\(\int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx\) [916]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 31, antiderivative size = 82 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=-\frac {i a^3 c^4 \sec ^6(e+f x)}{6 f}+\frac {a^3 c^4 \tan (e+f x)}{f}+\frac {2 a^3 c^4 \tan ^3(e+f x)}{3 f}+\frac {a^3 c^4 \tan ^5(e+f x)}{5 f} \]

[Out]

-1/6*I*a^3*c^4*sec(f*x+e)^6/f+a^3*c^4*tan(f*x+e)/f+2/3*a^3*c^4*tan(f*x+e)^3/f+1/5*a^3*c^4*tan(f*x+e)^5/f

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 82, 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} \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=\frac {a^3 c^4 \tan ^5(e+f x)}{5 f}+\frac {2 a^3 c^4 \tan ^3(e+f x)}{3 f}+\frac {a^3 c^4 \tan (e+f x)}{f}-\frac {i a^3 c^4 \sec ^6(e+f x)}{6 f} \]

[In]

Int[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^4,x]

[Out]

((-1/6*I)*a^3*c^4*Sec[e + f*x]^6)/f + (a^3*c^4*Tan[e + f*x])/f + (2*a^3*c^4*Tan[e + f*x]^3)/(3*f) + (a^3*c^4*T
an[e + f*x]^5)/(5*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^3 c^3\right ) \int \sec ^6(e+f x) (c-i c \tan (e+f x)) \, dx \\ & = -\frac {i a^3 c^4 \sec ^6(e+f x)}{6 f}+\left (a^3 c^4\right ) \int \sec ^6(e+f x) \, dx \\ & = -\frac {i a^3 c^4 \sec ^6(e+f x)}{6 f}-\frac {\left (a^3 c^4\right ) \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{f} \\ & = -\frac {i a^3 c^4 \sec ^6(e+f x)}{6 f}+\frac {a^3 c^4 \tan (e+f x)}{f}+\frac {2 a^3 c^4 \tan ^3(e+f x)}{3 f}+\frac {a^3 c^4 \tan ^5(e+f x)}{5 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=\frac {a^3 c^4 \tan (e+f x) \left (30-15 i \tan (e+f x)+20 \tan ^2(e+f x)-15 i \tan ^3(e+f x)+6 \tan ^4(e+f x)-5 i \tan ^5(e+f x)\right )}{30 f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^3*c^4*Tan[e + f*x]*(30 - (15*I)*Tan[e + f*x] + 20*Tan[e + f*x]^2 - (15*I)*Tan[e + f*x]^3 + 6*Tan[e + f*x]^4
 - (5*I)*Tan[e + f*x]^5))/(30*f)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.61

method result size
risch \(\frac {16 i a^{3} c^{4} \left (15 \,{\mathrm e}^{4 i \left (f x +e \right )}+6 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{6}}\) \(50\)
derivativedivides \(-\frac {i a^{3} c^{4} \left (\frac {\left (\tan ^{6}\left (f x +e \right )\right )}{6}+\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{2}+\frac {i \left (\tan ^{5}\left (f x +e \right )\right )}{5}+\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}+\frac {2 i \left (\tan ^{3}\left (f x +e \right )\right )}{3}+i \tan \left (f x +e \right )\right )}{f}\) \(75\)
default \(-\frac {i a^{3} c^{4} \left (\frac {\left (\tan ^{6}\left (f x +e \right )\right )}{6}+\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{2}+\frac {i \left (\tan ^{5}\left (f x +e \right )\right )}{5}+\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}+\frac {2 i \left (\tan ^{3}\left (f x +e \right )\right )}{3}+i \tan \left (f x +e \right )\right )}{f}\) \(75\)
parallelrisch \(-\frac {5 i a^{3} c^{4} \left (\tan ^{6}\left (f x +e \right )\right )+15 i a^{3} c^{4} \left (\tan ^{4}\left (f x +e \right )\right )-6 \left (\tan ^{5}\left (f x +e \right )\right ) a^{3} c^{4}+15 i a^{3} c^{4} \left (\tan ^{2}\left (f x +e \right )\right )-20 \left (\tan ^{3}\left (f x +e \right )\right ) a^{3} c^{4}-30 \tan \left (f x +e \right ) a^{3} c^{4}}{30 f}\) \(104\)
norman \(\frac {a^{3} c^{4} \tan \left (f x +e \right )}{f}+\frac {2 a^{3} c^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{3 f}+\frac {a^{3} c^{4} \left (\tan ^{5}\left (f x +e \right )\right )}{5 f}-\frac {i a^{3} c^{4} \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}-\frac {i a^{3} c^{4} \left (\tan ^{4}\left (f x +e \right )\right )}{2 f}-\frac {i a^{3} c^{4} \left (\tan ^{6}\left (f x +e \right )\right )}{6 f}\) \(116\)
parts \(a^{3} c^{4} x +\frac {a^{3} c^{4} \left (\frac {\left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {\left (\tan ^{3}\left (f x +e \right )\right )}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}-\frac {3 i a^{3} c^{4} \left (\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}-\frac {3 i a^{3} c^{4} \left (\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}-\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}-\frac {i a^{3} c^{4} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}-\frac {i a^{3} c^{4} \left (\frac {\left (\tan ^{6}\left (f x +e \right )\right )}{6}-\frac {\left (\tan ^{4}\left (f x +e \right )\right )}{4}+\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}+\frac {3 a^{3} c^{4} \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {3 a^{3} c^{4} \left (\frac {\left (\tan ^{3}\left (f x +e \right )\right )}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\) \(281\)

[In]

int((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

16/15*I*a^3*c^4*(15*exp(4*I*(f*x+e))+6*exp(2*I*(f*x+e))+1)/f/(exp(2*I*(f*x+e))+1)^6

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.46 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=-\frac {16 \, {\left (-15 i \, a^{3} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 6 i \, a^{3} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c^{4}\right )}}{15 \, {\left (f e^{\left (12 i \, f x + 12 i \, e\right )} + 6 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

-16/15*(-15*I*a^3*c^4*e^(4*I*f*x + 4*I*e) - 6*I*a^3*c^4*e^(2*I*f*x + 2*I*e) - I*a^3*c^4)/(f*e^(12*I*f*x + 12*I
*e) + 6*f*e^(10*I*f*x + 10*I*e) + 15*f*e^(8*I*f*x + 8*I*e) + 20*f*e^(6*I*f*x + 6*I*e) + 15*f*e^(4*I*f*x + 4*I*
e) + 6*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. 175 vs. \(2 (73) = 146\).

Time = 0.35 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.13 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=\frac {240 i a^{3} c^{4} e^{4 i e} e^{4 i f x} + 96 i a^{3} c^{4} e^{2 i e} e^{2 i f x} + 16 i a^{3} c^{4}}{15 f e^{12 i e} e^{12 i f x} + 90 f e^{10 i e} e^{10 i f x} + 225 f e^{8 i e} e^{8 i f x} + 300 f e^{6 i e} e^{6 i f x} + 225 f e^{4 i e} e^{4 i f x} + 90 f e^{2 i e} e^{2 i f x} + 15 f} \]

[In]

integrate((a+I*a*tan(f*x+e))**3*(c-I*c*tan(f*x+e))**4,x)

[Out]

(240*I*a**3*c**4*exp(4*I*e)*exp(4*I*f*x) + 96*I*a**3*c**4*exp(2*I*e)*exp(2*I*f*x) + 16*I*a**3*c**4)/(15*f*exp(
12*I*e)*exp(12*I*f*x) + 90*f*exp(10*I*e)*exp(10*I*f*x) + 225*f*exp(8*I*e)*exp(8*I*f*x) + 300*f*exp(6*I*e)*exp(
6*I*f*x) + 225*f*exp(4*I*e)*exp(4*I*f*x) + 90*f*exp(2*I*e)*exp(2*I*f*x) + 15*f)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.22 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=-\frac {5 i \, a^{3} c^{4} \tan \left (f x + e\right )^{6} - 6 \, a^{3} c^{4} \tan \left (f x + e\right )^{5} + 15 i \, a^{3} c^{4} \tan \left (f x + e\right )^{4} - 20 \, a^{3} c^{4} \tan \left (f x + e\right )^{3} + 15 i \, a^{3} c^{4} \tan \left (f x + e\right )^{2} - 30 \, a^{3} c^{4} \tan \left (f x + e\right )}{30 \, f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

-1/30*(5*I*a^3*c^4*tan(f*x + e)^6 - 6*a^3*c^4*tan(f*x + e)^5 + 15*I*a^3*c^4*tan(f*x + e)^4 - 20*a^3*c^4*tan(f*
x + e)^3 + 15*I*a^3*c^4*tan(f*x + e)^2 - 30*a^3*c^4*tan(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.46 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=-\frac {16 \, {\left (-15 i \, a^{3} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 6 i \, a^{3} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c^{4}\right )}}{15 \, {\left (f e^{\left (12 i \, f x + 12 i \, e\right )} + 6 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-16/15*(-15*I*a^3*c^4*e^(4*I*f*x + 4*I*e) - 6*I*a^3*c^4*e^(2*I*f*x + 2*I*e) - I*a^3*c^4)/(f*e^(12*I*f*x + 12*I
*e) + 6*f*e^(10*I*f*x + 10*I*e) + 15*f*e^(8*I*f*x + 8*I*e) + 20*f*e^(6*I*f*x + 6*I*e) + 15*f*e^(4*I*f*x + 4*I*
e) + 6*f*e^(2*I*f*x + 2*I*e) + f)

Mupad [B] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.43 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^4 \, dx=\frac {a^3\,c^4\,\sin \left (e+f\,x\right )\,\left (30\,{\cos \left (e+f\,x\right )}^5-{\cos \left (e+f\,x\right )}^4\,\sin \left (e+f\,x\right )\,15{}\mathrm {i}+20\,{\cos \left (e+f\,x\right )}^3\,{\sin \left (e+f\,x\right )}^2-{\cos \left (e+f\,x\right )}^2\,{\sin \left (e+f\,x\right )}^3\,15{}\mathrm {i}+6\,\cos \left (e+f\,x\right )\,{\sin \left (e+f\,x\right )}^4-{\sin \left (e+f\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,f\,{\cos \left (e+f\,x\right )}^6} \]

[In]

int((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^4,x)

[Out]

(a^3*c^4*sin(e + f*x)*(6*cos(e + f*x)*sin(e + f*x)^4 - cos(e + f*x)^4*sin(e + f*x)*15i + 30*cos(e + f*x)^5 - s
in(e + f*x)^5*5i - cos(e + f*x)^2*sin(e + f*x)^3*15i + 20*cos(e + f*x)^3*sin(e + f*x)^2))/(30*f*cos(e + f*x)^6
)