3.175 \(\int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=48 \[ \frac {F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{6 b}-\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

-1/6*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b-1/3*cos(b*x+a)^2/b/s
in(2*b*x+2*a)^(3/2)

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4295, 2641} \[ \frac {F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{6 b}-\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

EllipticF[a - Pi/4 + b*x, 2]/(6*b) - Cos[a + b*x]^2/(3*b*Sin[2*a + 2*b*x]^(3/2))

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 4295

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[((e*Cos[a + b
*x])^m*(g*Sin[c + d*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(e^2*(m + 2*p + 2))/(4*g^2*(p + 1)), Int[(e*Cos[a
+ b*x])^(m - 2)*(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, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && Inte
gersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx &=-\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {1}{6} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{6 b}-\frac {\cos ^2(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\\ \end {align*}

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Mathematica [A]  time = 1.00, size = 82, normalized size = 1.71 \[ -\frac {\sqrt {\sin (2 (a+b x))} \csc ^2(a+b x)+\frac {\sqrt {2} (\sin (a+b x)+\cos (a+b x)) F\left (\sin ^{-1}(\cos (a+b x)-\sin (a+b x))|\frac {1}{2}\right )}{\sqrt {\sin (2 (a+b x))+1}}}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

-1/12*(Csc[a + b*x]^2*Sqrt[Sin[2*(a + b*x)]] + (Sqrt[2]*EllipticF[ArcSin[Cos[a + b*x] - Sin[a + b*x]], 1/2]*(C
os[a + b*x] + Sin[a + b*x]))/Sqrt[1 + Sin[2*(a + b*x)]])/b

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\cos \left (b x + a\right )^{2}}{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} - 1\right )} \sqrt {\sin \left (2 \, b x + 2 \, a\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")

[Out]

integral(-cos(b*x + a)^2/((cos(2*b*x + 2*a)^2 - 1)*sqrt(sin(2*b*x + 2*a))), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^2/sin(2*b*x + 2*a)^(5/2), x)

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maple [A]  time = 59.05, size = 123, normalized size = 2.56 \[ \frac {\sqrt {1+\sin \left (2 b x +2 a \right )}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \EllipticF \left (\sqrt {1+\sin \left (2 b x +2 a \right )}, \frac {\sqrt {2}}{2}\right ) \sin \left (2 b x +2 a \right )-2 \left (\cos ^{2}\left (2 b x +2 a \right )\right )-2 \cos \left (2 b x +2 a \right )}{12 \sin \left (2 b x +2 a \right )^{\frac {3}{2}} \cos \left (2 b x +2 a \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x)

[Out]

1/12/sin(2*b*x+2*a)^(3/2)/cos(2*b*x+2*a)*((1+sin(2*b*x+2*a))^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a
))^(1/2)*EllipticF((1+sin(2*b*x+2*a))^(1/2),1/2*2^(1/2))*sin(2*b*x+2*a)-2*cos(2*b*x+2*a)^2-2*cos(2*b*x+2*a))/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^2/sin(2*b*x + 2*a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\cos \left (a+b\,x\right )}^2}{{\sin \left (2\,a+2\,b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/sin(2*a + 2*b*x)^(5/2),x)

[Out]

int(cos(a + b*x)^2/sin(2*a + 2*b*x)^(5/2), x)

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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(cos(b*x+a)**2/sin(2*b*x+2*a)**(5/2),x)

[Out]

Timed out

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