3.129 \(\int (c+d x)^2 C(a+b x) \, dx\)

Optimal. Leaf size=194 \[ -\frac {(b c-a d)^3 C(a+b x)}{3 b^3 d}+\frac {d (b c-a d) S(a+b x)}{\pi b^3}-\frac {(b c-a d)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {d (a+b x) (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {d^2 (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac {2 d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {(c+d x)^3 C(a+b x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351, 3296, 2638} \[ -\frac {(b c-a d)^3 \text {FresnelC}(a+b x)}{3 b^3 d}+\frac {d (b c-a d) S(a+b x)}{\pi b^3}-\frac {(b c-a d)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {d (a+b x) (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {d^2 (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac {2 d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {(c+d x)^3 \text {FresnelC}(a+b x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*FresnelC[a + b*x],x]

[Out]

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

Rule 2637

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.54, size = 237, normalized size = 1.22 \[ \frac {-\pi a^2 d^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi ^2 C(a+b x) \left (a^3 d^2-3 a^2 b c d+3 a b^2 c^2+b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )\right )-3 \pi b^2 c^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-3 \pi b^2 c d x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi b^2 \left (-d^2\right ) x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+3 \pi d (b c-a d) S(a+b x)+3 \pi a b c d \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi a b d^2 x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-2 d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*FresnelC[a + b*x],x]

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} {\rm fresnelc}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*fresnelc(b*x+a),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} {\rm fresnelc}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*fresnelc(b*x+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 249, normalized size = 1.28 \[ \frac {\frac {\FresnelC \left (b x +a \right ) \left (\left (b x +a \right ) d -a d +b c \right )^{3}}{3 b^{2} d}-\frac {\frac {d^{3} \left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 d^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (-3 a \,d^{3}+3 b c \,d^{2}\right ) \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\left (-3 a \,d^{3}+3 b c \,d^{2}\right ) \mathrm {S}\left (b x +a \right )}{\pi }+\frac {\left (3 a^{2} d^{3}-6 a b c \,d^{2}+3 b^{2} c^{2} d \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-a^{3} d^{3} \FresnelC \left (b x +a \right )+3 a^{2} b c \,d^{2} \FresnelC \left (b x +a \right )-3 a \,b^{2} c^{2} d \FresnelC \left (b x +a \right )+b^{3} c^{3} \FresnelC \left (b x +a \right )}{3 b^{2} d}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*FresnelC(b*x+a),x)

[Out]

1/b*(1/3*FresnelC(b*x+a)*((b*x+a)*d-a*d+b*c)^3/b^2/d-1/3/b^2/d*(d^3/Pi*(b*x+a)^2*sin(1/2*Pi*(b*x+a)^2)+2*d^3/P
i^2*cos(1/2*Pi*(b*x+a)^2)+(-3*a*d^3+3*b*c*d^2)/Pi*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)-(-3*a*d^3+3*b*c*d^2)/Pi*Fresne
lS(b*x+a)+(3*a^2*d^3-6*a*b*c*d^2+3*b^2*c^2*d)/Pi*sin(1/2*Pi*(b*x+a)^2)-a^3*d^3*FresnelC(b*x+a)+3*a^2*b*c*d^2*F
resnelC(b*x+a)-3*a*b^2*c^2*d*FresnelC(b*x+a)+b^3*c^3*FresnelC(b*x+a)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} {\rm fresnelc}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*fresnelc(b*x+a),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(a + b*x)*(c + d*x)^2,x)

[Out]

int(FresnelC(a + b*x)*(c + d*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} C\left (a + b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*fresnelc(b*x+a),x)

[Out]

Integral((c + d*x)**2*fresnelc(a + b*x), x)

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