3.447 \(\int \frac{\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=89 \[ \frac{(2 a+b) \csc ^2(c+d x)}{2 a^2 d}-\frac{(a+b)^2 \log \left (a+b \sin ^2(c+d x)\right )}{2 a^3 d}+\frac{(a+b)^2 \log (\sin (c+d x))}{a^3 d}-\frac{\csc ^4(c+d x)}{4 a d} \]

[Out]

((2*a + b)*Csc[c + d*x]^2)/(2*a^2*d) - Csc[c + d*x]^4/(4*a*d) + ((a + b)^2*Log[Sin[c + d*x]])/(a^3*d) - ((a +
b)^2*Log[a + b*Sin[c + d*x]^2])/(2*a^3*d)

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Rubi [A]  time = 0.0889625, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3194, 88} \[ \frac{(2 a+b) \csc ^2(c+d x)}{2 a^2 d}-\frac{(a+b)^2 \log \left (a+b \sin ^2(c+d x)\right )}{2 a^3 d}+\frac{(a+b)^2 \log (\sin (c+d x))}{a^3 d}-\frac{\csc ^4(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a + b)*Csc[c + d*x]^2)/(2*a^2*d) - Csc[c + d*x]^4/(4*a*d) + ((a + b)^2*Log[Sin[c + d*x]])/(a^3*d) - ((a +
b)^2*Log[a + b*Sin[c + d*x]^2])/(2*a^3*d)

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^3 (a+b x)} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a x^3}+\frac{-2 a-b}{a^2 x^2}+\frac{(a+b)^2}{a^3 x}-\frac{b (a+b)^2}{a^3 (a+b x)}\right ) \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac{(2 a+b) \csc ^2(c+d x)}{2 a^2 d}-\frac{\csc ^4(c+d x)}{4 a d}+\frac{(a+b)^2 \log (\sin (c+d x))}{a^3 d}-\frac{(a+b)^2 \log \left (a+b \sin ^2(c+d x)\right )}{2 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.563815, size = 72, normalized size = 0.81 \[ \frac{-a^2 \csc ^4(c+d x)+2 a (2 a+b) \csc ^2(c+d x)+2 (a+b)^2 \left (2 \log (\sin (c+d x))-\log \left (a+b \sin ^2(c+d x)\right )\right )}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(2*a + b)*Csc[c + d*x]^2 - a^2*Csc[c + d*x]^4 + 2*(a + b)^2*(2*Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]
^2]))/(4*a^3*d)

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Maple [B]  time = 0.099, size = 302, normalized size = 3.4 \begin{align*} -{\frac{1}{16\,da \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7}{16\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{b}{4\,{a}^{2}d \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{2\,da}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{{a}^{2}d}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ){b}^{2}}{2\,d{a}^{3}}}-{\frac{1}{16\,da \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{7}{16\,da \left ( 1+\cos \left ( dx+c \right ) \right ) }}+{\frac{b}{4\,{a}^{2}d \left ( 1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{2\,da}}+{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) b}{{a}^{2}d}}+{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ){b}^{2}}{2\,d{a}^{3}}}-{\frac{\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }{2\,da}}-{\frac{\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) b}{{a}^{2}d}}-{\frac{\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ){b}^{2}}{2\,d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5/(a+sin(d*x+c)^2*b),x)

[Out]

-1/16/d/a/(-1+cos(d*x+c))^2-7/16/d/a/(-1+cos(d*x+c))-1/4/d/a^2/(-1+cos(d*x+c))*b+1/2/d/a*ln(-1+cos(d*x+c))+1/d
/a^2*ln(-1+cos(d*x+c))*b+1/2/d/a^3*ln(-1+cos(d*x+c))*b^2-1/16/a/d/(1+cos(d*x+c))^2+7/16/a/d/(1+cos(d*x+c))+1/4
/d/a^2/(1+cos(d*x+c))*b+1/2/d/a*ln(1+cos(d*x+c))+1/d/a^2*ln(1+cos(d*x+c))*b+1/2/d/a^3*ln(1+cos(d*x+c))*b^2-1/2
/d/a*ln(b*cos(d*x+c)^2-a-b)-1/d/a^2*ln(b*cos(d*x+c)^2-a-b)*b-1/2/d/a^3*ln(b*cos(d*x+c)^2-a-b)*b^2

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Maxima [A]  time = 0.998426, size = 124, normalized size = 1.39 \begin{align*} -\frac{\frac{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{3}} - \frac{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right )^{2}\right )}{a^{3}} - \frac{2 \,{\left (2 \, a + b\right )} \sin \left (d x + c\right )^{2} - a}{a^{2} \sin \left (d x + c\right )^{4}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(a^2 + 2*a*b + b^2)*log(b*sin(d*x + c)^2 + a)/a^3 - 2*(a^2 + 2*a*b + b^2)*log(sin(d*x + c)^2)/a^3 - (2
*(2*a + b)*sin(d*x + c)^2 - a)/(a^2*sin(d*x + c)^4))/d

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Fricas [B]  time = 2.28817, size = 491, normalized size = 5.52 \begin{align*} -\frac{2 \,{\left (2 \, a^{2} + a b\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 2 \, a b + 2 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 4 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right )}{4 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(2*a^2 + a*b)*cos(d*x + c)^2 - 3*a^2 - 2*a*b + 2*((a^2 + 2*a*b + b^2)*cos(d*x + c)^4 - 2*(a^2 + 2*a*b
+ b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*log(-b*cos(d*x + c)^2 + a + b) - 4*((a^2 + 2*a*b + b^2)*cos(d*x + c
)^4 - 2*(a^2 + 2*a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*log(1/2*sin(d*x + c)))/(a^3*d*cos(d*x + c)^4 -
2*a^3*d*cos(d*x + c)^2 + a^3*d)

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

[Out]

Timed out

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Giac [B]  time = 1.22261, size = 277, normalized size = 3.11 \begin{align*} -\frac{\frac{a{\left (\frac{\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )}^{2} + 12 \, a{\left (\frac{\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 8 \, b{\left (\frac{\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )}}{a^{2}} + \frac{32 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | -a{\left (\frac{\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 2 \, a + 4 \, b \right |}\right )}{a^{3}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*((a*((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2 + 12*a*((cos(d*x +
c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1)) + 8*b*((cos(d*x + c) + 1)/(cos(d*x + c) -
1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/a^2 + 32*(a^2 + 2*a*b + b^2)*log(abs(-a*((cos(d*x + c) + 1)/(cos
(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1)) + 2*a + 4*b))/a^3)/d