3.2.30 \(\int (c+d x) \text {FresnelC}(a+b x) \, dx\) [130]

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

[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 = 0.08, 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 = {6564, 3515, 3433, 3461, 2717, 3467, 3432} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*FresnelC[a + b*x],x]

[Out]

-1/2*((b*c - a*d)^2*FresnelC[a + b*x])/(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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3461

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 3467

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 3515

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 6564

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/2)*(a + b*x)^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 {\text {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 \text {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) \text {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 \text {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) \text {Subst}\left (\int \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^2}+\frac {d \text {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 = 0.17, size = 74, normalized size = 0.61 \begin {gather*} \frac {-\pi (a+b x) (a d-b (2 c+d x)) \text {FresnelC}(a+b x)+d S(a+b x)+(-2 b c+a d-b d x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi } \end {gather*}

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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Maple [A]
time = 0.44, size = 108, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {\FresnelC \left (b x +a \right ) \left (d a \left (b x +a \right )-c b \left (b x +a \right )-\frac {d \left (b x +a \right )^{2}}{2}\right )}{b}+\frac {-\frac {d \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {d \,\mathrm {S}\left (b x +a \right )}{\pi }+\frac {\left (2 a d -2 c b \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }}{2 b}}{b}\)	\(108\)
default	\(\frac {-\frac {\FresnelC \left (b x +a \right ) \left (d a \left (b x +a \right )-c b \left (b x +a \right )-\frac {d \left (b x +a \right )^{2}}{2}\right )}{b}+\frac {-\frac {d \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {d \,\mathrm {S}\left (b x +a \right )}{\pi }+\frac {\left (2 a d -2 c b \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }}{2 b}}{b}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-FresnelC(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)*sin(1/2*Pi*(b*x+a)^2)+d
/Pi*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 132, normalized size = 1.08 \begin {gather*} \frac {\pi {\left (2 \, a b c - a^{2} d\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + \sqrt {b^{2}} d \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (\pi b^{3} d x^{2} + 2 \, \pi b^{3} c x\right )} \operatorname {C}\left (b x + a\right ) - {\left (b^{2} d x + 2 \, b^{2} c - a b d\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{2 \, \pi b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(pi*(2*a*b*c - a^2*d)*sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x + a)/b) + sqrt(b^2)*d*fresnel_sin(sqrt(b^2)*(b*
x + a)/b) + (pi*b^3*d*x^2 + 2*pi*b^3*c*x)*fresnel_cos(b*x + a) - (b^2*d*x + 2*b^2*c - a*b*d)*sin(1/2*pi*b^2*x^
2 + pi*a*b*x + 1/2*pi*a^2))/(pi*b^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right ) C\left (a + b x\right )\, dx \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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