3.152 \(\int (c+d x)^2 (a+b \sin (e+f x)) \, dx\)

Optimal. Leaf size=68 \[ \frac{a (c+d x)^3}{3 d}+\frac{2 b d (c+d x) \sin (e+f x)}{f^2}-\frac{b (c+d x)^2 \cos (e+f x)}{f}+\frac{2 b d^2 \cos (e+f x)}{f^3} \]

[Out]

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

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Rubi [A]  time = 0.0856717, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3317, 3296, 2638} \[ \frac{a (c+d x)^3}{3 d}+\frac{2 b d (c+d x) \sin (e+f x)}{f^2}-\frac{b (c+d x)^2 \cos (e+f x)}{f}+\frac{2 b d^2 \cos (e+f x)}{f^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*(a + b*Sin[e + f*x]),x]

[Out]

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

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 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 2638

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

Rubi steps

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

Mathematica [A]  time = 0.313986, size = 84, normalized size = 1.24 \[ \frac{1}{3} a x \left (3 c^2+3 c d x+d^2 x^2\right )-\frac{b \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2-2\right )\right ) \cos (e+f x)}{f^3}+\frac{2 b d (c+d x) \sin (e+f x)}{f^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*(a + b*Sin[e + f*x]),x]

[Out]

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

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Maple [B]  time = 0.008, size = 241, normalized size = 3.5 \begin{align*}{\frac{1}{f} \left ({\frac{a{d}^{2} \left ( fx+e \right ) ^{3}}{3\,{f}^{2}}}+{\frac{acd \left ( fx+e \right ) ^{2}}{f}}-{\frac{a{d}^{2}e \left ( fx+e \right ) ^{2}}{{f}^{2}}}+a{c}^{2} \left ( fx+e \right ) -2\,{\frac{acde \left ( fx+e \right ) }{f}}+{\frac{a{d}^{2}{e}^{2} \left ( fx+e \right ) }{{f}^{2}}}+{\frac{b{d}^{2} \left ( - \left ( fx+e \right ) ^{2}\cos \left ( fx+e \right ) +2\,\cos \left ( fx+e \right ) +2\, \left ( fx+e \right ) \sin \left ( fx+e \right ) \right ) }{{f}^{2}}}+2\,{\frac{cbd \left ( \sin \left ( fx+e \right ) - \left ( fx+e \right ) \cos \left ( fx+e \right ) \right ) }{f}}-2\,{\frac{b{d}^{2}e \left ( \sin \left ( fx+e \right ) - \left ( fx+e \right ) \cos \left ( fx+e \right ) \right ) }{{f}^{2}}}-{c}^{2}b\cos \left ( fx+e \right ) +2\,{\frac{cbde\cos \left ( fx+e \right ) }{f}}-{\frac{b{d}^{2}{e}^{2}\cos \left ( fx+e \right ) }{{f}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.994145, size = 323, normalized size = 4.75 \begin{align*} \frac{3 \,{\left (f x + e\right )} a c^{2} + \frac{{\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac{3 \,{\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac{3 \,{\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac{3 \,{\left (f x + e\right )}^{2} a c d}{f} - \frac{6 \,{\left (f x + e\right )} a c d e}{f} - 3 \, b c^{2} \cos \left (f x + e\right ) - \frac{3 \, b d^{2} e^{2} \cos \left (f x + e\right )}{f^{2}} + \frac{6 \, b c d e \cos \left (f x + e\right )}{f} + \frac{6 \,{\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} b d^{2} e}{f^{2}} - \frac{6 \,{\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} b c d}{f} - \frac{3 \,{\left ({\left ({\left (f x + e\right )}^{2} - 2\right )} \cos \left (f x + e\right ) - 2 \,{\left (f x + e\right )} \sin \left (f x + e\right )\right )} b d^{2}}{f^{2}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68959, size = 228, normalized size = 3.35 \begin{align*} \frac{a d^{2} f^{3} x^{3} + 3 \, a c d f^{3} x^{2} + 3 \, a c^{2} f^{3} x - 3 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2} - 2 \, b d^{2}\right )} \cos \left (f x + e\right ) + 6 \,{\left (b d^{2} f x + b c d f\right )} \sin \left (f x + e\right )}{3 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.813716, size = 151, normalized size = 2.22 \begin{align*} \begin{cases} a c^{2} x + a c d x^{2} + \frac{a d^{2} x^{3}}{3} - \frac{b c^{2} \cos{\left (e + f x \right )}}{f} - \frac{2 b c d x \cos{\left (e + f x \right )}}{f} + \frac{2 b c d \sin{\left (e + f x \right )}}{f^{2}} - \frac{b d^{2} x^{2} \cos{\left (e + f x \right )}}{f} + \frac{2 b d^{2} x \sin{\left (e + f x \right )}}{f^{2}} + \frac{2 b d^{2} \cos{\left (e + f x \right )}}{f^{3}} & \text{for}\: f \neq 0 \\\left (a + b \sin{\left (e \right )}\right ) \left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.49884, size = 128, normalized size = 1.88 \begin{align*} \frac{1}{3} \, a d^{2} x^{3} + a c d x^{2} + a c^{2} x - \frac{{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2} - 2 \, b d^{2}\right )} \cos \left (f x + e\right )}{f^{3}} + \frac{2 \,{\left (b d^{2} f x + b c d f\right )} \sin \left (f x + e\right )}{f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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