3.96 \(\int \frac {\sin ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=107 \[ \frac {\sin (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {\sin ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}}-\frac {4 \cos (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

1/9*sin(b*x+a)^3/b/sin(2*b*x+2*a)^(9/2)+1/15*sin(b*x+a)/b/sin(2*b*x+2*a)^(5/2)-4/45*cos(b*x+a)/b/sin(2*b*x+2*a
)^(3/2)+8/45*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4296, 4304, 4303, 4292} \[ \frac {\sin ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}+\frac {\sin (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}}-\frac {4 \cos (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/Sin[2*a + 2*b*x]^(11/2),x]

[Out]

Sin[a + b*x]^3/(9*b*Sin[2*a + 2*b*x]^(9/2)) + Sin[a + b*x]/(15*b*Sin[2*a + 2*b*x]^(5/2)) - (4*Cos[a + b*x])/(4
5*b*Sin[2*a + 2*b*x]^(3/2)) + (8*Sin[a + b*x])/(45*b*Sqrt[Sin[2*a + 2*b*x]])

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]

Rule 4296

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] && EqQ[
d/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && Int
egersQ[2*m, 2*p]

Rule 4303

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(Cos[a + b*x]*(g*Sin[c + d
*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x
], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

Rule 4304

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[(Sin[a + b*x]*(g*Sin[c +
d*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1),
x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Inte
gerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\sin ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx &=\frac {\sin ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}+\frac {1}{3} \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}+\frac {\sin (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4}{15} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}+\frac {\sin (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8}{45} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}+\frac {\sin (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 62, normalized size = 0.58 \[ \frac {\sqrt {\sin (2 (a+b x))} \left (5 \sec ^5(a+b x)+17 \sec ^3(a+b x)+113 \sec (a+b x)-15 \cot (a+b x) \csc (a+b x)\right )}{1440 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3/Sin[2*a + 2*b*x]^(11/2),x]

[Out]

((-15*Cot[a + b*x]*Csc[a + b*x] + 113*Sec[a + b*x] + 17*Sec[a + b*x]^3 + 5*Sec[a + b*x]^5)*Sqrt[Sin[2*(a + b*x
)]])/(1440*b)

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fricas [A]  time = 0.45, size = 98, normalized size = 0.92 \[ \frac {128 \, \cos \left (b x + a\right )^{7} - 128 \, \cos \left (b x + a\right )^{5} + \sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 96 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} - 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{1440 \, {\left (b \cos \left (b x + a\right )^{7} - b \cos \left (b x + a\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="fricas")

[Out]

1/1440*(128*cos(b*x + a)^7 - 128*cos(b*x + a)^5 + sqrt(2)*(128*cos(b*x + a)^6 - 96*cos(b*x + a)^4 - 12*cos(b*x
+ a)^2 - 5)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^7 - b*cos(b*x + a)^5)

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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)^(11/2),x, algorithm="giac")

[Out]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{3}\left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{\frac {11}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x)

[Out]

int(sin(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^3/sin(2*b*x + 2*a)^(11/2), x)

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mupad [B]  time = 5.16, size = 383, normalized size = 3.58 \[ -\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{60\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^3}-\frac {2\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^4}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^5}+\frac {{\mathrm {e}}^{a\,3{}\mathrm {i}+b\,x\,3{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,8{}\mathrm {i}}{45\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {49}{180\,b}-\frac {19\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}}{180\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3/sin(2*a + 2*b*x)^(11/2),x)

[Out]

(exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*1i)/(9*b*(exp(a*2i + b*x*2
i)*1i + 1i)^5) - (2*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(9*b*(
exp(a*2i + b*x*2i)*1i + 1i)^4) - (exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)
^(1/2)*1i)/(60*b*(exp(a*2i + b*x*2i)*1i + 1i)^3) + (exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2
i + b*x*2i)*1i)/2)^(1/2)*8i)/(45*b*(exp(a*2i + b*x*2i) - 1)*(exp(a*2i + b*x*2i)*1i + 1i)) - (exp(a*1i + b*x*1i
)*(49/(180*b) - (19*exp(a*2i + b*x*2i))/(180*b))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/
2))/((exp(a*2i + b*x*2i) - 1)^2*(exp(a*2i + b*x*2i)*1i + 1i)^2)

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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)**(11/2),x)

[Out]

Timed out

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