Integrand size = 29, antiderivative size = 97 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}-\frac {3 a^3 \log (1-\sin (c+d x))}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {2 a^5}{d \left (a^2-a^2 \sin (c+d x)\right )} \] Output:
-a^3*csc(d*x+c)/d-3*a^3*ln(1-sin(d*x+c))/d+3*a^3*ln(sin(d*x+c))/d+1/2*a^5/ d/(a-a*sin(d*x+c))^2+2*a^5/d/(a^2-a^2*sin(d*x+c))
Time = 0.34 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.65 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (-2 \csc (c+d x)-6 \log (1-\sin (c+d x))+6 \log (\sin (c+d x))+\frac {1}{(-1+\sin (c+d x))^2}-\frac {4}{-1+\sin (c+d x)}\right )}{2 d} \] Input:
Integrate[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]
Output:
(a^3*(-2*Csc[c + d*x] - 6*Log[1 - Sin[c + d*x]] + 6*Log[Sin[c + d*x]] + (- 1 + Sin[c + d*x])^(-2) - 4/(-1 + Sin[c + d*x])))/(2*d)
Time = 0.34 (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, 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.
| \(\displaystyle \int \csc ^2(c+d x) \sec ^5(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^3}{\sin (c+d x)^2 \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {\csc ^2(c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^7 \int \frac {\csc ^2(c+d x)}{a^2 (a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^7 \int \left (\frac {\csc ^2(c+d x)}{a^5}+\frac {3 \csc (c+d x)}{a^5}+\frac {3}{a^4 (a-a \sin (c+d x))}+\frac {2}{a^3 (a-a \sin (c+d x))^2}+\frac {1}{a^2 (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^7 \left (-\frac {\csc (c+d x)}{a^4}+\frac {3 \log (a \sin (c+d x))}{a^4}-\frac {3 \log (a-a \sin (c+d x))}{a^4}+\frac {2}{a^3 (a-a \sin (c+d x))}+\frac {1}{2 a^2 (a-a \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]
Output:
(a^7*(-(Csc[c + d*x]/a^4) + (3*Log[a*Sin[c + d*x]])/a^4 - (3*Log[a - a*Sin [c + d*x]])/a^4 + 1/(2*a^2*(a - a*Sin[c + d*x])^2) + 2/(a^3*(a - a*Sin[c + d*x]))))/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_.), 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[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]
Result contains complex when optimal does not.
Time = 0.84 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.41
| method | result | size |
| risch | \(-\frac {2 i a^{3} \left (-9 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+9 i {\mathrm e}^{2 i \left (d x +c \right )}-10 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {6 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(137\) |
| parallelrisch | \(\frac {\left (\frac {9}{2}+6 \left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\cos \left (d x +c \right )-1\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {9 \cos \left (2 d x +2 c \right )}{2}\right ) a^{3}}{d \left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right )}\) | \(146\) |
| derivativedivides | \(\frac {\frac {a^{3}}{4 \cos \left (d x +c \right )^{4}}+3 a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(168\) |
| default | \(\frac {\frac {a^{3}}{4 \cos \left (d x +c \right )^{4}}+3 a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(168\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-2*I*a^3/(exp(2*I*(d*x+c))-1)/(exp(I*(d*x+c))-I)^4/d*(-9*I*exp(4*I*(d*x+c) )+3*exp(5*I*(d*x+c))+9*I*exp(2*I*(d*x+c))-10*exp(3*I*(d*x+c))+3*exp(I*(d*x +c)))-6*a^3/d*ln(exp(I*(d*x+c))-I)+3*a^3/d*ln(exp(2*I*(d*x+c))-1)
Time = 0.09 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.91 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) - 8 \, a^{3} + 6 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (2 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
1/2*(6*a^3*cos(d*x + c)^2 + 9*a^3*sin(d*x + c) - 8*a^3 + 6*(2*a^3*cos(d*x + c)^2 - 2*a^3 - (a^3*cos(d*x + c)^2 - 2*a^3)*sin(d*x + c))*log(1/2*sin(d* x + c)) - 6*(2*a^3*cos(d*x + c)^2 - 2*a^3 - (a^3*cos(d*x + c)^2 - 2*a^3)*s in(d*x + c))*log(-sin(d*x + c) + 1))/(2*d*cos(d*x + c)^2 - (d*cos(d*x + c) ^2 - 2*d)*sin(d*x + c) - 2*d)
Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**5*(a+a*sin(d*x+c))**3,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {6 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - 6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + \frac {6 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3}}{\sin \left (d x + c\right )^{3} - 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )}}{2 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
-1/2*(6*a^3*log(sin(d*x + c) - 1) - 6*a^3*log(sin(d*x + c)) + (6*a^3*sin(d *x + c)^2 - 9*a^3*sin(d*x + c) + 2*a^3)/(sin(d*x + c)^3 - 2*sin(d*x + c)^2 + sin(d*x + c)))/d
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {1}{2} \, a^{3} {\left (\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{d} + \frac {6 \, \sin \left (d x + c\right )^{2} - 9 \, \sin \left (d x + c\right ) + 2}{d {\left (\sin \left (d x + c\right ) - 1\right )}^{2} \sin \left (d x + c\right )}\right )} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-1/2*a^3*(6*log(abs(sin(d*x + c) - 1))/d - 6*log(abs(sin(d*x + c)))/d + (6 *sin(d*x + c)^2 - 9*sin(d*x + c) + 2)/(d*(sin(d*x + c) - 1)^2*sin(d*x + c) ))
Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6\,a^3\,\mathrm {atanh}\left (2\,\sin \left (c+d\,x\right )-1\right )}{d}-\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^2-\frac {9\,a^3\,\sin \left (c+d\,x\right )}{2}+a^3}{d\,\left ({\sin \left (c+d\,x\right )}^3-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\right )} \] Input:
int((a + a*sin(c + d*x))^3/(cos(c + d*x)^5*sin(c + d*x)^2),x)
Output:
(6*a^3*atanh(2*sin(c + d*x) - 1))/d - (a^3 - (9*a^3*sin(c + d*x))/2 + 3*a^ 3*sin(c + d*x)^2)/(d*(sin(c + d*x) - 2*sin(c + d*x)^2 + sin(c + d*x)^3))
Time = 0.19 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.94 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-15 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}+21 \sin \left (d x +c \right )-8\right )}{8 \sin \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x)
Output:
(a**3*( - 48*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 96*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**2 - 48*log(tan((c + d*x)/2) - 1)*sin(c + d*x) + 24*log(tan((c + d*x)/2))*sin(c + d*x)**3 - 48*log(tan((c + d*x)/2))*sin(c + d*x)**2 + 24*log(tan((c + d*x)/2))*sin(c + d*x) - 15*sin(c + d*x)**3 + 6 *sin(c + d*x)**2 + 21*sin(c + d*x) - 8))/(8*sin(c + d*x)*d*(sin(c + d*x)** 2 - 2*sin(c + d*x) + 1))