Integrand size = 26, antiderivative size = 129 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {4 i f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3} \] Output:
I*(f*x+e)^2/a/d+1/3*(f*x+e)^3/a/f+(f*x+e)^2*cot(1/2*c+1/4*Pi+1/2*d*x)/a/d- 4*f*(f*x+e)*ln(1-I*exp(I*(d*x+c)))/a/d^2+4*I*f^2*polylog(2,I*exp(I*(d*x+c) ))/a/d^3
Time = 2.01 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.65 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {12 f (\cos (c)+i \sin (c)) \left (\frac {(e+f x)^2 (\cos (c)-i \sin (c))}{2 f}-\frac {(e+f x) \log (1+i \cos (c+d x)+\sin (c+d x)) (1+i \cos (c)+\sin (c))}{d}+\frac {f \operatorname {PolyLog}(2,-i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i (1+\sin (c)))}{d^2}\right )}{d (\cos (c)+i (1+\sin (c)))}-\frac {6 (e+f x)^2 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{3 a} \] Input:
Integrate[((e + f*x)^2*Sin[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
(x*(3*e^2 + 3*e*f*x + f^2*x^2) + (12*f*(Cos[c] + I*Sin[c])*(((e + f*x)^2*( Cos[c] - I*Sin[c]))/(2*f) - ((e + f*x)*Log[1 + I*Cos[c + d*x] + Sin[c + d* x]]*(1 + I*Cos[c] + Sin[c]))/d + (f*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])))/d^2))/(d*(Cos[c] + I*(1 + Sin[c]))) - (6 *(e + f*x)^2*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Si n[(c + d*x)/2])))/(3*a)
Time = 0.74 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5026, 17, 3042, 3799, 3042, 4672, 3042, 25, 4202, 2620, 2715, 2838}
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.
| \(\displaystyle \int \frac {(e+f x)^2 \sin (c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 5026 |
| \(\displaystyle \frac {\int (e+f x)^2dx}{a}-\int \frac {(e+f x)^2}{\sin (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\int \frac {(e+f x)^2}{\sin (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\int \frac {(e+f x)^2}{\sin (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 3799 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {\int (e+f x)^2 \csc ^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {\int (e+f x)^2 \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )^2dx}{2 a}\) |
| \(\Big \downarrow \) 4672 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {\frac {4 f \int (e+f x) \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{d}-\frac {2 (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {\frac {4 f \int -\left ((e+f x) \tan \left (\frac {c}{2}+\frac {d x}{2}+\frac {3 \pi }{4}\right )\right )dx}{d}-\frac {2 (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {-\frac {4 f \int (e+f x) \tan \left (\frac {1}{4} (2 c+3 \pi )+\frac {d x}{2}\right )dx}{d}-\frac {2 (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {-\frac {2 (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {4 f \left (\frac {i (e+f x)^2}{2 f}-2 i \int \frac {e^{\frac {1}{2} i (2 c+2 d x+3 \pi )} (e+f x)}{1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}}dx\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {-\frac {2 (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {4 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (\frac {i f \int \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )dx}{d}-\frac {i (e+f x) \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}\right )\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {-\frac {2 (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {4 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (\frac {f \int e^{-\frac {1}{2} i (2 c+2 d x+3 \pi )} \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )de^{\frac {1}{2} i (2 c+2 d x+3 \pi )}}{d^2}-\frac {i (e+f x) \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}\right )\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {(e+f x)^3}{3 a f}-\frac {-\frac {2 (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {4 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (-\frac {f \operatorname {PolyLog}\left (2,-e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d^2}-\frac {i (e+f x) \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}\right )\right )}{d}}{2 a}\) |
Input:
Int[((e + f*x)^2*Sin[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
(e + f*x)^3/(3*a*f) - ((-2*(e + f*x)^2*Cot[c/2 + Pi/4 + (d*x)/2])/d - (4*f *(((I/2)*(e + f*x)^2)/f - (2*I)*(((-I)*(e + f*x)*Log[1 + E^((I/2)*(2*c + 3 *Pi + 2*d*x))])/d - (f*PolyLog[2, -E^((I/2)*(2*c + 3*Pi + 2*d*x))])/d^2))) /d)/(2*a)
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Simp[(2*a)^n Int[(c + d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^ 2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
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]
Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_. )*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/b Int[(e + f*x)^m*Sin[c + d*x]^(n - 1), x], x] - Simp[a/b Int[(e + f*x)^m*(Sin[c + d*x]^(n - 1)/(a + b*Sin[c + d*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] & & IGtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (113 ) = 226\).
Time = 0.86 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.60
| method | result | size |
| risch | \(\frac {f^{2} x^{3}}{3 a}+\frac {f e \,x^{2}}{a}+\frac {e^{2} x}{a}+\frac {e^{3}}{3 a f}+\frac {2 x^{2} f^{2}+4 e f x +2 e^{2}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {4 f e \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}-\frac {2 f e \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a \,d^{2}}+\frac {2 i f^{2} x^{2}}{a d}-\frac {4 i f^{2} c \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}+\frac {4 i f e \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}+\frac {4 i f^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {2 f^{2} c \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a \,d^{3}}+\frac {4 i f^{2} c x}{a \,d^{2}}\) | \(335\) |
Input:
int((f*x+e)^2*sin(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/3/a*f^2*x^3+1/a*f*e*x^2+1/a*e^2*x+1/3/a/f*e^3+2*(f^2*x^2+2*e*f*x+e^2)/d/ a/(exp(I*(d*x+c))+I)+4/a/d^2*f*e*ln(exp(I*(d*x+c)))-2/a/d^2*f*e*ln(exp(2*I *(d*x+c))+1)+2*I/a/d*f^2*x^2-4*I/a/d^3*f^2*c*arctan(exp(I*(d*x+c)))+2*I/a/ d^3*f^2*c^2+4*I/a/d^2*f*e*arctan(exp(I*(d*x+c)))-4/a/d^2*f^2*ln(1-I*exp(I* (d*x+c)))*x-4/a/d^3*f^2*ln(1-I*exp(I*(d*x+c)))*c+4*I*f^2*polylog(2,I*exp(I *(d*x+c)))/a/d^3-4/a/d^3*f^2*c*ln(exp(I*(d*x+c)))+2/a/d^3*f^2*c*ln(exp(2*I *(d*x+c))+1)+4*I/a/d^2*f^2*c*x
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (108) = 216\).
Time = 0.11 (sec) , antiderivative size = 583, normalized size of antiderivative = 4.52 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {d^{3} f^{2} x^{3} + 3 \, d^{2} e^{2} + 3 \, {\left (d^{3} e f + d^{2} f^{2}\right )} x^{2} + 3 \, {\left (d^{3} e^{2} + 2 \, d^{2} e f\right )} x + {\left (d^{3} f^{2} x^{3} + 3 \, d^{2} e^{2} + 3 \, {\left (d^{3} e f + d^{2} f^{2}\right )} x^{2} + 3 \, {\left (d^{3} e^{2} + 2 \, d^{2} e f\right )} x\right )} \cos \left (d x + c\right ) - 6 \, {\left (-i \, f^{2} \cos \left (d x + c\right ) - i \, f^{2} \sin \left (d x + c\right ) - i \, f^{2}\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 6 \, {\left (i \, f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (d x + c\right ) + i \, f^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 6 \, {\left (d e f - c f^{2} + {\left (d e f - c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d e f - c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - 6 \, {\left (d f^{2} x + c f^{2} + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (d f^{2} x + c f^{2} + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (d e f - c f^{2} + {\left (d e f - c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d e f - c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + {\left (d^{3} f^{2} x^{3} - 3 \, d^{2} e^{2} + 3 \, {\left (d^{3} e f - d^{2} f^{2}\right )} x^{2} + 3 \, {\left (d^{3} e^{2} - 2 \, d^{2} e f\right )} x\right )} \sin \left (d x + c\right )}{3 \, {\left (a d^{3} \cos \left (d x + c\right ) + a d^{3} \sin \left (d x + c\right ) + a d^{3}\right )}} \] Input:
integrate((f*x+e)^2*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/3*(d^3*f^2*x^3 + 3*d^2*e^2 + 3*(d^3*e*f + d^2*f^2)*x^2 + 3*(d^3*e^2 + 2* d^2*e*f)*x + (d^3*f^2*x^3 + 3*d^2*e^2 + 3*(d^3*e*f + d^2*f^2)*x^2 + 3*(d^3 *e^2 + 2*d^2*e*f)*x)*cos(d*x + c) - 6*(-I*f^2*cos(d*x + c) - I*f^2*sin(d*x + c) - I*f^2)*dilog(I*cos(d*x + c) - sin(d*x + c)) - 6*(I*f^2*cos(d*x + c ) + I*f^2*sin(d*x + c) + I*f^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) - 6* (d*e*f - c*f^2 + (d*e*f - c*f^2)*cos(d*x + c) + (d*e*f - c*f^2)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - 6*(d*f^2*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d*x + c) + (d*f^2*x + c*f^2)*sin(d*x + c))*log(I*cos(d*x + c ) + sin(d*x + c) + 1) - 6*(d*f^2*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d*x + c ) + (d*f^2*x + c*f^2)*sin(d*x + c))*log(-I*cos(d*x + c) + sin(d*x + c) + 1 ) - 6*(d*e*f - c*f^2 + (d*e*f - c*f^2)*cos(d*x + c) + (d*e*f - c*f^2)*sin( d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) + (d^3*f^2*x^3 - 3*d^2*e ^2 + 3*(d^3*e*f - d^2*f^2)*x^2 + 3*(d^3*e^2 - 2*d^2*e*f)*x)*sin(d*x + c))/ (a*d^3*cos(d*x + c) + a*d^3*sin(d*x + c) + a*d^3)
\[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {e^{2} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate((f*x+e)**2*sin(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
(Integral(e**2*sin(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**2*x**2*si n(c + d*x)/(sin(c + d*x) + 1), x) + Integral(2*e*f*x*sin(c + d*x)/(sin(c + d*x) + 1), x))/a
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (108) = 216\).
Time = 0.22 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.12 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x - 6 i \, d^{2} e^{2} - 12 \, {\left (d e f \cos \left (d x + c\right ) + i \, d e f \sin \left (d x + c\right ) + i \, d e f\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) + 12 \, {\left (d f^{2} x \cos \left (d x + c\right ) + i \, d f^{2} x \sin \left (d x + c\right ) + i \, d f^{2} x\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - {\left (i \, d^{3} f^{2} x^{3} - 3 \, {\left (-i \, d^{3} e f + 2 \, d^{2} f^{2}\right )} x^{2} - 3 \, {\left (-i \, d^{3} e^{2} + 4 \, d^{2} e f\right )} x\right )} \cos \left (d x + c\right ) + 12 \, {\left (f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (d x + c\right ) + i \, f^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) - 6 \, {\left (d f^{2} x + d e f - {\left (i \, d f^{2} x + i \, d e f\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + d e f\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + {\left (d^{3} f^{2} x^{3} + 3 \, {\left (d^{3} e f + 2 i \, d^{2} f^{2}\right )} x^{2} + 3 \, {\left (d^{3} e^{2} + 4 i \, d^{2} e f\right )} x\right )} \sin \left (d x + c\right )}{-3 i \, a d^{3} \cos \left (d x + c\right ) + 3 \, a d^{3} \sin \left (d x + c\right ) + 3 \, a d^{3}} \] Input:
integrate((f*x+e)^2*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
(d^3*f^2*x^3 + 3*d^3*e*f*x^2 + 3*d^3*e^2*x - 6*I*d^2*e^2 - 12*(d*e*f*cos(d *x + c) + I*d*e*f*sin(d*x + c) + I*d*e*f)*arctan2(sin(d*x + c) + 1, cos(d* x + c)) + 12*(d*f^2*x*cos(d*x + c) + I*d*f^2*x*sin(d*x + c) + I*d*f^2*x)*a rctan2(cos(d*x + c), sin(d*x + c) + 1) - (I*d^3*f^2*x^3 - 3*(-I*d^3*e*f + 2*d^2*f^2)*x^2 - 3*(-I*d^3*e^2 + 4*d^2*e*f)*x)*cos(d*x + c) + 12*(f^2*cos( d*x + c) + I*f^2*sin(d*x + c) + I*f^2)*dilog(I*e^(I*d*x + I*c)) - 6*(d*f^2 *x + d*e*f - (I*d*f^2*x + I*d*e*f)*cos(d*x + c) + (d*f^2*x + d*e*f)*sin(d* x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + (d^3*f ^2*x^3 + 3*(d^3*e*f + 2*I*d^2*f^2)*x^2 + 3*(d^3*e^2 + 4*I*d^2*e*f)*x)*sin( d*x + c))/(-3*I*a*d^3*cos(d*x + c) + 3*a*d^3*sin(d*x + c) + 3*a*d^3)
\[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sin \left (d x + c\right )}{a \sin \left (d x + c\right ) + a} \,d x } \] Input:
integrate((f*x+e)^2*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
integrate((f*x + e)^2*sin(d*x + c)/(a*sin(d*x + c) + a), x)
Timed out. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\sin \left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:
int((sin(c + d*x)*(e + f*x)^2)/(a + a*sin(c + d*x)),x)
Output:
int((sin(c + d*x)*(e + f*x)^2)/(a + a*sin(c + d*x)), x)
\[ \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) x}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}d x \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d \,f^{2}+12 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) x}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}d x \right ) d \,f^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) e f +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) e f -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) e f -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) e f +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} e^{2} x +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} e f \,x^{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} f^{2} x^{3}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d \,e^{2}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d e f x -6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d \,f^{2} x^{2}+3 d^{2} e^{2} x +3 d^{2} e f \,x^{2}+d^{2} f^{2} x^{3}+6 d e f x}{3 a \,d^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \] Input:
int((f*x+e)^2*sin(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
(12*int((tan((c + d*x)/2)*x)/(tan((c + d*x)/2) + 1),x)*tan((c + d*x)/2)*d* f**2 + 12*int((tan((c + d*x)/2)*x)/(tan((c + d*x)/2) + 1),x)*d*f**2 + 6*lo g(tan((c + d*x)/2)**2 + 1)*tan((c + d*x)/2)*e*f + 6*log(tan((c + d*x)/2)** 2 + 1)*e*f - 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)*e*f - 12*log(ta n((c + d*x)/2) + 1)*e*f + 3*tan((c + d*x)/2)*d**2*e**2*x + 3*tan((c + d*x) /2)*d**2*e*f*x**2 + tan((c + d*x)/2)*d**2*f**2*x**3 - 6*tan((c + d*x)/2)*d *e**2 - 6*tan((c + d*x)/2)*d*e*f*x - 6*tan((c + d*x)/2)*d*f**2*x**2 + 3*d* *2*e**2*x + 3*d**2*e*f*x**2 + d**2*f**2*x**3 + 6*d*e*f*x)/(3*a*d**2*(tan(( c + d*x)/2) + 1))