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

Optimal. Leaf size=53 \[ -\frac{a \cos (c+d x)}{d}+\frac{2 b x \sin (c+d x)}{d^2}+\frac{2 b \cos (c+d x)}{d^3}-\frac{b x^2 \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0570932, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3329, 2638, 3296} \[ -\frac{a \cos (c+d x)}{d}+\frac{2 b x \sin (c+d x)}{d^2}+\frac{2 b \cos (c+d x)}{d^3}-\frac{b x^2 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)*Sin[c + d*x],x]

[Out]

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

Rule 3329

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegrand[Sin[c + d*x], (a
+ b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]

Rule 2638

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0829764, size = 41, normalized size = 0.77 \[ \frac{2 b d x \sin (c+d x)-\left (a d^2+b \left (d^2 x^2-2\right )\right ) \cos (c+d x)}{d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)*Sin[c + d*x],x]

[Out]

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

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Maple [A]  time = 0.007, size = 99, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({\frac{b \left ( - \left ( dx+c \right ) ^{2}\cos \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) +2\, \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) }{{d}^{2}}}-2\,{\frac{cb \left ( \sin \left ( dx+c \right ) - \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) }{{d}^{2}}}-\cos \left ( dx+c \right ) a-{\frac{{c}^{2}b\cos \left ( dx+c \right ) }{{d}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00949, size = 123, normalized size = 2.32 \begin{align*} -\frac{a \cos \left (d x + c\right ) + \frac{b c^{2} \cos \left (d x + c\right )}{d^{2}} - \frac{2 \,{\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b c}{d^{2}} + \frac{{\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \,{\left (d x + c\right )} \sin \left (d x + c\right )\right )} b}{d^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.36923, size = 93, normalized size = 1.75 \begin{align*} \frac{2 \, b d x \sin \left (d x + c\right ) -{\left (b d^{2} x^{2} + a d^{2} - 2 \, b\right )} \cos \left (d x + c\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.699871, size = 65, normalized size = 1.23 \begin{align*} \begin{cases} - \frac{a \cos{\left (c + d x \right )}}{d} - \frac{b x^{2} \cos{\left (c + d x \right )}}{d} + \frac{2 b x \sin{\left (c + d x \right )}}{d^{2}} + \frac{2 b \cos{\left (c + d x \right )}}{d^{3}} & \text{for}\: d \neq 0 \\\left (a x + \frac{b x^{3}}{3}\right ) \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10924, size = 57, normalized size = 1.08 \begin{align*} \frac{2 \, b x \sin \left (d x + c\right )}{d^{2}} - \frac{{\left (b d^{2} x^{2} + a d^{2} - 2 \, b\right )} \cos \left (d x + c\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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