3.1.57 \(\int x^5 (a+b \sin (c+d x^3)) \, dx\) [57]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {14, 3460, 3377, 2717} \begin {gather*} \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} \end {gather*}

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 2717

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

Rule 3377

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 3460

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]))

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 \text {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 \text {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*}

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Mathematica [A]
time = 0.01, size = 44, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 = 0.04, size = 39, normalized size = 0.89

method	result	size
risch	\(\frac {a \,x^{6}}{6}-\frac {b \,x^{3} \cos \left (d \,x^{3}+c \right )}{3 d}+\frac {b \sin \left (d \,x^{3}+c \right )}{3 d^{2}}\)	\(39\)
norman	\(\frac {\frac {a \,x^{6}}{6}+\frac {a \,x^{6} \left (\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}{6}+\frac {2 b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d^{2}}-\frac {b \,x^{3}}{3 d}+\frac {b \,x^{3} \left (\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}{3 d}}{1+\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 37, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 0.36, size = 40, normalized size = 0.91 \begin {gather*} \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 {gather*}

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 = 0.39, size = 49, normalized size = 1.11 \begin {gather*} \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 {gather*}

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 = 4.90, size = 75, normalized size = 1.70 \begin {gather*} \frac {{\left (d x^{3} + c\right )}^{2} a - 2 \, {\left (d x^{3} + c\right )} b \cos \left (d x^{3} + c\right ) + 2 \, b \sin \left (d x^{3} + c\right )}{6 \, d^{2}} - \frac {{\left (d x^{3} + c\right )} a c - b c \cos \left (d x^{3} + c\right )}{3 \, d^{2}} \end {gather*}

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

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Mupad [B]
time = 0.17, size = 38, normalized size = 0.86 \begin {gather*} \frac {a\,x^6}{6}+\frac {\frac {b\,\sin \left (d\,x^3+c\right )}{3}-\frac {b\,d\,x^3\,\cos \left (d\,x^3+c\right )}{3}}{d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*sin(c + d*x^3)),x)

[Out]

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

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