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

Optimal. Leaf size=71 \[ \frac {12 i 2^{5/6} a^2 (d \sec (e+f x))^{5/3} \, _2F_1\left (-\frac {11}{6},\frac {5}{6};\frac {11}{6};\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}} \]

[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)

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Rubi [A]  time = 0.18, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3505, 3523, 70, 69} \[ \frac {12 i 2^{5/6} a^2 (d \sec (e+f x))^{5/3} \text {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}} \]

Antiderivative was successfully verified.

[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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 70

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 3505

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 3523

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*} \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx &=\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 ) \operatorname {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 ) \operatorname {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 \, _2F_1\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*}

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Mathematica [B]  time = 2.82, size = 267, normalized size = 3.76 \[ \frac {(a+i a \tan (e+f x))^2 (d \sec (e+f x))^{5/3} \left (\frac {3}{4} \csc (e) (\cos (2 e)-i \sin (2 e)) \sec ^{\frac {8}{3}}(e+f x) (64 i \sin (2 e+f x)+75 \cos (2 e+f x)+55 \cos (2 e+3 f x)-64 i \sin (f x)+90 \cos (f x))-\frac {33 i 2^{2/3} \left (5 \sqrt [3]{1+e^{2 i (e+f x)}}-\left (-1+e^{2 i e}\right ) e^{2 i f x} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i (e+f x)}\right )\right )}{\left (-1+e^{2 i e}\right ) \sqrt [3]{\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt [3]{1+e^{2 i (e+f x)}}}\right )}{80 f \sec ^{\frac {11}{3}}(e+f x) (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sec[e + f*x])^(5/3)*(((-33*I)*2^(2/3)*(5*(1 + E^((2*I)*(e + f*x)))^(1/3) - E^((2*I)*f*x)*(-1 + E^((2*I)*e)
)*Hypergeometric2F1[2/3, 5/6, 11/6, -E^((2*I)*(e + f*x))]))/((-1 + E^((2*I)*e))*(E^(I*(e + f*x))/(1 + E^((2*I)
*(e + f*x))))^(1/3)*(1 + E^((2*I)*(e + f*x)))^(1/3)) + (3*Csc[e]*Sec[e + f*x]^(8/3)*(Cos[2*e] - I*Sin[2*e])*(9
0*Cos[f*x] + 75*Cos[2*e + f*x] + 55*Cos[2*e + 3*f*x] - (64*I)*Sin[f*x] + (64*I)*Sin[2*e + f*x]))/4)*(a + I*a*T
an[e + f*x])^2)/(80*f*Sec[e + f*x]^(11/3)*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ \frac {2^{\frac {2}{3}} {\left (-165 i \, a^{2} d e^{\left (5 i \, f x + 5 i \, e\right )} - 78 i \, a^{2} d e^{\left (3 i \, f x + 3 i \, e\right )} - 33 i \, a^{2} d e^{\left (i \, f x + i \, e\right )}\right )} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} e^{\left (\frac {2}{3} i \, f x + \frac {2}{3} i \, e\right )} + 80 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} {\rm integral}\left (\frac {11 i \cdot 2^{\frac {2}{3}} a^{2} d \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} e^{\left (\frac {2}{3} i \, f x + \frac {2}{3} i \, e\right )}}{16 \, f}, x\right )}{80 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(2^(2/3)*(-165*I*a^2*d*e^(5*I*f*x + 5*I*e) - 78*I*a^2*d*e^(3*I*f*x + 3*I*e) - 33*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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