3.449 \(\int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx\)

Optimal. Leaf size=163 \[ \frac {486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {81 i a^3 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f} \]

[Out]

486/35*I*a^4*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(1/3)+81/35*I*a^3*(d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e
))^(2/3)/f+27/35*I*a^2*(d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3)/f+3/10*I*a*(d*sec(f*x+e))^(2/3)*(a+I*a*ta
n(f*x+e))^(8/3)/f

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Rubi [A]  time = 0.32, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3494, 3493} \[ \frac {486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {81 i a^3 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(11/3),x]

[Out]

(((486*I)/35)*a^4*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(1/3)) + (((81*I)/35)*a^3*(d*Sec[e + f*x])
^(2/3)*(a + I*a*Tan[e + f*x])^(2/3))/f + (((27*I)/35)*a^2*(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3))
/f + (((3*I)/10)*a*(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(8/3))/f

Rule 3493

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

Rule 3494

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

Rubi steps

\begin {align*} \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx &=\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac {1}{5} (9 a) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx\\ &=\frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac {1}{35} \left (108 a^2\right ) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx\\ &=\frac {81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac {1}{35} \left (162 a^3\right ) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3} \, dx\\ &=\frac {486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}\\ \end {align*}

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Mathematica [A]  time = 1.15, size = 116, normalized size = 0.71 \[ \frac {3 a^3 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{5/3} (\sin (e-2 f x)+i \cos (e-2 f x)) (442 \cos (2 (e+f x))+45 i \tan (e+f x)+59 i \sin (3 (e+f x)) \sec (e+f x)+364)}{140 d f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(11/3),x]

[Out]

(3*a^3*(d*Sec[e + f*x])^(5/3)*(I*Cos[e - 2*f*x] + Sin[e - 2*f*x])*(364 + 442*Cos[2*(e + f*x)] + (59*I)*Sec[e +
f*x]*Sin[3*(e + f*x)] + (45*I)*Tan[e + f*x])*(a + I*a*Tan[e + f*x])^(2/3))/(140*d*f*(Cos[f*x] + I*Sin[f*x])^3
)

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fricas [A]  time = 0.68, size = 133, normalized size = 0.82 \[ \frac {2 \cdot 2^{\frac {1}{3}} {\left (420 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 945 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 810 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 243 i \, a^{3}\right )} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}}{35 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x, algorithm="fricas")

[Out]

2/35*2^(1/3)*(420*I*a^3*e^(6*I*f*x + 6*I*e) + 945*I*a^3*e^(4*I*f*x + 4*I*e) + 810*I*a^3*e^(2*I*f*x + 2*I*e) +
243*I*a^3)*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2*I*f*x + 2*I*e)/(f*e^(6
*I*f*x + 6*I*e) + 2*f*e^(4*I*f*x + 4*I*e) + f*e^(2*I*f*x + 2*I*e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {11}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(2/3)*(I*a*tan(f*x + e) + a)^(11/3), x)

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maple [F]  time = 0.77, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{\frac {11}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x)

[Out]

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x)

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maxima [B]  time = 1.08, size = 977, normalized size = 5.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x, algorithm="maxima")

[Out]

-1/35*((84*I*2^(1/3)*a^3*cos(10/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 84*2^(1/3)*a^3*sin(10/3*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (630*I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 630*I*2^(1/3)*a^3*si
n(2*f*x + 2*e)^2 + 1260*I*2^(1/3)*a^3*cos(2*f*x + 2*e) + 630*I*2^(1/3)*a^3)*cos(4/3*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e) + 1)) + 630*(2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 2^(1/3)*a^3*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a^3*
cos(2*f*x + 2*e) + 2^(1/3)*a^3)*sin(4/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*sqrt(cos(2*f*x + 2*e
)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*a^(2/3)*d^(2/3) + ((-360*I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 -
360*I*2^(1/3)*a^3*sin(2*f*x + 2*e)^2 - 720*I*2^(1/3)*a^3*cos(2*f*x + 2*e) - 360*I*2^(1/3)*a^3)*cos(7/3*arctan
2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (-840*I*2^(1/3)*a^3*cos(2*f*x + 2*e)^4 - 840*I*2^(1/3)*a^3*sin(2*
f*x + 2*e)^4 - 3360*I*2^(1/3)*a^3*cos(2*f*x + 2*e)^3 - 5040*I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 - 3360*I*2^(1/3)*
a^3*cos(2*f*x + 2*e) - 840*I*2^(1/3)*a^3 + (-1680*I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 - 3360*I*2^(1/3)*a^3*cos(2*
f*x + 2*e) - 1680*I*2^(1/3)*a^3)*sin(2*f*x + 2*e)^2)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))
- 360*(2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 2^(1/3)*a^3*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a^3*cos(2*f*x + 2*e) + 2^(1
/3)*a^3)*sin(7/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 840*(2^(1/3)*a^3*cos(2*f*x + 2*e)^4 + 2^(1
/3)*a^3*sin(2*f*x + 2*e)^4 + 4*2^(1/3)*a^3*cos(2*f*x + 2*e)^3 + 6*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 4*2^(1/3)*a
^3*cos(2*f*x + 2*e) + 2^(1/3)*a^3 + 2*(2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 2*2^(1/3)*a^3*cos(2*f*x + 2*e) + 2^(1/
3)*a^3)*sin(2*f*x + 2*e)^2)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*a^(2/3)*d^(2/3))/((cos(2
*f*x + 2*e)^4 + sin(2*f*x + 2*e)^4 + 4*cos(2*f*x + 2*e)^3 + 2*(cos(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*si
n(2*f*x + 2*e)^2 + 6*cos(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e) + 1)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2
*cos(2*f*x + 2*e) + 1)^(1/6)*f)

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mupad [B]  time = 7.88, size = 303, normalized size = 1.86 \[ \frac {{\left (-\frac {d}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,243{}\mathrm {i}}{35\,f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+1\right )\,162{}\mathrm {i}}{7\,f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,27{}\mathrm {i}}{f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (3\,e+3\,f\,x\right )}^2+\sin \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}+1\right )\,12{}\mathrm {i}}{f}\right )}{4\,\left ({\sin \left (e+f\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(2/3)*(a + a*tan(e + f*x)*1i)^(11/3),x)

[Out]

((-d/(2*sin(e/2 + (f*x)/2)^2 - 1))^(2/3)*(sin(4*e + 4*f*x)*1i + 2*sin(2*e + 2*f*x)^2 - 1)*((a^3*(a - (a*sin(e
+ f*x)*1i)/(2*sin(e/2 + (f*x)/2)^2 - 1))^(2/3)*243i)/(35*f) + (a^3*(a - (a*sin(e + f*x)*1i)/(2*sin(e/2 + (f*x)
/2)^2 - 1))^(2/3)*(sin(2*e + 2*f*x)*1i - 2*sin(e + f*x)^2 + 1)*162i)/(7*f) + (a^3*(a - (a*sin(e + f*x)*1i)/(2*
sin(e/2 + (f*x)/2)^2 - 1))^(2/3)*(sin(4*e + 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1)*27i)/f + (a^3*(a - (a*sin(e
+ f*x)*1i)/(2*sin(e/2 + (f*x)/2)^2 - 1))^(2/3)*(sin(6*e + 6*f*x)*1i - 2*sin(3*e + 3*f*x)^2 + 1)*12i)/f))/(4*(s
in(e + f*x)^2 - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(2/3)*(a+I*a*tan(f*x+e))**(11/3),x)

[Out]

Timed out

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