3.24 \(\int \frac {1}{\sin ^{\frac {7}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=70 \[ -\frac {6 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}-\frac {6 \cos (a+b x)}{5 b \sqrt {\sin (a+b x)}} \]

[Out]

6/5*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))
/b-2/5*cos(b*x+a)/b/sin(b*x+a)^(5/2)-6/5*cos(b*x+a)/b/sin(b*x+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2636, 2639} \[ -\frac {6 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}-\frac {6 \cos (a+b x)}{5 b \sqrt {\sin (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^(-7/2),x]

[Out]

(-6*EllipticE[(a - Pi/2 + b*x)/2, 2])/(5*b) - (2*Cos[a + b*x])/(5*b*Sin[a + b*x]^(5/2)) - (6*Cos[a + b*x])/(5*
b*Sqrt[Sin[a + b*x]])

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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}{\sin ^{\frac {7}{2}}(a+b x)} \, dx &=-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}+\frac {3}{5} \int \frac {1}{\sin ^{\frac {3}{2}}(a+b x)} \, dx\\ &=-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}-\frac {6 \cos (a+b x)}{5 b \sqrt {\sin (a+b x)}}-\frac {3}{5} \int \sqrt {\sin (a+b x)} \, dx\\ &=-\frac {6 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(a+b x)}-\frac {6 \cos (a+b x)}{5 b \sqrt {\sin (a+b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 55, normalized size = 0.79 \[ \frac {2 \left (3 E\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )-\frac {\left (3 \sin ^2(a+b x)+1\right ) \cos (a+b x)}{\sin ^{\frac {5}{2}}(a+b x)}\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^(-7/2),x]

[Out]

(2*(3*EllipticE[(-2*a + Pi - 2*b*x)/4, 2] - (Cos[a + b*x]*(1 + 3*Sin[a + b*x]^2))/Sin[a + b*x]^(5/2)))/(5*b)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\sin \left (b x + a\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sin(b*x + a))/(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(-7/2), x)

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maple [A]  time = 0.05, size = 160, normalized size = 2.29 \[ \frac {6 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \left (\sin ^{2}\left (b x +a \right )\right ) \EllipticE \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \left (\sin ^{2}\left (b x +a \right )\right ) \EllipticF \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{4}\left (b x +a \right )\right )-4 \left (\sin ^{2}\left (b x +a \right )\right )-2}{5 \sin \left (b x +a \right )^{\frac {5}{2}} \cos \left (b x +a \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(b*x+a)^(7/2),x)

[Out]

1/5/sin(b*x+a)^(5/2)*(6*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*sin(b*x+a)^2*Elliptic
E((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))-3*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*sin(b*x
+a)^2*EllipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+6*sin(b*x+a)^4-4*sin(b*x+a)^2-2)/cos(b*x+a)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(-7/2), x)

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mupad [B]  time = 0.59, size = 42, normalized size = 0.60 \[ -\frac {\cos \left (a+b\,x\right )\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {3}{2};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\sin \left (a+b\,x\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(a + b*x)^(7/2),x)

[Out]

-(cos(a + b*x)*(sin(a + b*x)^2)^(5/4)*hypergeom([1/2, 9/4], 3/2, cos(a + b*x)^2))/(b*sin(a + b*x)^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sin ^{\frac {7}{2}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(b*x+a)**(7/2),x)

[Out]

Integral(sin(a + b*x)**(-7/2), x)

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