\(\int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx\) [20]

Optimal result

Integrand size = 16, antiderivative size = 49 \[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {\cot (x)}{a^2}-\frac {2 b \log (\tan (x))}{a^3}+\frac {2 b \log (a+b \tan (x))}{a^3}-\frac {\frac {1}{b}+\frac {b}{a^2}}{a+b \tan (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3166, 908} \[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {2 b \log (\tan (x))}{a^3}+\frac {2 b \log (a+b \tan (x))}{a^3}-\frac {\frac {b}{a^2}+\frac {1}{b}}{a+b \tan (x)}-\frac {\cot (x)}{a^2} \]

[In]

Int[Csc[x]^2/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

-(Cot[x]/a^2) - (2*b*Log[Tan[x]])/a^3 + (2*b*Log[a + b*Tan[x]])/a^3 - (b^(-1) + b/a^2)/(a + b*Tan[x])

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3166

Int[sin[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[1/d, Subst[Int[x^m*((a + b*x)^n/(1 + x^2)^((m + n + 2)/2)), x], x, Tan[c + d*x]], x] /; FreeQ[{a,
b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1+x^2}{x^2 (a+b x)^2} \, dx,x,\tan (x)\right ) \\ & = \text {Subst}\left (\int \left (\frac {1}{a^2 x^2}-\frac {2 b}{a^3 x}+\frac {a^2+b^2}{a^2 (a+b x)^2}+\frac {2 b^2}{a^3 (a+b x)}\right ) \, dx,x,\tan (x)\right ) \\ & = -\frac {\cot (x)}{a^2}-\frac {2 b \log (\tan (x))}{a^3}+\frac {2 b \log (a+b \tan (x))}{a^3}-\frac {\frac {1}{b}+\frac {b}{a^2}}{a+b \tan (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.43 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.55 \[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {a^2+b^2-a^2 \cot ^2(x)-2 b^2 \log (\sin (x))-a b \cot (x) (1+2 \log (\sin (x))-2 \log (a \cos (x)+b \sin (x)))+2 b^2 \log (a \cos (x)+b \sin (x))}{a^3 (b+a \cot (x))} \]

[In]

Integrate[Csc[x]^2/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

(a^2 + b^2 - a^2*Cot[x]^2 - 2*b^2*Log[Sin[x]] - a*b*Cot[x]*(1 + 2*Log[Sin[x]] - 2*Log[a*Cos[x] + b*Sin[x]]) +
2*b^2*Log[a*Cos[x] + b*Sin[x]])/(a^3*(b + a*Cot[x]))

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.14

method	result	size
default	\(-\frac {a^{2}+b^{2}}{a^{2} b \left (a +b \tan \left (x \right )\right )}+\frac {2 b \ln \left (a +b \tan \left (x \right )\right )}{a^{3}}-\frac {1}{a^{2} \tan \left (x \right )}-\frac {2 b \ln \left (\tan \left (x \right )\right )}{a^{3}}\)	\(56\)
risch	\(-\frac {4 \left (b \,{\mathrm e}^{2 i x}-b +i a \right )}{\left ({\mathrm e}^{2 i x}-1\right ) \left (-i b \,{\mathrm e}^{2 i x}+a \,{\mathrm e}^{2 i x}+i b +a \right ) a^{2}}+\frac {2 b \ln \left ({\mathrm e}^{2 i x}-\frac {i b +a}{i b -a}\right )}{a^{3}}-\frac {2 b \ln \left ({\mathrm e}^{2 i x}-1\right )}{a^{3}}\)	\(100\)
norman	\(\frac {\frac {1}{2 a}+\frac {\tan \left (\frac {x}{2}\right )^{4}}{2 a}-\frac {\left (3 a^{2}+4 b^{2}\right ) \tan \left (\frac {x}{2}\right )^{2}}{a^{3}}}{\tan \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a \right )}-\frac {2 b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{3}}+\frac {2 b \ln \left (\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a \right )}{a^{3}}\)	\(106\)
parallelrisch	\(\frac {-2 \cot \left (x \right ) \cos \left (x \right ) a^{2}+2 a b \cos \left (x \right ) \ln \left (\frac {-2 a \cos \left (x \right )-2 b \sin \left (x \right )}{\cos \left (x \right )+1}\right )-2 \ln \left (\csc \left (x \right )-\cot \left (x \right )\right ) a b \cos \left (x \right )+2 \sin \left (x \right ) b^{2} \ln \left (\frac {-2 a \cos \left (x \right )-2 b \sin \left (x \right )}{\cos \left (x \right )+1}\right )-2 \sin \left (x \right ) \ln \left (\csc \left (x \right )-\cot \left (x \right )\right ) b^{2}+\csc \left (x \right ) a^{2}+2 \sin \left (x \right ) b^{2}}{\left (a \cos \left (x \right )+b \sin \left (x \right )\right ) a^{3}}\)	\(119\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (49) = 98\).

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.73 \[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 \, a^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) \sin \left (x\right ) - a^{2} + {\left (b^{2} \cos \left (x\right )^{2} - a b \cos \left (x\right ) \sin \left (x\right ) - b^{2}\right )} \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}\right ) - {\left (b^{2} \cos \left (x\right )^{2} - a b \cos \left (x\right ) \sin \left (x\right ) - b^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (x\right )^{2} + \frac {1}{4}\right )}{a^{3} b \cos \left (x\right )^{2} - a^{4} \cos \left (x\right ) \sin \left (x\right ) - a^{3} b} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=\int \frac {\csc ^{2}{\left (x \right )}}{\left (a \cos {\left (x \right )} + b \sin {\left (x \right )}\right )^{2}}\, dx \]

[In]

integrate(csc(x)**2/(a*cos(x)+b*sin(x))**2,x)

[Out]

Integral(csc(x)**2/(a*cos(x) + b*sin(x))**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.27 \[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {a b + {\left (a^{2} + 2 \, b^{2}\right )} \tan \left (x\right )}{a^{2} b^{2} \tan \left (x\right )^{2} + a^{3} b \tan \left (x\right )} + \frac {2 \, b \log \left (b \tan \left (x\right ) + a\right )}{a^{3}} - \frac {2 \, b \log \left (\tan \left (x\right )\right )}{a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.29 \[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 \, b \log \left ({\left | b \tan \left (x\right ) + a \right |}\right )}{a^{3}} - \frac {2 \, b \log \left ({\left | \tan \left (x\right ) \right |}\right )}{a^{3}} - \frac {a^{2} \tan \left (x\right ) + 2 \, b^{2} \tan \left (x\right ) + a b}{{\left (b \tan \left (x\right )^{2} + a \tan \left (x\right )\right )} a^{2} b} \]

[In]

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

[Out]

2*b*log(abs(b*tan(x) + a))/a^3 - 2*b*log(abs(tan(x)))/a^3 - (a^2*tan(x) + 2*b^2*tan(x) + a*b)/((b*tan(x)^2 + a
*tan(x))*a^2*b)

Mupad [B] (verification not implemented)

Time = 21.71 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.33 \[ \int \frac {\csc ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a^2}-\frac {a+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (5\,a^2+4\,b^2\right )}{a}}{-2\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+2\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )+4\,b\,a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}+\frac {2\,b\,\ln \left (-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a\right )}{a^3}-\frac {2\,b\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^3} \]

[In]

int(1/(sin(x)^2*(a*cos(x) + b*sin(x))^2),x)

[Out]

tan(x/2)/(2*a^2) - (a + 2*b*tan(x/2) - (tan(x/2)^2*(5*a^2 + 4*b^2))/a)/(2*a^3*tan(x/2) - 2*a^3*tan(x/2)^3 + 4*
a^2*b*tan(x/2)^2) + (2*b*log(a + 2*b*tan(x/2) - a*tan(x/2)^2))/a^3 - (2*b*log(tan(x/2)))/a^3