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 \[ -\frac {3 i \sqrt [6]{1+i \tan (e+f x)} \, _2F_1\left (-\frac {1}{6},\frac {19}{6};\frac {5}{6};\frac {1}{2} (1-i \tan (e+f x))\right )}{4 \sqrt [6]{2} a^2 f \sqrt [3]{d \sec (e+f x)}} \]

[Out]

-3/8*I*hypergeom([-1/6, 19/6],[5/6],1/2-1/2*I*tan(f*x+e))*(1+I*tan(f*x+e))^(1/6)*2^(5/6)/a^2/f/(d*sec(f*x+e))^
(1/3)

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Rubi [A]  time = 0.18, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 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]))

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 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]

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*}

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Mathematica [A]  time = 1.45, size = 141, normalized size = 1.99 \[ \frac {(d \sec (e+f x))^{2/3} (-3 \sin (2 (e+f x))-3 i \cos (2 (e+f x))) \left (16 e^{3 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i (e+f x)}\right )-10 \left (7 \cos (e+f x)+5 \cos (3 (e+f x))+18 i \sin (e+f x) \cos ^2(e+f x)\right )\right )}{260 a^2 d f} \]

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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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \frac {2^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (-39 i \, e^{\left (7 i \, f x + 7 i \, e\right )} - 57 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 27 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 69 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 15 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 15 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, e^{\left (i \, f x + i \, e\right )} - 3 i\right )} e^{\left (\frac {2}{3} i \, f x + \frac {2}{3} i \, e\right )} + 104 \, {\left (a^{2} d f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d f e^{\left (5 i \, f x + 5 i \, e\right )}\right )} {\rm integral}\left (\frac {2^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (-8 i \, e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, e^{\left (i \, f x + i \, e\right )} - 8 i\right )} e^{\left (\frac {2}{3} i \, f x + \frac {2}{3} i \, e\right )}}{13 \, {\left (a^{2} d f e^{\left (3 i \, f x + 3 i \, e\right )} - 2 \, a^{2} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d f e^{\left (i \, f x + i \, e\right )}\right )}}, x\right )}{104 \, {\left (a^{2} d f e^{\left (6 i \, f x + 6 i \, e\right )} - a^{2} d f e^{\left (5 i \, f x + 5 i \, e\right )}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

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)

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maple [F]  time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{\sqrt [3]{d \sec {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )} - 2 i \sqrt [3]{d \sec {\left (e + f x \right )}} \tan {\left (e + f x \right )} - \sqrt [3]{d \sec {\left (e + f x \right )}}}\, dx}{a^{2}} \]

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]

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

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