Integrand size = 19, antiderivative size = 74 \[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {1}{2} b \left (6 a^2+b^2\right ) x-\frac {a^3 \text {arctanh}(\cos (e+f x))}{f}-\frac {5 a b^2 \cos (e+f x)}{2 f}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f} \] Output:
1/2*b*(6*a^2+b^2)*x-a^3*arctanh(cos(f*x+e))/f-5/2*a*b^2*cos(f*x+e)/f-1/2*b ^2*cos(f*x+e)*(a+b*sin(f*x+e))/f
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=3 a^2 b x+\frac {b^3 x}{2}-\frac {a^3 \text {arctanh}(\cos (e+f x))}{f}-\frac {3 a b^2 \cos (e+f x)}{f}-\frac {b^3 \cos (e+f x) \sin (e+f x)}{2 f} \] Input:
Integrate[Csc[e + f*x]*(a + b*Sin[e + f*x])^3,x]
Output:
3*a^2*b*x + (b^3*x)/2 - (a^3*ArcTanh[Cos[e + f*x]])/f - (3*a*b^2*Cos[e + f *x])/f - (b^3*Cos[e + f*x]*Sin[e + f*x])/(2*f)
Time = 0.53 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3272, 3042, 3502, 3042, 3214, 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.
| \(\displaystyle \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^3}{\sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle \frac {1}{2} \int \csc (e+f x) \left (2 a^3+5 b^2 \sin ^2(e+f x) a+b \left (6 a^2+b^2\right ) \sin (e+f x)\right )dx-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {2 a^3+5 b^2 \sin (e+f x)^2 a+b \left (6 a^2+b^2\right ) \sin (e+f x)}{\sin (e+f x)}dx-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \csc (e+f x) \left (2 a^3+b \left (6 a^2+b^2\right ) \sin (e+f x)\right )dx-\frac {5 a b^2 \cos (e+f x)}{f}\right )-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {2 a^3+b \left (6 a^2+b^2\right ) \sin (e+f x)}{\sin (e+f x)}dx-\frac {5 a b^2 \cos (e+f x)}{f}\right )-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (2 a^3 \int \csc (e+f x)dx+b x \left (6 a^2+b^2\right )-\frac {5 a b^2 \cos (e+f x)}{f}\right )-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (2 a^3 \int \csc (e+f x)dx+b x \left (6 a^2+b^2\right )-\frac {5 a b^2 \cos (e+f x)}{f}\right )-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (-\frac {2 a^3 \text {arctanh}(\cos (e+f x))}{f}+b x \left (6 a^2+b^2\right )-\frac {5 a b^2 \cos (e+f x)}{f}\right )-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f}\) |
Input:
Int[Csc[e + f*x]*(a + b*Sin[e + f*x])^3,x]
Output:
(b*(6*a^2 + b^2)*x - (2*a^3*ArcTanh[Cos[e + f*x]])/f - (5*a*b^2*Cos[e + f* x])/f)/2 - (b^2*Cos[e + f*x]*(a + b*Sin[e + f*x]))/(2*f)
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.46 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93
| method | result | size |
| parallelrisch | \(\frac {12 a^{2} b f x +2 b^{3} f x -12 a \,b^{2}+4 a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\sin \left (2 f x +2 e \right ) b^{3}-12 \cos \left (f x +e \right ) a \,b^{2}}{4 f}\) | \(69\) |
| derivativedivides | \(\frac {a^{3} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )+3 a^{2} b \left (f x +e \right )-3 \cos \left (f x +e \right ) a \,b^{2}+b^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\) | \(75\) |
| default | \(\frac {a^{3} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )+3 a^{2} b \left (f x +e \right )-3 \cos \left (f x +e \right ) a \,b^{2}+b^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\) | \(75\) |
| risch | \(3 a^{2} b x +\frac {b^{3} x}{2}-\frac {3 a \,b^{2} {\mathrm e}^{i \left (f x +e \right )}}{2 f}-\frac {3 a \,b^{2} {\mathrm e}^{-i \left (f x +e \right )}}{2 f}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{f}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{f}-\frac {b^{3} \sin \left (2 f x +2 e \right )}{4 f}\) | \(107\) |
| norman | \(\frac {\left (3 a^{2} b +\frac {1}{2} b^{3}\right ) x +\frac {b^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}+\left (3 a^{2} b +\frac {1}{2} b^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (9 a^{2} b +\frac {3}{2} b^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (9 a^{2} b +\frac {3}{2} b^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\frac {6 a \,b^{2}}{f}-\frac {b^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {6 a \,b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{f}-\frac {12 a \,b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}\) | \(209\) |
Input:
int(csc(f*x+e)*(a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
1/4*(12*a^2*b*f*x+2*b^3*f*x-12*a*b^2+4*a^3*ln(tan(1/2*f*x+1/2*e))-sin(2*f* x+2*e)*b^3-12*cos(f*x+e)*a*b^2)/f
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=-\frac {b^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 6 \, a b^{2} \cos \left (f x + e\right ) + a^{3} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - a^{3} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - {\left (6 \, a^{2} b + b^{3}\right )} f x}{2 \, f} \] Input:
integrate(csc(f*x+e)*(a+b*sin(f*x+e))^3,x, algorithm="fricas")
Output:
-1/2*(b^3*cos(f*x + e)*sin(f*x + e) + 6*a*b^2*cos(f*x + e) + a^3*log(1/2*c os(f*x + e) + 1/2) - a^3*log(-1/2*cos(f*x + e) + 1/2) - (6*a^2*b + b^3)*f* x)/f
\[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=\int \left (a + b \sin {\left (e + f x \right )}\right )^{3} \csc {\left (e + f x \right )}\, dx \] Input:
integrate(csc(f*x+e)*(a+b*sin(f*x+e))**3,x)
Output:
Integral((a + b*sin(e + f*x))**3*csc(e + f*x), x)
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {12 \, {\left (f x + e\right )} a^{2} b + {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b^{3} - 12 \, a b^{2} \cos \left (f x + e\right ) - 4 \, a^{3} \log \left (\cot \left (f x + e\right ) + \csc \left (f x + e\right )\right )}{4 \, f} \] Input:
integrate(csc(f*x+e)*(a+b*sin(f*x+e))^3,x, algorithm="maxima")
Output:
1/4*(12*(f*x + e)*a^2*b + (2*f*x + 2*e - sin(2*f*x + 2*e))*b^3 - 12*a*b^2* cos(f*x + e) - 4*a^3*log(cot(f*x + e) + csc(f*x + e)))/f
Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.46 \[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {2 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) + {\left (6 \, a^{2} b + b^{3}\right )} {\left (f x + e\right )} + \frac {2 \, {\left (b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2}}}{2 \, f} \] Input:
integrate(csc(f*x+e)*(a+b*sin(f*x+e))^3,x, algorithm="giac")
Output:
1/2*(2*a^3*log(abs(tan(1/2*f*x + 1/2*e))) + (6*a^2*b + b^3)*(f*x + e) + 2* (b^3*tan(1/2*f*x + 1/2*e)^3 - 6*a*b^2*tan(1/2*f*x + 1/2*e)^2 - b^3*tan(1/2 *f*x + 1/2*e) - 6*a*b^2)/(tan(1/2*f*x + 1/2*e)^2 + 1)^2)/f
Time = 17.19 (sec) , antiderivative size = 259, normalized size of antiderivative = 3.50 \[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {a^3\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f}-\frac {b^3\,\mathrm {atan}\left (\frac {2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^3+6\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^2\,b+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,b^3}{-2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^3+6\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^2\,b+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,b^3}\right )}{f}-\frac {b^3\,\sin \left (2\,e+2\,f\,x\right )}{4\,f}-\frac {6\,a^2\,b\,\mathrm {atan}\left (\frac {2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^3+6\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^2\,b+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,b^3}{-2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^3+6\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^2\,b+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,b^3}\right )}{f}-\frac {3\,a\,b^2\,\cos \left (e+f\,x\right )}{f} \] Input:
int((a + b*sin(e + f*x))^3/sin(e + f*x),x)
Output:
(a^3*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/f - (b^3*atan((b^3*cos(e/ 2 + (f*x)/2) + 2*a^3*sin(e/2 + (f*x)/2) + 6*a^2*b*cos(e/2 + (f*x)/2))/(b^3 *sin(e/2 + (f*x)/2) - 2*a^3*cos(e/2 + (f*x)/2) + 6*a^2*b*sin(e/2 + (f*x)/2 ))))/f - (b^3*sin(2*e + 2*f*x))/(4*f) - (6*a^2*b*atan((b^3*cos(e/2 + (f*x) /2) + 2*a^3*sin(e/2 + (f*x)/2) + 6*a^2*b*cos(e/2 + (f*x)/2))/(b^3*sin(e/2 + (f*x)/2) - 2*a^3*cos(e/2 + (f*x)/2) + 6*a^2*b*sin(e/2 + (f*x)/2))))/f - (3*a*b^2*cos(e + f*x))/f
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int \csc (e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {-\cos \left (f x +e \right ) \sin \left (f x +e \right ) b^{3}-6 \cos \left (f x +e \right ) a \,b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{3}+6 a^{2} b e +6 a^{2} b f x +6 a \,b^{2}+b^{3} e +b^{3} f x}{2 f} \] Input:
int(csc(f*x+e)*(a+b*sin(f*x+e))^3,x)
Output:
( - cos(e + f*x)*sin(e + f*x)*b**3 - 6*cos(e + f*x)*a*b**2 + 2*log(tan((e + f*x)/2))*a**3 + 6*a**2*b*e + 6*a**2*b*f*x + 6*a*b**2 + b**3*e + b**3*f*x )/(2*f)