3.5.47 \(\int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx\) [447]

Optimal. Leaf size=81 \[ \frac {9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{4 f} \]

[Out]

9/2*I*a^2*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(1/3)+3/4*I*a*(d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(2/3
)/f

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Rubi [A]
time = 0.12, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3575, 3574} \begin {gather*} \frac {9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3),x]

[Out]

(((9*I)/2)*a^2*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(1/3)) + (((3*I)/4)*a*(d*Sec[e + f*x])^(2/3)*
(a + I*a*Tan[e + f*x])^(2/3))/f

Rule 3574

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 3575

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))^{5/3} \, dx &=\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{4 f}+\frac {1}{2} (3 a) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3} \, dx\\ &=\frac {9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{4 f}\\ \end {align*}

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Mathematica [A]
time = 0.59, size = 70, normalized size = 0.86 \begin {gather*} -\frac {3 a d (\cos (e)-i \sin (e)) (\cos (f x)-i \sin (f x)) (-7 i+\tan (e+f x)) (a+i a \tan (e+f x))^{2/3}}{4 f \sqrt [3]{d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3),x]

[Out]

(-3*a*d*(Cos[e] - I*Sin[e])*(Cos[f*x] - I*Sin[f*x])*(-7*I + Tan[e + f*x])*(a + I*a*Tan[e + f*x])^(2/3))/(4*f*(
d*Sec[e + f*x])^(1/3))

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Maple [F]
time = 0.37, size = 0, normalized size = 0.00 \[\int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{\frac {5}{3}}\, dx\]

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))^(5/3),x)

[Out]

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (65) = 130\).
time = 0.56, size = 336, normalized size = 4.15 \begin {gather*} \frac {3 \, {\left ({\left (-i \cdot 2^{\frac {1}{3}} a \cos \left (\frac {4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) - 2^{\frac {1}{3}} a \sin \left (\frac {4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} \sqrt {\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1} a^{\frac {2}{3}} d^{\frac {2}{3}} + 4 \, {\left ({\left (i \cdot 2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right )^{2} + i \cdot 2^{\frac {1}{3}} a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 i \cdot 2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right ) + i \cdot 2^{\frac {1}{3}} a\right )} \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + {\left (2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right )^{2} + 2^{\frac {1}{3}} a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \cdot 2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right ) + 2^{\frac {1}{3}} a\right )} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} a^{\frac {2}{3}} d^{\frac {2}{3}}\right )}}{2 \, {\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {7}{6}} f} \end {gather*}

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))^(5/3),x, algorithm="maxima")

[Out]

3/2*((-I*2^(1/3)*a*cos(4/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2^(1/3)*a*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) + 4*((I*2^(1/3)*a*cos(2*f*x + 2*e)^2 + I*2^(1/3)*a*sin(2*f*x + 2*e)^2 + 2*I*2^(1/3)*a*cos(2*f*x +
2*e) + I*2^(1/3)*a)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (2^(1/3)*a*cos(2*f*x + 2*e)^2
+ 2^(1/3)*a*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a*cos(2*f*x + 2*e) + 2^(1/3)*a)*sin(1/3*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e) + 1)))*a^(2/3)*d^(2/3))/((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(7
/6)*f)

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Fricas [A]
time = 0.38, size = 61, normalized size = 0.75 \begin {gather*} -\frac {3 \cdot 2^{\frac {1}{3}} {\left (-4 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a\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}}}{2 \, f} \end {gather*}

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))^(5/3),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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))**(5/3),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8569 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(5/3),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(2/3)*(I*a*tan(f*x + e) + a)^(5/3), x)

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Mupad [B]
time = 4.13, size = 90, normalized size = 1.11 \begin {gather*} \frac {3\,a\,{\left (\frac {d}{2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left ({\cos \left (e+f\,x\right )}^2\,6{}\mathrm {i}+3\,\sin \left (2\,e+2\,f\,x\right )+1{}\mathrm {i}\right )\,{\left (\frac {a\,\left (2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}\right )}^{2/3}}{4\,f} \end {gather*}

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)^(5/3),x)

[Out]

(3*a*(d/(2*cos(e/2 + (f*x)/2)^2 - 1))^(2/3)*(3*sin(2*e + 2*f*x) + cos(e + f*x)^2*6i + 1i)*((a*(sin(2*e + 2*f*x
)*1i + 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(2/3))/(4*f)

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