3.1.25 \(\int x^3 S(a+b x) \, dx\) [25]

Optimal. Leaf size=229 \[ -\frac {a^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {3 a^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {a (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {(a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {3 a^2 \text {FresnelC}(a+b x)}{2 b^4 \pi }-\frac {a^4 S(a+b x)}{4 b^4}+\frac {3 S(a+b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 S(a+b x)+\frac {2 a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6563, 3514, 3432, 3460, 2718, 3466, 3433, 3377, 2717, 3467} \begin {gather*} -\frac {a^4 S(a+b x)}{4 b^4}-\frac {a^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {3 a^2 \text {FresnelC}(a+b x)}{2 \pi b^4}+\frac {3 a^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac {3 S(a+b x)}{4 \pi ^2 b^4}+\frac {2 a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac {3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}-\frac {a (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}+\frac {(a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}+\frac {1}{4} x^4 S(a+b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*FresnelS[a + b*x],x]

[Out]

-((a^3*Cos[(Pi*(a + b*x)^2)/2])/(b^4*Pi)) + (3*a^2*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/(2*b^4*Pi) - (a*(a + b*x
)^2*Cos[(Pi*(a + b*x)^2)/2])/(b^4*Pi) + ((a + b*x)^3*Cos[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi) - (3*a^2*FresnelC[a +
b*x])/(2*b^4*Pi) - (a^4*FresnelS[a + b*x])/(4*b^4) + (3*FresnelS[a + b*x])/(4*b^4*Pi^2) + (x^4*FresnelS[a + b
*x])/4 + (2*a*Sin[(Pi*(a + b*x)^2)/2])/(b^4*Pi^2) - (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 x^3 S(a+b x) \, dx &=\frac {1}{4} x^4 S(a+b x)-\frac {1}{4} b \int x^4 \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) \, dx\\ &=\frac {1}{4} x^4 S(a+b x)-\frac {\text {Subst}\left (\int \left (a^4 \sin \left (\frac {\pi x^2}{2}\right )-4 a^3 x \sin \left (\frac {\pi x^2}{2}\right )+6 a^2 x^2 \sin \left (\frac {\pi x^2}{2}\right )-4 a x^3 \sin \left (\frac {\pi x^2}{2}\right )+x^4 \sin \left (\frac {\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac {1}{4} x^4 S(a+b x)-\frac {\text {Subst}\left (\int x^4 \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4}+\frac {a \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 a^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 {a^3 \text {Subst}\left (\int x \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac {a^4 \text {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac {3 a^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }+\frac {(a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {a^4 S(a+b x)}{4 b^4}+\frac {1}{4} x^4 S(a+b x)+\frac {a \text {Subst}\left (\int x \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}+\frac {a^3 \text {Subst}\left (\int \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}-\frac {3 \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 a^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 {a^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {3 a^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {a (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {(a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {3 a^2 C(a+b x)}{2 b^4 \pi }-\frac {a^4 S(a+b x)}{4 b^4}+\frac {1}{4} x^4 S(a+b x)-\frac {3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}+\frac {3 \text {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi ^2}+\frac {a \text {Subst}\left (\int \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{b^4 \pi }\\ &=-\frac {a^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {3 a^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {a (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {(a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {3 a^2 C(a+b x)}{2 b^4 \pi }-\frac {a^4 S(a+b x)}{4 b^4}+\frac {3 S(a+b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 S(a+b x)+\frac {2 a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {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 = 0.24, size = 166, normalized size = 0.72 \begin {gather*} \frac {-a^3 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+a^2 b \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a b^2 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+b^3 \pi x^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-6 a^2 \pi \text {FresnelC}(a+b x)+\left (3-a^4 \pi ^2+b^4 \pi ^2 x^4\right ) S(a+b x)+5 a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-3 b 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[x^3*FresnelS[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.40, size = 189, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\frac {\mathrm {S}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \mathrm {S}\left (b x +a \right )}{4}-\frac {a^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 a^{2} \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }-\frac {3 a^{2} \FresnelC \left (b x +a \right )}{2 \pi }-\frac {a \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 a \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }-\frac {3 \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\mathrm {S}\left (b x +a \right )}{\pi }\right )}{4 \pi }}{b^{4}}\)	\(189\)
default	\(\frac {\frac {\mathrm {S}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \mathrm {S}\left (b x +a \right )}{4}-\frac {a^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 a^{2} \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }-\frac {3 a^{2} \FresnelC \left (b x +a \right )}{2 \pi }-\frac {a \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 a \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }-\frac {3 \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\mathrm {S}\left (b x +a \right )}{\pi }\right )}{4 \pi }}{b^{4}}\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 1.16, size = 503, normalized size = 2.20 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {S}\left (b x + a\right ) - \frac {{\left (16 \, {\left (\pi ^{2} e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + \pi ^{2} e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a^{4} + 32 \, {\left (-i \, \pi \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + i \, \pi \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} a^{2} + 16 \, {\left ({\left (\pi ^{2} e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + \pi ^{2} e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a^{3} + 2 \, {\left (-i \, \pi \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + i \, \pi \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} a\right )} b x - {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \pi ^{\frac {5}{2}} {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )} + \left (i - 1\right ) \, \sqrt {2} \pi ^{\frac {5}{2}} {\left (\operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )}\right )} a^{4} - 12 \, {\left (\left (i - 1\right ) \, \sqrt {2} \pi \Gamma \left (\frac {3}{2}, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \pi \Gamma \left (\frac {3}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} a^{2} - \left (4 i + 4\right ) \, \sqrt {2} \Gamma \left (\frac {5}{2}, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + \left (4 i - 4\right ) \, \sqrt {2} \Gamma \left (\frac {5}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} \sqrt {2 \, \pi b^{2} x^{2} + 4 \, \pi a b x + 2 \, \pi a^{2}}\right )} b}{32 \, {\left (\pi ^{3} b^{6} x + \pi ^{3} a b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*fresnel_sin(b*x + a) - 1/32*(16*(pi^2*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + pi^2*e^(-1/2*
I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^4 + 32*(-I*pi*gamma(2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a
^2) + I*pi*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^2 + 16*((pi^2*e^(1/2*I*pi*b^2*x^2 + I*pi
*a*b*x + 1/2*I*pi*a^2) + pi^2*e^(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^3 + 2*(-I*pi*gamma(2, 1/2*I
*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + I*pi*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a)*b*
x - ((-(I + 1)*sqrt(2)*pi^(5/2)*(erf(sqrt(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2)) - 1) + (I - 1)*sqrt(2
)*pi^(5/2)*(erf(sqrt(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)) - 1))*a^4 - 12*((I - 1)*sqrt(2)*pi*gamma(
3/2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) - (I + 1)*sqrt(2)*pi*gamma(3/2, -1/2*I*pi*b^2*x^2 - I*pi*a*
b*x - 1/2*I*pi*a^2))*a^2 - (4*I + 4)*sqrt(2)*gamma(5/2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + (4*I -
4)*sqrt(2)*gamma(5/2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*sqrt(2*pi*b^2*x^2 + 4*pi*a*b*x + 2*pi*a
^2))*b/(pi^3*b^6*x + pi^3*a*b^5)

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Fricas [A]
time = 0.37, size = 175, normalized size = 0.76 \begin {gather*} \frac {\pi ^{2} b^{5} x^{4} \operatorname {S}\left (b x + a\right ) - 6 \, \pi a^{2} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi ^{2} a^{4} - 3\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (\pi b^{4} x^{3} - \pi a b^{3} x^{2} + \pi a^{2} b^{2} x - \pi a^{3} b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - {\left (3 \, b^{2} x - 5 \, a b\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(x^3*fresnel_sin(b*x+a),x, algorithm="fricas")

[Out]

1/4*(pi^2*b^5*x^4*fresnel_sin(b*x + a) - 6*pi*a^2*sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x + a)/b) - (pi^2*a^4 - 3
)*sqrt(b^2)*fresnel_sin(sqrt(b^2)*(b*x + a)/b) + (pi*b^4*x^3 - pi*a*b^3*x^2 + pi*a^2*b^2*x - pi*a^3*b)*cos(1/2
*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - (3*b^2*x - 5*a*b)*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 x^{3} S\left (a + b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^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 x^3\,\mathrm {FresnelS}\left (a+b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelS(a + b*x),x)

[Out]

int(x^3*FresnelS(a + b*x), x)

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