Integrand size = 20, antiderivative size = 127 \[ \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx=-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}+\frac {i d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3} \]
-2*(d*x+c)^2*arctanh(exp(2*I*(b*x+a)))/b+I*d*(d*x+c)*polylog(2,-exp(2*I*(b *x+a)))/b^2-I*d*(d*x+c)*polylog(2,exp(2*I*(b*x+a)))/b^2-1/2*d^2*polylog(3, -exp(2*I*(b*x+a)))/b^3+1/2*d^2*polylog(3,exp(2*I*(b*x+a)))/b^3
Time = 0.74 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.68 \[ \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx=\frac {-4 b^2 c^2 \text {arctanh}\left (e^{2 i (a+b x)}\right )+4 b^2 c d x \log \left (1-e^{2 i (a+b x)}\right )+2 b^2 d^2 x^2 \log \left (1-e^{2 i (a+b x)}\right )-4 b^2 c d x \log \left (1+e^{2 i (a+b x)}\right )-2 b^2 d^2 x^2 \log \left (1+e^{2 i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )-2 i b d (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )-d^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )+d^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3} \]
(-4*b^2*c^2*ArcTanh[E^((2*I)*(a + b*x))] + 4*b^2*c*d*x*Log[1 - E^((2*I)*(a + b*x))] + 2*b^2*d^2*x^2*Log[1 - E^((2*I)*(a + b*x))] - 4*b^2*c*d*x*Log[1 + E^((2*I)*(a + b*x))] - 2*b^2*d^2*x^2*Log[1 + E^((2*I)*(a + b*x))] + (2* I)*b*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))] - (2*I)*b*d*(c + d*x)*Po lyLog[2, E^((2*I)*(a + b*x))] - d^2*PolyLog[3, -E^((2*I)*(a + b*x))] + d^2 *PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3)
Time = 0.50 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4919, 3042, 4671, 3011, 2720, 7143}
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.
| \(\displaystyle \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx\) |
| \(\Big \downarrow \) 4919 |
| \(\displaystyle 2 \int (c+d x)^2 \csc (2 a+2 b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int (c+d x)^2 \csc (2 a+2 b x)dx\) |
| \(\Big \downarrow \) 4671 |
| \(\displaystyle 2 \left (-\frac {d \int (c+d x) \log \left (1-e^{2 i (a+b x)}\right )dx}{b}+\frac {d \int (c+d x) \log \left (1+e^{2 i (a+b x)}\right )dx}{b}-\frac {(c+d x)^2 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle 2 \left (\frac {d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \int \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \int \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {(c+d x)^2 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle 2 \left (\frac {d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b}-\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {(c+d x)^2 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle 2 \left (-\frac {(c+d x)^2 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}+\frac {d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}-\frac {d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b}-\frac {d \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}\right )\) |
2*(-(((c + d*x)^2*ArcTanh[E^((2*I)*(a + b*x))])/b) + (d*(((I/2)*(c + d*x)* PolyLog[2, -E^((2*I)*(a + b*x))])/b - (d*PolyLog[3, -E^((2*I)*(a + b*x))]) /(4*b^2)))/b - (d*(((I/2)*(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))])/b - ( d*PolyLog[3, E^((2*I)*(a + b*x))])/(4*b^2)))/b)
3.3.30.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x )^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG tQ[m, 0]
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Simp[2^n Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n , x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 468 vs. \(2 (111 ) = 222\).
Time = 1.21 (sec) , antiderivative size = 469, normalized size of antiderivative = 3.69
| method | result | size |
| risch | \(-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}-\frac {2 i d c \operatorname {polylog}\left (2, {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x^{2}}{b}+\frac {2 d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {d^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x^{2}}{b}-\frac {d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{2 b^{3}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) x^{2}}{b}+\frac {2 d^{2} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x}{b}-\frac {2 c d \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x}{b}+\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) x}{b}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{2}}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{b}-\frac {c^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b}+\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{2}}+\frac {i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right ) x}{b^{2}}+\frac {i c d \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{b^{2}}-\frac {2 i d c \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) a^{2}}{b^{3}}+\frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{3}}\) | \(469\) |
-2*I/b^2*d^2*polylog(2,exp(I*(b*x+a)))*x-2*I/b^2*d^2*polylog(2,-exp(I*(b*x +a)))*x-2*I/b^2*d*c*polylog(2,exp(I*(b*x+a)))+1/b*d^2*ln(exp(I*(b*x+a))+1) *x^2+2*d^2*polylog(3,-exp(I*(b*x+a)))/b^3-1/b*d^2*ln(exp(2*I*(b*x+a))+1)*x ^2-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3+1/b*d^2*ln(1-exp(I*(b*x+a)))*x ^2+2*d^2*polylog(3,exp(I*(b*x+a)))/b^3+2/b*d*c*ln(exp(I*(b*x+a))+1)*x-2/b* c*d*ln(exp(2*I*(b*x+a))+1)*x+2/b*d*c*ln(1-exp(I*(b*x+a)))*x-2/b^2*c*d*a*ln (exp(I*(b*x+a))-1)+1/b*c^2*ln(exp(I*(b*x+a))+1)-1/b*c^2*ln(exp(2*I*(b*x+a) )+1)+1/b*c^2*ln(exp(I*(b*x+a))-1)+2/b^2*d*c*ln(1-exp(I*(b*x+a)))*a+I/b^2*d ^2*polylog(2,-exp(2*I*(b*x+a)))*x+I/b^2*c*d*polylog(2,-exp(2*I*(b*x+a)))-2 *I/b^2*d*c*polylog(2,-exp(I*(b*x+a)))-1/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^2+1 /b^3*d^2*a^2*ln(exp(I*(b*x+a))-1)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1098 vs. \(2 (107) = 214\).
Time = 0.32 (sec) , antiderivative size = 1098, normalized size of antiderivative = 8.65 \[ \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx=\text {Too large to display} \]
1/2*(2*d^2*polylog(3, cos(b*x + a) + I*sin(b*x + a)) + 2*d^2*polylog(3, co s(b*x + a) - I*sin(b*x + a)) - 2*d^2*polylog(3, I*cos(b*x + a) + sin(b*x + a)) - 2*d^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 2*d^2*polylog(3, -I*cos(b*x + a) + sin(b*x + a)) - 2*d^2*polylog(3, -I*cos(b*x + a) - sin(b *x + a)) + 2*d^2*polylog(3, -cos(b*x + a) + I*sin(b*x + a)) + 2*d^2*polylo g(3, -cos(b*x + a) - I*sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d)*dilog(cos(b *x + a) + I*sin(b*x + a)) - 2*(-I*b*d^2*x - I*b*c*d)*dilog(cos(b*x + a) - I*sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) - 2*(-I*b*d^2*x - I*b*c*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 2*( -I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 2*(-I*b*d^2*x - I*b*c* d)*dilog(-cos(b*x + a) + I*sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d)*dilog(- cos(b*x + a) - I*sin(b*x + a)) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log (cos(b*x + a) + I*sin(b*x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log( cos(b*x + a) + I*sin(b*x + a) + I) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2) *log(cos(b*x + a) - I*sin(b*x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)* log(cos(b*x + a) - I*sin(b*x + a) + I) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a* b*c*d - a^2*d^2)*log(I*cos(b*x + a) + sin(b*x + a) + 1) - (b^2*d^2*x^2 + 2 *b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) ...
\[ \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx=\int \left (c + d x\right )^{2} \csc {\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 599 vs. \(2 (107) = 214\).
Time = 0.42 (sec) , antiderivative size = 599, normalized size of antiderivative = 4.72 \[ \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx=-\frac {c^{2} {\left (\log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )\right )} - \frac {2 \, a c d {\left (\log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )\right )}}{b} + \frac {a^{2} d^{2} {\left (\log \left (\sin \left (b x + a\right )^{2} - 1\right ) - \log \left (\sin \left (b x + a\right )^{2}\right )\right )}}{b^{2}} + \frac {d^{2} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 4 \, d^{2} {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 4 \, d^{2} {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) - 2 \, {\left (-i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (-i \, b c d + i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 2 \, {\left (i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (i \, b c d - i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 2 \, {\left (-i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (-i \, b c d + i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 2 \, {\left (i \, b c d + i \, {\left (b x + a\right )} d^{2} - i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 4 \, {\left (-i \, b c d - i \, {\left (b x + a\right )} d^{2} + i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 4 \, {\left (-i \, b c d - i \, {\left (b x + a\right )} d^{2} + i \, a d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right )}{b^{2}}}{2 \, b} \]
-1/2*(c^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2)) - 2*a*c*d*(log(s in(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b + a^2*d^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^2 + (d^2*polylog(3, -e^(2*I*b*x + 2*I*a)) - 4*d^2*polylog(3, -e^(I*b*x + I*a)) - 4*d^2*polylog(3, e^(I*b*x + I*a)) - 2 *(-I*(b*x + a)^2*d^2 + 2*(-I*b*c*d + I*a*d^2)*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 2*(I*(b*x + a)^2*d^2 + 2*(I*b*c*d - I*a*d ^2)*(b*x + a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 2*(-I*(b*x + a)^2 *d^2 + 2*(-I*b*c*d + I*a*d^2)*(b*x + a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 2*(I*b*c*d + I*(b*x + a)*d^2 - I*a*d^2)*dilog(-e^(2*I*b*x + 2*I* a)) - 4*(-I*b*c*d - I*(b*x + a)*d^2 + I*a*d^2)*dilog(-e^(I*b*x + I*a)) - 4 *(-I*b*c*d - I*(b*x + a)*d^2 + I*a*d^2)*dilog(e^(I*b*x + I*a)) + ((b*x + a )^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)* (b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - ((b *x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1))/b^2)/b
\[ \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx=\int { {\left (d x + c\right )}^{2} \csc \left (b x + a\right ) \sec \left (b x + a\right ) \,d x } \]
Timed out. \[ \int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx=\int \frac {{\left (c+d\,x\right )}^2}{\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )} \,d x \]