3.5.7 \(\int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx\) [407]

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

[Out]

-1/16*arcsin(cos(x)-sin(x))-1/16*ln(cos(x)+sin(x)+sin(2*x)^(1/2))-1/5*cos(x)^5/sin(2*x)^(5/2)+1/4*cos(x)/sin(2
*x)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 = {4378, 4392, 4391} \begin {gather*} -\frac {1}{16} \text {ArcSin}(\cos (x)-\sin (x))-\frac {\cos ^5(x)}{5 \sin ^{\frac {5}{2}}(2 x)}+\frac {\cos (x)}{4 \sqrt {\sin (2 x)}}-\frac {1}{16} \log \left (\sin (x)+\sqrt {\sin (2 x)}+\cos (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*ArcSin[Cos[x] - Sin[x]] - 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 4378

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 4391

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]

Rule 4392

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]

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*}

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Mathematica [A]
time = 0.06, size = 56, normalized size = 0.92 \begin {gather*} \frac {1}{16} \left (-\sin ^{-1}(\cos (x)-\sin (x))-\log \left (\cos (x)+\sin (x)+\sqrt {\sin (2 x)}\right )\right )+\left (\frac {3 \csc (x)}{20}-\frac {\csc ^3(x)}{40}\right ) \sqrt {\sin (2 x)} \end {gather*}

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] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.16, size = 1108, normalized size = 18.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^7/sin(2*x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/160*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/tan(1/2*x)^3*(192*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-ta
n(1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(
tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^6-96*(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))*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan(1/2*x)
*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^6-((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)
^2-1))^(1/2)*tan(1/2*x)^10+48*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)
*tan(1/2*x)^8+96*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^8-384*(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))*((1+tan(1
/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^4+192*(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))*((1+tan(1/2*x))*(t
an(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^4+3*((1+tan(1/2*x))*(tan(1/2*x)-
1)*tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^8-144*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*((1+
tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*tan(1/2*x)^6-192*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan
(1/2*x)^2-1))^(1/2)*tan(1/2*x)^6+192*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*Elliptic
E((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2
-1))^(1/2)*tan(1/2*x)^2-96*(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))*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)
*tan(1/2*x)^2+14*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*
x)^6+144*tan(1/2*x)^4*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)+96*tan(
1/2*x)^4*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+14*tan(1/2*x)^4*(tan(1/2*x)*(tan(
1/2*x)^2-1))^(1/2)*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)-48*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*((1+tan
(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*tan(1/2*x)^2+3*((1+tan(1/2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)*(tan
(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^2-(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*((1+tan(1/2*x))*(tan(1/2*x)-1
)*tan(1/2*x))^(1/2))/(tan(1/2*x)^2-1)/(tan(1/2*x)-1)/(1+tan(1/2*x))/(tan(1/2*x)^3-tan(1/2*x))^(1/2)/((1+tan(1/
2*x))*(tan(1/2*x)-1)*tan(1/2*x))^(1/2)

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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(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] Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (47) = 94\).
time = 1.25, size = 205, normalized size = 3.36 \begin {gather*} \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 {gather*}

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)] 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(cos(x)**7/sin(2*x)**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\cos \left (x\right )}^7}{{\sin \left (2\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(x)^7/sin(2*x)^(7/2), x)

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