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

Optimal. Leaf size=57 \[ \frac{a x^6}{6}+\frac{b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac{b \cos \left (c+d x^2\right )}{d^3}-\frac{b x^4 \cos \left (c+d x^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0730884, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {14, 3379, 3296, 2638} \[ \frac{a x^6}{6}+\frac{b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac{b \cos \left (c+d x^2\right )}{d^3}-\frac{b x^4 \cos \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sin[c + d*x^2]),x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0859434, size = 51, normalized size = 0.89 \[ \frac{a d^3 x^6-3 b \left (d^2 x^4-2\right ) \cos \left (c+d x^2\right )+6 b d x^2 \sin \left (c+d x^2\right )}{6 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 62, normalized size = 1.1 \begin{align*}{\frac{a{x}^{6}}{6}}+b \left ( -{\frac{{x}^{4}\cos \left ( d{x}^{2}+c \right ) }{2\,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)),x)

[Out]

1/6*a*x^6+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 = 0.980933, size = 63, normalized size = 1.11 \begin{align*} \frac{1}{6} \, a 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 )} b}{2 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.99113, size = 115, normalized size = 2.02 \begin{align*} \frac{a d^{3} x^{6} + 6 \, b d x^{2} \sin \left (d x^{2} + c\right ) - 3 \,{\left (b d^{2} x^{4} - 2 \, b\right )} \cos \left (d x^{2} + c\right )}{6 \, 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.95245, size = 65, normalized size = 1.14 \begin{align*} \begin{cases} \frac{a x^{6}}{6} - \frac{b x^{4} \cos{\left (c + d x^{2} \right )}}{2 d} + \frac{b x^{2} \sin{\left (c + d x^{2} \right )}}{d^{2}} + \frac{b \cos{\left (c + d x^{2} \right )}}{d^{3}} & \text{for}\: d \neq 0 \\\frac{x^{6} \left (a + b \sin{\left (c \right )}\right )}{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)),x)

[Out]

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

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Giac [A]  time = 1.17725, size = 93, normalized size = 1.63 \begin{align*} \frac{a d x^{6} + 3 \,{\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 )} b}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a*d*x^6 + 3*(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)*b)/
d