\(\int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))} \, dx\) [274]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))} \, dx=-\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {17}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}} \]

[Out]

-3/20*I*hypergeom([-5/6, 17/6],[1/6],1/2-1/2*I*tan(f*x+e))*(1+I*tan(f*x+e))^(5/6)*2^(1/6)/a/f/(d*sec(f*x+e))^(
5/3)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 71, 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} \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))} \, dx=-\frac {3 i (1+i \tan (e+f x))^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {17}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}} \]

[In]

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

[Out]

(((-3*I)/10)*Hypergeometric2F1[-5/6, 17/6, 1/6, (1 - I*Tan[e + f*x])/2]*(1 + I*Tan[e + f*x])^(5/6))/(2^(5/6)*a
*f*(d*Sec[e + f*x])^(5/3))

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 {1}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{11/6}} \, 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 {1}{(a-i a x)^{11/6} (a+i a x)^{17/6}} \, dx,x,\tan (e+f x)\right )}{f (d \sec (e+f x))^{5/3}} \\ & = \frac {\left ((a-i a \tan (e+f x))^{5/6} \left (\frac {a+i a \tan (e+f x)}{a}\right )^{5/6}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {1}{2}+\frac {i x}{2}\right )^{17/6} (a-i a x)^{11/6}} \, dx,x,\tan (e+f x)\right )}{4\ 2^{5/6} f (d \sec (e+f x))^{5/3}} \\ & = -\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {17}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.89 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.68 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))} \, dx=-\frac {3 \sec ^2(e+f x) \left (-26+6 \cos (2 (e+f x))+\frac {128 e^{2 i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-e^{2 i (e+f x)}\right )}{\left (1+e^{2 i (e+f x)}\right )^{2/3}}+16 i \sin (2 (e+f x))\right )}{220 a f (d \sec (e+f x))^{5/3} (-i+\tan (e+f x))} \]

[In]

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

[Out]

(-3*Sec[e + f*x]^2*(-26 + 6*Cos[2*(e + f*x)] + (128*E^((2*I)*(e + f*x))*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((
2*I)*(e + f*x))])/(1 + E^((2*I)*(e + f*x)))^(2/3) + (16*I)*Sin[2*(e + f*x)]))/(220*a*f*(d*Sec[e + f*x])^(5/3)*
(-I + Tan[e + f*x]))

Maple [F]

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

[In]

int(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e)),x)

[Out]

int(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e)),x)

Fricas [F]

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

[In]

integrate(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/440*(440*a*d^2*f*e^(4*I*f*x + 4*I*e)*integral(-16/55*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*d^2*f), x) - 3*2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*(11*I*e^(6*I*f*x + 6*I*e) - 15*I
*e^(4*I*f*x + 4*I*e) - 31*I*e^(2*I*f*x + 2*I*e) - 5*I)*e^(1/3*I*f*x + 1/3*I*e))*e^(-4*I*f*x - 4*I*e)/(a*d^2*f)

Sympy [F]

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

[In]

integrate(1/(d*sec(f*x+e))**(5/3)/(a+I*a*tan(f*x+e)),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

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

[In]

integrate(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate(1/((d*sec(f*x + e))^(5/3)*(I*a*tan(f*x + e) + a)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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