3.437 \(\int \frac {(e \sec (c+d x))^{7/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=86 \[ \frac {3 i 2^{2/3} a \sqrt [3]{1+i \tan (c+d x)} (e \sec (c+d x))^{7/3} \, _2F_1\left (\frac {1}{3},\frac {7}{6};\frac {13}{6};\frac {1}{2} (1-i \tan (c+d x))\right )}{7 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

3/7*I*2^(2/3)*a*hypergeom([1/3, 7/6],[13/6],1/2-1/2*I*tan(d*x+c))*(e*sec(d*x+c))^(7/3)*(1+I*tan(d*x+c))^(1/3)/
d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.21, 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 = {3505, 3523, 70, 69} \[ \frac {3 i 2^{2/3} a \sqrt [3]{1+i \tan (c+d x)} (e \sec (c+d x))^{7/3} \text {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{6},\frac {13}{6},\frac {1}{2} (1-i \tan (c+d x))\right )}{7 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(7/3)/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((3*I)/7)*2^(2/3)*a*Hypergeometric2F1[1/3, 7/6, 13/6, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(7/3)*(1 + I*T
an[c + d*x])^(1/3))/(d*(a + I*a*Tan[c + d*x])^(3/2))

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 \frac {(e \sec (c+d x))^{7/3}}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {(e \sec (c+d x))^{7/3} \int (a-i a \tan (c+d x))^{7/6} (a+i a \tan (c+d x))^{2/3} \, dx}{(a-i a \tan (c+d x))^{7/6} (a+i a \tan (c+d x))^{7/6}}\\ &=\frac {\left (a^2 (e \sec (c+d x))^{7/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [6]{a-i a x}}{\sqrt [3]{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{7/6} (a+i a \tan (c+d x))^{7/6}}\\ &=\frac {\left (a^2 (e \sec (c+d x))^{7/3} \sqrt [3]{\frac {a+i a \tan (c+d x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [6]{a-i a x}}{\sqrt [3]{\frac {1}{2}+\frac {i x}{2}}} \, dx,x,\tan (c+d x)\right )}{\sqrt [3]{2} d (a-i a \tan (c+d x))^{7/6} (a+i a \tan (c+d x))^{3/2}}\\ &=\frac {3 i 2^{2/3} a \, _2F_1\left (\frac {1}{3},\frac {7}{6};\frac {13}{6};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{7/3} \sqrt [3]{1+i \tan (c+d x)}}{7 d (a+i a \tan (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 118, normalized size = 1.37 \[ -\frac {3 i \sqrt [3]{2} e e^{i (c+d x)} \left (\frac {e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{4/3} \left (4+\left (1+e^{2 i (c+d x)}\right )^{5/6} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {5}{3};-e^{2 i (c+d x)}\right )\right )}{5 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(7/3)/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ \frac {-6 i \cdot 2^{\frac {5}{6}} e^{2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} + 5 \, a d {\rm integral}\left (-\frac {2 i \cdot 2^{\frac {5}{6}} e^{2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {1}{3} i \, d x + \frac {1}{3} i \, c\right )}}{5 \, a d}, x\right )}{5 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {7}{3}}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(7/3)/sqrt(I*a*tan(d*x + c) + a), x)

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maple [F]  time = 1.16, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x +c \right )\right )^{\frac {7}{3}}}{\sqrt {a +i a \tan \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(7/3)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

int((e*sec(d*x+c))^(7/3)/(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 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {7}{3}}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(7/3)/sqrt(I*a*tan(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/3}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(7/3)/(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int((e/cos(c + d*x))^(7/3)/(a + a*tan(c + d*x)*1i)^(1/2), 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((e*sec(d*x+c))**(7/3)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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