Integrand size = 23, antiderivative size = 63 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\csc ^2(c+d x)}{2 a d}-\frac {(a+b) \log (\sin (c+d x))}{a^2 d}+\frac {(a+b) \log \left (a+b \sin ^2(c+d x)\right )}{2 a^2 d} \] Output:
-1/2*csc(d*x+c)^2/a/d-(a+b)*ln(sin(d*x+c))/a^2/d+1/2*(a+b)*ln(a+b*sin(d*x+ c)^2)/a^2/d
Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.79 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a \csc ^2(c+d x)+(a+b) \left (2 \log (\sin (c+d x))-\log \left (a+b \sin ^2(c+d x)\right )\right )}{2 a^2 d} \] Input:
Integrate[Cot[c + d*x]^3/(a + b*Sin[c + d*x]^2),x]
Output:
-1/2*(a*Csc[c + d*x]^2 + (a + b)*(2*Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]^2]))/(a^2*d)
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3673, 86, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^3 \left (a+b \sin (c+d x)^2\right )}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\csc ^4(c+d x) \left (1-\sin ^2(c+d x)\right )}{b \sin ^2(c+d x)+a}d\sin ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {\csc ^4(c+d x)}{a}+\frac {(-a-b) \csc ^2(c+d x)}{a^2}+\frac {b (a+b)}{a^2 \left (b \sin ^2(c+d x)+a\right )}\right )d\sin ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {(a+b) \log \left (\sin ^2(c+d x)\right )}{a^2}+\frac {(a+b) \log \left (a+b \sin ^2(c+d x)\right )}{a^2}-\frac {\csc ^2(c+d x)}{a}}{2 d}\) |
Input:
Int[Cot[c + d*x]^3/(a + b*Sin[c + d*x]^2),x]
Output:
(-(Csc[c + d*x]^2/a) - ((a + b)*Log[Sin[c + d*x]^2])/a^2 + ((a + b)*Log[a + b*Sin[c + d*x]^2])/a^2)/(2*d)
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[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] && Integ erQ[(m - 1)/2]
Time = 1.83 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.60
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 a \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (-a -b \right ) \ln \left (\cos \left (d x +c \right )+1\right )}{2 a^{2}}+\frac {1}{4 a \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (-a -b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{2}}+\frac {\left (a +b \right ) \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 a^{2}}}{d}\) | \(101\) |
default | \(\frac {-\frac {1}{4 a \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (-a -b \right ) \ln \left (\cos \left (d x +c \right )+1\right )}{2 a^{2}}+\frac {1}{4 a \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (-a -b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{2}}+\frac {\left (a +b \right ) \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 a^{2}}}{d}\) | \(101\) |
risch | \(\frac {2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right ) b}{2 a^{2} d}\) | \(152\) |
Input:
int(cot(d*x+c)^3/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/4/a/(cos(d*x+c)+1)+1/2*(-a-b)/a^2*ln(cos(d*x+c)+1)+1/4/a/(cos(d*x+ c)-1)+1/2*(-a-b)/a^2*ln(cos(d*x+c)-1)+1/2*(a+b)/a^2*ln(a+b-b*cos(d*x+c)^2) )
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.44 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 2 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + a}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="fricas")
Output:
1/2*(((a + b)*cos(d*x + c)^2 - a - b)*log(-b*cos(d*x + c)^2 + a + b) - 2*( (a + b)*cos(d*x + c)^2 - a - b)*log(1/2*sin(d*x + c)) + a)/(a^2*d*cos(d*x + c)^2 - a^2*d)
\[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(cot(d*x+c)**3/(a+b*sin(d*x+c)**2),x)
Output:
Integral(cot(c + d*x)**3/(a + b*sin(c + d*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {{\left (a + b\right )} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{2}} - \frac {{\left (a + b\right )} \log \left (\sin \left (d x + c\right )^{2}\right )}{a^{2}} - \frac {1}{a \sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="maxima")
Output:
1/2*((a + b)*log(b*sin(d*x + c)^2 + a)/a^2 - (a + b)*log(sin(d*x + c)^2)/a ^2 - 1/(a*sin(d*x + c)^2))/d
Time = 0.52 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {{\left (a + b\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2} d} + \frac {{\left (a b + b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right )^{2} + a \right |}\right )}{2 \, a^{2} b d} - \frac {1}{2 \, a d \sin \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)^3/(a+b*sin(d*x+c)^2),x, algorithm="giac")
Output:
-(a + b)*log(abs(sin(d*x + c)))/(a^2*d) + 1/2*(a*b + b^2)*log(abs(b*sin(d* x + c)^2 + a))/(a^2*b*d) - 1/2/(a*d*sin(d*x + c)^2)
Time = 34.71 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\ln \left (a+a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )\,\left (a+b\right )}{2\,a^2\,d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2}{2\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a+b\right )}{a^2\,d} \] Input:
int(cot(c + d*x)^3/(a + b*sin(c + d*x)^2),x)
Output:
(log(a + a*tan(c + d*x)^2 + b*tan(c + d*x)^2)*(a + b))/(2*a^2*d) - cot(c + d*x)^2/(2*a*d) - (log(tan(c + d*x))*(a + b))/(a^2*d)
Time = 0.19 (sec) , antiderivative size = 316, normalized size of antiderivative = 5.02 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {2 \,\mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a +2 \,\mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b +2 \,\mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a +2 \,\mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b +2 \,\mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a +2 b \right ) \sin \left (d x +c \right )^{2} a +2 \,\mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a +2 b \right ) \sin \left (d x +c \right )^{2} b -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b +\sin \left (d x +c \right )^{2} a -2 a}{4 \sin \left (d x +c \right )^{2} a^{2} d} \] Input:
int(cot(d*x+c)^3/(a+b*sin(d*x+c)^2),x)
Output:
(2*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2) )*sin(c + d*x)**2*a + 2*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqr t(a)*tan((c + d*x)/2))*sin(c + d*x)**2*b + 2*log(sqrt(2*sqrt(b)*sqrt(a + b ) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2*a + 2*log(sqrt(2* sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2 *b + 2*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**2*a + 2*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2* b)*sin(c + d*x)**2*b - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*a - 4*log(t an((c + d*x)/2))*sin(c + d*x)**2*b + sin(c + d*x)**2*a - 2*a)/(4*sin(c + d *x)**2*a**2*d)