3.1.80 \(\int \frac {\sin (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx\) [80]

Optimal. Leaf size=105 \[ \frac {\sin (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {6 \cos (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

1/7*sin(b*x+a)/b/sin(2*b*x+2*a)^(7/2)-6/35*cos(b*x+a)/b/sin(2*b*x+2*a)^(5/2)+8/35*sin(b*x+a)/b/sin(2*b*x+2*a)^
(3/2)-16/35*cos(b*x+a)/b/sin(2*b*x+2*a)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4389, 4388, 4376} \begin {gather*} \frac {8 \sin (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {\sin (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {6 \cos (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

Sin[a + b*x]/(7*b*Sin[2*a + 2*b*x]^(7/2)) - (6*Cos[a + b*x])/(35*b*Sin[2*a + 2*b*x]^(5/2)) + (8*Sin[a + b*x])/
(35*b*Sin[2*a + 2*b*x]^(3/2)) - (16*Cos[a + b*x])/(35*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4376

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Cos[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] &&
EqQ[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4388

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 4389

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] && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\sin (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx &=\frac {\sin (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {6}{7} \int \frac {\cos (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {6 \cos (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {24}{35} \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {6 \cos (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {16}{35} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {6 \cos (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 67, normalized size = 0.64 \begin {gather*} \frac {(-5-10 \cos (2 (a+b x))+4 \cos (4 (a+b x))+4 \cos (6 (a+b x))) \csc ^3(a+b x) \sec ^4(a+b x) \sqrt {\sin (2 (a+b x))}}{560 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

((-5 - 10*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)] + 4*Cos[6*(a + b*x)])*Csc[a + b*x]^3*Sec[a + b*x]^4*Sqrt[Sin[2
*(a + b*x)]])/(560*b)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (x b +a \right )}{\sin \left (2 x b +2 a \right )^{\frac {9}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)/sin(2*b*x+2*a)^(9/2),x)

[Out]

int(sin(b*x+a)/sin(2*b*x+2*a)^(9/2),x)

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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)/sin(2*b*x+2*a)^(9/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)/sin(2*b*x + 2*a)^(9/2), x)

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Fricas [A]
time = 5.99, size = 113, normalized size = 1.08 \begin {gather*} -\frac {\sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 160 \, \cos \left (b x + a\right )^{4} + 20 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 128 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \sin \left (b x + a\right )}{560 \, {\left (b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )} \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/sin(2*b*x+2*a)^(9/2),x, algorithm="fricas")

[Out]

-1/560*(sqrt(2)*(128*cos(b*x + a)^6 - 160*cos(b*x + a)^4 + 20*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x +
a)) + 128*(cos(b*x + a)^6 - cos(b*x + a)^4)*sin(b*x + a))/((b*cos(b*x + a)^6 - b*cos(b*x + a)^4)*sin(b*x + a))

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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)/sin(2*b*x+2*a)**(9/2),x)

[Out]

Timed out

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Giac [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)/sin(2*b*x+2*a)^(9/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 4.34, size = 351, normalized size = 3.34 \begin {gather*} -\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}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^4}+\frac {16\,{\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}}}{35\,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 {1{}\mathrm {i}}{7\,b}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,8{}\mathrm {i}}{35\,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}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {16}{35\,b}-\frac {44\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}}{35\,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 )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/sin(2*a + 2*b*x)^(9/2),x)

[Out]

(16*exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(35*b*(exp(a*2i + b*x*
2i) - 1)*(exp(a*2i + b*x*2i)*1i + 1i)) - (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)/(7*b*(exp(a*2i + b*x*2i)*1i + 1i)^4) - (exp(a*1i + b*x*1i)*(1i/(7*b) + (exp(a*2i + b*x*2i)*
8i)/(35*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) - (exp(a*1i + b*x*1i)*(16/(35*b) - (44*exp(a*2i + b*x*2i))/(35*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)^3*(exp(a*2i + b*x*2i)*1i + 1i)^3)

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