3.128 \(\int (c+d x)^3 \text{FresnelC}(a+b x) \, dx\)

Optimal. Leaf size=298 \[ -\frac{d^2 (a+b x)^2 (b c-a d) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac{2 d^2 (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac{(b c-a d)^4 \text{FresnelC}(a+b x)}{4 b^4 d}+\frac{3 d (b c-a d)^2 S(a+b x)}{2 \pi b^4}-\frac{(b c-a d)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac{3 d (a+b x) (b c-a d)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac{3 d^3 \text{FresnelC}(a+b x)}{4 \pi ^2 b^4}-\frac{d^3 (a+b x)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}-\frac{3 d^3 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}+\frac{(c+d x)^4 \text{FresnelC}(a+b x)}{4 d} \]

[Out]

(-2*d^2*(b*c - a*d)*Cos[(Pi*(a + b*x)^2)/2])/(b^4*Pi^2) - (3*d^3*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi^
2) - ((b*c - a*d)^4*FresnelC[a + b*x])/(4*b^4*d) + (3*d^3*FresnelC[a + b*x])/(4*b^4*Pi^2) + ((c + d*x)^4*Fresn
elC[a + b*x])/(4*d) + (3*d*(b*c - a*d)^2*FresnelS[a + b*x])/(2*b^4*Pi) - ((b*c - a*d)^3*Sin[(Pi*(a + b*x)^2)/2
])/(b^4*Pi) - (3*d*(b*c - a*d)^2*(a + b*x)*Sin[(Pi*(a + b*x)^2)/2])/(2*b^4*Pi) - (d^2*(b*c - a*d)*(a + b*x)^2*
Sin[(Pi*(a + b*x)^2)/2])/(b^4*Pi) - (d^3*(a + b*x)^3*Sin[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi)

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Rubi [A]  time = 0.37306, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351, 3296, 2638, 3385} \[ -\frac{d^2 (a+b x)^2 (b c-a d) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac{2 d^2 (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac{(b c-a d)^4 \text{FresnelC}(a+b x)}{4 b^4 d}+\frac{3 d (b c-a d)^2 S(a+b x)}{2 \pi b^4}-\frac{(b c-a d)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac{3 d (a+b x) (b c-a d)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac{3 d^3 \text{FresnelC}(a+b x)}{4 \pi ^2 b^4}-\frac{d^3 (a+b x)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}-\frac{3 d^3 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}+\frac{(c+d x)^4 \text{FresnelC}(a+b x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*FresnelC[a + b*x],x]

[Out]

(-2*d^2*(b*c - a*d)*Cos[(Pi*(a + b*x)^2)/2])/(b^4*Pi^2) - (3*d^3*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi^
2) - ((b*c - a*d)^4*FresnelC[a + b*x])/(4*b^4*d) + (3*d^3*FresnelC[a + b*x])/(4*b^4*Pi^2) + ((c + d*x)^4*Fresn
elC[a + b*x])/(4*d) + (3*d*(b*c - a*d)^2*FresnelS[a + b*x])/(2*b^4*Pi) - ((b*c - a*d)^3*Sin[(Pi*(a + b*x)^2)/2
])/(b^4*Pi) - (3*d*(b*c - a*d)^2*(a + b*x)*Sin[(Pi*(a + b*x)^2)/2])/(2*b^4*Pi) - (d^2*(b*c - a*d)*(a + b*x)^2*
Sin[(Pi*(a + b*x)^2)/2])/(b^4*Pi) - (d^3*(a + b*x)^3*Sin[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi)

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]

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 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 2637

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

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 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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

Rule 3385

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]

Rubi steps

\begin{align*} \int (c+d x)^3 C(a+b x) \, dx &=\frac{(c+d x)^4 C(a+b x)}{4 d}-\frac{b \int (c+d x)^4 \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) \, dx}{4 d}\\ &=\frac{(c+d x)^4 C(a+b x)}{4 d}-\frac{\operatorname{Subst}\left (\int \left (b^4 c^4 \left (1+\frac{a d \left (-4 b^3 c^3+6 a b^2 c^2 d-4 a^2 b c d^2+a^3 d^3\right )}{b^4 c^4}\right ) \cos \left (\frac{\pi x^2}{2}\right )+4 b^3 c^3 d \left (1-\frac{a d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right )}{b^3 c^3}\right ) x \cos \left (\frac{\pi x^2}{2}\right )+6 b^2 c^2 d^2 \left (1+\frac{a d (-2 b c+a d)}{b^2 c^2}\right ) x^2 \cos \left (\frac{\pi x^2}{2}\right )+4 b c d^3 \left (1-\frac{a d}{b c}\right ) x^3 \cos \left (\frac{\pi x^2}{2}\right )+d^4 x^4 \cos \left (\frac{\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{4 b^4 d}\\ &=\frac{(c+d x)^4 C(a+b x)}{4 d}-\frac{d^3 \operatorname{Subst}\left (\int x^4 \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4}-\frac{\left (d^2 (b c-a d)\right ) \operatorname{Subst}\left (\int x^3 \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac{\left (3 d (b c-a d)^2\right ) \operatorname{Subst}\left (\int x^2 \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int x \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac{(b c-a d)^4 \operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 d}\\ &=-\frac{(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac{(c+d x)^4 C(a+b x)}{4 d}-\frac{3 d (b c-a d)^2 (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac{d^3 (a+b x)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac{\left (d^2 (b c-a d)\right ) \operatorname{Subst}\left (\int x \cos \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \cos \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}+\frac{\left (3 d^3\right ) \operatorname{Subst}\left (\int x^2 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi }+\frac{\left (3 d (b c-a d)^2\right ) \operatorname{Subst}\left (\int \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4 \pi }\\ &=-\frac{3 d^3 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac{(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac{(c+d x)^4 C(a+b x)}{4 d}+\frac{3 d (b c-a d)^2 S(a+b x)}{2 b^4 \pi }-\frac{(b c-a d)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac{3 d (b c-a d)^2 (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac{d^2 (b c-a d) (a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac{d^3 (a+b x)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }+\frac{\left (3 d^3\right ) \operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi ^2}+\frac{\left (d^2 (b c-a d)\right ) \operatorname{Subst}\left (\int \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{b^4 \pi }\\ &=-\frac{2 d^2 (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac{3 d^3 (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac{(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac{3 d^3 C(a+b x)}{4 b^4 \pi ^2}+\frac{(c+d x)^4 C(a+b x)}{4 d}+\frac{3 d (b c-a d)^2 S(a+b x)}{2 b^4 \pi }-\frac{(b c-a d)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac{3 d (b c-a d)^2 (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac{d^2 (b c-a d) (a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac{d^3 (a+b x)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }\\ \end{align*}

Mathematica [A]  time = 0.816556, size = 424, normalized size = 1.42 \[ \frac{\text{FresnelC}(a+b x) \left (6 \pi ^2 b^2 c^2 d \left (b^2 x^2-a^2\right )+4 \pi ^2 b c d^2 \left (a^3+b^3 x^3\right )+d^3 \left (-\pi ^2 a^4+\pi ^2 b^4 x^4+3\right )+4 \pi ^2 b^3 c^3 (a+b x)\right )-4 \pi a^2 b c d^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-\pi a^2 b d^3 x \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi a^3 d^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-6 \pi b^3 c^2 d x \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+6 \pi a b^2 c^2 d \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-4 \pi b^3 c^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-4 \pi b^3 c d^2 x^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+4 \pi a b^2 c d^2 x \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi b^3 \left (-d^3\right ) x^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi a b^2 d^3 x^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-8 b c d^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+6 \pi d (b c-a d)^2 S(a+b x)-3 b d^3 x \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+5 a d^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*FresnelC[a + b*x],x]

[Out]

(-8*b*c*d^2*Cos[(Pi*(a + b*x)^2)/2] + 5*a*d^3*Cos[(Pi*(a + b*x)^2)/2] - 3*b*d^3*x*Cos[(Pi*(a + b*x)^2)/2] + (4
*b^3*c^3*Pi^2*(a + b*x) + 6*b^2*c^2*d*Pi^2*(-a^2 + b^2*x^2) + 4*b*c*d^2*Pi^2*(a^3 + b^3*x^3) + d^3*(3 - a^4*Pi
^2 + b^4*Pi^2*x^4))*FresnelC[a + b*x] + 6*d*(b*c - a*d)^2*Pi*FresnelS[a + b*x] - 4*b^3*c^3*Pi*Sin[(Pi*(a + b*x
)^2)/2] + 6*a*b^2*c^2*d*Pi*Sin[(Pi*(a + b*x)^2)/2] - 4*a^2*b*c*d^2*Pi*Sin[(Pi*(a + b*x)^2)/2] + a^3*d^3*Pi*Sin
[(Pi*(a + b*x)^2)/2] - 6*b^3*c^2*d*Pi*x*Sin[(Pi*(a + b*x)^2)/2] + 4*a*b^2*c*d^2*Pi*x*Sin[(Pi*(a + b*x)^2)/2] -
a^2*b*d^3*Pi*x*Sin[(Pi*(a + b*x)^2)/2] - 4*b^3*c*d^2*Pi*x^2*Sin[(Pi*(a + b*x)^2)/2] + a*b^2*d^3*Pi*x^2*Sin[(P
i*(a + b*x)^2)/2] - b^3*d^3*Pi*x^3*Sin[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi^2)

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Maple [A]  time = 0.056, size = 397, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{{\it FresnelC} \left ( bx+a \right ) \left ( d \left ( bx+a \right ) -ad+bc \right ) ^{4}}{4\,d{b}^{3}}}-{\frac{1}{4\,d{b}^{3}} \left ({\frac{{d}^{4} \left ( bx+a \right ) ^{3}}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-3\,{\frac{{d}^{4}}{\pi } \left ( -{\frac{ \left ( bx+a \right ) \cos \left ( 1/2\,\pi \, \left ( bx+a \right ) ^{2} \right ) }{\pi }}+{\frac{{\it FresnelC} \left ( bx+a \right ) }{\pi }} \right ) }+{\frac{ \left ( -4\,a{d}^{4}+4\,bc{d}^{3} \right ) \left ( bx+a \right ) ^{2}}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+2\,{\frac{ \left ( -4\,a{d}^{4}+4\,bc{d}^{3} \right ) \cos \left ( 1/2\,\pi \, \left ( bx+a \right ) ^{2} \right ) }{{\pi }^{2}}}+{\frac{ \left ( 6\,{a}^{2}{d}^{4}-12\,abc{d}^{3}+6\,{b}^{2}{c}^{2}{d}^{2} \right ) \left ( bx+a \right ) }{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{ \left ( 6\,{a}^{2}{d}^{4}-12\,abc{d}^{3}+6\,{b}^{2}{c}^{2}{d}^{2} \right ){\it FresnelS} \left ( bx+a \right ) }{\pi }}+{\frac{-4\,{a}^{3}{d}^{4}+12\,{a}^{2}bc{d}^{3}-12\,a{b}^{2}{c}^{2}{d}^{2}+4\,{b}^{3}{c}^{3}d}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+{a}^{4}{d}^{4}{\it FresnelC} \left ( bx+a \right ) -4\,{a}^{3}bc{d}^{3}{\it FresnelC} \left ( bx+a \right ) +6\,{a}^{2}{b}^{2}{c}^{2}{d}^{2}{\it FresnelC} \left ( bx+a \right ) -4\,a{b}^{3}{c}^{3}d{\it FresnelC} \left ( bx+a \right ) +{b}^{4}{c}^{4}{\it FresnelC} \left ( bx+a \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*FresnelC(b*x+a),x)

[Out]

1/b*(1/4*FresnelC(b*x+a)*(d*(b*x+a)-a*d+b*c)^4/b^3/d-1/4/b^3/d*(d^4/Pi*(b*x+a)^3*sin(1/2*Pi*(b*x+a)^2)-3*d^4/P
i*(-1/Pi*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)+1/Pi*FresnelC(b*x+a))+(-4*a*d^4+4*b*c*d^3)/Pi*(b*x+a)^2*sin(1/2*Pi*(b*x
+a)^2)+2*(-4*a*d^4+4*b*c*d^3)/Pi^2*cos(1/2*Pi*(b*x+a)^2)+(6*a^2*d^4-12*a*b*c*d^3+6*b^2*c^2*d^2)/Pi*(b*x+a)*sin
(1/2*Pi*(b*x+a)^2)-(6*a^2*d^4-12*a*b*c*d^3+6*b^2*c^2*d^2)/Pi*FresnelS(b*x+a)+(-4*a^3*d^4+12*a^2*b*c*d^3-12*a*b
^2*c^2*d^2+4*b^3*c^3*d)/Pi*sin(1/2*Pi*(b*x+a)^2)+a^4*d^4*FresnelC(b*x+a)-4*a^3*b*c*d^3*FresnelC(b*x+a)+6*a^2*b
^2*c^2*d^2*FresnelC(b*x+a)-4*a*b^3*c^3*d*FresnelC(b*x+a)+b^4*c^4*FresnelC(b*x+a)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3}{\rm fresnelc}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*fresnelc(b*x+a),x, algorithm="maxima")

[Out]

integrate((d*x + c)^3*fresnelc(b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )}{\rm fresnelc}\left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*fresnelc(b*x+a),x, algorithm="fricas")

[Out]

integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*fresnelc(b*x + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{3} C\left (a + b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*fresnelc(b*x+a),x)

[Out]

Integral((c + d*x)**3*fresnelc(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3}{\rm fresnelc}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*fresnelc(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*fresnelc(b*x + a), x)