Integrand size = 21, antiderivative size = 72 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\cot (c+d x)}{a^2 d}-\frac {2 b \log (\tan (c+d x))}{a^3 d}+\frac {2 b \log (a+b \tan (c+d x))}{a^3 d}-\frac {b}{a^2 d (a+b \tan (c+d x))} \] Output:
-cot(d*x+c)/a^2/d-2*b*ln(tan(d*x+c))/a^3/d+2*b*ln(a+b*tan(d*x+c))/a^3/d-b/ a^2/d/(a+b*tan(d*x+c))
Time = 0.72 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.51 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {-a^2 \cot ^2(c+d x)-a b \cot (c+d x) (1+2 \log (\sin (c+d x))-2 \log (a \cos (c+d x)+b \sin (c+d x)))+b^2 (1-2 \log (\sin (c+d x))+2 \log (a \cos (c+d x)+b \sin (c+d x)))}{a^3 d (b+a \cot (c+d x))} \] Input:
Integrate[Csc[c + d*x]^2/(a + b*Tan[c + d*x])^2,x]
Output:
(-(a^2*Cot[c + d*x]^2) - a*b*Cot[c + d*x]*(1 + 2*Log[Sin[c + d*x]] - 2*Log [a*Cos[c + d*x] + b*Sin[c + d*x]]) + b^2*(1 - 2*Log[Sin[c + d*x]] + 2*Log[ a*Cos[c + d*x] + b*Sin[c + d*x]]))/(a^3*d*(b + a*Cot[c + d*x]))
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3999, 54, 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.
\(\displaystyle \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^2 (a+b \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 3999 |
\(\displaystyle \frac {b \int \frac {\cot ^2(c+d x)}{b^2 (a+b \tan (c+d x))^2}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {b \int \left (\frac {\cot ^2(c+d x)}{a^2 b^2}-\frac {2 \cot (c+d x)}{a^3 b}+\frac {2}{a^3 (a+b \tan (c+d x))}+\frac {1}{a^2 (a+b \tan (c+d x))^2}\right )d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (-\frac {2 \log (b \tan (c+d x))}{a^3}+\frac {2 \log (a+b \tan (c+d x))}{a^3}-\frac {1}{a^2 (a+b \tan (c+d x))}-\frac {\cot (c+d x)}{a^2 b}\right )}{d}\) |
Input:
Int[Csc[c + d*x]^2/(a + b*Tan[c + d*x])^2,x]
Output:
(b*(-(Cot[c + d*x]/(a^2*b)) - (2*Log[b*Tan[c + d*x]])/a^3 + (2*Log[a + b*T an[c + d*x]])/a^3 - 1/(a^2*(a + b*Tan[c + d*x]))))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[b/f Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
Time = 1.54 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}-\frac {2 b \ln \left (\tan \left (d x +c \right )\right )}{a^{3}}-\frac {b}{a^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(67\) |
default | \(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}-\frac {2 b \ln \left (\tan \left (d x +c \right )\right )}{a^{3}}-\frac {b}{a^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(67\) |
risch | \(-\frac {2 i \left (2 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2}-2 b^{2}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (i a +b \right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right ) a^{2} d}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{3} d}\) | \(178\) |
Input:
int(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/a^2/tan(d*x+c)-2/a^3*b*ln(tan(d*x+c))-b/a^2/(a+b*tan(d*x+c))+2/a^3 *b*ln(a+b*tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (72) = 144\).
Time = 0.10 (sec) , antiderivative size = 293, normalized size of antiderivative = 4.07 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {a^{2} b^{2} - {\left (a^{4} + 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4} - {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (a^{2} b^{2} + b^{4} - {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right )}{{\left (a^{5} b + a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} + a^{4} b^{2}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{5} b + a^{3} b^{3}\right )} d} \] Input:
integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
-(a^2*b^2 - (a^4 + 2*a^2*b^2)*cos(d*x + c)^2 - (a^3*b + 2*a*b^3)*cos(d*x + c)*sin(d*x + c) + (a^2*b^2 + b^4 - (a^2*b^2 + b^4)*cos(d*x + c)^2 + (a^3* b + a*b^3)*cos(d*x + c)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - (a^2*b^2 + b^4 - (a^2*b^2 + b^4)*cos (d*x + c)^2 + (a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4))/((a^5*b + a^3*b^3)*d*cos(d*x + c)^2 - (a^6 + a^4*b^2)*d*cos (d*x + c)*sin(d*x + c) - (a^5*b + a^3*b^3)*d)
\[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(csc(d*x+c)**2/(a+b*tan(d*x+c))**2,x)
Output:
Integral(csc(c + d*x)**2/(a + b*tan(c + d*x))**2, x)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, b \tan \left (d x + c\right ) + a}{a^{2} b \tan \left (d x + c\right )^{2} + a^{3} \tan \left (d x + c\right )} - \frac {2 \, b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{3}} + \frac {2 \, b \log \left (\tan \left (d x + c\right )\right )}{a^{3}}}{d} \] Input:
integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-((2*b*tan(d*x + c) + a)/(a^2*b*tan(d*x + c)^2 + a^3*tan(d*x + c)) - 2*b*l og(b*tan(d*x + c) + a)/a^3 + 2*b*log(tan(d*x + c))/a^3)/d
Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \, b \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{3} d} - \frac {2 \, b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3} d} - \frac {2 \, b \tan \left (d x + c\right ) + a}{{\left (b \tan \left (d x + c\right )^{2} + a \tan \left (d x + c\right )\right )} a^{2} d} \] Input:
integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
2*b*log(abs(b*tan(d*x + c) + a))/(a^3*d) - 2*b*log(abs(tan(d*x + c)))/(a^3 *d) - (2*b*tan(d*x + c) + a)/((b*tan(d*x + c)^2 + a*tan(d*x + c))*a^2*d)
Time = 0.94 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2\,b\,\ln \left (\frac {a+b\,\mathrm {tan}\left (c+d\,x\right )}{\mathrm {tan}\left (c+d\,x\right )}\right )}{a^3\,d}-\frac {2\,b}{a^2\,d\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {1}{a\,d\,\mathrm {tan}\left (c+d\,x\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \] Input:
int(1/(sin(c + d*x)^2*(a + b*tan(c + d*x))^2),x)
Output:
(2*b*log((a + b*tan(c + d*x))/tan(c + d*x)))/(a^3*d) - (2*b)/(a^2*d*(a + b *tan(c + d*x))) - 1/(a*d*tan(c + d*x)*(a + b*tan(c + d*x)))
Time = 0.22 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.26 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) \sin \left (d x +c \right ) a \,b^{2}-4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a \,b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3}-4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) \sin \left (d x +c \right )^{2} b^{3}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{3}+\sin \left (d x +c \right )^{2} a^{2} b -2 a^{2} b}{2 \sin \left (d x +c \right ) a^{3} b d \left (\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b \right )} \] Input:
int(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x)
Output:
(4*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin( c + d*x)*a*b**2 - 4*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a*b**2 - cos(c + d*x)*sin(c + d*x)*a**3 - 4*cos(c + d*x)*sin(c + d*x)*a*b**2 + 4 *log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**2*b** 3 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**3 + sin(c + d*x)**2*a**2*b - 2*a**2*b)/(2*sin(c + d*x)*a**3*b*d*(cos(c + d*x)*a + sin(c + d*x)*b))