3.722 \(\int \frac{(a+b \cot (x))^3 \csc ^2(x)}{c+d \cot (x)} \, dx\)

Optimal. Leaf size=78 \[ -\frac{b \cot (x) (b c-a d)^2}{d^3}+\frac{(b c-a d) (a+b \cot (x))^2}{2 d^2}+\frac{(b c-a d)^3 \log (c+d \cot (x))}{d^4}-\frac{(a+b \cot (x))^3}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.138954, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4344, 43} \[ -\frac{b \cot (x) (b c-a d)^2}{d^3}+\frac{(b c-a d) (a+b \cot (x))^2}{2 d^2}+\frac{(b c-a d)^3 \log (c+d \cot (x))}{d^4}-\frac{(a+b \cot (x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cot[x])^3*Csc[x]^2)/(c + d*Cot[x]),x]

[Out]

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

Rule 4344

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Cot[c*(a + b*x)], x]}, -Dist[d
/(b*c), Subst[Int[SubstFor[1, Cot[c*(a + b*x)]/d, u, x], x], x, Cot[c*(a + b*x)]/d], x] /; FunctionOfQ[Cot[c*(
a + b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Csc] || EqQ[F, csc])

Rule 43

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])

Rubi steps

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

Mathematica [A]  time = 1.26379, size = 135, normalized size = 1.73 \[ \frac{(a+b \cot (x))^3 (c \sin (x)+d \cos (x)) \left (b d \left (\sin (2 x) \left (-9 a^2 d^2+9 a b c d+b^2 \left (d^2-3 c^2\right )\right )+3 b d (b c-3 a d)\right )-6 \sin ^2(x) (b c-a d)^3 (\log (\sin (x))-\log (c \sin (x)+d \cos (x)))-2 b^3 d^3 \cot (x)\right )}{6 d^4 (c+d \cot (x)) (a \sin (x)+b \cos (x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cot[x])^3*Csc[x]^2)/(c + d*Cot[x]),x]

[Out]

((a + b*Cot[x])^3*(d*Cos[x] + c*Sin[x])*(-2*b^3*d^3*Cot[x] - 6*(b*c - a*d)^3*(Log[Sin[x]] - Log[d*Cos[x] + c*S
in[x]])*Sin[x]^2 + b*d*(3*b*d*(b*c - 3*a*d) + (9*a*b*c*d - 9*a^2*d^2 + b^2*(-3*c^2 + d^2))*Sin[2*x])))/(6*d^4*
(c + d*Cot[x])*(b*Cos[x] + a*Sin[x])^3)

________________________________________________________________________________________

Maple [B]  time = 0.078, size = 202, normalized size = 2.6 \begin{align*} -{\frac{{b}^{3}}{3\,d \left ( \tan \left ( x \right ) \right ) ^{3}}}+{\frac{\ln \left ( \tan \left ( x \right ) \right ){a}^{3}}{d}}-3\,{\frac{\ln \left ( \tan \left ( x \right ) \right ){a}^{2}bc}{{d}^{2}}}+3\,{\frac{\ln \left ( \tan \left ( x \right ) \right ) a{b}^{2}{c}^{2}}{{d}^{3}}}-{\frac{\ln \left ( \tan \left ( x \right ) \right ){b}^{3}{c}^{3}}{{d}^{4}}}-3\,{\frac{{a}^{2}b}{d\tan \left ( x \right ) }}+3\,{\frac{a{b}^{2}c}{{d}^{2}\tan \left ( x \right ) }}-{\frac{{b}^{3}{c}^{2}}{{d}^{3}\tan \left ( x \right ) }}-{\frac{3\,a{b}^{2}}{2\,d \left ( \tan \left ( x \right ) \right ) ^{2}}}+{\frac{{b}^{3}c}{2\,{d}^{2} \left ( \tan \left ( x \right ) \right ) ^{2}}}-{\frac{\ln \left ( c\tan \left ( x \right ) +d \right ){a}^{3}}{d}}+3\,{\frac{\ln \left ( c\tan \left ( x \right ) +d \right ){a}^{2}bc}{{d}^{2}}}-3\,{\frac{\ln \left ( c\tan \left ( x \right ) +d \right ) a{b}^{2}{c}^{2}}{{d}^{3}}}+{\frac{\ln \left ( c\tan \left ( x \right ) +d \right ){b}^{3}{c}^{3}}{{d}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cot(x))^3*csc(x)^2/(c+d*cot(x)),x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 0.990708, size = 217, normalized size = 2.78 \begin{align*} \frac{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (c \tan \left (x\right ) + d\right )}{d^{4}} - \frac{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\tan \left (x\right )\right )}{d^{4}} - \frac{2 \, b^{3} d^{2} + 6 \,{\left (b^{3} c^{2} - 3 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} \tan \left (x\right )^{2} - 3 \,{\left (b^{3} c d - 3 \, a b^{2} d^{2}\right )} \tan \left (x\right )}{6 \, d^{3} \tan \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x))^3*csc(x)^2/(c+d*cot(x)),x, algorithm="maxima")

[Out]

(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(c*tan(x) + d)/d^4 - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b
*c*d^2 - a^3*d^3)*log(tan(x))/d^4 - 1/6*(2*b^3*d^2 + 6*(b^3*c^2 - 3*a*b^2*c*d + 3*a^2*b*d^2)*tan(x)^2 - 3*(b^3
*c*d - 3*a*b^2*d^2)*tan(x))/(d^3*tan(x)^3)

________________________________________________________________________________________

Fricas [B]  time = 3.42416, size = 705, normalized size = 9.04 \begin{align*} -\frac{2 \,{\left (3 \, b^{3} c^{2} d - 9 \, a b^{2} c d^{2} +{\left (9 \, a^{2} b - b^{3}\right )} d^{3}\right )} \cos \left (x\right )^{3} + 3 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3} -{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (2 \, c d \cos \left (x\right ) \sin \left (x\right ) -{\left (c^{2} - d^{2}\right )} \cos \left (x\right )^{2} + c^{2}\right ) \sin \left (x\right ) - 3 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3} -{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{4} \, \cos \left (x\right )^{2} + \frac{1}{4}\right ) \sin \left (x\right ) - 6 \,{\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} \cos \left (x\right ) + 3 \,{\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} \sin \left (x\right )}{6 \,{\left (d^{4} \cos \left (x\right )^{2} - d^{4}\right )} \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x))^3*csc(x)^2/(c+d*cot(x)),x, algorithm="fricas")

[Out]

-1/6*(2*(3*b^3*c^2*d - 9*a*b^2*c*d^2 + (9*a^2*b - b^3)*d^3)*cos(x)^3 + 3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*
d^2 - a^3*d^3 - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*cos(x)^2)*log(2*c*d*cos(x)*sin(x) - (c^2 -
d^2)*cos(x)^2 + c^2)*sin(x) - 3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3 - (b^3*c^3 - 3*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 - a^3*d^3)*cos(x)^2)*log(-1/4*cos(x)^2 + 1/4)*sin(x) - 6*(b^3*c^2*d - 3*a*b^2*c*d^2 + 3*a^2*b
*d^3)*cos(x) + 3*(b^3*c*d^2 - 3*a*b^2*d^3)*sin(x))/((d^4*cos(x)^2 - d^4)*sin(x))

________________________________________________________________________________________

Sympy [A]  time = 75.3299, size = 97, normalized size = 1.24 \begin{align*} - \frac{b^{3} \cot ^{3}{\left (x \right )}}{3 d} - \frac{\left (3 a b^{2} d - b^{3} c\right ) \cot ^{2}{\left (x \right )}}{2 d^{2}} - \frac{\left (a d - b c\right )^{3} \left (\begin{cases} \frac{\cot{\left (x \right )}}{c} & \text{for}\: d = 0 \\\frac{\log{\left (c + d \cot{\left (x \right )} \right )}}{d} & \text{otherwise} \end{cases}\right )}{d^{3}} - \frac{\left (3 a^{2} b d^{2} - 3 a b^{2} c d + b^{3} c^{2}\right ) \cot{\left (x \right )}}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x))**3*csc(x)**2/(c+d*cot(x)),x)

[Out]

-b**3*cot(x)**3/(3*d) - (3*a*b**2*d - b**3*c)*cot(x)**2/(2*d**2) - (a*d - b*c)**3*Piecewise((cot(x)/c, Eq(d, 0
)), (log(c + d*cot(x))/d, True))/d**3 - (3*a**2*b*d**2 - 3*a*b**2*c*d + b**3*c**2)*cot(x)/d**3

________________________________________________________________________________________

Giac [B]  time = 1.17287, size = 313, normalized size = 4.01 \begin{align*} -\frac{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | \tan \left (x\right ) \right |}\right )}{d^{4}} + \frac{{\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \log \left ({\left | c \tan \left (x\right ) + d \right |}\right )}{c d^{4}} + \frac{11 \, b^{3} c^{3} \tan \left (x\right )^{3} - 33 \, a b^{2} c^{2} d \tan \left (x\right )^{3} + 33 \, a^{2} b c d^{2} \tan \left (x\right )^{3} - 11 \, a^{3} d^{3} \tan \left (x\right )^{3} - 6 \, b^{3} c^{2} d \tan \left (x\right )^{2} + 18 \, a b^{2} c d^{2} \tan \left (x\right )^{2} - 18 \, a^{2} b d^{3} \tan \left (x\right )^{2} + 3 \, b^{3} c d^{2} \tan \left (x\right ) - 9 \, a b^{2} d^{3} \tan \left (x\right ) - 2 \, b^{3} d^{3}}{6 \, d^{4} \tan \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(x))^3*csc(x)^2/(c+d*cot(x)),x, algorithm="giac")

[Out]

-(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(abs(tan(x)))/d^4 + (b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b
*c^2*d^2 - a^3*c*d^3)*log(abs(c*tan(x) + d))/(c*d^4) + 1/6*(11*b^3*c^3*tan(x)^3 - 33*a*b^2*c^2*d*tan(x)^3 + 33
*a^2*b*c*d^2*tan(x)^3 - 11*a^3*d^3*tan(x)^3 - 6*b^3*c^2*d*tan(x)^2 + 18*a*b^2*c*d^2*tan(x)^2 - 18*a^2*b*d^3*ta
n(x)^2 + 3*b^3*c*d^2*tan(x) - 9*a*b^2*d^3*tan(x) - 2*b^3*d^3)/(d^4*tan(x)^3)