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

Optimal. Leaf size=53 \[ -\frac {(b c-a d)^2 \log (c+d \cot (x))}{d^3}+\frac {b \cot (x) (b c-a d)}{d^2}-\frac {(a+b \cot (x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 53, normalized size of antiderivative = 1.00, 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} \[ \frac {b \cot (x) (b c-a d)}{d^2}-\frac {(b c-a d)^2 \log (c+d \cot (x))}{d^3}-\frac {(a+b \cot (x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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])

Rubi steps

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

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Mathematica [A]  time = 0.56, size = 62, normalized size = 1.17 \[ \frac {2 b d \cot (x) (b c-2 a d)+2 (b c-a d)^2 (\log (\sin (x))-\log (c \sin (x)+d \cos (x)))-b^2 d^2 \csc ^2(x)}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.47, size = 182, normalized size = 3.43 \[ \frac {b^{2} d^{2} - 2 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} \cos \relax (x) \sin \relax (x) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (2 \, c d \cos \relax (x) \sin \relax (x) - {\left (c^{2} - d^{2}\right )} \cos \relax (x)^{2} + c^{2}\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{4} \, \cos \relax (x)^{2} + \frac {1}{4}\right )}{2 \, {\left (d^{3} \cos \relax (x)^{2} - d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 139, normalized size = 2.62 \[ \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | \tan \relax (x) \right |}\right )}{d^{3}} - \frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \log \left ({\left | c \tan \relax (x) + d \right |}\right )}{c d^{3}} - \frac {3 \, b^{2} c^{2} \tan \relax (x)^{2} - 6 \, a b c d \tan \relax (x)^{2} + 3 \, a^{2} d^{2} \tan \relax (x)^{2} - 2 \, b^{2} c d \tan \relax (x) + 4 \, a b d^{2} \tan \relax (x) + b^{2} d^{2}}{2 \, d^{3} \tan \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.13, size = 119, normalized size = 2.25 \[ -\frac {b^{2}}{2 d \tan \relax (x )^{2}}+\frac {\ln \left (\tan \relax (x )\right ) a^{2}}{d}-\frac {2 \ln \left (\tan \relax (x )\right ) a b c}{d^{2}}+\frac {\ln \left (\tan \relax (x )\right ) b^{2} c^{2}}{d^{3}}-\frac {2 b a}{d \tan \relax (x )}+\frac {b^{2} c}{d^{2} \tan \relax (x )}-\frac {\ln \left (c \tan \relax (x )+d \right ) a^{2}}{d}+\frac {2 \ln \left (c \tan \relax (x )+d \right ) a b c}{d^{2}}-\frac {\ln \left (c \tan \relax (x )+d \right ) b^{2} c^{2}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 92, normalized size = 1.74 \[ -\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (c \tan \relax (x) + d\right )}{d^{3}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\tan \relax (x)\right )}{d^{3}} - \frac {b^{2} d - 2 \, {\left (b^{2} c - 2 \, a b d\right )} \tan \relax (x)}{2 \, d^{2} \tan \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.07, size = 92, normalized size = 1.74 \[ -\frac {\frac {b^2}{2\,d}+\frac {b\,\mathrm {tan}\relax (x)\,\left (2\,a\,d-b\,c\right )}{d^2}}{{\mathrm {tan}\relax (x)}^2}-\frac {2\,\mathrm {atanh}\left (\frac {\left (d+2\,c\,\mathrm {tan}\relax (x)\right )\,{\left (a\,d-b\,c\right )}^2}{d\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 55.64, size = 58, normalized size = 1.09 \[ - \frac {b^{2} \cot ^{2}{\relax (x )}}{2 d} - \frac {\left (a d - b c\right )^{2} \left (\begin {cases} \frac {\cot {\relax (x )}}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d \cot {\relax (x )} \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {\left (2 a b d - b^{2} c\right ) \cot {\relax (x )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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