3.12 \(\int x^5 (a+b \sin (c+d x^2))^2 \, dx\)

Optimal. Leaf size=163 \[ \frac{a^2 x^6}{6}+\frac{2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac{2 a b \cos \left (c+d x^2\right )}{d^3}-\frac{a b x^4 \cos \left (c+d x^2\right )}{d}+\frac{b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}+\frac{b^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{8 d^3}-\frac{b^2 x^4 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{4 d}-\frac{b^2 x^2}{8 d^2}+\frac{b^2 x^6}{12} \]

[Out]

-(b^2*x^2)/(8*d^2) + (a^2*x^6)/6 + (b^2*x^6)/12 + (2*a*b*Cos[c + d*x^2])/d^3 - (a*b*x^4*Cos[c + d*x^2])/d + (2
*a*b*x^2*Sin[c + d*x^2])/d^2 + (b^2*Cos[c + d*x^2]*Sin[c + d*x^2])/(8*d^3) - (b^2*x^4*Cos[c + d*x^2]*Sin[c + d
*x^2])/(4*d) + (b^2*x^2*Sin[c + d*x^2]^2)/(4*d^2)

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Rubi [A]  time = 0.246997, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3379, 3317, 3296, 2638, 3311, 30, 2635, 8} \[ \frac{a^2 x^6}{6}+\frac{2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac{2 a b \cos \left (c+d x^2\right )}{d^3}-\frac{a b x^4 \cos \left (c+d x^2\right )}{d}+\frac{b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}+\frac{b^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{8 d^3}-\frac{b^2 x^4 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{4 d}-\frac{b^2 x^2}{8 d^2}+\frac{b^2 x^6}{12} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sin[c + d*x^2])^2,x]

[Out]

-(b^2*x^2)/(8*d^2) + (a^2*x^6)/6 + (b^2*x^6)/12 + (2*a*b*Cos[c + d*x^2])/d^3 - (a*b*x^4*Cos[c + d*x^2])/d + (2
*a*b*x^2*Sin[c + d*x^2])/d^2 + (b^2*Cos[c + d*x^2]*Sin[c + d*x^2])/(8*d^3) - (b^2*x^4*Cos[c + d*x^2]*Sin[c + d
*x^2])/(4*d) + (b^2*x^2*Sin[c + d*x^2]^2)/(4*d^2)

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 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

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 2638

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

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 30

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 (a+b \sin (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \sin (c+d x)+b^2 x^2 \sin ^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 x^6}{6}+(a b) \operatorname{Subst}\left (\int x^2 \sin (c+d x) \, dx,x,x^2\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int x^2 \sin ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac{a^2 x^6}{6}-\frac{a b x^4 \cos \left (c+d x^2\right )}{d}-\frac{b^2 x^4 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac{b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}+\frac{1}{4} b^2 \operatorname{Subst}\left (\int x^2 \, dx,x,x^2\right )-\frac{b^2 \operatorname{Subst}\left (\int \sin ^2(c+d x) \, dx,x,x^2\right )}{4 d^2}+\frac{(2 a b) \operatorname{Subst}\left (\int x \cos (c+d x) \, dx,x,x^2\right )}{d}\\ &=\frac{a^2 x^6}{6}+\frac{b^2 x^6}{12}-\frac{a b x^4 \cos \left (c+d x^2\right )}{d}+\frac{2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac{b^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{8 d^3}-\frac{b^2 x^4 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac{b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}-\frac{(2 a b) \operatorname{Subst}\left (\int \sin (c+d x) \, dx,x,x^2\right )}{d^2}-\frac{b^2 \operatorname{Subst}\left (\int 1 \, dx,x,x^2\right )}{8 d^2}\\ &=-\frac{b^2 x^2}{8 d^2}+\frac{a^2 x^6}{6}+\frac{b^2 x^6}{12}+\frac{2 a b \cos \left (c+d x^2\right )}{d^3}-\frac{a b x^4 \cos \left (c+d x^2\right )}{d}+\frac{2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac{b^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{8 d^3}-\frac{b^2 x^4 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac{b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.389886, size = 122, normalized size = 0.75 \[ \frac{8 a^2 d^3 x^6-48 a b \left (d^2 x^4-2\right ) \cos \left (c+d x^2\right )+96 a b d x^2 \sin \left (c+d x^2\right )-6 b^2 d^2 x^4 \sin \left (2 \left (c+d x^2\right )\right )+3 b^2 \sin \left (2 \left (c+d x^2\right )\right )-6 b^2 d x^2 \cos \left (2 \left (c+d x^2\right )\right )+4 b^2 d^3 x^6}{48 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*Sin[c + d*x^2])^2,x]

[Out]

(8*a^2*d^3*x^6 + 4*b^2*d^3*x^6 - 48*a*b*(-2 + d^2*x^4)*Cos[c + d*x^2] - 6*b^2*d*x^2*Cos[2*(c + d*x^2)] + 96*a*
b*d*x^2*Sin[c + d*x^2] + 3*b^2*Sin[2*(c + d*x^2)] - 6*b^2*d^2*x^4*Sin[2*(c + d*x^2)])/(48*d^3)

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Maple [A]  time = 0.016, size = 140, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}{x}^{6}}{6}}+{\frac{{b}^{2}{x}^{6}}{12}}-{\frac{{b}^{2}}{2} \left ({\frac{{x}^{4}\sin \left ( 2\,d{x}^{2}+2\,c \right ) }{4\,d}}-{\frac{1}{d} \left ( -{\frac{{x}^{2}\cos \left ( 2\,d{x}^{2}+2\,c \right ) }{4\,d}}+{\frac{\sin \left ( 2\,d{x}^{2}+2\,c \right ) }{8\,{d}^{2}}} \right ) } \right ) }+2\,ab \left ( -1/2\,{\frac{{x}^{4}\cos \left ( d{x}^{2}+c \right ) }{d}}+2\,{\frac{1}{d} \left ( 1/2\,{\frac{{x}^{2}\sin \left ( d{x}^{2}+c \right ) }{d}}+1/2\,{\frac{\cos \left ( d{x}^{2}+c \right ) }{{d}^{2}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*sin(d*x^2+c))^2,x)

[Out]

1/6*a^2*x^6+1/12*b^2*x^6-1/2*b^2*(1/4/d*x^4*sin(2*d*x^2+2*c)-1/d*(-1/4/d*x^2*cos(2*d*x^2+2*c)+1/8/d^2*sin(2*d*
x^2+2*c)))+2*a*b*(-1/2/d*x^4*cos(d*x^2+c)+2/d*(1/2/d*x^2*sin(d*x^2+c)+1/2/d^2*cos(d*x^2+c)))

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Maxima [A]  time = 1.03411, size = 143, normalized size = 0.88 \begin{align*} \frac{1}{6} \, a^{2} x^{6} + \frac{{\left (2 \, d x^{2} \sin \left (d x^{2} + c\right ) -{\left (d^{2} x^{4} - 2\right )} \cos \left (d x^{2} + c\right )\right )} a b}{d^{3}} + \frac{{\left (4 \, d^{3} x^{6} - 6 \, d x^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 3 \,{\left (2 \, d^{2} x^{4} - 1\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2}}{48 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^2+c))^2,x, algorithm="maxima")

[Out]

1/6*a^2*x^6 + (2*d*x^2*sin(d*x^2 + c) - (d^2*x^4 - 2)*cos(d*x^2 + c))*a*b/d^3 + 1/48*(4*d^3*x^6 - 6*d*x^2*cos(
2*d*x^2 + 2*c) - 3*(2*d^2*x^4 - 1)*sin(2*d*x^2 + 2*c))*b^2/d^3

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Fricas [A]  time = 2.05386, size = 265, normalized size = 1.63 \begin{align*} \frac{2 \,{\left (2 \, a^{2} + b^{2}\right )} d^{3} x^{6} - 6 \, b^{2} d x^{2} \cos \left (d x^{2} + c\right )^{2} + 3 \, b^{2} d x^{2} - 24 \,{\left (a b d^{2} x^{4} - 2 \, a b\right )} \cos \left (d x^{2} + c\right ) + 3 \,{\left (16 \, a b d x^{2} -{\left (2 \, b^{2} d^{2} x^{4} - b^{2}\right )} \cos \left (d x^{2} + c\right )\right )} \sin \left (d x^{2} + c\right )}{24 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^2+c))^2,x, algorithm="fricas")

[Out]

1/24*(2*(2*a^2 + b^2)*d^3*x^6 - 6*b^2*d*x^2*cos(d*x^2 + c)^2 + 3*b^2*d*x^2 - 24*(a*b*d^2*x^4 - 2*a*b)*cos(d*x^
2 + c) + 3*(16*a*b*d*x^2 - (2*b^2*d^2*x^4 - b^2)*cos(d*x^2 + c))*sin(d*x^2 + c))/d^3

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Sympy [A]  time = 7.29998, size = 209, normalized size = 1.28 \begin{align*} \begin{cases} \frac{a^{2} x^{6}}{6} - \frac{a b x^{4} \cos{\left (c + d x^{2} \right )}}{d} + \frac{2 a b x^{2} \sin{\left (c + d x^{2} \right )}}{d^{2}} + \frac{2 a b \cos{\left (c + d x^{2} \right )}}{d^{3}} + \frac{b^{2} x^{6} \sin ^{2}{\left (c + d x^{2} \right )}}{12} + \frac{b^{2} x^{6} \cos ^{2}{\left (c + d x^{2} \right )}}{12} - \frac{b^{2} x^{4} \sin{\left (c + d x^{2} \right )} \cos{\left (c + d x^{2} \right )}}{4 d} + \frac{b^{2} x^{2} \sin ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} - \frac{b^{2} x^{2} \cos ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} + \frac{b^{2} \sin{\left (c + d x^{2} \right )} \cos{\left (c + d x^{2} \right )}}{8 d^{3}} & \text{for}\: d \neq 0 \\\frac{x^{6} \left (a + b \sin{\left (c \right )}\right )^{2}}{6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*sin(d*x**2+c))**2,x)

[Out]

Piecewise((a**2*x**6/6 - a*b*x**4*cos(c + d*x**2)/d + 2*a*b*x**2*sin(c + d*x**2)/d**2 + 2*a*b*cos(c + d*x**2)/
d**3 + b**2*x**6*sin(c + d*x**2)**2/12 + b**2*x**6*cos(c + d*x**2)**2/12 - b**2*x**4*sin(c + d*x**2)*cos(c + d
*x**2)/(4*d) + b**2*x**2*sin(c + d*x**2)**2/(8*d**2) - b**2*x**2*cos(c + d*x**2)**2/(8*d**2) + b**2*sin(c + d*
x**2)*cos(c + d*x**2)/(8*d**3), Ne(d, 0)), (x**6*(a + b*sin(c))**2/6, True))

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Giac [A]  time = 1.11926, size = 244, normalized size = 1.5 \begin{align*} \frac{8 \, a^{2} d x^{6} + 48 \,{\left (\frac{2 \, x^{2} \sin \left (d x^{2} + c\right )}{d} - \frac{{\left ({\left (d x^{2} + c\right )}^{2} - 2 \,{\left (d x^{2} + c\right )} c + c^{2} - 2\right )} \cos \left (d x^{2} + c\right )}{d^{2}}\right )} a b -{\left (\frac{6 \, x^{2} \cos \left (2 \, d x^{2} + 2 \, c\right )}{d} + \frac{3 \,{\left (2 \,{\left (d x^{2} + c\right )}^{2} - 4 \,{\left (d x^{2} + c\right )} c + 2 \, c^{2} - 1\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{d^{2}} - \frac{4 \,{\left ({\left (d x^{2} + c\right )}^{3} - 3 \,{\left (d x^{2} + c\right )}^{2} c + 3 \,{\left (d x^{2} + c\right )} c^{2}\right )}}{d^{2}}\right )} b^{2}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^2+c))^2,x, algorithm="giac")

[Out]

1/48*(8*a^2*d*x^6 + 48*(2*x^2*sin(d*x^2 + c)/d - ((d*x^2 + c)^2 - 2*(d*x^2 + c)*c + c^2 - 2)*cos(d*x^2 + c)/d^
2)*a*b - (6*x^2*cos(2*d*x^2 + 2*c)/d + 3*(2*(d*x^2 + c)^2 - 4*(d*x^2 + c)*c + 2*c^2 - 1)*sin(2*d*x^2 + 2*c)/d^
2 - 4*((d*x^2 + c)^3 - 3*(d*x^2 + c)^2*c + 3*(d*x^2 + c)*c^2)/d^2)*b^2)/d