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

Optimal. Leaf size=198 \[ \frac{1}{6} x^3 \left (2 a^2+b^2\right )+\frac{\sqrt{\frac{\pi }{2}} a b \cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )}{d^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} a b \sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )}{d^{3/2}}-\frac{a b x \cos \left (c+d x^2\right )}{d}+\frac{\sqrt{\pi } b^2 \sin (2 c) \text{FresnelC}\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )}{16 d^{3/2}}+\frac{\sqrt{\pi } b^2 \cos (2 c) S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )}{16 d^{3/2}}-\frac{b^2 x \sin \left (2 c+2 d x^2\right )}{8 d} \]

[Out]

((2*a^2 + b^2)*x^3)/6 - (a*b*x*Cos[c + d*x^2])/d + (a*b*Sqrt[Pi/2]*Cos[c]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x])/d^(3
/2) + (b^2*Sqrt[Pi]*Cos[2*c]*FresnelS[(2*Sqrt[d]*x)/Sqrt[Pi]])/(16*d^(3/2)) - (a*b*Sqrt[Pi/2]*FresnelS[Sqrt[d]
*Sqrt[2/Pi]*x]*Sin[c])/d^(3/2) + (b^2*Sqrt[Pi]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c])/(16*d^(3/2)) - (b^2*
x*Sin[2*c + 2*d*x^2])/(8*d)

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Rubi [A]  time = 0.15663, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3403, 6, 3386, 3353, 3352, 3351, 3385, 3354} \[ \frac{1}{6} x^3 \left (2 a^2+b^2\right )+\frac{\sqrt{\frac{\pi }{2}} a b \cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )}{d^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} a b \sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )}{d^{3/2}}-\frac{a b x \cos \left (c+d x^2\right )}{d}+\frac{\sqrt{\pi } b^2 \sin (2 c) \text{FresnelC}\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )}{16 d^{3/2}}+\frac{\sqrt{\pi } b^2 \cos (2 c) S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )}{16 d^{3/2}}-\frac{b^2 x \sin \left (2 c+2 d x^2\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*Sin[c + d*x^2])^2,x]

[Out]

((2*a^2 + b^2)*x^3)/6 - (a*b*x*Cos[c + d*x^2])/d + (a*b*Sqrt[Pi/2]*Cos[c]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x])/d^(3
/2) + (b^2*Sqrt[Pi]*Cos[2*c]*FresnelS[(2*Sqrt[d]*x)/Sqrt[Pi]])/(16*d^(3/2)) - (a*b*Sqrt[Pi/2]*FresnelS[Sqrt[d]
*Sqrt[2/Pi]*x]*Sin[c])/d^(3/2) + (b^2*Sqrt[Pi]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c])/(16*d^(3/2)) - (b^2*
x*Sin[2*c + 2*d*x^2])/(8*d)

Rule 3403

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 3353

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

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]

Rubi steps

\begin{align*} \int x^2 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx &=\int \left (a^2 x^2+\frac{b^2 x^2}{2}-\frac{1}{2} b^2 x^2 \cos \left (2 c+2 d x^2\right )+2 a b x^2 \sin \left (c+d x^2\right )\right ) \, dx\\ &=\int \left (\left (a^2+\frac{b^2}{2}\right ) x^2-\frac{1}{2} b^2 x^2 \cos \left (2 c+2 d x^2\right )+2 a b x^2 \sin \left (c+d x^2\right )\right ) \, dx\\ &=\frac{1}{6} \left (2 a^2+b^2\right ) x^3+(2 a b) \int x^2 \sin \left (c+d x^2\right ) \, dx-\frac{1}{2} b^2 \int x^2 \cos \left (2 c+2 d x^2\right ) \, dx\\ &=\frac{1}{6} \left (2 a^2+b^2\right ) x^3-\frac{a b x \cos \left (c+d x^2\right )}{d}-\frac{b^2 x \sin \left (2 c+2 d x^2\right )}{8 d}+\frac{(a b) \int \cos \left (c+d x^2\right ) \, dx}{d}+\frac{b^2 \int \sin \left (2 c+2 d x^2\right ) \, dx}{8 d}\\ &=\frac{1}{6} \left (2 a^2+b^2\right ) x^3-\frac{a b x \cos \left (c+d x^2\right )}{d}-\frac{b^2 x \sin \left (2 c+2 d x^2\right )}{8 d}+\frac{(a b \cos (c)) \int \cos \left (d x^2\right ) \, dx}{d}+\frac{\left (b^2 \cos (2 c)\right ) \int \sin \left (2 d x^2\right ) \, dx}{8 d}-\frac{(a b \sin (c)) \int \sin \left (d x^2\right ) \, dx}{d}+\frac{\left (b^2 \sin (2 c)\right ) \int \cos \left (2 d x^2\right ) \, dx}{8 d}\\ &=\frac{1}{6} \left (2 a^2+b^2\right ) x^3-\frac{a b x \cos \left (c+d x^2\right )}{d}+\frac{a b \sqrt{\frac{\pi }{2}} \cos (c) C\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )}{d^{3/2}}+\frac{b^2 \sqrt{\pi } \cos (2 c) S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )}{16 d^{3/2}}-\frac{a b \sqrt{\frac{\pi }{2}} S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right ) \sin (c)}{d^{3/2}}+\frac{b^2 \sqrt{\pi } C\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right ) \sin (2 c)}{16 d^{3/2}}-\frac{b^2 x \sin \left (2 c+2 d x^2\right )}{8 d}\\ \end{align*}

Mathematica [A]  time = 0.53196, size = 191, normalized size = 0.96 \[ \frac{16 a^2 d^{3/2} x^3+24 \sqrt{2 \pi } a b \cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-24 \sqrt{2 \pi } a b \sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-48 a b \sqrt{d} x \cos \left (c+d x^2\right )+3 \sqrt{\pi } b^2 \sin (2 c) \text{FresnelC}\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )+3 \sqrt{\pi } b^2 \cos (2 c) S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )-6 b^2 \sqrt{d} x \sin \left (2 \left (c+d x^2\right )\right )+8 b^2 d^{3/2} x^3}{48 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2*d^(3/2)*x^3 + 8*b^2*d^(3/2)*x^3 - 48*a*b*Sqrt[d]*x*Cos[c + d*x^2] + 24*a*b*Sqrt[2*Pi]*Cos[c]*FresnelC[
Sqrt[d]*Sqrt[2/Pi]*x] + 3*b^2*Sqrt[Pi]*Cos[2*c]*FresnelS[(2*Sqrt[d]*x)/Sqrt[Pi]] - 24*a*b*Sqrt[2*Pi]*FresnelS[
Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c] + 3*b^2*Sqrt[Pi]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c] - 6*b^2*Sqrt[d]*x*Sin[
2*(c + d*x^2)])/(48*d^(3/2))

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Maple [A]  time = 0.016, size = 142, normalized size = 0.7 \begin{align*}{\frac{{x}^{3}{a}^{2}}{3}}+{\frac{{x}^{3}{b}^{2}}{6}}-{\frac{{b}^{2}}{2} \left ({\frac{x\sin \left ( 2\,d{x}^{2}+2\,c \right ) }{4\,d}}-{\frac{\sqrt{\pi }}{8} \left ( \cos \left ( 2\,c \right ){\it FresnelS} \left ( 2\,{\frac{x\sqrt{d}}{\sqrt{\pi }}} \right ) +\sin \left ( 2\,c \right ){\it FresnelC} \left ( 2\,{\frac{x\sqrt{d}}{\sqrt{\pi }}} \right ) \right ){d}^{-{\frac{3}{2}}}} \right ) }+2\,ab \left ( -1/2\,{\frac{x\cos \left ( d{x}^{2}+c \right ) }{d}}+1/4\,{\frac{\sqrt{2}\sqrt{\pi }}{{d}^{3/2}} \left ( \cos \left ( c \right ){\it FresnelC} \left ({\frac{x\sqrt{d}\sqrt{2}}{\sqrt{\pi }}} \right ) -\sin \left ( c \right ){\it FresnelS} \left ({\frac{x\sqrt{d}\sqrt{2}}{\sqrt{\pi }}} \right ) \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*a^2+1/6*x^3*b^2-1/2*b^2*(1/4/d*x*sin(2*d*x^2+2*c)-1/8/d^(3/2)*Pi^(1/2)*(cos(2*c)*FresnelS(2*x*d^(1/2)/
Pi^(1/2))+sin(2*c)*FresnelC(2*x*d^(1/2)/Pi^(1/2))))+2*a*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))-sin(c)*FresnelS(x*d^(1/2)*2^(1/2)/Pi^(1/2))))

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Maxima [C]  time = 1.81792, size = 743, normalized size = 3.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*x^3 - 1/8*(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*
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*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*arc
tan2(0, d)))*sin(c))*erf(sqrt(-I*d)*x))*sqrt(abs(d)))*a*b/(d*abs(d)) + 1/384*(64*d*x^3*abs(d) - 48*x*abs(d)*si
n(2*d*x^2 + 2*c) - sqrt(2)*sqrt(pi)*(((-3*I*cos(1/4*pi + 1/2*arctan2(0, d)) - 3*I*cos(-1/4*pi + 1/2*arctan2(0,
d)) - 3*sin(1/4*pi + 1/2*arctan2(0, d)) + 3*sin(-1/4*pi + 1/2*arctan2(0, d)))*cos(2*c) - (3*cos(1/4*pi + 1/2*
arctan2(0, d)) + 3*cos(-1/4*pi + 1/2*arctan2(0, d)) - 3*I*sin(1/4*pi + 1/2*arctan2(0, d)) + 3*I*sin(-1/4*pi +
1/2*arctan2(0, d)))*sin(2*c))*erf(sqrt(2*I*d)*x) + ((3*I*cos(1/4*pi + 1/2*arctan2(0, d)) + 3*I*cos(-1/4*pi + 1
/2*arctan2(0, d)) - 3*sin(1/4*pi + 1/2*arctan2(0, d)) + 3*sin(-1/4*pi + 1/2*arctan2(0, d)))*cos(2*c) - (3*cos(
1/4*pi + 1/2*arctan2(0, d)) + 3*cos(-1/4*pi + 1/2*arctan2(0, d)) + 3*I*sin(1/4*pi + 1/2*arctan2(0, d)) - 3*I*s
in(-1/4*pi + 1/2*arctan2(0, d)))*sin(2*c))*erf(sqrt(-2*I*d)*x))*sqrt(abs(d)))*b^2/(d*abs(d))

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Fricas [A]  time = 2.30346, size = 498, normalized size = 2.52 \begin{align*} \frac{8 \,{\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{3} + 24 \, \sqrt{2} \pi a b \sqrt{\frac{d}{\pi }} \cos \left (c\right ) \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) - 12 \, b^{2} d x \cos \left (d x^{2} + c\right ) \sin \left (d x^{2} + c\right ) - 24 \, \sqrt{2} \pi a b \sqrt{\frac{d}{\pi }} \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) \sin \left (c\right ) + 3 \, \pi b^{2} \sqrt{\frac{d}{\pi }} \cos \left (2 \, c\right ) \operatorname{S}\left (2 \, x \sqrt{\frac{d}{\pi }}\right ) + 3 \, \pi b^{2} \sqrt{\frac{d}{\pi }} \operatorname{C}\left (2 \, x \sqrt{\frac{d}{\pi }}\right ) \sin \left (2 \, c\right ) - 48 \, a b d x \cos \left (d x^{2} + c\right )}{48 \, 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))^2,x, algorithm="fricas")

[Out]

1/48*(8*(2*a^2 + b^2)*d^2*x^3 + 24*sqrt(2)*pi*a*b*sqrt(d/pi)*cos(c)*fresnel_cos(sqrt(2)*x*sqrt(d/pi)) - 12*b^2
*d*x*cos(d*x^2 + c)*sin(d*x^2 + c) - 24*sqrt(2)*pi*a*b*sqrt(d/pi)*fresnel_sin(sqrt(2)*x*sqrt(d/pi))*sin(c) + 3
*pi*b^2*sqrt(d/pi)*cos(2*c)*fresnel_sin(2*x*sqrt(d/pi)) + 3*pi*b^2*sqrt(d/pi)*fresnel_cos(2*x*sqrt(d/pi))*sin(
2*c) - 48*a*b*d*x*cos(d*x^2 + c))/d^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (a + b \sin{\left (c + d x^{2} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.52749, size = 382, normalized size = 1.93 \begin{align*} \frac{1}{3} \, a^{2} x^{3} + \frac{1}{6} \, b^{2} x^{3} + \frac{i \, b^{2} x e^{\left (2 i \, d x^{2} + 2 i \, c\right )}}{16 \, d} - \frac{a b x e^{\left (i \, d x^{2} + i \, c\right )}}{2 \, d} - \frac{a b x e^{\left (-i \, d x^{2} - i \, c\right )}}{2 \, d} - \frac{i \, b^{2} x e^{\left (-2 i \, d x^{2} - 2 i \, c\right )}}{16 \, d} - \frac{\sqrt{2} \sqrt{\pi } a 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 )}}{4 \, d{\left (-\frac{i \, d}{{\left | d \right |}} + 1\right )} \sqrt{{\left | d \right |}}} - \frac{\sqrt{2} \sqrt{\pi } a 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 )}}{4 \, d{\left (\frac{i \, d}{{\left | d \right |}} + 1\right )} \sqrt{{\left | d \right |}}} + \frac{i \, \sqrt{\pi } b^{2} \operatorname{erf}\left (-\sqrt{d} x{\left (-\frac{i \, d}{{\left | d \right |}} + 1\right )}\right ) e^{\left (2 i \, c\right )}}{32 \, d^{\frac{3}{2}}{\left (-\frac{i \, d}{{\left | d \right |}} + 1\right )}} - \frac{i \, \sqrt{\pi } b^{2} \operatorname{erf}\left (-\sqrt{d} x{\left (\frac{i \, d}{{\left | d \right |}} + 1\right )}\right ) e^{\left (-2 i \, c\right )}}{32 \, d^{\frac{3}{2}}{\left (\frac{i \, d}{{\left | d \right |}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*x^3 + 1/6*b^2*x^3 + 1/16*I*b^2*x*e^(2*I*d*x^2 + 2*I*c)/d - 1/2*a*b*x*e^(I*d*x^2 + I*c)/d - 1/2*a*b*x*e
^(-I*d*x^2 - I*c)/d - 1/16*I*b^2*x*e^(-2*I*d*x^2 - 2*I*c)/d - 1/4*sqrt(2)*sqrt(pi)*a*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))) - 1/4*sqrt(2)*sqrt(pi)*a*b*erf(-1/2*sqr
t(2)*x*(I*d/abs(d) + 1)*sqrt(abs(d)))*e^(-I*c)/(d*(I*d/abs(d) + 1)*sqrt(abs(d))) + 1/32*I*sqrt(pi)*b^2*erf(-sq
rt(d)*x*(-I*d/abs(d) + 1))*e^(2*I*c)/(d^(3/2)*(-I*d/abs(d) + 1)) - 1/32*I*sqrt(pi)*b^2*erf(-sqrt(d)*x*(I*d/abs
(d) + 1))*e^(-2*I*c)/(d^(3/2)*(I*d/abs(d) + 1))