\(\int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx\) [1083]

Optimal result

Integrand size = 28, antiderivative size = 115 \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=\frac {4 a^3 x}{c-i d}-\frac {a^3 (i c-3 d) \log (\cos (e+f x))}{d^2 f}-\frac {a^3 (c+i d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 (i c+d) f}-\frac {a^3+i a^3 \tan (e+f x)}{d f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3637, 3670, 3556, 3612, 3611} \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=-\frac {a^3 (-3 d+i c) \log (\cos (e+f x))}{d^2 f}-\frac {a^3 (c+i d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 f (d+i c)}+\frac {4 a^3 x}{c-i d}-\frac {a^3+i a^3 \tan (e+f x)}{d f} \]

[In]

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

[Out]

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

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3612

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

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3670

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[B*(d/b), Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^3+i a^3 \tan (e+f x)}{d f}+\frac {a \int \frac {(a+i a \tan (e+f x)) (a (i c+d)+a (c+3 i d) \tan (e+f x))}{c+d \tan (e+f x)} \, dx}{d} \\ & = -\frac {a^3+i a^3 \tan (e+f x)}{d f}+\frac {a \int \frac {a^2 d (i c+d)-a^2 \left (i c^2-3 c d-4 i d^2\right ) \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{d^2}+\frac {\left (a^3 (i c-3 d)\right ) \int \tan (e+f x) \, dx}{d^2} \\ & = \frac {4 a^3 x}{c-i d}-\frac {a^3 (i c-3 d) \log (\cos (e+f x))}{d^2 f}-\frac {a^3+i a^3 \tan (e+f x)}{d f}-\frac {\left (a^3 (c+i d)^2\right ) \int \frac {d-c \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{d^2 (i c+d)} \\ & = \frac {4 a^3 x}{c-i d}-\frac {a^3 (i c-3 d) \log (\cos (e+f x))}{d^2 f}-\frac {a^3 (c+i d)^2 \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 (i c+d) f}-\frac {a^3+i a^3 \tan (e+f x)}{d f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=-\frac {a^3 \left (-d^2+8 d^2 \log (i+\tan (e+f x))+2 c^2 \log (c+d \tan (e+f x))+4 i c d \log (c+d \tan (e+f x))-2 d^2 \log (c+d \tan (e+f x))-2 (c-i d) d \tan (e+f x)\right )}{2 d^2 (i c+d) f} \]

[In]

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

[Out]

-1/2*(a^3*(-d^2 + 8*d^2*Log[I + Tan[e + f*x]] + 2*c^2*Log[c + d*Tan[e + f*x]] + (4*I)*c*d*Log[c + d*Tan[e + f*
x]] - 2*d^2*Log[c + d*Tan[e + f*x]] - 2*(c - I*d)*d*Tan[e + f*x]))/(d^2*(I*c + d)*f)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93

method	result	size
norman	\(\frac {4 a^{3} x}{-i d +c}-\frac {i a^{3} \tan \left (f x +e \right )}{d f}+\frac {i a^{3} \left (2 i c d +c^{2}-d^{2}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{d^{2} f \left (-i d +c \right )}+\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f \left (-i d +c \right )}\)	\(107\)
derivativedivides	\(\frac {a^{3} \left (-\frac {i \tan \left (f x +e \right )}{d}+\frac {\left (i c^{3}-3 i c \,d^{2}-3 c^{2} d +d^{3}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{d^{2} \left (c^{2}+d^{2}\right )}+\frac {\frac {\left (4 i c -4 d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (4 i d +4 c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{c^{2}+d^{2}}\right )}{f}\)	\(116\)
default	\(\frac {a^{3} \left (-\frac {i \tan \left (f x +e \right )}{d}+\frac {\left (i c^{3}-3 i c \,d^{2}-3 c^{2} d +d^{3}\right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{d^{2} \left (c^{2}+d^{2}\right )}+\frac {\frac {\left (4 i c -4 d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (4 i d +4 c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{c^{2}+d^{2}}\right )}{f}\)	\(116\)
parallelrisch	\(\frac {4 i x \,a^{3} d^{3} f +2 i \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a^{3} c \,d^{2}+i \ln \left (c +d \tan \left (f x +e \right )\right ) a^{3} c^{3}-3 i \ln \left (c +d \tan \left (f x +e \right )\right ) a^{3} c \,d^{2}-i \tan \left (f x +e \right ) a^{3} c^{2} d -i \tan \left (f x +e \right ) a^{3} d^{3}+4 x \,a^{3} c \,d^{2} f -2 \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a^{3} d^{3}-3 \ln \left (c +d \tan \left (f x +e \right )\right ) a^{3} c^{2} d +\ln \left (c +d \tan \left (f x +e \right )\right ) a^{3} d^{3}}{d^{2} f \left (c^{2}+d^{2}\right )}\)	\(190\)
risch	\(-\frac {8 a^{3} x}{i d -c}-\frac {6 i a^{3} x}{d}-\frac {i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) c}{d^{2} f}-\frac {2 a^{3} c x}{d^{2}}-\frac {2 a^{3} c e}{d^{2} f}-\frac {4 a^{3} c x}{d \left (i c +d \right )}-\frac {4 a^{3} c e}{d f \left (i c +d \right )}+\frac {2 i a^{3} c^{2} e}{d^{2} f \left (i c +d \right )}-\frac {2 i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) c}{d f \left (i c +d \right )}-\frac {2 i a^{3} e}{f \left (i c +d \right )}+\frac {2 i a^{3} c^{2} x}{d^{2} \left (i c +d \right )}+\frac {2 a^{3}}{f d \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{d f}-\frac {2 i a^{3} x}{i c +d}-\frac {6 i a^{3} e}{d f}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) c^{2}}{d^{2} f \left (i c +d \right )}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right )}{f \left (i c +d \right )}\)	\(398\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.83 \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=\frac {2 i \, a^{3} c d + 2 \, a^{3} d^{2} - {\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} d^{2} + {\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right ) + {\left (a^{3} c^{2} + 2 i \, a^{3} c d + 3 \, a^{3} d^{2} + {\left (a^{3} c^{2} + 2 i \, a^{3} c d + 3 \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{{\left (i \, c d^{2} + d^{3}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c d^{2} + d^{3}\right )} f} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (94) = 188\).

Time = 8.69 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.23 \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=\frac {2 a^{3}}{d f e^{2 i e} e^{2 i f x} + d f} - \frac {i a^{3} \left (c + 3 i d\right ) \log {\left (e^{2 i f x} + \frac {a^{3} c^{2} + 3 i a^{3} c d - 2 a^{3} d^{2} - i a^{3} d \left (c + 3 i d\right )}{a^{3} c^{2} e^{2 i e} + 2 i a^{3} c d e^{2 i e} + a^{3} d^{2} e^{2 i e}} \right )}}{d^{2} f} + \frac {i a^{3} \left (c + i d\right )^{2} \log {\left (e^{2 i f x} + \frac {a^{3} c^{2} + 3 i a^{3} c d - 2 a^{3} d^{2} + \frac {i a^{3} d \left (c + i d\right )^{2}}{c - i d}}{a^{3} c^{2} e^{2 i e} + 2 i a^{3} c d e^{2 i e} + a^{3} d^{2} e^{2 i e}} \right )}}{d^{2} f \left (c - i d\right )} \]

[In]

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

[Out]

2*a**3/(d*f*exp(2*I*e)*exp(2*I*f*x) + d*f) - I*a**3*(c + 3*I*d)*log(exp(2*I*f*x) + (a**3*c**2 + 3*I*a**3*c*d -
2*a**3*d**2 - I*a**3*d*(c + 3*I*d))/(a**3*c**2*exp(2*I*e) + 2*I*a**3*c*d*exp(2*I*e) + a**3*d**2*exp(2*I*e)))/
(d**2*f) + I*a**3*(c + I*d)**2*log(exp(2*I*f*x) + (a**3*c**2 + 3*I*a**3*c*d - 2*a**3*d**2 + I*a**3*d*(c + I*d)
**2/(c - I*d))/(a**3*c**2*exp(2*I*e) + 2*I*a**3*c*d*exp(2*I*e) + a**3*d**2*exp(2*I*e)))/(d**2*f*(c - I*d))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.22 \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=\frac {-\frac {i \, a^{3} \tan \left (f x + e\right )}{d} + \frac {4 \, {\left (a^{3} c + i \, a^{3} d\right )} {\left (f x + e\right )}}{c^{2} + d^{2}} + \frac {{\left (i \, a^{3} c^{3} - 3 \, a^{3} c^{2} d - 3 i \, a^{3} c d^{2} + a^{3} d^{3}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{2} d^{2} + d^{4}} - \frac {2 \, {\left (-i \, a^{3} c + a^{3} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{2} + d^{2}}}{f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (106) = 212\).

Time = 0.49 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.07 \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=-\frac {-\frac {8 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c - i \, d} + \frac {{\left (-i \, a^{3} c^{2} + 2 \, a^{3} c d + i \, a^{3} d^{2}\right )} \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}{c d^{2} - i \, d^{3}} - \frac {{\left (-i \, a^{3} c + 3 \, a^{3} d\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{d^{2}} + \frac {{\left (i \, a^{3} c - 3 \, a^{3} d\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{d^{2}} + \frac {-i \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 i \, a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i \, a^{3} c - 3 \, a^{3} d}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} d^{2}}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.65 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \frac {(a+i a \tan (e+f x))^3}{c+d \tan (e+f x)} \, dx=\frac {a^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{f\,\left (c-d\,1{}\mathrm {i}\right )}-\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{d\,f}-\frac {\ln \left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (-a^3\,c^2\,1{}\mathrm {i}+2\,a^3\,c\,d+a^3\,d^2\,1{}\mathrm {i}\right )}{d^2\,f\,\left (c-d\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

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