3.1.19 \(\int (c+d x)^3 S(a+b x) \, dx\) [19]

Optimal. Leaf size=296 \[ \frac {(b c-a d)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {3 d (b c-a d)^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }+\frac {d^2 (b c-a d) (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {d^3 (a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {3 d (b c-a d)^2 \text {FresnelC}(a+b x)}{2 b^4 \pi }-\frac {(b c-a d)^4 S(a+b x)}{4 b^4 d}+\frac {3 d^3 S(a+b x)}{4 b^4 \pi ^2}+\frac {(c+d x)^4 S(a+b x)}{4 d}-\frac {2 d^2 (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 d^3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \]

[Out]

(-a*d+b*c)^3*cos(1/2*Pi*(b*x+a)^2)/b^4/Pi+3/2*d*(-a*d+b*c)^2*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)/b^4/Pi+d^2*(-a*d+b*
c)*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)/b^4/Pi+1/4*d^3*(b*x+a)^3*cos(1/2*Pi*(b*x+a)^2)/b^4/Pi-3/2*d*(-a*d+b*c)^2*Fr
esnelC(b*x+a)/b^4/Pi-1/4*(-a*d+b*c)^4*FresnelS(b*x+a)/b^4/d+3/4*d^3*FresnelS(b*x+a)/b^4/Pi^2+1/4*(d*x+c)^4*Fre
snelS(b*x+a)/d-2*d^2*(-a*d+b*c)*sin(1/2*Pi*(b*x+a)^2)/b^4/Pi^2-3/4*d^3*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)/b^4/Pi^2

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Rubi [A]
time = 0.28, antiderivative size = 296, normalized size of antiderivative = 1.00, 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 = {6563, 3514, 3432, 3460, 2718, 3466, 3433, 3377, 2717, 3467} \begin {gather*} -\frac {2 d^2 (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}+\frac {d^2 (a+b x)^2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {3 d (b c-a d)^2 \text {FresnelC}(a+b x)}{2 \pi b^4}-\frac {(b c-a d)^4 S(a+b x)}{4 b^4 d}+\frac {(b c-a d)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}+\frac {3 d (a+b x) (b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac {3 d^3 S(a+b x)}{4 \pi ^2 b^4}-\frac {3 d^3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}+\frac {d^3 (a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}+\frac {(c+d x)^4 S(a+b x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*FresnelS[a + b*x],x]

[Out]

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

Rule 2717

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

Rule 2718

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

Rule 3377

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

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Mathematica [A]
time = 0.56, size = 424, normalized size = 1.43 \begin {gather*} \frac {4 b^3 c^3 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-6 a b^2 c^2 d \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+4 a^2 b c d^2 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a^3 d^3 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+6 b^3 c^2 d \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-4 a b^2 c d^2 \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+a^2 b d^3 \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+4 b^3 c d^2 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a b^2 d^3 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+b^3 d^3 \pi x^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-6 d (b c-a d)^2 \pi \text {FresnelC}(a+b x)+\left (4 b^3 c^3 \pi ^2 (a+b x)+6 b^2 c^2 d \pi ^2 \left (-a^2+b^2 x^2\right )+4 b c d^2 \pi ^2 \left (a^3+b^3 x^3\right )+d^3 \left (3-a^4 \pi ^2+b^4 \pi ^2 x^4\right )\right ) S(a+b x)-8 b c d^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+5 a d^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-3 b d^3 x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.52, size = 401, normalized size = 1.35

method	result	size
derivativedivides	\(\frac {\frac {\mathrm {S}\left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{4}}{4 b^{3} d}-\frac {-\frac {d^{4} \left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 d^{4} \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\mathrm {S}\left (b x +a \right )}{\pi }\right )}{\pi }-\frac {\left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 \left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \FresnelC \left (b x +a \right )}{\pi }-\frac {\left (-4 a^{3} d^{4}+12 a^{2} b c \,d^{3}-12 a \,b^{2} c^{2} d^{2}+4 b^{3} c^{3} d \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+a^{4} d^{4} \mathrm {S}\left (b x +a \right )-4 a^{3} b c \,d^{3} \mathrm {S}\left (b x +a \right )+6 a^{2} b^{2} c^{2} d^{2} \mathrm {S}\left (b x +a \right )-4 a \,b^{3} c^{3} d \,\mathrm {S}\left (b x +a \right )+b^{4} c^{4} \mathrm {S}\left (b x +a \right )}{4 b^{3} d}}{b}\)	\(401\)
default	\(\frac {\frac {\mathrm {S}\left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{4}}{4 b^{3} d}-\frac {-\frac {d^{4} \left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 d^{4} \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\mathrm {S}\left (b x +a \right )}{\pi }\right )}{\pi }-\frac {\left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 \left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \FresnelC \left (b x +a \right )}{\pi }-\frac {\left (-4 a^{3} d^{4}+12 a^{2} b c \,d^{3}-12 a \,b^{2} c^{2} d^{2}+4 b^{3} c^{3} d \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+a^{4} d^{4} \mathrm {S}\left (b x +a \right )-4 a^{3} b c \,d^{3} \mathrm {S}\left (b x +a \right )+6 a^{2} b^{2} c^{2} d^{2} \mathrm {S}\left (b x +a \right )-4 a \,b^{3} c^{3} d \,\mathrm {S}\left (b x +a \right )+b^{4} c^{4} \mathrm {S}\left (b x +a \right )}{4 b^{3} d}}{b}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*FresnelS(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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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)^3*fresnel_sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 376, normalized size = 1.27 \begin {gather*} -\frac {6 \, \pi {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi ^{2} {\left (4 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 4 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} + 3 \, d^{3}\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi b^{4} d^{3} x^{3} + \pi {\left (4 \, b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + \pi {\left (6 \, b^{4} c^{2} d - 4 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}\right )} x + \pi {\left (4 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d + 4 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - {\left (\pi ^{2} b^{5} d^{3} x^{4} + 4 \, \pi ^{2} b^{5} c d^{2} x^{3} + 6 \, \pi ^{2} b^{5} c^{2} d x^{2} + 4 \, \pi ^{2} b^{5} c^{3} x\right )} \operatorname {S}\left (b x + a\right ) + {\left (3 \, b^{2} d^{3} x + 8 \, b^{2} c d^{2} - 5 \, a b d^{3}\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{4 \, \pi ^{2} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(6*pi*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x + a)/b) - (pi^2*(4*a*b^3*c
^3 - 6*a^2*b^2*c^2*d + 4*a^3*b*c*d^2 - a^4*d^3) + 3*d^3)*sqrt(b^2)*fresnel_sin(sqrt(b^2)*(b*x + a)/b) - (pi*b^
4*d^3*x^3 + pi*(4*b^4*c*d^2 - a*b^3*d^3)*x^2 + pi*(6*b^4*c^2*d - 4*a*b^3*c*d^2 + a^2*b^2*d^3)*x + pi*(4*b^4*c^
3 - 6*a*b^3*c^2*d + 4*a^2*b^2*c*d^2 - a^3*b*d^3))*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - (pi^2*b^5*d^3*
x^4 + 4*pi^2*b^5*c*d^2*x^3 + 6*pi^2*b^5*c^2*d*x^2 + 4*pi^2*b^5*c^3*x)*fresnel_sin(b*x + a) + (3*b^2*d^3*x + 8*
b^2*c*d^2 - 5*a*b*d^3)*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2))/(pi^2*b^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right )^{3} S\left (a + b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)**3*fresnels(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)^3*fresnel_sin(b*x+a),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \mathrm {FresnelS}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(a + b*x)*(c + d*x)^3,x)

[Out]

int(FresnelS(a + b*x)*(c + d*x)^3, x)

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