Integrand size = 29, antiderivative size = 97 \[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \csc (c+d x)}{a d}+\frac {\csc ^2(c+d x)}{2 a d}-\frac {\csc ^3(c+d x)}{3 a d}+\frac {2 \log (\sin (c+d x))}{a d}+\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d} \] Output:
2*csc(d*x+c)/a/d+1/2*csc(d*x+c)^2/a/d-1/3*csc(d*x+c)^3/a/d+2*ln(sin(d*x+c) )/a/d+sin(d*x+c)/a/d-1/2*sin(d*x+c)^2/a/d
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \csc (c+d x)+3 \csc ^2(c+d x)-2 \csc ^3(c+d x)+12 \log (\sin (c+d x))+6 \sin (c+d x)-3 \sin ^2(c+d x)}{6 a d} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(12*Csc[c + d*x] + 3*Csc[c + d*x]^2 - 2*Csc[c + d*x]^3 + 12*Log[Sin[c + d* x]] + 6*Sin[c + d*x] - 3*Sin[c + d*x]^2)/(6*a*d)
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x)^4 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^4(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^4(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}{a^4}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (a \csc ^4(c+d x)-a \csc ^3(c+d x)-2 a \csc ^2(c+d x)+2 a \csc (c+d x)+a-a \sin (c+d x)\right )d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} a^2 \sin ^2(c+d x)+a^2 \sin (c+d x)-\frac {1}{3} a^2 \csc ^3(c+d x)+\frac {1}{2} a^2 \csc ^2(c+d x)+2 a^2 \csc (c+d x)+2 a^2 \log (a \sin (c+d x))}{a^3 d}\) |
Input:
Int[(Cos[c + d*x]^3*Cot[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(2*a^2*Csc[c + d*x] + (a^2*Csc[c + d*x]^2)/2 - (a^2*Csc[c + d*x]^3)/3 + 2* a^2*Log[a*Sin[c + d*x]] + a^2*Sin[c + d*x] - (a^2*Sin[c + d*x]^2)/2)/(a^3* d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 1.74 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )^{2}}{2}+\sin \left (d x +c \right )+2 \ln \left (\sin \left (d x +c \right )\right )+\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {2}{\sin \left (d x +c \right )}}{d a}\) | \(64\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )^{2}}{2}+\sin \left (d x +c \right )+2 \ln \left (\sin \left (d x +c \right )\right )+\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {2}{\sin \left (d x +c \right )}}{d a}\) | \(64\) |
risch | \(-\frac {2 i x}{a}+\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d a}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}-\frac {4 i c}{a d}+\frac {2 i \left (6 \,{\mathrm e}^{5 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 i {\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}-3 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(189\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/2*sin(d*x+c)^2+sin(d*x+c)+2*ln(sin(d*x+c))+1/2/sin(d*x+c)^2-1/3/ sin(d*x+c)^3+2/sin(d*x+c))
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \, \cos \left (d x + c\right )^{4} - 24 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 48 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (2 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 32}{12 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/12*(12*cos(d*x + c)^4 - 24*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c))*s in(d*x + c) - 48*cos(d*x + c)^2 - 3*(2*cos(d*x + c)^4 - 3*cos(d*x + c)^2 - 1)*sin(d*x + c) + 32)/((a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**4/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {3 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )}}{a} - \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {12 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 2}{a \sin \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/6*(3*(sin(d*x + c)^2 - 2*sin(d*x + c))/a - 12*log(sin(d*x + c))/a - (12 *sin(d*x + c)^2 + 3*sin(d*x + c) - 2)/(a*sin(d*x + c)^3))/d
Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \sin \left (d x + c\right )^{2} - \frac {12 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 2}{\sin \left (d x + c\right )^{3}} - 12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 6 \, \sin \left (d x + c\right )}{6 \, a d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/6*(3*sin(d*x + c)^2 - (12*sin(d*x + c)^2 + 3*sin(d*x + c) - 2)/sin(d*x + c)^3 - 12*log(abs(sin(d*x + c))) - 6*sin(d*x + c))/(a*d)
Time = 31.48 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.28 \[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}+\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}+\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {89\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{3}}{d\,\left (8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \] Input:
int((cos(c + d*x)^3*cot(c + d*x)^4)/(a + a*sin(c + d*x)),x)
Output:
tan(c/2 + (d*x)/2)^2/(8*a*d) - tan(c/2 + (d*x)/2)^3/(24*a*d) + (2*log(tan( c/2 + (d*x)/2)))/(a*d) + (7*tan(c/2 + (d*x)/2))/(8*a*d) + (tan(c/2 + (d*x) /2) + (19*tan(c/2 + (d*x)/2)^2)/3 + 2*tan(c/2 + (d*x)/2)^3 + (89*tan(c/2 + (d*x)/2)^4)/3 - 15*tan(c/2 + (d*x)/2)^5 + 23*tan(c/2 + (d*x)/2)^6 - 1/3)/ (d*(8*a*tan(c/2 + (d*x)/2)^3 + 16*a*tan(c/2 + (d*x)/2)^5 + 8*a*tan(c/2 + ( d*x)/2)^7)) - (2*log(tan(c/2 + (d*x)/2)^2 + 1))/(a*d)
\[ \int \frac {\cos ^3(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{3} \cot \left (d x +c \right )^{4}}{\sin \left (d x +c \right ) a +a}d x \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^4/(a+a*sin(d*x+c)),x)
Output:
int(cos(d*x+c)^3*cot(d*x+c)^4/(a+a*sin(d*x+c)),x)