\(\int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx\) [267]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {12 i 2^{5/6} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3}}{5 f (1+i \tan (e+f x))^{5/6}} \]

[Out]

12/5*I*2^(5/6)*a^2*hypergeom([-11/6, 5/6],[11/6],1/2-1/2*I*tan(f*x+e))*(d*sec(f*x+e))^(5/3)/f/(1+I*tan(f*x+e))
^(5/6)

Rubi [A] (verified)

Time = 0.23 (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} \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {12 i 2^{5/6} a^2 (d \sec (e+f x))^{5/3} \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}} \]

[In]

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

[Out]

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

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 {(d \sec (e+f x))^{5/3} \int (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{17/6} \, 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 {(a+i a x)^{11/6}}{\sqrt [6]{a-i a x}} \, 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 (2\ 2^{5/6} a^3 (d \sec (e+f x))^{5/3}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{11/6}}{\sqrt [6]{a-i a x}} \, dx,x,\tan (e+f x)\right )}{f (a-i a \tan (e+f x))^{5/6} \left (\frac {a+i a \tan (e+f x)}{a}\right )^{5/6}} \\ & = \frac {12 i 2^{5/6} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3}}{5 f (1+i \tan (e+f x))^{5/6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.56 \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {3 i a^2 (d \sec (e+f x))^{5/3} \left (i \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+i \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 \sqrt {-\tan ^2(e+f x)}\right )}{5 f \sqrt {-\tan ^2(e+f x)}} \]

[In]

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

[Out]

(((3*I)/5)*a^2*(d*Sec[e + f*x])^(5/3)*(I*Hypergeometric2F1[-1/2, 5/6, 11/6, Sec[e + f*x]^2]*Tan[e + f*x] + I*H
ypergeometric2F1[1/2, 5/6, 11/6, Sec[e + f*x]^2]*Tan[e + f*x] + 2*Sqrt[-Tan[e + f*x]^2]))/(f*Sqrt[-Tan[e + f*x
]^2])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

-1/80*(3*2^(2/3)*(55*I*a^2*d*e^(5*I*f*x + 5*I*e) + 26*I*a^2*d*e^(3*I*f*x + 3*I*e) + 11*I*a^2*d*e^(I*f*x + I*e)
)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e) - 80*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2
*I*e) + f)*integral(11/16*I*2^(2/3)*a^2*d*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e)/f, x))/(
f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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