Integrand size = 14, antiderivative size = 29 \[ \int (c+d x) \csc ^2(a+b x) \, dx=-\frac {(c+d x) \cot (a+b x)}{b}+\frac {d \log (\sin (a+b x))}{b^2} \] Output:
-(d*x+c)*cot(b*x+a)/b+d*ln(sin(b*x+a))/b^2
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79 \[ \int (c+d x) \csc ^2(a+b x) \, dx=-\frac {d x \cot (a)}{b}-\frac {c \cot (a+b x)}{b}+\frac {d \log (\sin (a+b x))}{b^2}+\frac {d x \csc (a) \csc (a+b x) \sin (b x)}{b} \] Input:
Integrate[(c + d*x)*Csc[a + b*x]^2,x]
Output:
-((d*x*Cot[a])/b) - (c*Cot[a + b*x])/b + (d*Log[Sin[a + b*x]])/b^2 + (d*x* Csc[a]*Csc[a + b*x]*Sin[b*x])/b
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4672, 3042, 25, 3956}
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.
\(\displaystyle \int (c+d x) \csc ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d x) \csc (a+b x)^2dx\) |
\(\Big \downarrow \) 4672 |
\(\displaystyle \frac {d \int \cot (a+b x)dx}{b}-\frac {(c+d x) \cot (a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d \int -\tan \left (a+b x+\frac {\pi }{2}\right )dx}{b}-\frac {(c+d x) \cot (a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {d \int \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )dx}{b}-\frac {(c+d x) \cot (a+b x)}{b}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {d \log (-\sin (a+b x))}{b^2}-\frac {(c+d x) \cot (a+b x)}{b}\) |
Input:
Int[(c + d*x)*Csc[a + b*x]^2,x]
Output:
-(((c + d*x)*Cot[a + b*x])/b) + (d*Log[-Sin[a + b*x]])/b^2
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp [(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1) *Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Time = 0.87 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83
method | result | size |
derivativedivides | \(\frac {\frac {d a \cot \left (b x +a \right )}{b}-c \cot \left (b x +a \right )+\frac {d \left (-\left (b x +a \right ) \cot \left (b x +a \right )+\ln \left (\sin \left (b x +a \right )\right )\right )}{b}}{b}\) | \(53\) |
default | \(\frac {\frac {d a \cot \left (b x +a \right )}{b}-c \cot \left (b x +a \right )+\frac {d \left (-\left (b x +a \right ) \cot \left (b x +a \right )+\ln \left (\sin \left (b x +a \right )\right )\right )}{b}}{b}\) | \(53\) |
risch | \(-\frac {2 i d x}{b}-\frac {2 i d a}{b^{2}}-\frac {2 i \left (d x +c \right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {d \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b^{2}}\) | \(59\) |
parallelrisch | \(\frac {-2 \ln \left (\sec \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right ) d +2 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -b \left (\cot \left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (d x +c \right )}{2 b^{2}}\) | \(64\) |
norman | \(\frac {-\frac {c}{2 b}-\frac {d x}{2 b}+\frac {c \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{2 b}+\frac {d x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{2 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}+\frac {d \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b^{2}}-\frac {d \ln \left (1+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}{b^{2}}\) | \(98\) |
Input:
int((d*x+c)*csc(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(1/b*d*a*cot(b*x+a)-c*cot(b*x+a)+1/b*d*(-(b*x+a)*cot(b*x+a)+ln(sin(b*x +a))))
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int (c+d x) \csc ^2(a+b x) \, dx=\frac {d \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - {\left (b d x + b c\right )} \cos \left (b x + a\right )}{b^{2} \sin \left (b x + a\right )} \] Input:
integrate((d*x+c)*csc(b*x+a)^2,x, algorithm="fricas")
Output:
(d*log(1/2*sin(b*x + a))*sin(b*x + a) - (b*d*x + b*c)*cos(b*x + a))/(b^2*s in(b*x + a))
\[ \int (c+d x) \csc ^2(a+b x) \, dx=\int \left (c + d x\right ) \csc ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate((d*x+c)*csc(b*x+a)**2,x)
Output:
Integral((c + d*x)*csc(a + b*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (29) = 58\).
Time = 0.04 (sec) , antiderivative size = 217, normalized size of antiderivative = 7.48 \[ \int (c+d x) \csc ^2(a+b x) \, dx=\frac {\frac {{\left ({\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + {\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - 4 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d}{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} b} - \frac {2 \, c}{\tan \left (b x + a\right )} + \frac {2 \, a d}{b \tan \left (b x + a\right )}}{2 \, b} \] Input:
integrate((d*x+c)*csc(b*x+a)^2,x, algorithm="maxima")
Output:
1/2*(((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*l og(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (cos(2*b*x + 2* a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + s in(b*x + a)^2 - 2*cos(b*x + a) + 1) - 4*(b*x + a)*sin(2*b*x + 2*a))*d/((co s(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*b) - 2*c/t an(b*x + a) + 2*a*d/(b*tan(b*x + a)))/b
Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (29) = 58\).
Time = 0.57 (sec) , antiderivative size = 1027, normalized size of antiderivative = 35.41 \[ \int (c+d x) \csc ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)*csc(b*x+a)^2,x, algorithm="giac")
Output:
1/2*(b*d*x*tan(1/2*b*x)^2*tan(1/2*a)^2 + b*c*tan(1/2*b*x)^2*tan(1/2*a)^2 - b*d*x*tan(1/2*b*x)^2 - 4*b*d*x*tan(1/2*b*x)*tan(1/2*a) + d*log(16*(tan(1/ 2*b*x)^4*tan(1/2*a)^2 + 2*tan(1/2*b*x)^3*tan(1/2*a)^3 + tan(1/2*b*x)^2*tan (1/2*a)^4 - 2*tan(1/2*b*x)^3*tan(1/2*a) - 4*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*b*x)^2 + 2*tan(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2)/(tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*tan(1/2*b*x)^4*tan(1/2*a)^ 2 + 2*tan(1/2*b*x)^2*tan(1/2*a)^4 + tan(1/2*b*x)^4 + 4*tan(1/2*b*x)^2*tan( 1/2*a)^2 + tan(1/2*a)^4 + 2*tan(1/2*b*x)^2 + 2*tan(1/2*a)^2 + 1))*tan(1/2* b*x)^2*tan(1/2*a) - b*d*x*tan(1/2*a)^2 + d*log(16*(tan(1/2*b*x)^4*tan(1/2* a)^2 + 2*tan(1/2*b*x)^3*tan(1/2*a)^3 + tan(1/2*b*x)^2*tan(1/2*a)^4 - 2*tan (1/2*b*x)^3*tan(1/2*a) - 4*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*tan(1/2*b*x)*ta n(1/2*a)^3 + tan(1/2*b*x)^2 + 2*tan(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2)/(t an(1/2*b*x)^4*tan(1/2*a)^4 + 2*tan(1/2*b*x)^4*tan(1/2*a)^2 + 2*tan(1/2*b*x )^2*tan(1/2*a)^4 + tan(1/2*b*x)^4 + 4*tan(1/2*b*x)^2*tan(1/2*a)^2 + tan(1/ 2*a)^4 + 2*tan(1/2*b*x)^2 + 2*tan(1/2*a)^2 + 1))*tan(1/2*b*x)*tan(1/2*a)^2 - b*c*tan(1/2*b*x)^2 - 4*b*c*tan(1/2*b*x)*tan(1/2*a) - b*c*tan(1/2*a)^2 + b*d*x - d*log(16*(tan(1/2*b*x)^4*tan(1/2*a)^2 + 2*tan(1/2*b*x)^3*tan(1/2* a)^3 + tan(1/2*b*x)^2*tan(1/2*a)^4 - 2*tan(1/2*b*x)^3*tan(1/2*a) - 4*tan(1 /2*b*x)^2*tan(1/2*a)^2 - 2*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*b*x)^2 + 2* tan(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2)/(tan(1/2*b*x)^4*tan(1/2*a)^4 + ...
Time = 35.54 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int (c+d x) \csc ^2(a+b x) \, dx=\frac {d\,\ln \left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}-1\right )}{b^2}-\frac {\left (c+d\,x\right )\,2{}\mathrm {i}}{b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}-\frac {d\,x\,2{}\mathrm {i}}{b} \] Input:
int((c + d*x)/sin(a + b*x)^2,x)
Output:
(d*log(exp(a*2i)*exp(b*x*2i) - 1))/b^2 - ((c + d*x)*2i)/(b*(exp(a*2i + b*x *2i) - 1)) - (d*x*2i)/b
Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.59 \[ \int (c+d x) \csc ^2(a+b x) \, dx=\frac {-\cos \left (b x +a \right ) b c -\cos \left (b x +a \right ) b d x -\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) \sin \left (b x +a \right ) d +\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right ) d}{\sin \left (b x +a \right ) b^{2}} \] Input:
int((d*x+c)*csc(b*x+a)^2,x)
Output:
( - cos(a + b*x)*b*c - cos(a + b*x)*b*d*x - log(tan((a + b*x)/2)**2 + 1)*s in(a + b*x)*d + log(tan((a + b*x)/2))*sin(a + b*x)*d)/(sin(a + b*x)*b**2)