3.57 \(\int x^5 (a+b \sin (c+d x^3)) \, dx\)

Optimal. Leaf size=44 \[ \frac{a x^6}{6}+\frac{b \sin \left (c+d x^3\right )}{3 d^2}-\frac{b x^3 \cos \left (c+d x^3\right )}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0519132, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {14, 3379, 3296, 2637} \[ \frac{a x^6}{6}+\frac{b \sin \left (c+d x^3\right )}{3 d^2}-\frac{b x^3 \cos \left (c+d x^3\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sin[c + d*x^3]),x]

[Out]

(a*x^6)/6 - (b*x^3*Cos[c + d*x^3])/(3*d) + (b*Sin[c + d*x^3])/(3*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 2637

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

Rubi steps

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

Mathematica [A]  time = 0.0084399, size = 44, normalized size = 1. \[ \frac{a x^6}{6}+\frac{b \sin \left (c+d x^3\right )}{3 d^2}-\frac{b x^3 \cos \left (c+d x^3\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*Sin[c + d*x^3]),x]

[Out]

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

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Maple [A]  time = 0.019, size = 73, normalized size = 1.7 \begin{align*}{\frac{a{x}^{6}}{6}}+{b \left ( -{\frac{{x}^{3}}{3\,d}}+{\frac{2}{3\,{d}^{2}}\tan \left ({\frac{d{x}^{3}}{2}}+{\frac{c}{2}} \right ) }+{\frac{{x}^{3}}{3\,d} \left ( \tan \left ({\frac{d{x}^{3}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{d{x}^{3}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*sin(d*x^3+c)),x)

[Out]

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

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Maxima [A]  time = 0.97892, size = 50, normalized size = 1.14 \begin{align*} \frac{1}{6} \, a x^{6} - \frac{{\left (d x^{3} \cos \left (d x^{3} + c\right ) - \sin \left (d x^{3} + c\right )\right )} b}{3 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^3+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75342, size = 93, normalized size = 2.11 \begin{align*} \frac{a d^{2} x^{6} - 2 \, b d x^{3} \cos \left (d x^{3} + c\right ) + 2 \, b \sin \left (d x^{3} + c\right )}{6 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^3+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.5573, size = 49, normalized size = 1.11 \begin{align*} \begin{cases} \frac{a x^{6}}{6} - \frac{b x^{3} \cos{\left (c + d x^{3} \right )}}{3 d} + \frac{b \sin{\left (c + d x^{3} \right )}}{3 d^{2}} & \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**3+c)),x)

[Out]

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

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Giac [A]  time = 1.09509, size = 82, normalized size = 1.86 \begin{align*} \frac{\frac{{\left ({\left (d x^{3} + c\right )}^{2} - 2 \,{\left (d x^{3} + c\right )} c\right )} a}{d} - \frac{2 \,{\left (d x^{3} \cos \left (d x^{3} + c\right ) - \sin \left (d x^{3} + c\right )\right )} b}{d}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^3+c)),x, algorithm="giac")

[Out]

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