3.202 \(\int x^6 C(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\)

Optimal. Leaf size=185 \[ -\frac {15 C(b x)^2}{2 \pi ^3 b^7}-\frac {15 x^2}{4 \pi ^3 b^5}-\frac {x^5 C(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {11 \sin \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7}+\frac {15 x C(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac {7 x^2 \cos \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}+\frac {5 x^3 C(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}+\frac {x^4 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {x^6}{12 \pi b} \]

[Out]

-15/4*x^2/b^5/Pi^3+1/12*x^6/b/Pi+7/4*x^2*cos(b^2*Pi*x^2)/b^5/Pi^3+15*x*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/b^6/P
i^3-x^5*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/b^2/Pi-15/2*FresnelC(b*x)^2/b^7/Pi^3+5*x^3*FresnelC(b*x)*sin(1/2*b^2
*Pi*x^2)/b^4/Pi^2-11/2*sin(b^2*Pi*x^2)/b^7/Pi^4+1/4*x^4*sin(b^2*Pi*x^2)/b^3/Pi^2

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Rubi [A]  time = 0.25, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6463, 6455, 6441, 30, 3380, 2634, 3379, 3296, 2637, 3309} \[ \frac {5 x^3 \text {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^5 \text {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {15 x \text {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}-\frac {15 \text {FresnelC}(b x)^2}{2 \pi ^3 b^7}-\frac {15 x^2}{4 \pi ^3 b^5}+\frac {x^4 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac {11 \sin \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7}+\frac {7 x^2 \cos \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}+\frac {x^6}{12 \pi b} \]

Antiderivative was successfully verified.

[In]

Int[x^6*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

(-15*x^2)/(4*b^5*Pi^3) + x^6/(12*b*Pi) + (7*x^2*Cos[b^2*Pi*x^2])/(4*b^5*Pi^3) + (15*x*Cos[(b^2*Pi*x^2)/2]*Fres
nelC[b*x])/(b^6*Pi^3) - (x^5*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^2*Pi) - (15*FresnelC[b*x]^2)/(2*b^7*Pi^3) +
(5*x^3*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^4*Pi^2) - (11*Sin[b^2*Pi*x^2])/(2*b^7*Pi^4) + (x^4*Sin[b^2*Pi*x^
2])/(4*b^3*Pi^2)

Rule 30

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

Rule 2634

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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 3309

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + ((f_.)*(x_))/2]^2, x_Symbol] :> Dist[1/2, Int[(c + d*x)^m, x], x] -
Dist[1/2, Int[(c + d*x)^m*Cos[2*e + f*x], x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 3379

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 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 6441

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]^(n_.), x_Symbol] :> Dist[(Pi*b)/(2*d), Subst[Int[x^n, x], x, Fresne
lC[b*x]], x] /; FreeQ[{b, d, n}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 6455

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] + (-Dist[(m - 1)/(2*d), Int[x^(m - 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Dist[b/(4*d), Int[x^(m - 1)*
Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rule 6463

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] + (Dist[(m - 1)/(2*d), Int[x^(m - 2)*Cos[d*x^2]*FresnelC[b*x], x], x] + Dist[b/(2*d), Int[x^(m - 1)*
Cos[d*x^2]^2, x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rubi steps

\begin {align*} \int x^6 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx &=-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac {5 \int x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx}{b^2 \pi }+\frac {\int x^5 \cos ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac {5 x^3 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {15 \int x^2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^4 \pi ^2}-\frac {5 \int x^3 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b^3 \pi ^2}+\frac {\operatorname {Subst}\left (\int x^2 \cos ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b \pi }\\ &=\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac {5 x^3 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {15 \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx}{b^6 \pi ^3}-\frac {15 \int x \cos ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}-\frac {5 \operatorname {Subst}\left (\int x \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^3 \pi ^2}+\frac {\operatorname {Subst}\left (\int x^2 \, dx,x,x^2\right )}{4 b \pi }+\frac {\operatorname {Subst}\left (\int x^2 \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b \pi }\\ &=\frac {x^6}{12 b \pi }+\frac {5 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac {5 x^3 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {15 \operatorname {Subst}(\int x \, dx,x,C(b x))}{b^7 \pi ^3}-\frac {5 \operatorname {Subst}\left (\int \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^5 \pi ^3}-\frac {15 \operatorname {Subst}\left (\int \cos ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^5 \pi ^3}-\frac {\operatorname {Subst}\left (\int x \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^3 \pi ^2}\\ &=-\frac {15 x^2}{4 b^5 \pi ^3}+\frac {x^6}{12 b \pi }+\frac {7 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }-\frac {15 C(b x)^2}{2 b^7 \pi ^3}+\frac {5 x^3 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {5 \sin \left (b^2 \pi x^2\right )}{b^7 \pi ^4}+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {\operatorname {Subst}\left (\int \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^5 \pi ^3}\\ &=-\frac {15 x^2}{4 b^5 \pi ^3}+\frac {x^6}{12 b \pi }+\frac {7 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {15 x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^6 \pi ^3}-\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }-\frac {15 C(b x)^2}{2 b^7 \pi ^3}+\frac {5 x^3 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {11 \sin \left (b^2 \pi x^2\right )}{2 b^7 \pi ^4}+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 185, normalized size = 1.00 \[ -\frac {15 C(b x)^2}{2 \pi ^3 b^7}-\frac {15 x^2}{4 \pi ^3 b^5}-\frac {x^5 C(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {11 \sin \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7}+\frac {15 x C(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac {7 x^2 \cos \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}+\frac {5 x^3 C(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}+\frac {x^4 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {x^6}{12 \pi b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

(-15*x^2)/(4*b^5*Pi^3) + x^6/(12*b*Pi) + (7*x^2*Cos[b^2*Pi*x^2])/(4*b^5*Pi^3) + (15*x*Cos[(b^2*Pi*x^2)/2]*Fres
nelC[b*x])/(b^6*Pi^3) - (x^5*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^2*Pi) - (15*FresnelC[b*x]^2)/(2*b^7*Pi^3) +
(5*x^3*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^4*Pi^2) - (11*Sin[b^2*Pi*x^2])/(2*b^7*Pi^4) + (x^4*Sin[b^2*Pi*x^
2])/(4*b^3*Pi^2)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{6} {\rm fresnelc}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")

[Out]

integral(x^6*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{6} {\rm fresnelc}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")

[Out]

integrate(x^6*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int x^{6} \FresnelC \left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)

[Out]

int(x^6*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{6} {\rm fresnelc}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*fresnelc(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")

[Out]

integrate(x^6*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^6\,\mathrm {FresnelC}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*FresnelC(b*x)*sin((Pi*b^2*x^2)/2),x)

[Out]

int(x^6*FresnelC(b*x)*sin((Pi*b^2*x^2)/2), x)

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sympy [A]  time = 77.91, size = 264, normalized size = 1.43 \[ \begin {cases} \frac {x^{6} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{12 \pi b} + \frac {x^{6} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{12 \pi b} - \frac {x^{5} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )}{\pi b^{2}} + \frac {x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{2} b^{3}} + \frac {5 x^{3} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )}{\pi ^{2} b^{4}} - \frac {11 x^{2} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{3} b^{5}} - \frac {2 x^{2} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{\pi ^{3} b^{5}} + \frac {15 x \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )}{\pi ^{3} b^{6}} - \frac {11 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{\pi ^{4} b^{7}} - \frac {15 C^{2}\left (b x\right )}{2 \pi ^{3} b^{7}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*fresnelc(b*x)*sin(1/2*b**2*pi*x**2),x)

[Out]

Piecewise((x**6*sin(pi*b**2*x**2/2)**2/(12*pi*b) + x**6*cos(pi*b**2*x**2/2)**2/(12*pi*b) - x**5*cos(pi*b**2*x*
*2/2)*fresnelc(b*x)/(pi*b**2) + x**4*sin(pi*b**2*x**2/2)*cos(pi*b**2*x**2/2)/(2*pi**2*b**3) + 5*x**3*sin(pi*b*
*2*x**2/2)*fresnelc(b*x)/(pi**2*b**4) - 11*x**2*sin(pi*b**2*x**2/2)**2/(2*pi**3*b**5) - 2*x**2*cos(pi*b**2*x**
2/2)**2/(pi**3*b**5) + 15*x*cos(pi*b**2*x**2/2)*fresnelc(b*x)/(pi**3*b**6) - 11*sin(pi*b**2*x**2/2)*cos(pi*b**
2*x**2/2)/(pi**4*b**7) - 15*fresnelc(b*x)**2/(2*pi**3*b**7), Ne(b, 0)), (0, True))

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