3.8 \(\int x^2 (a+b \sin (c+d x^2)) \, dx\)

Optimal. Leaf size=102 \[ \frac{a x^3}{3}+\frac{\sqrt{\frac{\pi }{2}} b \cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )}{2 d^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} b \sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )}{2 d^{3/2}}-\frac{b x \cos \left (c+d x^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.068195, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {14, 3385, 3354, 3352, 3351} \[ \frac{a x^3}{3}+\frac{\sqrt{\frac{\pi }{2}} b \cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )}{2 d^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} b \sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )}{2 d^{3/2}}-\frac{b x \cos \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*Sin[c + d*x^2]),x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3385

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 3354

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

Rubi steps

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

Mathematica [A]  time = 0.194966, size = 104, normalized size = 1.02 \[ \frac{a x^3}{3}+\frac{\sqrt{\frac{\pi }{2}} b \left (\cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )\right )}{2 d^{3/2}}+\frac{b x \sin (c) \sin \left (d x^2\right )}{2 d}-\frac{b x \cos (c) \cos \left (d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 68, normalized size = 0.7 \begin{align*}{\frac{a{x}^{3}}{3}}+b \left ( -{\frac{x\cos \left ( d{x}^{2}+c \right ) }{2\,d}}+{\frac{\sqrt{2}\sqrt{\pi }}{4} \left ( \cos \left ( c \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{d}} \right ) -\sin \left ( c \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{d}} \right ) \right ){d}^{-{\frac{3}{2}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.64559, size = 350, normalized size = 3.43 \begin{align*} \frac{1}{3} \, a x^{3} - \frac{{\left (8 \, x{\left | d \right |} \cos \left (d x^{2} + c\right ) - \sqrt{\pi }{\left ({\left ({\left (\cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) + \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) - i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) + i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) -{\left (i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) + i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) + \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) - \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} \operatorname{erf}\left (\sqrt{i \, d} x\right ) +{\left ({\left (\cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) + \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) + i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) - i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) -{\left (-i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) - i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) + \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) - \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} \operatorname{erf}\left (\sqrt{-i \, d} x\right )\right )} \sqrt{{\left | d \right |}}\right )} b}{16 \, d{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 - 1/16*(8*x*abs(d)*cos(d*x^2 + c) - sqrt(pi)*(((cos(1/4*pi + 1/2*arctan2(0, d)) + cos(-1/4*pi + 1/2*
arctan2(0, d)) - I*sin(1/4*pi + 1/2*arctan2(0, d)) + I*sin(-1/4*pi + 1/2*arctan2(0, d)))*cos(c) - (I*cos(1/4*p
i + 1/2*arctan2(0, d)) + I*cos(-1/4*pi + 1/2*arctan2(0, d)) + sin(1/4*pi + 1/2*arctan2(0, d)) - sin(-1/4*pi +
1/2*arctan2(0, d)))*sin(c))*erf(sqrt(I*d)*x) + ((cos(1/4*pi + 1/2*arctan2(0, d)) + cos(-1/4*pi + 1/2*arctan2(0
, d)) + I*sin(1/4*pi + 1/2*arctan2(0, d)) - I*sin(-1/4*pi + 1/2*arctan2(0, d)))*cos(c) - (-I*cos(1/4*pi + 1/2*
arctan2(0, d)) - I*cos(-1/4*pi + 1/2*arctan2(0, d)) + sin(1/4*pi + 1/2*arctan2(0, d)) - sin(-1/4*pi + 1/2*arct
an2(0, d)))*sin(c))*erf(sqrt(-I*d)*x))*sqrt(abs(d)))*b/(d*abs(d))

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Fricas [A]  time = 2.08909, size = 252, normalized size = 2.47 \begin{align*} \frac{4 \, a d^{2} x^{3} + 3 \, \sqrt{2} \pi b \sqrt{\frac{d}{\pi }} \cos \left (c\right ) \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) - 3 \, \sqrt{2} \pi b \sqrt{\frac{d}{\pi }} \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) \sin \left (c\right ) - 6 \, b d x \cos \left (d x^{2} + c\right )}{12 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*a*d^2*x^3 + 3*sqrt(2)*pi*b*sqrt(d/pi)*cos(c)*fresnel_cos(sqrt(2)*x*sqrt(d/pi)) - 3*sqrt(2)*pi*b*sqrt(d
/pi)*fresnel_sin(sqrt(2)*x*sqrt(d/pi))*sin(c) - 6*b*d*x*cos(d*x^2 + c))/d^2

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Sympy [B]  time = 2.9315, size = 223, normalized size = 2.19 \begin{align*} \frac{a x^{3}}{3} - \frac{b d^{\frac{3}{2}} x^{5} \sqrt{\frac{1}{d}} \cos{\left (c \right )} \Gamma \left (\frac{3}{4}\right ) \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle |{- \frac{d^{2} x^{4}}{4}} \right )}}{8 \Gamma \left (\frac{7}{4}\right ) \Gamma \left (\frac{9}{4}\right )} - \frac{b \sqrt{d} x^{3} \sqrt{\frac{1}{d}} \sin{\left (c \right )} \Gamma \left (\frac{1}{4}\right ) \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} \\ \frac{1}{2}, \frac{5}{4}, \frac{7}{4} \end{matrix}\middle |{- \frac{d^{2} x^{4}}{4}} \right )}}{8 \Gamma \left (\frac{5}{4}\right ) \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{2} \sqrt{\pi } b x^{2} \sqrt{\frac{1}{d}} \sin{\left (c \right )} C\left (\frac{\sqrt{2} \sqrt{d} x}{\sqrt{\pi }}\right )}{2} + \frac{\sqrt{2} \sqrt{\pi } b x^{2} \sqrt{\frac{1}{d}} \cos{\left (c \right )} S\left (\frac{\sqrt{2} \sqrt{d} x}{\sqrt{\pi }}\right )}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 - b*d**(3/2)*x**5*sqrt(1/d)*cos(c)*gamma(3/4)*gamma(5/4)*hyper((3/4, 5/4), (3/2, 7/4, 9/4), -d**2*x**
4/4)/(8*gamma(7/4)*gamma(9/4)) - b*sqrt(d)*x**3*sqrt(1/d)*sin(c)*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (1/2,
5/4, 7/4), -d**2*x**4/4)/(8*gamma(5/4)*gamma(7/4)) + sqrt(2)*sqrt(pi)*b*x**2*sqrt(1/d)*sin(c)*fresnelc(sqrt(2
)*sqrt(d)*x/sqrt(pi))/2 + sqrt(2)*sqrt(pi)*b*x**2*sqrt(1/d)*cos(c)*fresnels(sqrt(2)*sqrt(d)*x/sqrt(pi))/2

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Giac [C]  time = 1.15799, size = 196, normalized size = 1.92 \begin{align*} \frac{1}{3} \, a x^{3} - \frac{b x e^{\left (i \, d x^{2} + i \, c\right )}}{4 \, d} - \frac{b x e^{\left (-i \, d x^{2} - i \, c\right )}}{4 \, d} - \frac{\sqrt{2} \sqrt{\pi } b \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{2} x{\left (-\frac{i \, d}{{\left | d \right |}} + 1\right )} \sqrt{{\left | d \right |}}\right ) e^{\left (i \, c\right )}}{8 \, d{\left (-\frac{i \, d}{{\left | d \right |}} + 1\right )} \sqrt{{\left | d \right |}}} - \frac{\sqrt{2} \sqrt{\pi } b \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{2} x{\left (\frac{i \, d}{{\left | d \right |}} + 1\right )} \sqrt{{\left | d \right |}}\right ) e^{\left (-i \, c\right )}}{8 \, d{\left (\frac{i \, d}{{\left | d \right |}} + 1\right )} \sqrt{{\left | d \right |}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 - 1/4*b*x*e^(I*d*x^2 + I*c)/d - 1/4*b*x*e^(-I*d*x^2 - I*c)/d - 1/8*sqrt(2)*sqrt(pi)*b*erf(-1/2*sqrt(
2)*x*(-I*d/abs(d) + 1)*sqrt(abs(d)))*e^(I*c)/(d*(-I*d/abs(d) + 1)*sqrt(abs(d))) - 1/8*sqrt(2)*sqrt(pi)*b*erf(-
1/2*sqrt(2)*x*(I*d/abs(d) + 1)*sqrt(abs(d)))*e^(-I*c)/(d*(I*d/abs(d) + 1)*sqrt(abs(d)))