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3.2.50 \(\int \cos ^2(a+b x) \csc ^5(2 a+2 b x) \, dx\) [150]

Optimal. Leaf size=60 \[ -\frac {3 \cot ^2(a+b x)}{64 b}-\frac {\cot ^4(a+b x)}{128 b}+\frac {3 \log (\tan (a+b x))}{32 b}+\frac {\tan ^2(a+b x)}{64 b} \]

[Out]

-3/64*cot(b*x+a)^2/b-1/128*cot(b*x+a)^4/b+3/32*ln(tan(b*x+a))/b+1/64*tan(b*x+a)^2/b

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Rubi [A]
time = 0.05, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4372, 2700, 272, 45} \begin {gather*} \frac {\tan ^2(a+b x)}{64 b}-\frac {\cot ^4(a+b x)}{128 b}-\frac {3 \cot ^2(a+b x)}{64 b}+\frac {3 \log (\tan (a+b x))}{32 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Csc[2*a + 2*b*x]^5,x]

[Out]

(-3*Cot[a + b*x]^2)/(64*b) - Cot[a + b*x]^4/(128*b) + (3*Log[Tan[a + b*x]])/(32*b) + Tan[a + b*x]^2/(64*b)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 4372

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rubi steps

\begin {align*} \int \cos ^2(a+b x) \csc ^5(2 a+2 b x) \, dx &=\frac {1}{32} \int \csc ^5(a+b x) \sec ^3(a+b x) \, dx\\ &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^5} \, dx,x,\tan (a+b x)\right )}{32 b}\\ &=\frac {\text {Subst}\left (\int \frac {(1+x)^3}{x^3} \, dx,x,\tan ^2(a+b x)\right )}{64 b}\\ &=\frac {\text {Subst}\left (\int \left (1+\frac {1}{x^3}+\frac {3}{x^2}+\frac {3}{x}\right ) \, dx,x,\tan ^2(a+b x)\right )}{64 b}\\ &=-\frac {3 \cot ^2(a+b x)}{64 b}-\frac {\cot ^4(a+b x)}{128 b}+\frac {3 \log (\tan (a+b x))}{32 b}+\frac {\tan ^2(a+b x)}{64 b}\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 54, normalized size = 0.90 \begin {gather*} -\frac {4 \csc ^2(a+b x)+\csc ^4(a+b x)+12 \log (\cos (a+b x))-12 \log (\sin (a+b x))-2 \sec ^2(a+b x)}{128 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Csc[2*a + 2*b*x]^5,x]

[Out]

-1/128*(4*Csc[a + b*x]^2 + Csc[a + b*x]^4 + 12*Log[Cos[a + b*x]] - 12*Log[Sin[a + b*x]] - 2*Sec[a + b*x]^2)/b

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Maple [A]
time = 0.22, size = 62, normalized size = 1.03

method result size
default \(\frac {-\frac {1}{4 \sin \left (x b +a \right )^{4} \cos \left (x b +a \right )^{2}}+\frac {3}{4 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{2}}-\frac {3}{2 \sin \left (x b +a \right )^{2}}+3 \ln \left (\tan \left (x b +a \right )\right )}{32 b}\) \(62\)
risch \(\frac {3 \,{\mathrm e}^{10 i \left (x b +a \right )}-6 \,{\mathrm e}^{8 i \left (x b +a \right )}-2 \,{\mathrm e}^{6 i \left (x b +a \right )}-6 \,{\mathrm e}^{4 i \left (x b +a \right )}+3 \,{\mathrm e}^{2 i \left (x b +a \right )}}{16 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{32 b}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{32 b}\) \(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/b*(-1/4/sin(b*x+a)^4/cos(b*x+a)^2+3/4/sin(b*x+a)^2/cos(b*x+a)^2-3/2/sin(b*x+a)^2+3*ln(tan(b*x+a)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 3188 vs. \(2 (52) = 104\).
time = 0.36, size = 3188, normalized size = 53.13 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(4*(3*cos(10*b*x + 10*a) - 6*cos(8*b*x + 8*a) - 2*cos(6*b*x + 6*a) - 6*cos(4*b*x + 4*a) + 3*cos(2*b*x + 2
*a))*cos(12*b*x + 12*a) + 4*(9*cos(8*b*x + 8*a) + 16*cos(6*b*x + 6*a) + 9*cos(4*b*x + 4*a) - 12*cos(2*b*x + 2*
a) + 3)*cos(10*b*x + 10*a) - 24*cos(10*b*x + 10*a)^2 - 4*(22*cos(6*b*x + 6*a) - 12*cos(4*b*x + 4*a) - 9*cos(2*
b*x + 2*a) + 6)*cos(8*b*x + 8*a) + 24*cos(8*b*x + 8*a)^2 - 8*(11*cos(4*b*x + 4*a) - 8*cos(2*b*x + 2*a) + 1)*co
s(6*b*x + 6*a) - 32*cos(6*b*x + 6*a)^2 + 12*(3*cos(2*b*x + 2*a) - 2)*cos(4*b*x + 4*a) + 24*cos(4*b*x + 4*a)^2
- 24*cos(2*b*x + 2*a)^2 + 3*(2*(2*cos(10*b*x + 10*a) + cos(8*b*x + 8*a) - 4*cos(6*b*x + 6*a) + cos(4*b*x + 4*a
) + 2*cos(2*b*x + 2*a) - 1)*cos(12*b*x + 12*a) - cos(12*b*x + 12*a)^2 - 4*(cos(8*b*x + 8*a) - 4*cos(6*b*x + 6*
a) + cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(10*b*x + 10*a) - 4*cos(10*b*x + 10*a)^2 + 2*(4*cos(6*b*x +
 6*a) - cos(4*b*x + 4*a) - 2*cos(2*b*x + 2*a) + 1)*cos(8*b*x + 8*a) - cos(8*b*x + 8*a)^2 + 8*(cos(4*b*x + 4*a)
 + 2*cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - 16*cos(6*b*x + 6*a)^2 - 2*(2*cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4
*a) - cos(4*b*x + 4*a)^2 - 4*cos(2*b*x + 2*a)^2 + 2*(2*sin(10*b*x + 10*a) + sin(8*b*x + 8*a) - 4*sin(6*b*x + 6
*a) + sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(12*b*x + 12*a) - sin(12*b*x + 12*a)^2 - 4*(sin(8*b*x + 8*a) -
 4*sin(6*b*x + 6*a) + sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - 4*sin(10*b*x + 10*a)^2 + 2*(
4*sin(6*b*x + 6*a) - sin(4*b*x + 4*a) - 2*sin(2*b*x + 2*a))*sin(8*b*x + 8*a) - sin(8*b*x + 8*a)^2 + 8*(sin(4*b
*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(6*b*x + 6*a) - 16*sin(6*b*x + 6*a)^2 - sin(4*b*x + 4*a)^2 - 4*sin(4*b*x +
4*a)*sin(2*b*x + 2*a) - 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) - 1)*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a
) + cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2) - 3*(2*(2*cos(10*b*x + 10*a) + cos(8*b*x +
 8*a) - 4*cos(6*b*x + 6*a) + cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(12*b*x + 12*a) - cos(12*b*x + 12*a
)^2 - 4*(cos(8*b*x + 8*a) - 4*cos(6*b*x + 6*a) + cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(10*b*x + 10*a)
 - 4*cos(10*b*x + 10*a)^2 + 2*(4*cos(6*b*x + 6*a) - cos(4*b*x + 4*a) - 2*cos(2*b*x + 2*a) + 1)*cos(8*b*x + 8*a
) - cos(8*b*x + 8*a)^2 + 8*(cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - 16*cos(6*b*x + 6*a)^
2 - 2*(2*cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x + 4*a)^2 - 4*cos(2*b*x + 2*a)^2 + 2*(2*sin(10*b*x
+ 10*a) + sin(8*b*x + 8*a) - 4*sin(6*b*x + 6*a) + sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(12*b*x + 12*a) -
sin(12*b*x + 12*a)^2 - 4*(sin(8*b*x + 8*a) - 4*sin(6*b*x + 6*a) + sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(1
0*b*x + 10*a) - 4*sin(10*b*x + 10*a)^2 + 2*(4*sin(6*b*x + 6*a) - sin(4*b*x + 4*a) - 2*sin(2*b*x + 2*a))*sin(8*
b*x + 8*a) - sin(8*b*x + 8*a)^2 + 8*(sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(6*b*x + 6*a) - 16*sin(6*b*x +
6*a)^2 - sin(4*b*x + 4*a)^2 - 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a)
- 1)*log(cos(b*x)^2 + 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - 3*(2*(2*cos(
10*b*x + 10*a) + cos(8*b*x + 8*a) - 4*cos(6*b*x + 6*a) + cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(12*b*x
 + 12*a) - cos(12*b*x + 12*a)^2 - 4*(cos(8*b*x + 8*a) - 4*cos(6*b*x + 6*a) + cos(4*b*x + 4*a) + 2*cos(2*b*x +
2*a) - 1)*cos(10*b*x + 10*a) - 4*cos(10*b*x + 10*a)^2 + 2*(4*cos(6*b*x + 6*a) - cos(4*b*x + 4*a) - 2*cos(2*b*x
 + 2*a) + 1)*cos(8*b*x + 8*a) - cos(8*b*x + 8*a)^2 + 8*(cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(6*b*x +
 6*a) - 16*cos(6*b*x + 6*a)^2 - 2*(2*cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x + 4*a)^2 - 4*cos(2*b*x
 + 2*a)^2 + 2*(2*sin(10*b*x + 10*a) + sin(8*b*x + 8*a) - 4*sin(6*b*x + 6*a) + sin(4*b*x + 4*a) + 2*sin(2*b*x +
 2*a))*sin(12*b*x + 12*a) - sin(12*b*x + 12*a)^2 - 4*(sin(8*b*x + 8*a) - 4*sin(6*b*x + 6*a) + sin(4*b*x + 4*a)
 + 2*sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - 4*sin(10*b*x + 10*a)^2 + 2*(4*sin(6*b*x + 6*a) - sin(4*b*x + 4*a)
- 2*sin(2*b*x + 2*a))*sin(8*b*x + 8*a) - sin(8*b*x + 8*a)^2 + 8*(sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(6*
b*x + 6*a) - 16*sin(6*b*x + 6*a)^2 - sin(4*b*x + 4*a)^2 - 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 4*sin(2*b*x +
2*a)^2 + 4*cos(2*b*x + 2*a) - 1)*log(cos(b*x)^2 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a
) + sin(a)^2) + 4*(3*sin(10*b*x + 10*a) - 6*sin(8*b*x + 8*a) - 2*sin(6*b*x + 6*a) - 6*sin(4*b*x + 4*a) + 3*sin
(2*b*x + 2*a))*sin(12*b*x + 12*a) + 4*(9*sin(8*b*x + 8*a) + 16*sin(6*b*x + 6*a) + 9*sin(4*b*x + 4*a) - 12*sin(
2*b*x + 2*a))*sin(10*b*x + 10*a) - 24*sin(10*b*x + 10*a)^2 - 4*(22*sin(6*b*x + 6*a) - 12*sin(4*b*x + 4*a) - 9*
sin(2*b*x + 2*a))*sin(8*b*x + 8*a) + 24*sin(8*b*x + 8*a)^2 - 8*(11*sin(4*b*x + 4*a) - 8*sin(2*b*x + 2*a))*sin(
6*b*x + 6*a) - 32*sin(6*b*x + 6*a)^2 + 24*sin(4*b*x + 4*a)^2 + 36*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 24*sin(2
*b*x + 2*a)^2 + 12*cos(2*b*x + 2*a))/(b*cos(12*b*x + 12*a)^2 + 4*b*cos(10*b*x + 10*a)^2 + b*cos(8*b*x + 8*a)^2
 + 16*b*cos(6*b*x + 6*a)^2 + b*cos(4*b*x + 4*a)...

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (52) = 104\).
time = 2.47, size = 138, normalized size = 2.30 \begin {gather*} \frac {6 \, \cos \left (b x + a\right )^{4} - 9 \, \cos \left (b x + a\right )^{2} - 6 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 6 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) + 2}{128 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(6*cos(b*x + a)^4 - 9*cos(b*x + a)^2 - 6*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*log(cos(b*
x + a)^2) + 6*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*log(-1/4*cos(b*x + a)^2 + 1/4) + 2)/(b*cos(
b*x + a)^6 - 2*b*cos(b*x + a)^4 + b*cos(b*x + a)^2)

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

[Out]

Timed out

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Giac [A]
time = 0.44, size = 74, normalized size = 1.23 \begin {gather*} \frac {\frac {6 \, \cos \left (b x + a\right )^{4} - 9 \, \cos \left (b x + a\right )^{2} + 2}{{\left (\cos \left (b x + a\right )^{2} - 1\right )}^{2} \cos \left (b x + a\right )^{2}} + 6 \, \log \left (-\cos \left (b x + a\right )^{2} + 1\right ) - 12 \, \log \left ({\left | \cos \left (b x + a\right ) \right |}\right )}{128 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*((6*cos(b*x + a)^4 - 9*cos(b*x + a)^2 + 2)/((cos(b*x + a)^2 - 1)^2*cos(b*x + a)^2) + 6*log(-cos(b*x + a)
^2 + 1) - 12*log(abs(cos(b*x + a))))/b

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Mupad [B]
time = 0.14, size = 82, normalized size = 1.37 \begin {gather*} \frac {3\,\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{64\,b}-\frac {3\,\ln \left (\cos \left (a+b\,x\right )\right )}{32\,b}+\frac {\frac {3\,{\cos \left (a+b\,x\right )}^4}{64}-\frac {9\,{\cos \left (a+b\,x\right )}^2}{128}+\frac {1}{64}}{b\,\left ({\cos \left (a+b\,x\right )}^6-2\,{\cos \left (a+b\,x\right )}^4+{\cos \left (a+b\,x\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*log(sin(a + b*x)^2))/(64*b) - (3*log(cos(a + b*x)))/(32*b) + ((3*cos(a + b*x)^4)/64 - (9*cos(a + b*x)^2)/12
8 + 1/64)/(b*(cos(a + b*x)^2 - 2*cos(a + b*x)^4 + cos(a + b*x)^6))

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