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]

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]])

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Rubi [A]  time = 0.0586601, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, 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 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]

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 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 (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*}

Mathematica [A]  time = 0.189159, 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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Maple [C]  time = 257.353, size = 308, normalized size = 3.9 \begin{align*} -{\frac{1}{48\,b}\sqrt{-{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) ^{-1}}} \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) \left ( 5\,\sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1}\sqrt{-2\,\tan \left ( 1/2\,bx+a/2 \right ) +2}\sqrt{-\tan \left ( 1/2\,bx+a/2 \right ) }{\it EllipticF} \left ( \sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1},1/2\,\sqrt{2} \right ) \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}- \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{6}+5\,\sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1}\sqrt{-2\,\tan \left ( 1/2\,bx+a/2 \right ) +2}\sqrt{-\tan \left ( 1/2\,bx+a/2 \right ) }{\it EllipticF} \left ( \sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1},1/2\,\sqrt{2} \right ) \tan \left ( 1/2\,bx+a/2 \right ) -7\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+7\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1 \right ) \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) }}}{\frac{1}{\sqrt{ \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{3}-\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) }}} \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}+1 \right ) ^{-1}} \end{align*}

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]

-1/48/b*(-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)/tan(1/2*b*x+1/2*a)*(5*(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)*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)^3-tan(1/2*b*x+1/2*a)^6+5*(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)-7*tan(1/2*b*x+1/2*a)^4+7*tan(1/2*b*x+1/2*a)^2+1)/(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)^3-tan(1/2*b*x+1/2*a))^(1/2)/(tan(1/2*b*x+1/2*a)^2+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

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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Fricas [A]  time = 0.514501, size = 231, normalized size = 2.92 \begin{align*} \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 )}} \end{align*}

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

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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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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