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

Optimal. Leaf size=25 \[ \frac {a x^2}{2}-\frac {b \cos \left (c+d x^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {14, 3379, 2638} \[ \frac {a x^2}{2}-\frac {b \cos \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sin[c + d*x^2]),x]

[Out]

(a*x^2)/2 - (b*Cos[c + d*x^2])/(2*d)

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 2638

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

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]))

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 41, normalized size = 1.64 \[ \frac {a x^2}{2}+\frac {b \sin (c) \sin \left (d x^2\right )}{2 d}-\frac {b \cos (c) \cos \left (d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*Sin[c + d*x^2]),x]

[Out]

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

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fricas [A]  time = 0.76, size = 23, normalized size = 0.92 \[ \frac {a d x^{2} - b \cos \left (d x^{2} + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*d*x^2 - b*cos(d*x^2 + c))/d

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giac [A]  time = 0.71, size = 26, normalized size = 1.04 \[ \frac {{\left (d x^{2} + c\right )} a - b \cos \left (d x^{2} + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 27, normalized size = 1.08 \[ \frac {a \left (d \,x^{2}+c \right )-b \cos \left (d \,x^{2}+c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 21, normalized size = 0.84 \[ \frac {1}{2} \, a x^{2} - \frac {b \cos \left (d x^{2} + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 - 1/2*b*cos(d*x^2 + c)/d

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mupad [B]  time = 4.61, size = 21, normalized size = 0.84 \[ \frac {a\,x^2}{2}-\frac {b\,\cos \left (d\,x^2+c\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*sin(c + d*x^2)),x)

[Out]

(a*x^2)/2 - (b*cos(c + d*x^2))/(2*d)

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sympy [A]  time = 0.24, size = 31, normalized size = 1.24 \[ \begin {cases} \frac {a x^{2}}{2} - \frac {b \cos {\left (c + d x^{2} \right )}}{2 d} & \text {for}\: d \neq 0 \\\frac {x^{2} \left (a + b \sin {\relax (c )}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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