\(\int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx\) [268]

Optimal result

Integrand size = 28, antiderivative size = 69 \[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=\frac {12 i \sqrt [6]{2} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{6},\frac {7}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [3]{d \sec (e+f x)}}{f \sqrt [6]{1+i \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3586, 3604, 72, 71} \[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=\frac {12 i \sqrt [6]{2} a^2 \sqrt [3]{d \sec (e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{6},\frac {7}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [6]{1+i \tan (e+f x)}} \]

[In]

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

[Out]

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

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*} \text {integral}& = \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))^{13/6} \, 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 {(a+i a x)^{7/6}}{(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 (2 \sqrt [6]{2} a^3 \sqrt [3]{d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{7/6}}{(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 {12 i \sqrt [6]{2} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {7}{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] (verified)

Time = 0.37 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.48 \[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=\frac {3 a^2 \sqrt [3]{d \sec (e+f x)} \left (2 i+\cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{6},\frac {7}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}+\cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{2}d x\]

[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)

Fricas [F]

\[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]

[In]

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

[Out]

-1/4*(3*2^(1/3)*(-9*I*a^2*e^(2*I*f*x + 2*I*e) - 7*I*a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(1/3*I*f*x + 1/
3*I*e) - 4*(f*e^(2*I*f*x + 2*I*e) + f)*integral(-7/4*I*2^(1/3)*a^2*(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*e^(2*I*f*x + 2*I*e) + f)

Sympy [F]

\[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=- a^{2} \left (\int \left (- \sqrt [3]{d \sec {\left (e + f x \right )}}\right )\, dx + \int \sqrt [3]{d \sec {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i \sqrt [3]{d \sec {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx\right ) \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]

[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)