3.454 \(\int \frac{\cot ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=71 \[ \frac{(a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{(a+b) \cot (c+d x)}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d} \]

[Out]

((a + b)^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(5/2)*d) + ((a + b)*Cot[c + d*x])/(a^2*d) - Cot[
c + d*x]^3/(3*a*d)

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Rubi [A]  time = 0.0793176, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3195, 325, 205} \[ \frac{(a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{(a+b) \cot (c+d x)}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

((a + b)^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(5/2)*d) + ((a + b)*Cot[c + d*x])/(a^2*d) - Cot[
c + d*x]^3/(3*a*d)

Rule 3195

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.292023, size = 72, normalized size = 1.01 \[ \frac{3 (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )+\sqrt{a} \cot (c+d x) \left (-a \csc ^2(c+d x)+4 a+3 b\right )}{3 a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

(3*(a + b)^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*Cot[c + d*x]*(4*a + 3*b - a*Csc[c + d*x]
^2))/(3*a^(5/2)*d)

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Maple [B]  time = 0.121, size = 147, normalized size = 2.1 \begin{align*}{\frac{1}{d}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}}+2\,{\frac{b}{da\sqrt{a \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a+b \right ) \tan \left ( dx+c \right ) }{\sqrt{a \left ( a+b \right ) }}} \right ) }+{\frac{{b}^{2}}{{a}^{2}d}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}}-{\frac{1}{3\,da \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{1}{da\tan \left ( dx+c \right ) }}+{\frac{b}{{a}^{2}d\tan \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4/(a+sin(d*x+c)^2*b),x)

[Out]

1/d/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))+2/d/a/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*
(a+b))^(1/2))*b+1/d/a^2*b^2/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/3/d/a/tan(d*x+c)^3+1/d/
a/tan(d*x+c)+1/d/a^2/tan(d*x+c)*b

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82973, size = 973, normalized size = 13.7 \begin{align*} \left [\frac{4 \,{\left (4 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{-\frac{a + b}{a}} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left ({\left (2 \, a^{2} + a b\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + a b\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{a + b}{a}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) - 12 \,{\left (a + b\right )} \cos \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )}, \frac{2 \,{\left (4 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{a + b}{a}} \arctan \left (\frac{{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{a + b}{a}}}{2 \,{\left (a + b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 6 \,{\left (a + b\right )} \cos \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/12*(4*(4*a + 3*b)*cos(d*x + c)^3 + 3*((a + b)*cos(d*x + c)^2 - a - b)*sqrt(-(a + b)/a)*log(((8*a^2 + 8*a*b
+ b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 - 4*((2*a^2 + a*b)*cos(d*x + c)^3 - (a^2 + a*b)
*cos(d*x + c))*sqrt(-(a + b)/a)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x
+ c)^2 + a^2 + 2*a*b + b^2))*sin(d*x + c) - 12*(a + b)*cos(d*x + c))/((a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x +
c)), 1/6*(2*(4*a + 3*b)*cos(d*x + c)^3 - 3*((a + b)*cos(d*x + c)^2 - a - b)*sqrt((a + b)/a)*arctan(1/2*((2*a
+ b)*cos(d*x + c)^2 - a - b)*sqrt((a + b)/a)/((a + b)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c) - 6*(a + b)*cos
(d*x + c))/((a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{4}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4/(a+b*sin(d*x+c)**2),x)

[Out]

Integral(cot(c + d*x)**4/(a + b*sin(c + d*x)**2), x)

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Giac [A]  time = 1.17035, size = 162, normalized size = 2.28 \begin{align*} \frac{\frac{3 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )}{\left (a^{2} + 2 \, a b + b^{2}\right )}}{\sqrt{a^{2} + a b} a^{2}} + \frac{3 \, a \tan \left (d x + c\right )^{2} + 3 \, b \tan \left (d x + c\right )^{2} - a}{a^{2} \tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)
))*(a^2 + 2*a*b + b^2)/(sqrt(a^2 + a*b)*a^2) + (3*a*tan(d*x + c)^2 + 3*b*tan(d*x + c)^2 - a)/(a^2*tan(d*x + c)
^3))/d