3.3.76 \(\int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\) [276]

Optimal. Leaf size=87 \[ \frac {3 i \, _2F_1\left (\frac {1}{6},\frac {17}{6};\frac {7}{6};\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [3]{d \sec (e+f x)} (1+i \tan (e+f x))^{5/6}}{2\ 2^{5/6} f \left (a^2+i a^2 \tan (e+f x)\right )} \]

[Out]

3/4*I*hypergeom([1/6, 17/6],[7/6],1/2-1/2*I*tan(f*x+e))*(d*sec(f*x+e))^(1/3)*(1+I*tan(f*x+e))^(5/6)*2^(1/6)/f/
(a^2+I*a^2*tan(f*x+e))

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Rubi [A]
time = 0.13, antiderivative size = 87, 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 = {3586, 3604, 72, 71} \begin {gather*} \frac {3 i (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)} \, _2F_1\left (\frac {1}{6},\frac {17}{6};\frac {7}{6};\frac {1}{2} (1-i \tan (e+f x))\right )}{2\ 2^{5/6} f \left (a^2+i a^2 \tan (e+f x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/2)*Hypergeometric2F1[1/6, 17/6, 7/6, (1 - I*Tan[e + f*x])/2]*(d*Sec[e + f*x])^(1/3)*(1 + I*Tan[e + f*x
])^(5/6))/(2^(5/6)*f*(a^2 + I*a^2*Tan[e + f*x]))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 3586

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 3604

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 {\sqrt [3]{d \sec (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {\sqrt [3]{d \sec (e+f x)} \int \frac {\sqrt [6]{a-i a \tan (e+f x)}}{(a+i a \tan (e+f x))^{11/6}} \, dx}{\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}}\\ &=\frac {\left (a^2 \sqrt [3]{d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{(a-i a x)^{5/6} (a+i a x)^{17/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}}\\ &=\frac {\left (\sqrt [3]{d \sec (e+f x)} \left (\frac {a+i a \tan (e+f x)}{a}\right )^{5/6}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {1}{2}+\frac {i x}{2}\right )^{17/6} (a-i a x)^{5/6}} \, dx,x,\tan (e+f x)\right )}{4\ 2^{5/6} f \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x))}\\ &=\frac {3 i \, _2F_1\left (\frac {1}{6},\frac {17}{6};\frac {7}{6};\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [3]{d \sec (e+f x)} (1+i \tan (e+f x))^{5/6}}{2\ 2^{5/6} f \left (a^2+i a^2 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.68, size = 121, normalized size = 1.39 \begin {gather*} \frac {3 \sec ^2(e+f x) \sqrt [3]{d \sec (e+f x)} \left (-2 i-2 i \cos (2 (e+f x))+4 i e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (e+f x)}\right )+\sin (2 (e+f x))\right )}{22 a^2 f (-i+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sec[e + f*x]^2*(d*Sec[e + f*x])^(1/3)*(-2*I - (2*I)*Cos[2*(e + f*x)] + (4*I)*E^((2*I)*(e + f*x))*(1 + E^((2
*I)*(e + f*x)))^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(e + f*x))] + Sin[2*(e + f*x)]))/(22*a^2*f*(-
I + Tan[e + f*x])^2)

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Maple [F]
time = 0.50, size = 0, normalized size = 0.00 \[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/44*(44*a^2*f*e^(4*I*f*x + 4*I*e)*integral(-2/11*I*2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(-2/3*I*f*x
- 2/3*I*e)/(a^2*f), x) - 3*2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*(-3*I*e^(4*I*f*x + 4*I*e) - 4*I*e^(2*I*
f*x + 2*I*e) - I)*e^(1/3*I*f*x + 1/3*I*e))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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