\(\int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx\) [917]

Optimal result

Integrand size = 31, antiderivative size = 58 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=\frac {i a^2 (c-i c \tan (e+f x))^4}{2 f}-\frac {i a^2 (c-i c \tan (e+f x))^5}{5 c f} \]

[Out]

1/2*I*a^2*(c-I*c*tan(f*x+e))^4/f-1/5*I*a^2*(c-I*c*tan(f*x+e))^5/c/f

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 58, 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, 3568, 45} \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=\frac {i a^2 (c-i c \tan (e+f x))^4}{2 f}-\frac {i a^2 (c-i c \tan (e+f x))^5}{5 c f} \]

[In]

Int[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^4,x]

[Out]

((I/2)*a^2*(c - I*c*Tan[e + f*x])^4)/f - ((I/5)*a^2*(c - I*c*Tan[e + f*x])^5)/(c*f)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

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]))

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \sec ^4(e+f x) (c-i c \tan (e+f x))^2 \, dx \\ & = \frac {\left (i a^2\right ) \text {Subst}\left (\int (c-x) (c+x)^3 \, dx,x,-i c \tan (e+f x)\right )}{c f} \\ & = \frac {\left (i a^2\right ) \text {Subst}\left (\int \left (2 c (c+x)^3-(c+x)^4\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f} \\ & = \frac {i a^2 (c-i c \tan (e+f x))^4}{2 f}-\frac {i a^2 (c-i c \tan (e+f x))^5}{5 c f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=\frac {a^2 c^4 \left (\tan (e+f x)-i \tan ^2(e+f x)-\frac {1}{2} i \tan ^4(e+f x)-\frac {1}{5} \tan ^5(e+f x)\right )}{f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^2*c^4*(Tan[e + f*x] - I*Tan[e + f*x]^2 - (I/2)*Tan[e + f*x]^4 - Tan[e + f*x]^5/5))/f

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {8 i a^{2} c^{4} \left (5 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{5 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\)	\(39\)
derivativedivides	\(\frac {a^{2} c^{4} \left (\tan \left (f x +e \right )-\frac {\left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {i \left (\tan ^{4}\left (f x +e \right )\right )}{2}-i \left (\tan ^{2}\left (f x +e \right )\right )\right )}{f}\)	\(50\)
default	\(\frac {a^{2} c^{4} \left (\tan \left (f x +e \right )-\frac {\left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {i \left (\tan ^{4}\left (f x +e \right )\right )}{2}-i \left (\tan ^{2}\left (f x +e \right )\right )\right )}{f}\)	\(50\)
parallelrisch	\(-\frac {5 i a^{2} c^{4} \left (\tan ^{4}\left (f x +e \right )\right )+2 \left (\tan ^{5}\left (f x +e \right )\right ) a^{2} c^{4}+10 i a^{2} c^{4} \left (\tan ^{2}\left (f x +e \right )\right )-10 \tan \left (f x +e \right ) a^{2} c^{4}}{10 f}\)	\(71\)
norman	\(\frac {a^{2} c^{4} \tan \left (f x +e \right )}{f}-\frac {a^{2} c^{4} \left (\tan ^{5}\left (f x +e \right )\right )}{5 f}-\frac {i a^{2} c^{4} \left (\tan ^{2}\left (f x +e \right )\right )}{f}-\frac {i a^{2} c^{4} \left (\tan ^{4}\left (f x +e \right )\right )}{2 f}\)	\(77\)
parts	\(a^{2} c^{4} x +\frac {a^{2} c^{4} \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}-\frac {4 i a^{2} c^{4} \left (\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}-\frac {i a^{2} c^{4} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f}-\frac {2 i a^{2} c^{4} \left (\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}-\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}-\frac {a^{2} c^{4} \left (\frac {\left (\tan ^{3}\left (f x +e \right )\right )}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}-\frac {a^{2} c^{4} \left (\frac {\left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {\left (\tan ^{3}\left (f x +e \right )\right )}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(225\)

[In]

int((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

8/5*I*a^2*c^4*(5*exp(2*I*(f*x+e))+1)/f/(exp(2*I*(f*x+e))+1)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.57 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=-\frac {8 \, {\left (-5 i \, a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c^{4}\right )}}{5 \, {\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

-8/5*(-5*I*a^2*c^4*e^(2*I*f*x + 2*I*e) - I*a^2*c^4)/(f*e^(10*I*f*x + 10*I*e) + 5*f*e^(8*I*f*x + 8*I*e) + 10*f*
e^(6*I*f*x + 6*I*e) + 10*f*e^(4*I*f*x + 4*I*e) + 5*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. 131 vs. \(2 (44) = 88\).

Time = 0.26 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.26 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=\frac {40 i a^{2} c^{4} e^{2 i e} e^{2 i f x} + 8 i a^{2} c^{4}}{5 f e^{10 i e} e^{10 i f x} + 25 f e^{8 i e} e^{8 i f x} + 50 f e^{6 i e} e^{6 i f x} + 50 f e^{4 i e} e^{4 i f x} + 25 f e^{2 i e} e^{2 i f x} + 5 f} \]

[In]

integrate((a+I*a*tan(f*x+e))**2*(c-I*c*tan(f*x+e))**4,x)

[Out]

(40*I*a**2*c**4*exp(2*I*e)*exp(2*I*f*x) + 8*I*a**2*c**4)/(5*f*exp(10*I*e)*exp(10*I*f*x) + 25*f*exp(8*I*e)*exp(
8*I*f*x) + 50*f*exp(6*I*e)*exp(6*I*f*x) + 50*f*exp(4*I*e)*exp(4*I*f*x) + 25*f*exp(2*I*e)*exp(2*I*f*x) + 5*f)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=-\frac {2 \, a^{2} c^{4} \tan \left (f x + e\right )^{5} + 5 i \, a^{2} c^{4} \tan \left (f x + e\right )^{4} + 10 i \, a^{2} c^{4} \tan \left (f x + e\right )^{2} - 10 \, a^{2} c^{4} \tan \left (f x + e\right )}{10 \, f} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

-1/10*(2*a^2*c^4*tan(f*x + e)^5 + 5*I*a^2*c^4*tan(f*x + e)^4 + 10*I*a^2*c^4*tan(f*x + e)^2 - 10*a^2*c^4*tan(f*
x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.57 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=-\frac {8 \, {\left (-5 i \, a^{2} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c^{4}\right )}}{5 \, {\left (f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-8/5*(-5*I*a^2*c^4*e^(2*I*f*x + 2*I*e) - I*a^2*c^4)/(f*e^(10*I*f*x + 10*I*e) + 5*f*e^(8*I*f*x + 8*I*e) + 10*f*
e^(6*I*f*x + 6*I*e) + 10*f*e^(4*I*f*x + 4*I*e) + 5*f*e^(2*I*f*x + 2*I*e) + f)

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.38 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4 \, dx=-\frac {a^2\,c^4\,\sin \left (e+f\,x\right )\,\left (-10\,{\cos \left (e+f\,x\right )}^4+{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )\,10{}\mathrm {i}+\cos \left (e+f\,x\right )\,{\sin \left (e+f\,x\right )}^3\,5{}\mathrm {i}+2\,{\sin \left (e+f\,x\right )}^4\right )}{10\,f\,{\cos \left (e+f\,x\right )}^5} \]

[In]

int((a + a*tan(e + f*x)*1i)^2*(c - c*tan(e + f*x)*1i)^4,x)

[Out]

-(a^2*c^4*sin(e + f*x)*(cos(e + f*x)*sin(e + f*x)^3*5i + cos(e + f*x)^3*sin(e + f*x)*10i - 10*cos(e + f*x)^4 +
2*sin(e + f*x)^4))/(10*f*cos(e + f*x)^5)