\(\int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx\) [442]

Optimal result

Integrand size = 30, antiderivative size = 88 \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {13}{6},\frac {1}{3},\frac {1}{2} (1-i \tan (c+d x))\right ) \sqrt [6]{1+i \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 \sqrt [6]{2} a d (e \sec (c+d x))^{4/3}} \]

[Out]

-3/16*I*hypergeom([-2/3, 13/6],[1/3],1/2-1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(1/2)*(1+I*tan(d*x+c))^(1/6)*2^(
5/6)/a/d/(e*sec(d*x+c))^(4/3)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 88, 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} \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 i \sqrt [6]{1+i \tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {13}{6},\frac {1}{3},\frac {1}{2} (1-i \tan (c+d x))\right )}{8 \sqrt [6]{2} a d (e \sec (c+d x))^{4/3}} \]

[In]

Int[1/((e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(((-3*I)/8)*Hypergeometric2F1[-2/3, 13/6, 1/3, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(1/6)*Sqrt[a + I*a
*Tan[c + d*x]])/(2^(1/6)*a*d*(e*Sec[c + d*x])^(4/3))

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 (c+d x))^{2/3} (a+i a \tan (c+d x))^{2/3}\right ) \int \frac {1}{(a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{7/6}} \, dx}{(e \sec (c+d x))^{4/3}} \\ & = \frac {\left (a^2 (a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{2/3}\right ) \text {Subst}\left (\int \frac {1}{(a-i a x)^{5/3} (a+i a x)^{13/6}} \, dx,x,\tan (c+d x)\right )}{d (e \sec (c+d x))^{4/3}} \\ & = \frac {\left ((a-i a \tan (c+d x))^{2/3} \sqrt {a+i a \tan (c+d x)} \sqrt [6]{\frac {a+i a \tan (c+d x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {1}{2}+\frac {i x}{2}\right )^{13/6} (a-i a x)^{5/3}} \, dx,x,\tan (c+d x)\right )}{4 \sqrt [6]{2} d (e \sec (c+d x))^{4/3}} \\ & = -\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {13}{6},\frac {1}{3},\frac {1}{2} (1-i \tan (c+d x))\right ) \sqrt [6]{1+i \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 \sqrt [6]{2} a d (e \sec (c+d x))^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 i \sec ^2(c+d x) \left (3+3 \cos (2 (c+d x))-55 \sqrt [6]{1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{6},\frac {5}{6},-e^{2 i (c+d x)}\right )+11 i \sin (2 (c+d x))\right )}{112 d (e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[1/((e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(((-3*I)/112)*Sec[c + d*x]^2*(3 + 3*Cos[2*(c + d*x)] - 55*(1 + E^((2*I)*(c + d*x)))^(1/6)*Hypergeometric2F1[-1
/6, 1/6, 5/6, -E^((2*I)*(c + d*x))] + (11*I)*Sin[2*(c + d*x)]))/(d*(e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c +
d*x]])

Maple [F]

\[\int \frac {1}{\left (e \sec \left (d x +c \right )\right )^{\frac {4}{3}} \sqrt {a +i a \tan \left (d x +c \right )}}d x\]

[In]

int(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

int(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x)

Fricas [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/112*(3*2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(7*I*e^(8*I*d*x + 8*I*
c) - 14*I*e^(7*I*d*x + 7*I*c) - 38*I*e^(6*I*d*x + 6*I*c) - 20*I*e^(5*I*d*x + 5*I*c) - 101*I*e^(4*I*d*x + 4*I*c
) + 2*I*e^(3*I*d*x + 3*I*c) - 60*I*e^(2*I*d*x + 2*I*c) + 8*I*e^(I*d*x + I*c) - 4*I)*e^(2/3*I*d*x + 2/3*I*c) -
112*(a*d*e^2*e^(5*I*d*x + 5*I*c) - 2*a*d*e^2*e^(4*I*d*x + 4*I*c) + a*d*e^2*e^(3*I*d*x + 3*I*c))*integral(-55/1
12*2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-I*e^(4*I*d*x + 4*I*c) - 7*I
*e^(3*I*d*x + 3*I*c) - 5*I*e^(2*I*d*x + 2*I*c) - 7*I*e^(I*d*x + I*c) - 4*I)*e^(2/3*I*d*x + 2/3*I*c)/(a*d*e^2*e
^(4*I*d*x + 4*I*c) - 3*a*d*e^2*e^(3*I*d*x + 3*I*c) + 3*a*d*e^2*e^(2*I*d*x + 2*I*c) - a*d*e^2*e^(I*d*x + I*c)),
x))/(a*d*e^2*e^(5*I*d*x + 5*I*c) - 2*a*d*e^2*e^(4*I*d*x + 4*I*c) + a*d*e^2*e^(3*I*d*x + 3*I*c))

Sympy [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {4}{3}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

[In]

integrate(1/(e*sec(d*x+c))**(4/3)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(1/((e*sec(c + d*x))**(4/3)*sqrt(I*a*(tan(c + d*x) - I))), x)

Maxima [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((e*sec(d*x + c))^(4/3)*sqrt(I*a*tan(d*x + c) + a)), x)

Giac [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(4/3)*sqrt(I*a*tan(d*x + c) + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

[In]

int(1/((e/cos(c + d*x))^(4/3)*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

int(1/((e/cos(c + d*x))^(4/3)*(a + a*tan(c + d*x)*1i)^(1/2)), x)