3.457 \(\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=91 \[ \frac {a^2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}} \]

[Out]

-a*b*(a*sin(f*x+e))^(1/2)/f/(b*sec(f*x+e))^(1/2)-1/2*a^2*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*Ellipti
cF(cos(e+1/4*Pi+f*x),2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2*f*x+2*e)^(1/2)/f/(a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2583, 2585, 2573, 2641} \[ \frac {a^2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[b*Sec[e + f*x]])) + (a^2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]
]*Sqrt[Sin[2*e + 2*f*x]])/(2*f*Sqrt[a*Sin[e + f*x]])

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 2583

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*b*(a*Sin[
e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(m - n)), x] + Dist[(a^2*(m - 1))/(m - n), Int[(a*Sin[e + f*x])
^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m - n, 0] && 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 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 \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx &=-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}+\frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}} \, dx\\ &=-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}+\frac {1}{2} \left (a^2 \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\\ &=-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}+\frac {\left (a^2 \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}{2 \sqrt {a \sin (e+f x)}}\\ &=-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}+\frac {a^2 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)}}{2 f \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 1.28, size = 66, normalized size = 0.73 \[ \frac {(a \sin (e+f x))^{5/2} (b \sec (e+f x))^{3/2} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {1}{2};\sec ^2(e+f x)\right )}{a b f \left (-\tan ^2(e+f x)\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[-1/2, -1/4, 1/2, Sec[e + f*x]^2]*(b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(5/2))/(a*b*f*(-Ta
n[e + f*x]^2)^(5/4))

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right )} a \sin \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))*a*sin(f*x + e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 184, normalized size = 2.02 \[ -\frac {\left (\sin \left (f x +e \right ) \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}-\cos \left (f x +e \right ) \sqrt {2}\right ) \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\frac {b}{\cos \left (f x +e \right )}}\, \sqrt {2}}{2 f \left (-1+\cos \left (f x +e \right )\right ) \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(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))+cos(f*x
+e)^2*2^(1/2)-cos(f*x+e)*2^(1/2))*(a*sin(f*x+e))^(3/2)*(b/cos(f*x+e))^(1/2)/(-1+cos(f*x+e))/sin(f*x+e)*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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