3.406 \(\int \frac{\sin ^7(x)}{\sin ^{\frac{7}{2}}(2 x)} \, dx\)

Optimal. Leaf size=61 \[ \frac{\sin ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{\sin (x)}{4 \sqrt{\sin (2 x)}}-\frac{1}{16} \sin ^{-1}(\cos (x)-\sin (x))+\frac{1}{16} \log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right ) \]

[Out]

-ArcSin[Cos[x] - Sin[x]]/16 + Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]]/16 + Sin[x]^5/(5*Sin[2*x]^(5/2)) - Sin[x]/
(4*Sqrt[Sin[2*x]])

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Rubi [A]  time = 0.0801925, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4294, 4308, 4305} \[ \frac{\sin ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{\sin (x)}{4 \sqrt{\sin (2 x)}}-\frac{1}{16} \sin ^{-1}(\cos (x)-\sin (x))+\frac{1}{16} \log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSin[Cos[x] - Sin[x]]/16 + Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]]/16 + Sin[x]^5/(5*Sin[2*x]^(5/2)) - Sin[x]/
(4*Sqrt[Sin[2*x]])

Rule 4294

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[(e^2*(e*Sin[
a + b*x])^(m - 2)*(g*Sin[c + d*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(e^4*(m + p - 1))/(4*g^2*(p + 1)), Int[
(e*Sin[a + b*x])^(m - 4)*(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, 2] && LtQ[p, -1] && (GtQ[m, 3] || EqQ[p, -3/2]) && IntegersQ[2*m, 2
*p]

Rule 4308

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

Rule 4305

Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> -Simp[ArcSin[Cos[a + b*x] - Sin[a + b*
x]]/d, x] + Simp[Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[
b*c - a*d, 0] && EqQ[d/b, 2]

Rubi steps

\begin{align*} \int \frac{\sin ^7(x)}{\sin ^{\frac{7}{2}}(2 x)} \, dx &=\frac{\sin ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{1}{4} \int \frac{\sin ^3(x)}{\sin ^{\frac{3}{2}}(2 x)} \, dx\\ &=\frac{\sin ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{\sin (x)}{4 \sqrt{\sin (2 x)}}+\frac{1}{16} \int \csc (x) \sqrt{\sin (2 x)} \, dx\\ &=\frac{\sin ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{\sin (x)}{4 \sqrt{\sin (2 x)}}+\frac{1}{8} \int \frac{\cos (x)}{\sqrt{\sin (2 x)}} \, dx\\ &=-\frac{1}{16} \sin ^{-1}(\cos (x)-\sin (x))+\frac{1}{16} \log \left (\cos (x)+\sin (x)+\sqrt{\sin (2 x)}\right )+\frac{\sin ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{\sin (x)}{4 \sqrt{\sin (2 x)}}\\ \end{align*}

Mathematica [A]  time = 0.0879294, size = 50, normalized size = 0.82 \[ \frac{1}{80} \left (2 \sqrt{\sin (2 x)} \sec (x) \left (\sec ^2(x)-6\right )+5 \left (\log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right )-\sin ^{-1}(\cos (x)-\sin (x))\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(-ArcSin[Cos[x] - Sin[x]] + Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]]) + 2*Sec[x]*(-6 + Sec[x]^2)*Sqrt[Sin[2*x]
])/80

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Maple [C]  time = 0.105, size = 510, normalized size = 8.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^7/sin(2*x)^(7/2),x)

[Out]

1/2688*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^2-1)*(5*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(
-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*tan(1/2*x)^14+35*(1+tan(1/2*x))^(1/2)*(-2*tan(1
/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*tan(1/2*x)^12+10*tan(1/2*x)^15+
105*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2
))*tan(1/2*x)^10+66*tan(1/2*x)^13+175*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*Ellipti
cF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*tan(1/2*x)^8-1014*tan(1/2*x)^11+175*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2
)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*tan(1/2*x)^6+2002*tan(1/2*x)^9+105*(1+
tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*tan(
1/2*x)^4-2002*tan(1/2*x)^7+35*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+ta
n(1/2*x))^(1/2),1/2*2^(1/2))*tan(1/2*x)^2+1014*tan(1/2*x)^5+5*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-t
an(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))-66*tan(1/2*x)^3-10*tan(1/2*x))/(tan(1/2*x)*(tan(1
/2*x)^2-1))^(1/2)/(tan(1/2*x)^3-tan(1/2*x))^(1/2)/(tan(1/2*x)^2+1)^7

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(x)^7/sin(2*x)^(7/2), x)

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Fricas [B]  time = 2.07311, size = 594, normalized size = 9.74 \begin{align*} \frac{10 \, \arctan \left (-\frac{\sqrt{2} \sqrt{\cos \left (x\right ) \sin \left (x\right )}{\left (\cos \left (x\right ) - \sin \left (x\right )\right )} + \cos \left (x\right ) \sin \left (x\right )}{\cos \left (x\right )^{2} + 2 \, \cos \left (x\right ) \sin \left (x\right ) - 1}\right ) \cos \left (x\right )^{3} - 10 \, \arctan \left (-\frac{2 \, \sqrt{2} \sqrt{\cos \left (x\right ) \sin \left (x\right )} - \cos \left (x\right ) - \sin \left (x\right )}{\cos \left (x\right ) - \sin \left (x\right )}\right ) \cos \left (x\right )^{3} - 5 \, \cos \left (x\right )^{3} \log \left (-32 \, \cos \left (x\right )^{4} + 4 \, \sqrt{2}{\left (4 \, \cos \left (x\right )^{3} -{\left (4 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right ) - 5 \, \cos \left (x\right )\right )} \sqrt{\cos \left (x\right ) \sin \left (x\right )} + 32 \, \cos \left (x\right )^{2} + 16 \, \cos \left (x\right ) \sin \left (x\right ) + 1\right ) - 48 \, \cos \left (x\right )^{3} - 8 \, \sqrt{2}{\left (6 \, \cos \left (x\right )^{2} - 1\right )} \sqrt{\cos \left (x\right ) \sin \left (x\right )}}{320 \, \cos \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(10*arctan(-(sqrt(2)*sqrt(cos(x)*sin(x))*(cos(x) - sin(x)) + cos(x)*sin(x))/(cos(x)^2 + 2*cos(x)*sin(x)
- 1))*cos(x)^3 - 10*arctan(-(2*sqrt(2)*sqrt(cos(x)*sin(x)) - cos(x) - sin(x))/(cos(x) - sin(x)))*cos(x)^3 - 5*
cos(x)^3*log(-32*cos(x)^4 + 4*sqrt(2)*(4*cos(x)^3 - (4*cos(x)^2 + 1)*sin(x) - 5*cos(x))*sqrt(cos(x)*sin(x)) +
32*cos(x)^2 + 16*cos(x)*sin(x) + 1) - 48*cos(x)^3 - 8*sqrt(2)*(6*cos(x)^2 - 1)*sqrt(cos(x)*sin(x)))/cos(x)^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(x)**7/sin(2*x)**(7/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^7/sin(2*x)^(7/2),x, algorithm="giac")

[Out]

integrate(sin(x)^7/sin(2*x)^(7/2), x)