3.465 \(\int \frac{1}{\sqrt{b \sec (e+f x)} \sin ^{\frac{7}{2}}(e+f x)} \, dx\)

Optimal. Leaf size=115 \[ -\frac{2 b}{5 f \sin ^{\frac{5}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{4 b}{5 f \sqrt{\sin (e+f x)} (b \sec (e+f x))^{3/2}}-\frac{4 \sqrt{\sin (e+f x)} E\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{5 f \sqrt{\sin (2 e+2 f x)} \sqrt{b \sec (e+f x)}} \]

[Out]

(-2*b)/(5*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(5/2)) - (4*b)/(5*f*(b*Sec[e + f*x])^(3/2)*Sqrt[Sin[e + f*x]])
- (4*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[Sin[e + f*x]])/(5*f*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])

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Rubi [A]  time = 0.162471, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2584, 2585, 2572, 2639} \[ -\frac{2 b}{5 f \sin ^{\frac{5}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{4 b}{5 f \sqrt{\sin (e+f x)} (b \sec (e+f x))^{3/2}}-\frac{4 \sqrt{\sin (e+f x)} E\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{5 f \sqrt{\sin (2 e+2 f x)} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(7/2)),x]

[Out]

(-2*b)/(5*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(5/2)) - (4*b)/(5*f*(b*Sec[e + f*x])^(3/2)*Sqrt[Sin[e + f*x]])
- (4*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[Sin[e + f*x]])/(5*f*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])

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

Mathematica [C]  time = 0.461951, size = 82, normalized size = 0.71 \[ \frac{2 b \left (2 \sin ^2(e+f x) \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{1}{2};\sec ^2(e+f x)\right )+\cos (2 (e+f x))-2\right )}{5 f \sin ^{\frac{5}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(7/2)),x]

[Out]

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

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Maple [B]  time = 0.159, size = 1030, normalized size = 9. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(f*x+e)^(7/2)/(b*sec(f*x+e))^(1/2),x)

[Out]

-16/5/f*2^(1/2)*(-1+cos(f*x+e))^4*(4*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)*EllipticE(((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)^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))+4*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)*EllipticE(((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)^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*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))+2*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*2^(1/2)*cos(f*x+e)^3-4*((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)*EllipticE(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+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))+2^(1/2)*cos(f*x+e)^2+2*2^(1/2)*cos(f*x+e))/sin(f*x+e)^(5/2)/
(b/cos(f*x+e))^(1/2)/(-1+cos(f*x+e)+sin(f*x+e))/(1-cos(f*x+e)+sin(f*x+e))/(sin(f*x+e)^2+cos(f*x+e)^2-2*cos(f*x
+e)+1)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(7/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(7/2)), 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{\sin \left (f x + e\right )}}{{\left (b \cos \left (f x + e\right )^{4} - 2 \, b \cos \left (f x + e\right )^{2} + b\right )} \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(7/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))*sqrt(sin(f*x + e))/((b*cos(f*x + e)^4 - 2*b*cos(f*x + e)^2 + 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(1/sin(f*x+e)**(7/2)/(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{1}{\sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(7/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(7/2)), x)