Integrand size = 12, antiderivative size = 121 \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\frac {(b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^2 \pi }+\frac {d (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi }-\frac {d \operatorname {FresnelC}(a+b x)}{2 b^2 \pi }-\frac {(b c-a d)^2 \operatorname {FresnelS}(a+b x)}{2 b^2 d}+\frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d} \]
(-a*d+b*c)*cos(1/2*Pi*(b*x+a)^2)/b^2/Pi+1/2*d*(b*x+a)*cos(1/2*Pi*(b*x+a)^2 )/b^2/Pi-1/2*d*FresnelC(b*x+a)/b^2/Pi-1/2*(-a*d+b*c)^2*FresnelS(b*x+a)/b^2 /d+1/2*(d*x+c)^2*FresnelS(b*x+a)/d
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.50 \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\frac {-d \operatorname {FresnelC}(a+b x)+(2 b c-a d+b d x) \left (\cos \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi (a+b x) \operatorname {FresnelS}(a+b x)\right )}{2 b^2 \pi } \]
(-(d*FresnelC[a + b*x]) + (2*b*c - a*d + b*d*x)*(Cos[(Pi*(a + b*x)^2)/2] + Pi*(a + b*x)*FresnelS[a + b*x]))/(2*b^2*Pi)
Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6982, 3914, 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.
\(\displaystyle \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx\) |
\(\Big \downarrow \) 6982 |
\(\displaystyle \frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d}-\frac {b \int (c+d x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )dx}{2 d}\) |
\(\Big \downarrow \) 3914 |
\(\displaystyle \frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d}-\frac {\int \left (\sin \left (\frac {1}{2} \pi (a+b x)^2\right ) (b c-a d)^2+2 d (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) (b c-a d)+d^2 (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )\right )d(a+b x)}{2 b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d}-\frac {(b c-a d)^2 \operatorname {FresnelS}(a+b x)-\frac {2 d (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }+\frac {d^2 \operatorname {FresnelC}(a+b x)}{\pi }-\frac {d^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }}{2 b^2 d}\) |
((c + d*x)^2*FresnelS[a + b*x])/(2*d) - ((-2*d*(b*c - a*d)*Cos[(Pi*(a + b* x)^2)/2])/Pi - (d^2*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/Pi + (d^2*FresnelC[ a + b*x])/Pi + (b*c - a*d)^2*FresnelS[a + b*x])/(2*b^2*d)
3.1.21.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f _.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat or[n], 1]}, Simp[k/f^(m + 1) Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x ^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x ]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]
Int[FresnelS[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> S imp[(c + d*x)^(m + 1)*(FresnelS[a + b*x]/(d*(m + 1))), x] - Simp[b/(d*(m + 1)) Int[(c + d*x)^(m + 1)*Sin[(Pi/2)*(a + b*x)^2], x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0]
Time = 0.54 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {\operatorname {FresnelS}\left (b x +a \right ) \left (d a \left (b x +a \right )-c b \left (b x +a \right )-\frac {d \left (b x +a \right )^{2}}{2}\right )}{b}+\frac {\frac {d \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {d \,\operatorname {FresnelC}\left (b x +a \right )}{\pi }-\frac {\left (2 a d -2 b c \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }}{2 b}}{b}\) | \(109\) |
default | \(\frac {-\frac {\operatorname {FresnelS}\left (b x +a \right ) \left (d a \left (b x +a \right )-c b \left (b x +a \right )-\frac {d \left (b x +a \right )^{2}}{2}\right )}{b}+\frac {\frac {d \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {d \,\operatorname {FresnelC}\left (b x +a \right )}{\pi }-\frac {\left (2 a d -2 b c \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }}{2 b}}{b}\) | \(109\) |
parts | \(\frac {\operatorname {FresnelS}\left (b x +a \right ) d \,x^{2}}{2}+\operatorname {FresnelS}\left (b x +a \right ) c x -\frac {b \left (-\frac {d x \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {d a \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}+\frac {d \,\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}-\frac {2 c \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {2 c \sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{2}\) | \(254\) |
1/b*(-FresnelS(b*x+a)/b*(d*a*(b*x+a)-c*b*(b*x+a)-1/2*d*(b*x+a)^2)+1/2/b*(d /Pi*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)-d/Pi*FresnelC(b*x+a)-(2*a*d-2*b*c)/Pi*co s(1/2*Pi*(b*x+a)^2)))
Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.09 \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\frac {\pi {\left (2 \, a b c - a^{2} d\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - \sqrt {b^{2}} d \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (b^{2} d x + 2 \, b^{2} c - a b d\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + {\left (\pi b^{3} d x^{2} + 2 \, \pi b^{3} c x\right )} \operatorname {S}\left (b x + a\right )}{2 \, \pi b^{3}} \]
1/2*(pi*(2*a*b*c - a^2*d)*sqrt(b^2)*fresnel_sin(sqrt(b^2)*(b*x + a)/b) - s qrt(b^2)*d*fresnel_cos(sqrt(b^2)*(b*x + a)/b) + (b^2*d*x + 2*b^2*c - a*b*d )*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) + (pi*b^3*d*x^2 + 2*pi*b^3*c *x)*fresnel_sin(b*x + a))/(pi*b^3)
\[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int \left (c + d x\right ) S\left (a + b x\right )\, dx \]
\[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {S}\left (b x + a\right ) \,d x } \]
\[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {S}\left (b x + a\right ) \,d x } \]
Timed out. \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int \mathrm {FresnelS}\left (a+b\,x\right )\,\left (c+d\,x\right ) \,d x \]