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

Optimal. Leaf size=67 \[ -\frac{3 i 2^{5/6} a \sqrt [6]{1+i \tan (e+f x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{6},\frac{5}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [3]{d \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.154474, 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} \[ -\frac{3 i 2^{5/6} a \sqrt [6]{1+i \tan (e+f x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{6},\frac{5}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [3]{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*2^(5/6)*a*Hypergeometric2F1[-1/6, 1/6, 5/6, (1 - I*Tan[e + f*x])/2]*(1 + I*Tan[e + f*x])^(1/6))/(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{a+i a \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx &=\frac{\left (\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}\right ) \int \frac{(a+i a \tan (e+f x))^{5/6}}{\sqrt [6]{a-i a \tan (e+f x)}} \, 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} \sqrt [6]{a+i a x}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)}}\\ &=\frac{\left (a^2 \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}{\sqrt [6]{\frac{1}{2}+\frac{i x}{2}} (a-i a x)^{7/6}} \, dx,x,\tan (e+f x)\right )}{\sqrt [6]{2} f \sqrt [3]{d \sec (e+f x)}}\\ &=-\frac{3 i 2^{5/6} a \, _2F_1\left (-\frac{1}{6},\frac{1}{6};\frac{5}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) \sqrt [6]{1+i \tan (e+f x)}}{f \sqrt [3]{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.440552, size = 98, normalized size = 1.46 \[ -\frac{3 i 2^{2/3} a e^{2 i (e+f x)} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (e+f x)}\right )}{5 f \sqrt [3]{1+e^{2 i (e+f x)}} \sqrt [3]{\frac{d e^{i (e+f x)}}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/5)*2^(2/3)*a*E^((2*I)*(e + f*x))*Hypergeometric2F1[2/3, 5/6, 11/6, -E^((2*I)*(e + f*x))])/(((d*E^(I*(
e + f*x)))/(1 + E^((2*I)*(e + f*x))))^(1/3)*(1 + E^((2*I)*(e + f*x)))^(1/3)*f)

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Maple [F]  time = 0.143, size = 0, normalized size = 0. \begin{align*} \int{(a+ia\tan \left ( fx+e \right ) ){\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((a+I*a*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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