3.130 \(\int (c+d x) C(a+b x) \, dx\)

Optimal. Leaf size=122 \[ -\frac {(b c-a d)^2 C(a+b x)}{2 b^2 d}-\frac {(b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^2}+\frac {d S(a+b x)}{2 \pi b^2}-\frac {d (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac {(c+d x)^2 C(a+b x)}{2 d} \]

[Out]

-1/2*(-a*d+b*c)^2*FresnelC(b*x+a)/b^2/d+1/2*(d*x+c)^2*FresnelC(b*x+a)/d+1/2*d*FresnelS(b*x+a)/b^2/Pi-(-a*d+b*c
)*sin(1/2*Pi*(b*x+a)^2)/b^2/Pi-1/2*d*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)/b^2/Pi

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Rubi [A]  time = 0.11, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351} \[ -\frac {(b c-a d)^2 \text {FresnelC}(a+b x)}{2 b^2 d}-\frac {(b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^2}+\frac {d S(a+b x)}{2 \pi b^2}-\frac {d (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac {(c+d x)^2 \text {FresnelC}(a+b x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*FresnelC[a + b*x],x]

[Out]

-((b*c - a*d)^2*FresnelC[a + b*x])/(2*b^2*d) + ((c + d*x)^2*FresnelC[a + b*x])/(2*d) + (d*FresnelS[a + b*x])/(
2*b^2*Pi) - ((b*c - a*d)*Sin[(Pi*(a + b*x)^2)/2])/(b^2*Pi) - (d*(a + b*x)*Sin[(Pi*(a + b*x)^2)/2])/(2*b^2*Pi)

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3351

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

Rule 3352

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

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 3386

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

Rule 3434

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Cos[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 6429

Int[FresnelC[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*FresnelC[a +
b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[(Pi*(a + b*x)^2)/2], x], x] /; FreeQ[{a
, b, c, d}, x] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 74, normalized size = 0.61 \[ \frac {-\pi (a+b x) C(a+b x) (a d-b (2 c+d x))+\sin \left (\frac {1}{2} \pi (a+b x)^2\right ) (a d-2 b c-b d x)+d S(a+b x)}{2 \pi b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Pi*(a + b*x)*(a*d - b*(2*c + d*x))*FresnelC[a + b*x]) + d*FresnelS[a + b*x] + (-2*b*c + a*d - b*d*x)*Sin[(P
i*(a + b*x)^2)/2])/(2*b^2*Pi)

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d x + c\right )} {\rm fresnelc}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x + c)*fresnelc(b*x + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} {\rm fresnelc}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 107, normalized size = 0.88 \[ \frac {\frac {\FresnelC \left (b x +a \right ) \left (\frac {\left (b x +a \right )^{2} d}{2}-a d \left (b x +a \right )+b c \left (b x +a \right )\right )}{b}-\frac {\frac {d \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {d \,\mathrm {S}\left (b x +a \right )}{\pi }+\frac {\left (-2 a d +2 b c \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }}{2 b}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*FresnelC(b*x+a),x)

[Out]

1/b*(FresnelC(b*x+a)/b*(1/2*(b*x+a)^2*d-a*d*(b*x+a)+b*c*(b*x+a))-1/2/b*(d/Pi*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)-d/P
i*FresnelS(b*x+a)+(-2*a*d+2*b*c)/Pi*sin(1/2*Pi*(b*x+a)^2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} {\rm fresnelc}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {FresnelC}\left (a+b\,x\right )\,\left (c+d\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(a + b*x)*(c + d*x),x)

[Out]

int(FresnelC(a + b*x)*(c + d*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) C\left (a + b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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