3.482 \(\int \frac{1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac{4 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{77 a^6 b^2 f \sqrt{a \sin (e+f x)}}+\frac{4}{77 a^5 b f (a \sin (e+f x))^{3/2} \sqrt{b \sec (e+f x)}}+\frac{2}{77 a^3 b f (a \sin (e+f x))^{7/2} \sqrt{b \sec (e+f x)}}-\frac{2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt{b \sec (e+f x)}} \]

[Out]

-2/(11*a*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(11/2)) + 2/(77*a^3*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x
])^(7/2)) + 4/(77*a^5*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2)) - (4*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[
b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(77*a^6*b^2*f*Sqrt[a*Sin[e + f*x]])

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Rubi [A]  time = 0.282325, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2581, 2584, 2585, 2573, 2641} \[ -\frac{4 \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{b \sec (e+f x)}}{77 a^6 b^2 f \sqrt{a \sin (e+f x)}}+\frac{4}{77 a^5 b f (a \sin (e+f x))^{3/2} \sqrt{b \sec (e+f x)}}+\frac{2}{77 a^3 b f (a \sin (e+f x))^{7/2} \sqrt{b \sec (e+f x)}}-\frac{2}{11 a b f (a \sin (e+f x))^{11/2} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(13/2)),x]

[Out]

-2/(11*a*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(11/2)) + 2/(77*a^3*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x
])^(7/2)) + 4/(77*a^5*b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2)) - (4*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[
b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(77*a^6*b^2*f*Sqrt[a*Sin[e + f*x]])

Rule 2581

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((a*Sin[e + f
*x])^(m + 1)*(b*Sec[e + f*x])^(n + 1))/(a*b*f*(m + 1)), x] - Dist[(n + 1)/(a^2*b^2*(m + 1)), Int[(a*Sin[e + f*
x])^(m + 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && LtQ[m, -1] && Integers
Q[2*m, 2*n]

Rule 2584

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(a*Sin[e +
f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + 1)), x] + Dist[(m - n + 2)/(a^2*(m + 1)), Int[(a*Sin[e + f*
x])^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2585

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(b*Cos[e + f*
x])^n*(b*Sec[e + f*x])^n, Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&
IntegerQ[m - 1/2] && IntegerQ[n - 1/2]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{13/2}} \, dx &=-\frac{2}{11 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{11/2}}-\frac{\int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{9/2}} \, dx}{11 a^2 b^2}\\ &=-\frac{2}{11 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{11/2}}+\frac{2}{77 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}-\frac{6 \int \frac{\sqrt{b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx}{77 a^4 b^2}\\ &=-\frac{2}{11 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{11/2}}+\frac{2}{77 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{4}{77 a^5 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{4 \int \frac{\sqrt{b \sec (e+f x)}}{\sqrt{a \sin (e+f x)}} \, dx}{77 a^6 b^2}\\ &=-\frac{2}{11 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{11/2}}+\frac{2}{77 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{4}{77 a^5 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{\left (4 \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{b \cos (e+f x)} \sqrt{a \sin (e+f x)}} \, dx}{77 a^6 b^2}\\ &=-\frac{2}{11 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{11/2}}+\frac{2}{77 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{4}{77 a^5 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{\left (4 \sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{77 a^6 b^2 \sqrt{a \sin (e+f x)}}\\ &=-\frac{2}{11 a b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{11/2}}+\frac{2}{77 a^3 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{7/2}}+\frac{4}{77 a^5 b f \sqrt{b \sec (e+f x)} (a \sin (e+f x))^{3/2}}-\frac{4 F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{b \sec (e+f x)} \sqrt{\sin (2 e+2 f x)}}{77 a^6 b^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.20343, size = 131, normalized size = 0.75 \[ \frac{2 \cot (2 (e+f x)) \csc (2 (e+f x)) \sqrt{a \sin (e+f x)} \left (8 \left (-\tan ^2(e+f x)\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\sec ^2(e+f x)\right )+(6 \cos (2 (e+f x))-\cos (4 (e+f x))+23) \csc ^4(e+f x)\right )}{77 a^7 b f \left (\sec ^2(e+f x)-2\right ) \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(13/2)),x]

[Out]

(2*Cot[2*(e + f*x)]*Csc[2*(e + f*x)]*Sqrt[a*Sin[e + f*x]]*((23 + 6*Cos[2*(e + f*x)] - Cos[4*(e + f*x)])*Csc[e
+ f*x]^4 + 8*Hypergeometric2F1[1/2, 3/4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(77*a^7*b*f*Sqrt[b*Sec
[e + f*x]]*(-2 + Sec[e + f*x]^2))

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Maple [B]  time = 0.168, size = 793, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x)

[Out]

-1/77/f*2^(1/2)*(4*sin(f*x+e)*cos(f*x+e)^5*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1
/2),1/2*2^(1/2))+4*sin(f*x+e)*cos(f*x+e)^4*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1
/2),1/2*2^(1/2))-8*sin(f*x+e)*cos(f*x+e)^3*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1
/2),1/2*2^(1/2))-8*sin(f*x+e)*cos(f*x+e)^2*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1
/2),1/2*2^(1/2))+4*sin(f*x+e)*cos(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+
e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2
),1/2*2^(1/2))-2*cos(f*x+e)^5*2^(1/2)+4*sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e
)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*
x+e))^(1/2),1/2*2^(1/2))+5*2^(1/2)*cos(f*x+e)^3+4*2^(1/2)*cos(f*x+e))*sin(f*x+e)/cos(f*x+e)^2/(b/cos(f*x+e))^(
3/2)/(a*sin(f*x+e))^(13/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (a \sin \left (f x + e\right )\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(f*x + e))^(3/2)*(a*sin(f*x + e))^(13/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sec \left (f x + e\right )} \sqrt{a \sin \left (f x + e\right )}}{{\left (a^{7} b^{2} \cos \left (f x + e\right )^{6} - 3 \, a^{7} b^{2} \cos \left (f x + e\right )^{4} + 3 \, a^{7} b^{2} \cos \left (f x + e\right )^{2} - a^{7} b^{2}\right )} \sec \left (f x + e\right )^{2} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))/((a^7*b^2*cos(f*x + e)^6 - 3*a^7*b^2*cos(f*x + e)^4 + 3*a^
7*b^2*cos(f*x + e)^2 - a^7*b^2)*sec(f*x + e)^2*sin(f*x + e)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))**(3/2)/(a*sin(f*x+e))**(13/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (a \sin \left (f x + e\right )\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(13/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(f*x + e))^(3/2)*(a*sin(f*x + e))^(13/2)), x)