3.277 \(\int \frac{1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=71 \[ -\frac{3 i \sqrt [6]{1+i \tan (e+f x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{19}{6},\frac{5}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{4 \sqrt [6]{2} a^2 f \sqrt [3]{d \sec (e+f x)}} \]

[Out]

(((-3*I)/4)*Hypergeometric2F1[-1/6, 19/6, 5/6, (1 - I*Tan[e + f*x])/2]*(1 + I*Tan[e + f*x])^(1/6))/(2^(1/6)*a^
2*f*(d*Sec[e + f*x])^(1/3))

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Rubi [A]  time = 0.179847, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3505, 3523, 70, 69} \[ -\frac{3 i \sqrt [6]{1+i \tan (e+f x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{19}{6},\frac{5}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{4 \sqrt [6]{2} a^2 f \sqrt [3]{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/4)*Hypergeometric2F1[-1/6, 19/6, 5/6, (1 - I*Tan[e + f*x])/2]*(1 + I*Tan[e + f*x])^(1/6))/(2^(1/6)*a^
2*f*(d*Sec[e + f*x])^(1/3))

Rule 3505

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

Rule 3523

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2} \, dx &=\frac{\left (\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}\right ) \int \frac{1}{\sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x))^{13/6}} \, dx}{\sqrt [3]{d \sec (e+f x)}}\\ &=\frac{\left (a^2 \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(a-i a x)^{7/6} (a+i a x)^{19/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)}}\\ &=\frac{\left (\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{\frac{a+i a \tan (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{1}{2}+\frac{i x}{2}\right )^{19/6} (a-i a x)^{7/6}} \, dx,x,\tan (e+f x)\right )}{8 \sqrt [6]{2} a f \sqrt [3]{d \sec (e+f x)}}\\ &=-\frac{3 i \, _2F_1\left (-\frac{1}{6},\frac{19}{6};\frac{5}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) \sqrt [6]{1+i \tan (e+f x)}}{4 \sqrt [6]{2} a^2 f \sqrt [3]{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.35768, size = 141, normalized size = 1.99 \[ \frac{(d \sec (e+f x))^{2/3} (-3 \sin (2 (e+f x))-3 i \cos (2 (e+f x))) \left (16 e^{3 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^{2/3} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (e+f x)}\right )-10 \left (7 \cos (e+f x)+5 \cos (3 (e+f x))+18 i \sin (e+f x) \cos ^2(e+f x)\right )\right )}{260 a^2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sec[e + f*x])^(2/3)*(16*E^((3*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(2/3)*Hypergeometric2F1[2/3, 5/6, 11
/6, -E^((2*I)*(e + f*x))] - 10*(7*Cos[e + f*x] + 5*Cos[3*(e + f*x)] + (18*I)*Cos[e + f*x]^2*Sin[e + f*x]))*((-
3*I)*Cos[2*(e + f*x)] - 3*Sin[2*(e + f*x)]))/(260*a^2*d*f)

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Maple [F]  time = 0.15, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt [3]{d\sec \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2^{\frac{2}{3}} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}}{\left (-39 i \, e^{\left (7 i \, f x + 7 i \, e\right )} - 57 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 27 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 69 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 15 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 15 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, e^{\left (i \, f x + i \, e\right )} - 3 i\right )} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )} + 104 \,{\left (a^{2} d f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d f e^{\left (5 i \, f x + 5 i \, e\right )}\right )}{\rm integral}\left (\frac{2^{\frac{2}{3}} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}}{\left (-8 i \, e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, e^{\left (i \, f x + i \, e\right )} - 8 i\right )} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )}}{13 \,{\left (a^{2} d f e^{\left (3 i \, f x + 3 i \, e\right )} - 2 \, a^{2} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d f e^{\left (i \, f x + i \, e\right )}\right )}}, x\right )}{104 \,{\left (a^{2} d f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d f e^{\left (5 i \, f x + 5 i \, e\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/104*(2^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(-39*I*e^(7*I*f*x + 7*I*e) - 57*I*e^(6*I*f*x + 6*I*e) - 27*
I*e^(5*I*f*x + 5*I*e) - 69*I*e^(4*I*f*x + 4*I*e) + 15*I*e^(3*I*f*x + 3*I*e) - 15*I*e^(2*I*f*x + 2*I*e) + 3*I*e
^(I*f*x + I*e) - 3*I)*e^(2/3*I*f*x + 2/3*I*e) + 104*(a^2*d*f*e^(6*I*f*x + 6*I*e) - a^2*d*f*e^(5*I*f*x + 5*I*e)
)*integral(1/13*2^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(-8*I*e^(2*I*f*x + 2*I*e) - 8*I*e^(I*f*x + I*e) -
8*I)*e^(2/3*I*f*x + 2/3*I*e)/(a^2*d*f*e^(3*I*f*x + 3*I*e) - 2*a^2*d*f*e^(2*I*f*x + 2*I*e) + a^2*d*f*e^(I*f*x +
I*e)), x))/(a^2*d*f*e^(6*I*f*x + 6*I*e) - a^2*d*f*e^(5*I*f*x + 5*I*e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))**(1/3)/(a+I*a*tan(f*x+e))**2,x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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