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]

-2/5*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(5/2)-4/5*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(1/2)+4/5*(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 = 0.16, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [C]  time = 0.47, 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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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right )} \sqrt {\sin \left (f x + e\right )}}{{\left (b \cos \left (f x + e\right )^{4} - 2 \, b \cos \left (f x + e\right )^{2} + b\right )} \sec \left (f x + e\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {7}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.18, size = 1030, normalized size = 8.96 \[ \text {result too large to display} \]

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*(-1+cos(f*x+e))^4*(4*cos(f*x+e)^3*((-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))*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))^(1/2)-2*cos(f*x+e)^3*((-1+cos(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))/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))^(1/2)*EllipticE(((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)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)-2*cos(
f*x+e)^2*((-1+cos(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))/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)+si
n(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)*E
llipticF(((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*2^(1/2)-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)*El
lipticE(((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)+s
in(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+cos(f*x+e)^2*2^(1/2)+2*cos(f*x+e)*2^(1/2))/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
*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {7}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (e+f\,x\right )}^{7/2}\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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