3.5.64 \(\int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {3}{2}}(e+f x)} \, dx\) [464]

Optimal. Leaf size=81 \[ -\frac {2 b}{f (b \sec (e+f x))^{3/2} \sqrt {\sin (e+f x)}}-\frac {2 E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (e+f x)}}{f \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}} \]

[Out]

-2*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(1/2)+2*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticE(cos(e+1
/4*Pi+f*x),2^(1/2))*sin(f*x+e)^(1/2)/f/(b*sec(f*x+e))^(1/2)/sin(2*f*x+2*e)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 81, normalized size of antiderivative = 1.00, 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 = {2664, 2665, 2652, 2719} \begin {gather*} -\frac {2 b}{f \sqrt {\sin (e+f x)} (b \sec (e+f x))^{3/2}}-\frac {2 \sqrt {\sin (e+f x)} E\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{f \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2652

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 2664

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 2665

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 2719

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.19, size = 63, normalized size = 0.78 \begin {gather*} \frac {2 b \left (-1+\, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};\sec ^2(e+f x)\right ) \sqrt [4]{-\tan ^2(e+f x)}\right )}{f (b \sec (e+f x))^{3/2} \sqrt {\sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(481\) vs. \(2(96)=192\).
time = 0.21, size = 482, normalized size = 5.95

method	result	size
default	\(-\frac {\left (-2 \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\sin \left (f x +e \right )-1+\cos \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 )}}\, \EllipticE \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 ) \cos \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 {\sin \left (f x +e \right )-1+\cos \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 ) \cos \left (f x +e \right )-2 \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\sin \left (f x +e \right )-1+\cos \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 )}}\, \EllipticE \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 )+\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\sin \left (f x +e \right )-1+\cos \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 )+\cos \left (f x +e \right ) \sqrt {2}\right ) \sqrt {2}}{f \cos \left (f x +e \right ) \sqrt {\sin \left (f x +e \right )}\, \sqrt {\frac {b}{\cos \left (f x +e \right )}}}\)	\(482\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*(-2*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(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))*cos(f*x+e)+((1-cos
(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(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*((1-cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))^(1/2)*((sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*Ellipt
icE(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((s
in(f*x+e)-1+cos(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^(1/2))/cos(f*x+e)/sin(f*x+e)^(1/2)/(b/cos(f*x+e))^(1/2)*2^(1/2
)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \sec {\left (e + f x \right )}} \sin ^{\frac {3}{2}}{\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(b*sec(e + f*x))*sin(e + f*x)**(3/2)), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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