\(\int x^5 \operatorname {FresnelC}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [203]

Optimal result

Integrand size = 20, antiderivative size = 167 \[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {4 x}{b^5 \pi ^3}+\frac {x^5}{10 b \pi }+\frac {11 x \cos \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^6 \pi ^3}-\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^2 \pi }-\frac {43 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{8 \sqrt {2} b^6 \pi ^3}+\frac {4 x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {x^3 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \] Output:

-4*x/b^5/Pi^3+1/10*x^5/b/Pi+11/8*x*cos(b^2*Pi*x^2)/b^5/Pi^3+8*cos(1/2*b^2* 
Pi*x^2)*FresnelC(b*x)/b^6/Pi^3-x^4*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/b^2/P 
i-43/16*FresnelC(2^(1/2)*b*x)*2^(1/2)/b^6/Pi^3+4*x^2*FresnelC(b*x)*sin(1/2 
*b^2*Pi*x^2)/b^4/Pi^2+1/4*x^3*sin(b^2*Pi*x^2)/b^3/Pi^2
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.75 \[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {-215 \sqrt {2} \operatorname {FresnelC}\left (\sqrt {2} b x\right )-80 \operatorname {FresnelC}(b x) \left (\left (-8+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )-4 b^2 \pi x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )\right )+2 b x \left (-160+4 b^4 \pi ^2 x^4+55 \cos \left (b^2 \pi x^2\right )+10 b^2 \pi x^2 \sin \left (b^2 \pi x^2\right )\right )}{80 b^6 \pi ^3} \] Input:

Integrate[x^5*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]
 

Output:

(-215*Sqrt[2]*FresnelC[Sqrt[2]*b*x] - 80*FresnelC[b*x]*((-8 + b^4*Pi^2*x^4 
)*Cos[(b^2*Pi*x^2)/2] - 4*b^2*Pi*x^2*Sin[(b^2*Pi*x^2)/2]) + 2*b*x*(-160 + 
4*b^4*Pi^2*x^4 + 55*Cos[b^2*Pi*x^2] + 10*b^2*Pi*x^2*Sin[b^2*Pi*x^2]))/(80* 
b^6*Pi^3)
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.69, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {7017, 3873, 15, 3867, 3866, 3833, 7009, 3866, 3833, 7015, 3839, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right ) \, dx\)

\(\Big \downarrow \) 7017

\(\displaystyle \frac {4 \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)dx}{\pi b^2}+\frac {\int x^4 \cos ^2\left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3873

\(\displaystyle \frac {4 \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)dx}{\pi b^2}+\frac {\frac {1}{2} \int x^4 \cos \left (b^2 \pi x^2\right )dx+\frac {\int x^4dx}{2}}{\pi b}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {4 \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)dx}{\pi b^2}+\frac {\frac {1}{2} \int x^4 \cos \left (b^2 \pi x^2\right )dx+\frac {x^5}{10}}{\pi b}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3867

\(\displaystyle \frac {4 \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)dx}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \int x^2 \sin \left (b^2 \pi x^2\right )dx}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3866

\(\displaystyle \frac {4 \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)dx}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\int \cos \left (b^2 \pi x^2\right )dx}{2 \pi b^2}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {4 \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)dx}{\pi b^2}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}\)

\(\Big \downarrow \) 7009

\(\displaystyle \frac {4 \left (-\frac {2 \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\int x^2 \sin \left (b^2 \pi x^2\right )dx}{2 \pi b}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}\)

\(\Big \downarrow \) 3866

\(\displaystyle \frac {4 \left (-\frac {2 \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\frac {\int \cos \left (b^2 \pi x^2\right )dx}{2 \pi b^2}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {4 \left (-\frac {2 \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}\right )}{\pi b^2}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}\)

\(\Big \downarrow \) 7015

\(\displaystyle \frac {4 \left (-\frac {2 \left (\frac {\int \cos ^2\left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b}-\frac {\operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}\right )}{\pi b^2}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}\)

\(\Big \downarrow \) 3839

\(\displaystyle \frac {4 \left (-\frac {2 \left (\frac {\int \left (\frac {1}{2} \cos \left (b^2 \pi x^2\right )+\frac {1}{2}\right )dx}{\pi b}-\frac {\operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}+\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}\right )}{\pi b^2}-\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {x^4 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {4 \left (\frac {x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {2 \left (\frac {\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} b}+\frac {x}{2}}{\pi b}-\frac {\operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}-\frac {\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}\right )}{\pi b^2}+\frac {\frac {1}{2} \left (\frac {x^3 \sin \left (\pi b^2 x^2\right )}{2 \pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^3}-\frac {x \cos \left (\pi b^2 x^2\right )}{2 \pi b^2}\right )}{2 \pi b^2}\right )+\frac {x^5}{10}}{\pi b}\)

Input:

Int[x^5*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]
 

Output:

-((x^4*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^2*Pi)) + (4*(-1/2*(-1/2*(x*Co 
s[b^2*Pi*x^2])/(b^2*Pi) + FresnelC[Sqrt[2]*b*x]/(2*Sqrt[2]*b^3*Pi))/(b*Pi) 
 - (2*(-((Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^2*Pi)) + (x/2 + FresnelC[S 
qrt[2]*b*x]/(2*Sqrt[2]*b))/(b*Pi)))/(b^2*Pi) + (x^2*FresnelC[b*x]*Sin[(b^2 
*Pi*x^2)/2])/(b^2*Pi)))/(b^2*Pi) + (x^5/10 + ((-3*(-1/2*(x*Cos[b^2*Pi*x^2] 
)/(b^2*Pi) + FresnelC[Sqrt[2]*b*x]/(2*Sqrt[2]*b^3*Pi)))/(2*b^2*Pi) + (x^3* 
Sin[b^2*Pi*x^2])/(2*b^2*Pi))/2)/(b*Pi)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3833

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 3839

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_), x_Sy 
mbol] :> Int[ExpandTrigReduce[(a + b*Cos[c + d*(e + f*x)^n])^p, x], x] /; F 
reeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]
 

rule 3866

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] + Simp[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 3867

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] - Simp[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 3873

Int[Cos[(a_.) + ((b_.)*(x_)^(n_))/2]^2*(x_)^(m_.), x_Symbol] :> Simp[1/2 
Int[x^m, x], x] + Simp[1/2   Int[x^m*Cos[2*a + b*x^n], x], x] /; FreeQ[{a, 
b, m, n}, x]
 

rule 7009

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[x^( 
m - 1)*Sin[d*x^2]*(FresnelC[b*x]/(2*d)), x] + (-Simp[(m - 1)/(2*d)   Int[x^ 
(m - 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Simp[b/(4*d)   Int[x^(m - 1)*Sin 
[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && IGtQ[m, 
 1]
 

rule 7015

Int[FresnelC[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(-Cos[d* 
x^2])*(FresnelC[b*x]/(2*d)), x] + Simp[b/(2*d)   Int[Cos[d*x^2]^2, x], x] / 
; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4]
 

rule 7017

Int[FresnelC[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(-x 
^(m - 1))*Cos[d*x^2]*(FresnelC[b*x]/(2*d)), x] + (Simp[(m - 1)/(2*d)   Int[ 
x^(m - 2)*Cos[d*x^2]*FresnelC[b*x], x], x] + Simp[b/(2*d)   Int[x^(m - 1)*C 
os[d*x^2]^2, x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && IGtQ[ 
m, 1]
 

Maple [A] (verified)

Time = 1.74 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.27

method	result	size
default	\(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) \left (-\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\frac {4 b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {8 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}}{\pi }\right )}{b^{5}}-\frac {-\frac {\frac {1}{5} \pi ^{2} b^{5} x^{5}-8 b x}{2 \pi ^{3}}+\frac {-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{\pi }+\frac {\sqrt {2}\, \operatorname {FresnelC}\left (\sqrt {2}\, b x \right )}{2 \pi }}{\pi ^{2}}-\frac {\frac {\pi \,b^{3} x^{3} \sin \left (b^{2} \pi \,x^{2}\right )}{2}-\frac {3 \pi \left (-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \operatorname {FresnelC}\left (\sqrt {2}\, b x \right )}{4 \pi }\right )}{2}-4 \sqrt {2}\, \operatorname {FresnelC}\left (\sqrt {2}\, b x \right )}{2 \pi ^{3}}}{b^{5}}}{b}\)	\(212\)

Input:

int(x^5*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x,method=_RETURNVERBOSE)
 

Output:

(FresnelC(b*x)/b^5*(-1/Pi*b^4*x^4*cos(1/2*b^2*Pi*x^2)+4/Pi*(1/Pi*b^2*x^2*s 
in(1/2*b^2*Pi*x^2)+2/Pi^2*cos(1/2*b^2*Pi*x^2)))-1/b^5*(-1/2/Pi^3*(1/5*Pi^2 
*b^5*x^5-8*b*x)+2/Pi^2*(-1/2/Pi*b*x*cos(b^2*Pi*x^2)+1/4/Pi*2^(1/2)*Fresnel 
C(2^(1/2)*b*x))-1/2/Pi^3*(1/2*Pi*b^3*x^3*sin(b^2*Pi*x^2)-3/2*Pi*(-1/2/Pi*b 
*x*cos(b^2*Pi*x^2)+1/4/Pi*2^(1/2)*FresnelC(2^(1/2)*b*x))-4*2^(1/2)*Fresnel 
C(2^(1/2)*b*x))))/b
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.83 \[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {8 \, \pi ^{2} b^{6} x^{5} + 220 \, b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} - 430 \, b^{2} x - 80 \, {\left (\pi ^{2} b^{5} x^{4} - 8 \, b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) - 215 \, \sqrt {2} \sqrt {b^{2}} \operatorname {C}\left (\sqrt {2} \sqrt {b^{2}} x\right ) + 40 \, {\left (\pi b^{4} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 8 \, \pi b^{3} x^{2} \operatorname {C}\left (b x\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{80 \, \pi ^{3} b^{7}} \] Input:

integrate(x^5*fresnel_cos(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")
 

Output:

1/80*(8*pi^2*b^6*x^5 + 220*b^2*x*cos(1/2*pi*b^2*x^2)^2 - 430*b^2*x - 80*(p 
i^2*b^5*x^4 - 8*b)*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x) - 215*sqrt(2)*sqrt 
(b^2)*fresnel_cos(sqrt(2)*sqrt(b^2)*x) + 40*(pi*b^4*x^3*cos(1/2*pi*b^2*x^2 
) + 8*pi*b^3*x^2*fresnel_cos(b*x))*sin(1/2*pi*b^2*x^2))/(pi^3*b^7)
 

Sympy [F]

\[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int x^{5} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )\, dx \] Input:

integrate(x**5*fresnelc(b*x)*sin(1/2*b**2*pi*x**2),x)
 

Output:

Integral(x**5*sin(pi*b**2*x**2/2)*fresnelc(b*x), x)
 

Maxima [F]

\[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { x^{5} \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \] Input:

integrate(x^5*fresnel_cos(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")
 

Output:

integrate(x^5*fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2), x)
 

Giac [F]

\[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { x^{5} \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \] Input:

integrate(x^5*fresnel_cos(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")
 

Output:

integrate(x^5*fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int x^5\,\mathrm {FresnelC}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \] Input:

int(x^5*FresnelC(b*x)*sin((Pi*b^2*x^2)/2),x)
 

Output:

int(x^5*FresnelC(b*x)*sin((Pi*b^2*x^2)/2), x)
 

Reduce [F]

\[ \int x^5 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int x^{5} \mathrm {FresnelC}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )d x \] Input:

int(x^5*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)
 

Output:

int(x^5*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)