Integrand size = 27, antiderivative size = 33 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=2 a A x-\frac {a A \text {arctanh}(\cos (c+d x))}{d}-\frac {a A \cos (c+d x)}{d} \] Output:
2*a*A*x-a*A*arctanh(cos(d*x+c))/d-a*A*cos(d*x+c)/d
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=2 a A x-\frac {a A \text {arctanh}(\cos (c+d x))}{d}-\frac {a A \cos (c) \cos (d x)}{d}+\frac {a A \sin (c) \sin (d x)}{d} \] Input:
Integrate[(a + a*Csc[c + d*x])*(A + A*Csc[c + d*x])*Sin[c + d*x],x]
Output:
2*a*A*x - (a*A*ArcTanh[Cos[c + d*x]])/d - (a*A*Cos[c]*Cos[d*x])/d + (a*A*S in[c]*Sin[d*x])/d
Time = 0.37 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2011, 3042, 4275, 24, 3042, 4533, 3042, 4257}
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.
\(\displaystyle \int \sin (c+d x) (a \csc (c+d x)+a) (A \csc (c+d x)+A) \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle \frac {A \int (\csc (c+d x) a+a)^2 \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \int \frac {(\csc (c+d x) a+a)^2}{\csc (c+d x)}dx}{a}\) |
\(\Big \downarrow \) 4275 |
\(\displaystyle \frac {A \left (\int \left (\csc ^2(c+d x) a^2+a^2\right ) \sin (c+d x)dx+2 a^2 \int 1dx\right )}{a}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {A \left (\int \left (\csc ^2(c+d x) a^2+a^2\right ) \sin (c+d x)dx+2 a^2 x\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \left (\int \frac {\csc (c+d x)^2 a^2+a^2}{\csc (c+d x)}dx+2 a^2 x\right )}{a}\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {A \left (a^2 \int \csc (c+d x)dx-\frac {a^2 \cos (c+d x)}{d}+2 a^2 x\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \left (a^2 \int \csc (c+d x)dx-\frac {a^2 \cos (c+d x)}{d}+2 a^2 x\right )}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {A \left (-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {a^2 \cos (c+d x)}{d}+2 a^2 x\right )}{a}\) |
Input:
Int[(a + a*Csc[c + d*x])*(A + A*Csc[c + d*x])*Sin[c + d*x],x]
Output:
(A*(2*a^2*x - (a^2*ArcTanh[Cos[c + d*x]])/d - (a^2*Cos[c + d*x])/d))/a
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(\frac {A a \left (2 d x +1+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\cos \left (d x +c \right )\right )}{d}\) | \(31\) |
derivativedivides | \(\frac {A a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+2 A a \left (d x +c \right )-A \cos \left (d x +c \right ) a}{d}\) | \(44\) |
default | \(\frac {A a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+2 A a \left (d x +c \right )-A \cos \left (d x +c \right ) a}{d}\) | \(44\) |
parts | \(-\frac {a A \cos \left (d x +c \right )}{d}+\frac {A a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {2 A a \left (d x +c \right )}{d}\) | \(49\) |
risch | \(2 a A x -\frac {A a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {A a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\) | \(76\) |
norman | \(\frac {\frac {2 A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+2 a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 a A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {A a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(94\) |
Input:
int((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/d*A*a*(2*d*x+1+ln(tan(1/2*d*x+1/2*c))-cos(d*x+c))
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=\frac {4 \, A a d x - 2 \, A a \cos \left (d x + c\right ) - A a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + A a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, d} \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c),x, algorithm="frica s")
Output:
1/2*(4*A*a*d*x - 2*A*a*cos(d*x + c) - A*a*log(1/2*cos(d*x + c) + 1/2) + A* a*log(-1/2*cos(d*x + c) + 1/2))/d
\[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=A a \left (\int 2 \sin {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c),x)
Output:
A*a*(Integral(2*sin(c + d*x)*csc(c + d*x), x) + Integral(sin(c + d*x)*csc( c + d*x)**2, x) + Integral(sin(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} A a - A a {\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, A a \cos \left (d x + c\right )}{2 \, d} \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c),x, algorithm="maxim a")
Output:
1/2*(4*(d*x + c)*A*a - A*a*(log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) - 2*A*a*cos(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} A a + A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, A a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \] Input:
integrate((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c),x, algorithm="giac" )
Output:
(2*(d*x + c)*A*a + A*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*A*a/(tan(1/2*d*x + 1/2*c)^2 + 1))/d
Time = 15.86 (sec) , antiderivative size = 129, normalized size of antiderivative = 3.91 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=\frac {A\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,A\,a}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {4\,A\,a\,\mathrm {atan}\left (\frac {16\,A^2\,a^2}{8\,A^2\,a^2-16\,A^2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,A^2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,A^2\,a^2-16\,A^2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \] Input:
int(sin(c + d*x)*(A + A/sin(c + d*x))*(a + a/sin(c + d*x)),x)
Output:
(A*a*log(tan(c/2 + (d*x)/2)))/d - (2*A*a)/(d*(tan(c/2 + (d*x)/2)^2 + 1)) + (4*A*a*atan((16*A^2*a^2)/(8*A^2*a^2 - 16*A^2*a^2*tan(c/2 + (d*x)/2)) + (8 *A^2*a^2*tan(c/2 + (d*x)/2))/(8*A^2*a^2 - 16*A^2*a^2*tan(c/2 + (d*x)/2)))) /d
Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int (a+a \csc (c+d x)) (A+A \csc (c+d x)) \sin (c+d x) \, dx=\frac {a^{2} \left (-\cos \left (d x +c \right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 d x \right )}{d} \] Input:
int((a+a*csc(d*x+c))*(A+A*csc(d*x+c))*sin(d*x+c),x)
Output:
(a**2*( - cos(c + d*x) + log(tan((c + d*x)/2)) + 2*d*x))/d