3.93 \(\int \frac {\sin ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {\sin ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

1/3*sin(b*x+a)^3/b/sin(2*b*x+2*a)^(3/2)

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Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {4292} \[ \frac {\sin ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

Sin[a + b*x]^3/(3*b*Sin[2*a + 2*b*x]^(3/2))

Rule 4292

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))/(b*g*m), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

\begin {align*} \int \frac {\sin ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx &=\frac {\sin ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 27, normalized size = 0.96 \[ \frac {\sin ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 (a+b x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

Sin[a + b*x]^3/(3*b*Sin[2*(a + b*x)]^(3/2))

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fricas [A]  time = 0.45, size = 48, normalized size = 1.71 \[ -\frac {\cos \left (b x + a\right )^{2} - \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right )}{12 \, b \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(cos(b*x + a)^2 - sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*sin(b*x + a))/(b*cos(b*x + a)^2)

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giac [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(sin(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 183.93, size = 727, normalized size = 25.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x)

[Out]

-1/48*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)^2-1)*(6*(tan(1/2*b*x+1/2*a)+1)^
(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2
^(1/2))*tan(1/2*b*x+1/2*a)^6-3*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*
a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^6+18*(tan(1/2*b*x+1/2*a)+1)^(
1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^
(1/2))*tan(1/2*b*x+1/2*a)^4-9*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a
))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^4+6*tan(1/2*b*x+1/2*a)^8+18*(t
an(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticE((tan(1/2*b*x+
1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2-9*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/
2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2-2*tan(
1/2*b*x+1/2*a)^6+6*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*El
lipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))-3*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/
2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))+10*tan(1/2*b*x+1/2*a)^4-14*
tan(1/2*b*x+1/2*a)^2)/(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)/(1+tan(1/2*b*x+1/2*a)^2)^3/(tan(1/2*
b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^3/sin(2*b*x + 2*a)^(5/2), x)

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mupad [B]  time = 2.19, size = 85, normalized size = 3.04 \[ -\frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}\,\left (2\,\sin \left (a+b\,x\right )+3\,\sin \left (3\,a+3\,b\,x\right )+\sin \left (5\,a+5\,b\,x\right )\right )}{6\,b\,\left (30\,{\sin \left (a+b\,x\right )}^2+12\,{\sin \left (2\,a+2\,b\,x\right )}^2+2\,{\sin \left (3\,a+3\,b\,x\right )}^2-32\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3/sin(2*a + 2*b*x)^(5/2),x)

[Out]

-(sin(2*a + 2*b*x)^(1/2)*(2*sin(a + b*x) + 3*sin(3*a + 3*b*x) + sin(5*a + 5*b*x)))/(6*b*(12*sin(2*a + 2*b*x)^2
+ 2*sin(3*a + 3*b*x)^2 + 30*sin(a + b*x)^2 - 32))

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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(sin(b*x+a)**3/sin(2*b*x+2*a)**(5/2),x)

[Out]

Timed out

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