3.97 \(\int (c+d x) (a+a \sin (e+f x)) \, dx\)

Optimal. Leaf size=45 \[ -\frac{a (c+d x) \cos (e+f x)}{f}+\frac{a (c+d x)^2}{2 d}+\frac{a d \sin (e+f x)}{f^2} \]

[Out]

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

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Rubi [A]  time = 0.0422202, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3317, 3296, 2637} \[ -\frac{a (c+d x) \cos (e+f x)}{f}+\frac{a (c+d x)^2}{2 d}+\frac{a d \sin (e+f x)}{f^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + a*Sin[e + f*x]),x]

[Out]

(a*(c + d*x)^2)/(2*d) - (a*(c + d*x)*Cos[e + f*x])/f + (a*d*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.356213, size = 51, normalized size = 1.13 \[ -\frac{a ((e+f x) (-2 c f+d e-d f x)+2 f (c+d x) \cos (e+f x)-2 d \sin (e+f x))}{2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 90, normalized size = 2. \begin{align*}{\frac{1}{f} \left ({\frac{da \left ( \sin \left ( fx+e \right ) - \left ( fx+e \right ) \cos \left ( fx+e \right ) \right ) }{f}}-ac\cos \left ( fx+e \right ) +{\frac{ade\cos \left ( fx+e \right ) }{f}}+{\frac{da \left ( fx+e \right ) ^{2}}{2\,f}}+ac \left ( fx+e \right ) -{\frac{ade \left ( fx+e \right ) }{f}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.969237, size = 126, normalized size = 2.8 \begin{align*} \frac{2 \,{\left (f x + e\right )} a c + \frac{{\left (f x + e\right )}^{2} a d}{f} - \frac{2 \,{\left (f x + e\right )} a d e}{f} - 2 \, a c \cos \left (f x + e\right ) + \frac{2 \, a d e \cos \left (f x + e\right )}{f} - \frac{2 \,{\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} a d}{f}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76993, size = 126, normalized size = 2.8 \begin{align*} \frac{a d f^{2} x^{2} + 2 \, a c f^{2} x + 2 \, a d \sin \left (f x + e\right ) - 2 \,{\left (a d f x + a c f\right )} \cos \left (f x + e\right )}{2 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.343821, size = 68, normalized size = 1.51 \begin{align*} \begin{cases} a c x - \frac{a c \cos{\left (e + f x \right )}}{f} + \frac{a d x^{2}}{2} - \frac{a d x \cos{\left (e + f x \right )}}{f} + \frac{a d \sin{\left (e + f x \right )}}{f^{2}} & \text{for}\: f \neq 0 \\\left (a \sin{\left (e \right )} + a\right ) \left (c x + \frac{d x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09574, size = 63, normalized size = 1.4 \begin{align*} \frac{1}{2} \, a d x^{2} + a c x + \frac{a d \sin \left (f x + e\right )}{f^{2}} - \frac{{\left (a d f x + a c f\right )} \cos \left (f x + e\right )}{f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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