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

Optimal. Leaf size=86 \[ \frac {3 i \sqrt [3]{2} a \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{5/3} (1+i \tan (c+d x))^{2/3}}{5 d (a+i a \tan (c+d x))^{3/2}} \]

[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)

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Rubi [A]
time = 0.14, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3586, 3604, 72, 71} \begin {gather*} \frac {3 i \sqrt [3]{2} a (1+i \tan (c+d x))^{2/3} (e \sec (c+d x))^{5/3} \, _2F_1\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}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {(e \sec (c+d x))^{5/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\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 \, _2F_1\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*}

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Mathematica [A]
time = 0.67, size = 116, normalized size = 1.35 \begin {gather*} \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)}} \, _2F_1\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)}} \end {gather*}

Antiderivative was successfully verified.

[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]])

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Maple [F]
time = 0.99, size = 0, normalized size = 0.00 \[\int \frac {\left (e \sec \left (d x +c \right )\right )^{\frac {5}{3}}}{\sqrt {a +i a \tan \left (d x +c \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*d*integral(2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(I*e^(5/3) + I*e^(2*I*d*x + 2*I*c + 5/3))*e^(-4/3*I*d*
x - 4/3*I*c)/(a*d*(e^(2*I*d*x + 2*I*c) + 1)^(2/3)), x) - 6*2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(I*e^(5/3
) + I*e^(2*I*d*x + 2*I*c + 5/3))*e^(2/3*I*d*x + 2/3*I*c)/(e^(2*I*d*x + 2*I*c) + 1)^(2/3))/(a*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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