\(\int \frac {(e \sec (c+d x))^{5/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [438]

Optimal result

Integrand size = 30, antiderivative size = 86 \[ \int \frac {(e \sec (c+d x))^{5/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 i \sqrt [3]{2} a \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{5/3} (1+i \tan (c+d x))^{2/3}}{5 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3586, 3604, 72, 71} \[ \int \frac {(e \sec (c+d x))^{5/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 i \sqrt [3]{2} a (1+i \tan (c+d x))^{2/3} (e \sec (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.35 \[ \int \frac {(e \sec (c+d x))^{5/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 i 2^{2/3} e e^{i (c+d x)} \left (\frac {e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (-2+\sqrt [6]{1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {4}{3},-e^{2 i (c+d x)}\right )\right )}{d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((3*I)*2^(2/3)*e*E^(I*(c + d*x))*((e*E^(I*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^(2/3)*(-2 + (1 + E^((2*I)*(c
+ d*x)))^(1/6)*Hypergeometric2F1[1/6, 1/3, 4/3, -E^((2*I)*(c + d*x))]))/(d*Sqrt[a + I*a*Tan[c + d*x]])

Maple [F]

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

[In]

int((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

int((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x)

Fricas [F]

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

[In]

integrate((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-(6*2^(1/6)*(I*e*e^(2*I*d*x + 2*I*c) + I*e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(2
/3)*e^(2/3*I*d*x + 2/3*I*c) - a*d*integral(2^(1/6)*(I*e*e^(2*I*d*x + 2*I*c) + I*e)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*e^(-4/3*I*d*x - 4/3*I*c)/(a*d), x))/(a*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{5/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((e*sec(d*x+c))**(5/3)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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