\,f\,x\,\sin \left (2\,e+2\,f\,x\right )+12\,c\,d^2\,f^2\,x\,\cos \left (2\,e+2\,f\,x\right )+12\,c^2\,d\,f^3\,x\,\sin \left (2\,e+2\,f\,x\right )+12\,c\,d^2\,f^3\,x^2\,\sin \left (2\,e+2\,f\,x\right )+d^3\,\sin \left (2\,e+2\,f\,x\right )\,3{}\mathrm {i}+c^3\,f^3\,\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-c^2\,d\,f^2\,\sin \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+d^3\,f^3\,x^3\,\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-d^3\,f^2\,x^2\,\sin \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}-c\,d^2\,f\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}-d^3\,f\,x\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+c^2\,d\,f^3\,x\,\cos \left (2\,e+2\,f\,x\right )\,12{}\mathrm {i}-c\,d^2\,f^2\,x\,\sin \left (2\,e+2\,f\,x\right )\,12{}\mathrm {i}+c\,d^2\,f^3\,x^2\,\cos \left (2\,e+2\,f\,x\right )\,12{}\mathrm {i}}{16\,a\,f^4} \] Input:

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

Output:

(d^3*sin(2*e + 2*f*x)*3i - 3*d^3*cos(2*e + 2*f*x) + 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*co 
s(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)*1 
2i + 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)
 

Reduce [F]

\[ \int \frac {(c+d x)^3}{a+i a \tan (e+f x)} \, dx=\frac {\left (\int \frac {x^{3}}{\tan \left (f x +e \right ) i +1}d x \right ) d^{3}+3 \left (\int \frac {x^{2}}{\tan \left (f x +e \right ) i +1}d x \right ) c \,d^{2}+3 \left (\int \frac {x}{\tan \left (f x +e \right ) i +1}d x \right ) c^{2} d +\left (\int \frac {1}{\tan \left (f x +e \right ) i +1}d x \right ) c^{3}}{a} \] Input:

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

Output:

(int(x**3/(tan(e + f*x)*i + 1),x)*d**3 + 3*int(x**2/(tan(e + f*x)*i + 1),x 
)*c*d**2 + 3*int(x/(tan(e + f*x)*i + 1),x)*c**2*d + int(1/(tan(e + f*x)*i 
+ 1),x)*c**3)/a