\(\int \frac {(d \sec (e+f x))^{5/3}}{a+i a \tan (e+f x)} \, dx\) [271]

Optimal result

Integrand size = 28, antiderivative size = 83 \[ \int \frac {(d \sec (e+f x))^{5/3}}{a+i a \tan (e+f x)} \, dx=\frac {3 i \operatorname {Hypergeometric2F1}\left (\frac {5}{6},\frac {7}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3} \sqrt [6]{1+i \tan (e+f x)}}{5 \sqrt [6]{2} f (a+i a \tan (e+f x))} \]

[Out]

3/10*I*hypergeom([5/6, 7/6],[11/6],1/2-1/2*I*tan(f*x+e))*(d*sec(f*x+e))^(5/3)*(1+I*tan(f*x+e))^(1/6)*2^(5/6)/f
/(a+I*a*tan(f*x+e))

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 83, 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} \[ \int \frac {(d \sec (e+f x))^{5/3}}{a+i a \tan (e+f x)} \, dx=\frac {3 i \sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {5}{6},\frac {7}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 \sqrt [6]{2} f (a+i a \tan (e+f x))} \]

[In]

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

[Out]

(((3*I)/5)*Hypergeometric2F1[5/6, 7/6, 11/6, (1 - I*Tan[e + f*x])/2]*(d*Sec[e + f*x])^(5/3)*(1 + I*Tan[e + f*x
])^(1/6))/(2^(1/6)*f*(a + I*a*Tan[e + f*x]))

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

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01 \[ \int \frac {(d \sec (e+f x))^{5/3}}{a+i a \tan (e+f x)} \, dx=\frac {6 d e^{i (e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {2}{3},\frac {5}{6},-e^{2 i (e+f x)}\right ) (d \sec (e+f x))^{2/3}}{a \sqrt [3]{1+e^{2 i (e+f x)}} f (-i+\tan (e+f x))} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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