\(\int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx\) [897]

Optimal result

Integrand size = 31, antiderivative size = 61 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=\frac {i a^3 c^2 \sec ^4(e+f x)}{4 f}+\frac {a^3 c^2 \tan (e+f x)}{f}+\frac {a^3 c^2 \tan ^3(e+f x)}{3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3567, 3852} \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=\frac {a^3 c^2 \tan ^3(e+f x)}{3 f}+\frac {a^3 c^2 \tan (e+f x)}{f}+\frac {i a^3 c^2 \sec ^4(e+f x)}{4 f} \]

[In]

Int[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2,x]

[Out]

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

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \sec ^4(e+f x) (a+i a \tan (e+f x)) \, dx \\ & = \frac {i a^3 c^2 \sec ^4(e+f x)}{4 f}+\left (a^3 c^2\right ) \int \sec ^4(e+f x) \, dx \\ & = \frac {i a^3 c^2 \sec ^4(e+f x)}{4 f}-\frac {\left (a^3 c^2\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{f} \\ & = \frac {i a^3 c^2 \sec ^4(e+f x)}{4 f}+\frac {a^3 c^2 \tan (e+f x)}{f}+\frac {a^3 c^2 \tan ^3(e+f x)}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.96 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.64 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=\frac {a^3 c^2 (-1-i \tan (e+f x))^3 (5 i+3 \tan (e+f x))}{12 f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^2,x]

[Out]

(a^3*c^2*(-1 - I*Tan[e + f*x])^3*(5*I + 3*Tan[e + f*x]))/(12*f)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {4 i a^{3} c^{2} \left (6 \,{\mathrm e}^{4 i \left (f x +e \right )}+4 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}\)	\(50\)
derivativedivides	\(\frac {i a^{3} c^{2} \left (\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}+\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {i \left (\tan ^{3}\left (f x +e \right )\right )}{3}-i \tan \left (f x +e \right )\right )}{f}\)	\(54\)
default	\(\frac {i a^{3} c^{2} \left (\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}+\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {i \left (\tan ^{3}\left (f x +e \right )\right )}{3}-i \tan \left (f x +e \right )\right )}{f}\)	\(54\)
parallelrisch	\(\frac {3 i a^{3} c^{2} \left (\tan ^{4}\left (f x +e \right )\right )+6 i a^{3} c^{2} \left (\tan ^{2}\left (f x +e \right )\right )+4 \left (\tan ^{3}\left (f x +e \right )\right ) a^{3} c^{2}+12 \tan \left (f x +e \right ) a^{3} c^{2}}{12 f}\)	\(71\)
norman	\(\frac {a^{3} c^{2} \tan \left (f x +e \right )}{f}+\frac {a^{3} c^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{3 f}+\frac {i a^{3} c^{2} \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {i a^{3} c^{2} \left (\tan ^{4}\left (f x +e \right )\right )}{4 f}\)	\(77\)
parts	\(a^{3} c^{2} x +\frac {a^{3} c^{2} \left (\frac {\left (\tan ^{3}\left (f x +e \right )\right )}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {i a^{3} c^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {i a^{3} c^{2} \left (\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}-\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}+\frac {2 i a^{3} c^{2} \left (\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}+\frac {2 a^{3} c^{2} \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(178\)

[In]

int((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

4/3*I*a^3*c^2*(6*exp(4*I*(f*x+e))+4*exp(2*I*(f*x+e))+1)/f/(exp(2*I*(f*x+e))+1)^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=-\frac {4 \, {\left (-6 i \, a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c^{2}\right )}}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

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

[Out]

-4/3*(-6*I*a^3*c^2*e^(4*I*f*x + 4*I*e) - 4*I*a^3*c^2*e^(2*I*f*x + 2*I*e) - I*a^3*c^2)/(f*e^(8*I*f*x + 8*I*e) +
4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + 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. 138 vs. \(2 (53) = 106\).

Time = 0.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.26 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=\frac {24 i a^{3} c^{2} e^{4 i e} e^{4 i f x} + 16 i a^{3} c^{2} e^{2 i e} e^{2 i f x} + 4 i a^{3} c^{2}}{3 f e^{8 i e} e^{8 i f x} + 12 f e^{6 i e} e^{6 i f x} + 18 f e^{4 i e} e^{4 i f x} + 12 f e^{2 i e} e^{2 i f x} + 3 f} \]

[In]

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

[Out]

(24*I*a**3*c**2*exp(4*I*e)*exp(4*I*f*x) + 16*I*a**3*c**2*exp(2*I*e)*exp(2*I*f*x) + 4*I*a**3*c**2)/(3*f*exp(8*I
*e)*exp(8*I*f*x) + 12*f*exp(6*I*e)*exp(6*I*f*x) + 18*f*exp(4*I*e)*exp(4*I*f*x) + 12*f*exp(2*I*e)*exp(2*I*f*x)
+ 3*f)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=\frac {3 i \, a^{3} c^{2} \tan \left (f x + e\right )^{4} + 4 \, a^{3} c^{2} \tan \left (f x + e\right )^{3} + 6 i \, a^{3} c^{2} \tan \left (f x + e\right )^{2} + 12 \, a^{3} c^{2} \tan \left (f x + e\right )}{12 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=-\frac {4 \, {\left (-6 i \, a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c^{2}\right )}}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.91 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2 \, dx=\frac {a^3\,c^2\,\sin \left (e+f\,x\right )\,\left (12\,{\cos \left (e+f\,x\right )}^3+{\cos \left (e+f\,x\right )}^2\,\sin \left (e+f\,x\right )\,6{}\mathrm {i}+4\,\cos \left (e+f\,x\right )\,{\sin \left (e+f\,x\right )}^2+{\sin \left (e+f\,x\right )}^3\,3{}\mathrm {i}\right )}{12\,f\,{\cos \left (e+f\,x\right )}^4} \]

[In]

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

[Out]

(a^3*c^2*sin(e + f*x)*(4*cos(e + f*x)*sin(e + f*x)^2 + cos(e + f*x)^2*sin(e + f*x)*6i + 12*cos(e + f*x)^3 + si
n(e + f*x)^3*3i))/(12*f*cos(e + f*x)^4)