\(\int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx\) [270]

Optimal result

Integrand size = 28, antiderivative size = 85 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=-\frac {6 i \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \left (a^2+i a^2 \tan (e+f x)\right )}{5 f (d \sec (e+f x))^{5/3} \sqrt [6]{1+i \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 85, 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} \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=-\frac {6 i \sqrt [6]{2} \left (a^2+i a^2 \tan (e+f x)\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f \sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3}} \]

[In]

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

[Out]

(((-6*I)/5)*2^(1/6)*Hypergeometric2F1[-5/6, -1/6, 1/6, (1 - I*Tan[e + f*x])/2]*(a^2 + I*a^2*Tan[e + f*x]))/(f*
(d*Sec[e + f*x])^(5/3)*(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 {\left ((a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \int \frac {(a+i a \tan (e+f x))^{7/6}}{(a-i a \tan (e+f x))^{5/6}} \, dx}{(d \sec (e+f x))^{5/3}} \\ & = \frac {\left (a^2 (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \text {Subst}\left (\int \frac {\sqrt [6]{a+i a x}}{(a-i a x)^{11/6}} \, dx,x,\tan (e+f x)\right )}{f (d \sec (e+f x))^{5/3}} \\ & = \frac {\left (\sqrt [6]{2} a^2 (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))\right ) \text {Subst}\left (\int \frac {\sqrt [6]{\frac {1}{2}+\frac {i x}{2}}}{(a-i a x)^{11/6}} \, dx,x,\tan (e+f x)\right )}{f (d \sec (e+f x))^{5/3} \sqrt [6]{\frac {a+i a \tan (e+f x)}{a}}} \\ & = -\frac {6 i \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \left (a^2+i a^2 \tan (e+f x)\right )}{5 f (d \sec (e+f x))^{5/3} \sqrt [6]{1+i \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=\frac {3 a^2 \left (\operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{2},\frac {1}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+\operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\sec ^2(e+f x)\right ) \tan (e+f x)-2 i \sqrt {-\tan ^2(e+f x)}\right )}{5 f (d \sec (e+f x))^{5/3} \sqrt {-\tan ^2(e+f x)}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{2}}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}d x\]

[In]

int((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(5/3),x)

[Out]

int((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(5/3),x)

Fricas [F]

\[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=- a^{2} \left (\int \left (- \frac {1}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\right )\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(f*x+e))**2/(d*sec(f*x+e))**(5/3),x)

[Out]

-a**2*(Integral(-1/(d*sec(e + f*x))**(5/3), x) + Integral(tan(e + f*x)**2/(d*sec(e + f*x))**(5/3), x) + Integr
al(-2*I*tan(e + f*x)/(d*sec(e + f*x))**(5/3), x))

Maxima [F]

\[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(5/3), x)

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(5/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}} \,d x \]

[In]

int((a + a*tan(e + f*x)*1i)^2/(d/cos(e + f*x))^(5/3),x)

[Out]

int((a + a*tan(e + f*x)*1i)^2/(d/cos(e + f*x))^(5/3), x)