3.462 \(\int \frac{\sin ^{\frac{5}{2}}(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\)

Optimal. Leaf size=85 \[ \frac{\sqrt{\sin (e+f x)} E\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{2 f \sqrt{\sin (2 e+2 f x)} \sqrt{b \sec (e+f x)}}-\frac{b \sin ^{\frac{3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}} \]

[Out]

-(b*Sin[e + f*x]^(3/2))/(3*f*(b*Sec[e + f*x])^(3/2)) + (EllipticE[e - Pi/4 + f*x, 2]*Sqrt[Sin[e + f*x]])/(2*f*
Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])

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Rubi [A]  time = 0.118849, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2583, 2585, 2572, 2639} \[ \frac{\sqrt{\sin (e+f x)} E\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{2 f \sqrt{\sin (2 e+2 f x)} \sqrt{b \sec (e+f x)}}-\frac{b \sin ^{\frac{3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^(5/2)/Sqrt[b*Sec[e + f*x]],x]

[Out]

-(b*Sin[e + f*x]^(3/2))/(3*f*(b*Sec[e + f*x])^(3/2)) + (EllipticE[e - Pi/4 + f*x, 2]*Sqrt[Sin[e + f*x]])/(2*f*
Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*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 2572

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

Rule 2639

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

Rubi steps

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

Mathematica [C]  time = 0.292999, size = 74, normalized size = 0.87 \[ \frac{b \left (-3 \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{1}{2};\sec ^2(e+f x)\right )+\cos (2 (e+f x))-1\right )}{6 f \sqrt{\sin (e+f x)} (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^(5/2)/Sqrt[b*Sec[e + f*x]],x]

[Out]

(b*(-1 + Cos[2*(e + f*x)] - 3*Hypergeometric2F1[-1/2, 1/4, 1/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(1/4)))/(6*f
*(b*Sec[e + f*x])^(3/2)*Sqrt[Sin[e + f*x]])

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Maple [B]  time = 0.136, size = 511, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x)

[Out]

1/12/f*2^(1/2)*(2*2^(1/2)*cos(f*x+e)^4+3*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))-6*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)*EllipticE(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),
1/2*2^(1/2))+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))-6*((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)*EllipticE(((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)^2+3*2^(1/2)*co
s(f*x+e))/cos(f*x+e)/sin(f*x+e)^(1/2)/(b/cos(f*x+e))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{\frac{5}{2}}}{\sqrt{b \sec \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^(5/2)/sqrt(b*sec(f*x + e)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt{b \sec \left (f x + e\right )} \sqrt{\sin \left (f x + e\right )}}{b \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-(cos(f*x + e)^2 - 1)*sqrt(b*sec(f*x + e))*sqrt(sin(f*x + e))/(b*sec(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(sin(f*x+e)**(5/2)/(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{\frac{5}{2}}}{\sqrt{b \sec \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(5/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^(5/2)/sqrt(b*sec(f*x + e)), x)