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

Optimal. Leaf size=79 \[ \frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4304, 4303, 4292} \[ \frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2)) - (4*Cos[a + b*x])/(15*b*Sin[2*a + 2*b*x]^(3/2)) + (8*Sin[a + b*x])/
(15*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 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 (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx &=\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4}{5} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8}{15} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 52, normalized size = 0.66 \[ \frac {\sqrt {\sin (2 (a+b x))} \left (3 \sec (a+b x) \left (\sec ^2(a+b x)+9\right )-5 \cot (a+b x) \csc (a+b x)\right )}{120 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*Cot[a + b*x]*Csc[a + b*x] + 3*Sec[a + b*x]*(9 + Sec[a + b*x]^2))*Sqrt[Sin[2*(a + b*x)]])/(120*b)

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fricas [A]  time = 0.43, size = 88, normalized size = 1.11 \[ \frac {32 \, \cos \left (b x + a\right )^{5} - 32 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{120 \, {\left (b \cos \left (b x + a\right )^{5} - b \cos \left (b x + a\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(32*cos(b*x + a)^5 - 32*cos(b*x + a)^3 + sqrt(2)*(32*cos(b*x + a)^4 - 24*cos(b*x + a)^2 - 3)*sqrt(cos(b*
x + a)*sin(b*x + a)))/(b*cos(b*x + a)^5 - b*cos(b*x + a)^3)

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

[Out]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.39, size = 131, normalized size = 1.66 \[ \frac {4\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left (2\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-3\,{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+2\,{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}+2\,{\mathrm {e}}^{a\,8{}\mathrm {i}+b\,x\,8{}\mathrm {i}}+2\right )}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)*(2*exp(a*2i + b*x*2i) -
3*exp(a*4i + b*x*4i) + 2*exp(a*6i + b*x*6i) + 2*exp(a*8i + b*x*8i) + 2))/(15*b*(exp(a*2i + b*x*2i) - 1)^2*(exp
(a*2i + b*x*2i) + 1)^3)

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

[Out]

Timed out

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