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

Optimal. Leaf size=67 \[ \frac{6 i \sqrt [6]{2} a \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [6]{1+i \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.14672, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3505, 3523, 70, 69} \[ \frac{6 i \sqrt [6]{2} a \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [6]{1+i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.555025, size = 92, normalized size = 1.37 \[ \frac{3 a d e^{-i e} (\tan (e+f x)-i) (\cos (f x)-i \sin (f x)) \left (-1+\sqrt [3]{1+e^{2 i (e+f x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (e+f x)}\right )\right )}{f (d \sec (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{d\sec \left ( fx+e \right ) } \left ( a+ia\tan \left ( fx+e \right ) \right ) \, dx \end{align*}

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

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]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 i \cdot 2^{\frac{1}{3}} a \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{1}{3} i \, f x + \frac{1}{3} i \, e\right )} + f{\rm integral}\left (-\frac{i \cdot 2^{\frac{1}{3}} a \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 )}}{f}, x\right )}{f} \end{align*}

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]

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

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

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]

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

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

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)