3.1.43 \(\int (a+b x^2) \sin (c+d x) \, dx\) [43]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 = {3410, 2718, 3377} \begin {gather*} -\frac {a \cos (c+d x)}{d}+\frac {2 b \cos (c+d x)}{d^3}+\frac {2 b x \sin (c+d x)}{d^2}-\frac {b x^2 \cos (c+d x)}{d} \end {gather*}

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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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 3410

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]

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*}

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Mathematica [A]
time = 0.06, size = 41, normalized size = 0.77 \begin {gather*} \frac {-\left (\left (a d^2+b \left (-2+d^2 x^2\right )\right ) \cos (c+d x)\right )+2 b d x \sin (c+d x)}{d^3} \end {gather*}

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 = 0.04, size = 99, normalized size = 1.87

method	result	size
risch	\(-\frac {\left (d^{2} x^{2} b +d^{2} a -2 b \right ) \cos \left (d x +c \right )}{d^{3}}+\frac {2 b x \sin \left (d x +c \right )}{d^{2}}\)	\(43\)
norman	\(\frac {\frac {b \,x^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 d^{2} a -4 b}{d^{3}}-\frac {b \,x^{2}}{d}+\frac {4 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d^{2}}}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\)	\(77\)
derivativedivides	\(\frac {-a \cos \left (d x +c \right )-\frac {b \,c^{2} \cos \left (d x +c \right )}{d^{2}}-\frac {2 b c \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}+\frac {b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}}{d}\)	\(99\)
default	\(\frac {-a \cos \left (d x +c \right )-\frac {b \,c^{2} \cos \left (d x +c \right )}{d^{2}}-\frac {2 b c \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}+\frac {b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}}{d}\)	\(99\)
meijerg	\(\frac {4 b \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {x \left (d^{2}\right )^{\frac {3}{2}} \cos \left (d x \right )}{2 \sqrt {\pi }\, d^{2}}-\frac {\left (d^{2}\right )^{\frac {3}{2}} \left (-\frac {3 d^{2} x^{2}}{2}+3\right ) \sin \left (d x \right )}{6 \sqrt {\pi }\, d^{3}}\right )}{d^{2} \sqrt {d^{2}}}+\frac {4 b \sqrt {\pi }\, \cos \left (c \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {d^{2} x^{2}}{2}+1\right ) \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {d x \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}+\frac {a \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {a \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 91, normalized size = 1.72 \begin {gather*} -\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 {gather*}

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 = 0.35, size = 41, normalized size = 0.77 \begin {gather*} \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 {gather*}

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 = 0.14, size = 65, normalized size = 1.23 \begin {gather*} \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 {gather*}

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 = 5.31, size = 42, normalized size = 0.79 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 4.69, size = 49, normalized size = 0.92 \begin {gather*} \frac {\cos \left (c+d\,x\right )\,\left (2\,b-a\,d^2\right )}{d^3}+\frac {2\,b\,x\,\sin \left (c+d\,x\right )}{d^2}-\frac {b\,x^2\,\cos \left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)*(a + b*x^2),x)

[Out]

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

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