3.2.52 \(\int (c+d x)^2 (a+b \sin (e+f x)) \, dx\) [152]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 = {3398, 3377, 2718} \begin {gather*} \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} \end {gather*}

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 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 3398

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])

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*}

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Mathematica [A]
time = 0.20, size = 84, normalized size = 1.24 \begin {gather*} \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 (-2+f^2 x^2\right )\right ) \cos (e+f x)}{f^3}+\frac {2 b d (c+d x) \sin (e+f x)}{f^2} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(240\) vs. \(2(66)=132\).
time = 0.05, size = 241, normalized size = 3.54

method	result	size
risch	\(\frac {a \,d^{2} x^{3}}{3}+a d c \,x^{2}+a \,c^{2} x +\frac {a \,c^{3}}{3 d}-\frac {b \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}-2 d^{2}\right ) \cos \left (f x +e \right )}{f^{3}}+\frac {2 b d \left (d x +c \right ) \sin \left (f x +e \right )}{f^{2}}\)	\(94\)
norman	\(\frac {\frac {\left (2 b \,c^{2} f^{2}-4 b \,d^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f^{3}}+\frac {c \left (a c f -2 b d \right ) x}{f}+\frac {d \left (a c f -b d \right ) x^{2}}{f}+\frac {c \left (a c f +2 b d \right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {d \left (a c f +b d \right ) x^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {a \,d^{2} x^{3}}{3}+\frac {a \,d^{2} x^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\frac {4 b c d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f^{2}}+\frac {4 b \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f^{2}}}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}\)	\(199\)
derivativedivides	\(\frac {a \,c^{2} \left (f x +e \right )-\frac {2 a c d e \left (f x +e \right )}{f}+\frac {a c d \left (f x +e \right )^{2}}{f}+\frac {a \,d^{2} e^{2} \left (f x +e \right )}{f^{2}}-\frac {a \,d^{2} e \left (f x +e \right )^{2}}{f^{2}}+\frac {a \,d^{2} \left (f x +e \right )^{3}}{3 f^{2}}-b \,c^{2} \cos \left (f x +e \right )+\frac {2 b c d e \cos \left (f x +e \right )}{f}+\frac {2 b c d \left (\sin \left (f x +e \right )-\left (f x +e \right ) \cos \left (f x +e \right )\right )}{f}-\frac {b \,d^{2} e^{2} \cos \left (f x +e \right )}{f^{2}}-\frac {2 b \,d^{2} e \left (\sin \left (f x +e \right )-\left (f x +e \right ) \cos \left (f x +e \right )\right )}{f^{2}}+\frac {b \,d^{2} \left (-\left (f x +e \right )^{2} \cos \left (f x +e \right )+2 \cos \left (f x +e \right )+2 \left (f x +e \right ) \sin \left (f x +e \right )\right )}{f^{2}}}{f}\)	\(241\)
default	\(\frac {a \,c^{2} \left (f x +e \right )-\frac {2 a c d e \left (f x +e \right )}{f}+\frac {a c d \left (f x +e \right )^{2}}{f}+\frac {a \,d^{2} e^{2} \left (f x +e \right )}{f^{2}}-\frac {a \,d^{2} e \left (f x +e \right )^{2}}{f^{2}}+\frac {a \,d^{2} \left (f x +e \right )^{3}}{3 f^{2}}-b \,c^{2} \cos \left (f x +e \right )+\frac {2 b c d e \cos \left (f x +e \right )}{f}+\frac {2 b c d \left (\sin \left (f x +e \right )-\left (f x +e \right ) \cos \left (f x +e \right )\right )}{f}-\frac {b \,d^{2} e^{2} \cos \left (f x +e \right )}{f^{2}}-\frac {2 b \,d^{2} e \left (\sin \left (f x +e \right )-\left (f x +e \right ) \cos \left (f x +e \right )\right )}{f^{2}}+\frac {b \,d^{2} \left (-\left (f x +e \right )^{2} \cos \left (f x +e \right )+2 \cos \left (f x +e \right )+2 \left (f x +e \right ) \sin \left (f x +e \right )\right )}{f^{2}}}{f}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (69) = 138\).
time = 0.37, size = 260, normalized size = 3.82 \begin {gather*} \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 c d}{f} - 3 \, b c^{2} \cos \left (f x + e\right ) - \frac {3 \, {\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} - \frac {6 \, {\left (f x + e\right )} a c d e}{f} + \frac {6 \, b c d \cos \left (f x + e\right ) e}{f} - \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 (f x + e\right )} a d^{2} e^{2}}{f^{2}} - \frac {3 \, b d^{2} \cos \left (f x + e\right ) e^{2}}{f^{2}} + \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 {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 {gather*}

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*c*d/f - 3*b*c^2*cos(f*x + e) - 3*(f*x + e)^2*
a*d^2*e/f^2 - 6*(f*x + e)*a*c*d*e/f + 6*b*c*d*cos(f*x + e)*e/f - 6*((f*x + e)*cos(f*x + e) - sin(f*x + e))*b*c
*d/f + 3*(f*x + e)*a*d^2*e^2/f^2 - 3*b*d^2*cos(f*x + e)*e^2/f^2 + 6*((f*x + e)*cos(f*x + e) - sin(f*x + e))*b*
d^2*e/f^2 - 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 = 0.37, size = 104, normalized size = 1.53 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (65) = 130\).
time = 0.15, size = 151, normalized size = 2.22 \begin {gather*} \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 {gather*}

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 = 5.39, size = 95, normalized size = 1.40 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.67, size = 112, normalized size = 1.65 \begin {gather*} \frac {a\,d^2\,x^3}{3}+\frac {\cos \left (e+f\,x\right )\,\left (2\,b\,d^2-b\,c^2\,f^2\right )}{f^3}+a\,c^2\,x+a\,c\,d\,x^2+\frac {2\,b\,d^2\,x\,\sin \left (e+f\,x\right )}{f^2}-\frac {b\,d^2\,x^2\,\cos \left (e+f\,x\right )}{f}+\frac {2\,b\,c\,d\,\sin \left (e+f\,x\right )}{f^2}-\frac {2\,b\,c\,d\,x\,\cos \left (e+f\,x\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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