3.407 \(\int \frac{\cos ^7(x)}{\sin ^{\frac{7}{2}}(2 x)} \, dx\)
Optimal. Leaf size=61 \[ -\frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{1}{16} \sin ^{-1}(\cos (x)-\sin (x))+\frac{\cos (x)}{4 \sqrt{\sin (2 x)}}-\frac{1}{16} \log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right ) \]
[Out]
-ArcSin[Cos[x] - Sin[x]]/16 - Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]]/16 - Cos[x]^5/(5*Sin[2*x]^(5/2)) + Cos[x]/
(4*Sqrt[Sin[2*x]])
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Rubi [A] time = 0.082215, 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 =
{4293, 4307, 4306} \[ -\frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{1}{16} \sin ^{-1}(\cos (x)-\sin (x))+\frac{\cos (x)}{4 \sqrt{\sin (2 x)}}-\frac{1}{16} \log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right ) \]
Antiderivative was successfully verified.
[In]
Int[Cos[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 - Cos[x]^5/(5*Sin[2*x]^(5/2)) + Cos[x]/
(4*Sqrt[Sin[2*x]])
Rule 4293
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e^2*(e*Cos[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*Cos[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 4307
Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/cos[(a_.) + (b_.)*(x_)], x_Symbol] :> Dist[2*g, Int[Sin[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 4306
Int[sin[(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{\cos ^7(x)}{\sin ^{\frac{7}{2}}(2 x)} \, dx &=-\frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{1}{4} \int \frac{\cos ^3(x)}{\sin ^{\frac{3}{2}}(2 x)} \, dx\\ &=-\frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}+\frac{\cos (x)}{4 \sqrt{\sin (2 x)}}+\frac{1}{16} \int \sec (x) \sqrt{\sin (2 x)} \, dx\\ &=-\frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}+\frac{\cos (x)}{4 \sqrt{\sin (2 x)}}+\frac{1}{8} \int \frac{\sin (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{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}+\frac{\cos (x)}{4 \sqrt{\sin (2 x)}}\\ \end{align*}
Mathematica [A] time = 0.0739581, size = 56, normalized size = 0.92 \[ \sqrt{\sin (2 x)} \left (\frac{3 \csc (x)}{20}-\frac{\csc ^3(x)}{40}\right )+\frac{1}{16} \left (-\sin ^{-1}(\cos (x)-\sin (x))-\log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[x]^7/Sin[2*x]^(7/2),x]
[Out]
(-ArcSin[Cos[x] - Sin[x]] - Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]])/16 + ((3*Csc[x])/20 - Csc[x]^3/40)*Sqrt[Sin
[2*x]]
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Maple [C] time = 0.058, size = 1108, normalized size = 18.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(x)^7/sin(2*x)^(7/2),x)
[Out]
1/160*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/tan(1/2*x)^3*(192*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x)
)^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(tan(1/2*x)*(t
an(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^6-96*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(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)*(tan(1/2*x)-
1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^6-(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/
2*x)))^(1/2)*tan(1/2*x)^10-384*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2
)*(-tan(1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(
1/2)*tan(1/2*x)^4+192*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(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)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(
1/2*x)^4+96*(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)^8+3*(tan(1/2*x)*(ta
n(1/2*x)^2-1))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^8+48*(tan(1/2*x)^3-tan(1/2*x)
)^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^8+192*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*
(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(
tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^2-96*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(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)
*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^2-192*(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)^6+14*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1
/2)*tan(1/2*x)^6-144*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*
x)^6+96*tan(1/2*x)^4*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)+14*tan(1/2*x)^4*(tan(
1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)+144*(tan(1/2*x)^3-tan(1/2*x))^
(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^4+3*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan
(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^2-48*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(
1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^2-(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*(tan(1/2*x)*(tan
(1/2*x)^2-1))^(1/2))/(tan(1/2*x)^2-1)/(tan(1/2*x)^3-tan(1/2*x))^(1/2)/(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)
))^(1/2)/(tan(1/2*x)-1)/(1+tan(1/2*x))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \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(cos(x)^7/sin(2*x)^(7/2),x, algorithm="maxima")
[Out]
integrate(cos(x)^7/sin(2*x)^(7/2), x)
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Fricas [B] time = 2.13289, size = 684, normalized size = 11.21 \begin{align*} \frac{10 \,{\left (\cos \left (x\right )^{2} - 1\right )} \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 ) \sin \left (x\right ) - 10 \,{\left (\cos \left (x\right )^{2} - 1\right )} \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 ) \sin \left (x\right ) + 5 \,{\left (\cos \left (x\right )^{2} - 1\right )} \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 ) \sin \left (x\right ) + 8 \, \sqrt{2}{\left (6 \, \cos \left (x\right )^{2} - 5\right )} \sqrt{\cos \left (x\right ) \sin \left (x\right )} + 48 \,{\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{320 \,{\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(x)^7/sin(2*x)^(7/2),x, algorithm="fricas")
[Out]
1/320*(10*(cos(x)^2 - 1)*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))*sin(x) - 10*(cos(x)^2 - 1)*arctan(-(2*sqrt(2)*sqrt(cos(x)*sin(x)) - cos(x) - sin(x))/(cos
(x) - sin(x)))*sin(x) + 5*(cos(x)^2 - 1)*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)*sin(x) + 8*sqrt(2)*(6*cos(x)^2 - 5)*sqrt(c
os(x)*sin(x)) + 48*(cos(x)^2 - 1)*sin(x))/((cos(x)^2 - 1)*sin(x))
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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(cos(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{\cos \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(cos(x)^7/sin(2*x)^(7/2),x, algorithm="giac")
[Out]
integrate(cos(x)^7/sin(2*x)^(7/2), x)