3.1.88 \(\int \frac {\sin ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) [88]

Optimal. Leaf size=77 \[ -\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}+\frac {\sin ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

3/10*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b+1/5*sin(b*x+a)^2/b/s
in(2*b*x+2*a)^(5/2)-3/10*cos(2*b*x+2*a)/b/sin(2*b*x+2*a)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4381, 2716, 2719} \begin {gather*} \frac {\sin ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{10 b}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

(-3*EllipticE[a - Pi/4 + b*x, 2])/(10*b) + Sin[a + b*x]^2/(5*b*Sin[2*a + 2*b*x]^(5/2)) - (3*Cos[2*a + 2*b*x])/
(10*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 2716

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 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]

Rule 4381

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Sin[a +
b*x])^m)*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Dist[e^2*((m + 2*p + 2)/(4*g^2*(p + 1))), Int[(e*Sin
[a + b*x])^(m - 2)*(g*Sin[c + d*x])^(p + 2), x], x] /; FreeQ[{a, b, c, d, e, g}, x] && EqQ[b*c - a*d, 0] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && I
ntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx &=\frac {\sin ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {3}{10} \int \frac {1}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}}-\frac {3}{10} \int \sqrt {\sin (2 a+2 b x)} \, dx\\ &=-\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}+\frac {\sin ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 66, normalized size = 0.86 \begin {gather*} -\frac {12 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\frac {4 (1+6 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 (a+b x))}}{40 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

-1/40*(12*EllipticE[a - Pi/4 + b*x, 2] + (4*(1 + 6*Cos[2*(a + b*x)] + 3*Cos[4*(a + b*x)])*Sin[a + b*x]^2)/Sin[
2*(a + b*x)]^(5/2))/b

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(92)=184\).
time = 247.57, size = 227, normalized size = 2.95

method	result	size
default	\(\frac {\sqrt {2}\, \left (\frac {8 \sqrt {2}}{5 \sin \left (2 x b +2 a \right )^{\frac {5}{2}}}+\frac {4 \sqrt {2}\, \left (6 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \left (\sin ^{2}\left (2 x b +2 a \right )\right ) \EllipticE \left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \left (\sin ^{2}\left (2 x b +2 a \right )\right ) \EllipticF \left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{4}\left (2 x b +2 a \right )\right )-4 \left (\sin ^{2}\left (2 x b +2 a \right )\right )-2\right )}{5 \sin \left (2 x b +2 a \right )^{\frac {5}{2}} \cos \left (2 x b +2 a \right )}\right )}{32 b}\)	\(227\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2/sin(2*b*x+2*a)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/32*2^(1/2)*(8/5*2^(1/2)/sin(2*b*x+2*a)^(5/2)+4/5*2^(1/2)/sin(2*b*x+2*a)^(5/2)*(6*(sin(2*b*x+2*a)+1)^(1/2)*(-
2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*sin(2*b*x+2*a)^2*EllipticE((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1
/2))-3*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*sin(2*b*x+2*a)^2*EllipticF
((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))+6*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-2)/cos(2*b*x+2*a))/b

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^2}{{\sin \left (2\,a+2\,b\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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