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

Optimal. Leaf size=81 \[ \frac {3 i (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)} \, _2F_1\left (\frac {1}{6},\frac {11}{6};\frac {7}{6};\frac {1}{2} (1-i \tan (e+f x))\right )}{2^{5/6} f (a+i a \tan (e+f x))} \]

[Out]

3/2*I*hypergeom([1/6, 11/6],[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+I*a*tan(f*x+e))

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Rubi [A]  time = 0.17, antiderivative size = 81, 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} \[ \frac {3 i (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)} \text {Hypergeometric2F1}\left (\frac {1}{6},\frac {11}{6},\frac {7}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{2^{5/6} f (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*I)*Hypergeometric2F1[1/6, 11/6, 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 + I*a*Tan[e + f*x]))

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

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Mathematica [A]  time = 0.49, size = 103, normalized size = 1.27 \[ -\frac {3 i e^{-2 i (e+f x)} \left (4 e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (e+f x)}\right )-e^{2 i (e+f x)}-1\right ) \sqrt [3]{d \sec (e+f x)}}{10 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/10)*(-1 - E^((2*I)*(e + f*x)) + 4*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*Hypergeometric2
F1[1/6, 1/3, 7/6, -E^((2*I)*(e + f*x))])*(d*Sec[e + f*x])^(1/3))/(a*E^((2*I)*(e + f*x))*f)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \frac {{\left (10 \, a f e^{\left (2 i \, f x + 2 i \, e\right )} {\rm integral}\left (-\frac {2 i \cdot 2^{\frac {1}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}} e^{\left (-\frac {2}{3} i \, f x - \frac {2}{3} i \, e\right )}}{5 \, a f}, x\right ) + 2^{\frac {1}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}} {\left (3 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (\frac {1}{3} i \, f x + \frac {1}{3} i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{10 \, a f} \]

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)),x, algorithm="fricas")

[Out]

1/10*(10*a*f*e^(2*I*f*x + 2*I*e)*integral(-2/5*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*f), x) + 2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*(3*I*e^(2*I*f*x + 2*I*e) + 3*I)*e^(1/3*I*f*x +
1/3*I*e))*e^(-2*I*f*x - 2*I*e)/(a*f)

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

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)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}{a +i a \tan \left (f x +e \right )}\, dx \]

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)),x)

[Out]

int((d*sec(f*x+e))^(1/3)/(a+I*a*tan(f*x+e)),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((d*sec(f*x+e))^(1/3)/(a+I*a*tan(f*x+e)),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 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}}{a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]

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),x)

[Out]

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

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

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)),x)

[Out]

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

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