\(\int (c+d x) \operatorname {FresnelS}(a+b x) \, dx\) [21]

Optimal result

Integrand size = 12, antiderivative size = 121 \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\frac {(b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^2 \pi }+\frac {d (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi }-\frac {d \operatorname {FresnelC}(a+b x)}{2 b^2 \pi }-\frac {(b c-a d)^2 \operatorname {FresnelS}(a+b x)}{2 b^2 d}+\frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d} \]

[Out]

(-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/P
i-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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6563, 3514, 3432, 3460, 2718, 3466, 3433} \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=-\frac {(b c-a d)^2 \operatorname {FresnelS}(a+b x)}{2 b^2 d}+\frac {(b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac {d \operatorname {FresnelC}(a+b x)}{2 \pi b^2}+\frac {d (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d} \]

[In]

Int[(c + d*x)*FresnelS[a + b*x],x]

[Out]

((b*c - a*d)*Cos[(Pi*(a + b*x)^2)/2])/(b^2*Pi) + (d*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/(2*b^2*Pi) - (d*Fresnel
C[a + b*x])/(2*b^2*Pi) - ((b*c - a*d)^2*FresnelS[a + b*x])/(2*b^2*d) + ((c + d*x)^2*FresnelS[a + b*x])/(2*d)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3466

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(-e^(n - 1))*(e*x)^(m - n + 1)*(Cos[c +
d*x^n]/(d*n)), x] + Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}
, x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3514

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[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]

Rule 6563

Int[FresnelS[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(FresnelS[a +
b*x]/(d*(m + 1))), x] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \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} \\ & = \frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d}-\frac {\text {Subst}\left (\int \left (b^2 c^2 \left (1+\frac {a d (-2 b c+a d)}{b^2 c^2}\right ) \sin \left (\frac {\pi x^2}{2}\right )+2 b c d \left (1-\frac {a d}{b c}\right ) x \sin \left (\frac {\pi x^2}{2}\right )+d^2 x^2 \sin \left (\frac {\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{2 b^2 d} \\ & = \frac {(c+d x)^2 \operatorname {FresnelS}(a+b x)}{2 d}-\frac {d \text {Subst}\left (\int x^2 \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2}-\frac {(b c-a d) \text {Subst}\left (\int x \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^2}-\frac {(b c-a d)^2 \text {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2 d} \\ & = \frac {d (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{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}-\frac {(b c-a d) \text {Subst}\left (\int \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^2}-\frac {d \text {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2 \pi } \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.50 \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\frac {-d \operatorname {FresnelC}(a+b x)+(2 b c-a d+b d x) \left (\cos \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi (a+b x) \operatorname {FresnelS}(a+b x)\right )}{2 b^2 \pi } \]

[In]

Integrate[(c + d*x)*FresnelS[a + b*x],x]

[Out]

(-(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)

Maple [A] (verified)

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	\(\frac {\operatorname {FresnelS}\left (b x +a \right ) d \,x^{2}}{2}+\operatorname {FresnelS}\left (b x +a \right ) c x -\frac {b \left (-\frac {d x \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {d a \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}+\frac {d \,\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}-\frac {2 c \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {2 c \sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{2}\)	\(254\)

[In]

int((d*x+c)*FresnelS(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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*cos(1/2*Pi*(b*x+a)^2)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.09 \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\frac {\pi {\left (2 \, a b c - a^{2} d\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - \sqrt {b^{2}} d \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (b^{2} d x + 2 \, b^{2} c - a b d\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + {\left (\pi b^{3} d x^{2} + 2 \, \pi b^{3} c x\right )} \operatorname {S}\left (b x + a\right )}{2 \, \pi b^{3}} \]

[In]

integrate((d*x+c)*fresnel_sin(b*x+a),x, algorithm="fricas")

[Out]

1/2*(pi*(2*a*b*c - a^2*d)*sqrt(b^2)*fresnel_sin(sqrt(b^2)*(b*x + a)/b) - sqrt(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)

Sympy [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int \left (c + d x\right ) S\left (a + b x\right )\, dx \]

[In]

integrate((d*x+c)*fresnels(b*x+a),x)

[Out]

Integral((c + d*x)*fresnels(a + b*x), x)

Maxima [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {S}\left (b x + a\right ) \,d x } \]

[In]

integrate((d*x+c)*fresnel_sin(b*x+a),x, algorithm="maxima")

[Out]

integrate((d*x + c)*fresnel_sin(b*x + a), x)

Giac [F]

\[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {S}\left (b x + a\right ) \,d x } \]

[In]

integrate((d*x+c)*fresnel_sin(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)*fresnel_sin(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \operatorname {FresnelS}(a+b x) \, dx=\int \mathrm {FresnelS}\left (a+b\,x\right )\,\left (c+d\,x\right ) \,d x \]

[In]

int(FresnelS(a + b*x)*(c + d*x),x)

[Out]

int(FresnelS(a + b*x)*(c + d*x), x)