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

Optimal. Leaf size=118 \[ -\frac{2 a^2 (c+d x) \cos (e+f x)}{f}-\frac{a^2 (c+d x) \sin (e+f x) \cos (e+f x)}{2 f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{1}{2} a^2 c x+\frac{a^2 d \sin ^2(e+f x)}{4 f^2}+\frac{2 a^2 d \sin (e+f x)}{f^2}+\frac{1}{4} a^2 d x^2 \]

[Out]

(a^2*c*x)/2 + (a^2*d*x^2)/4 + (a^2*(c + d*x)^2)/(2*d) - (2*a^2*(c + d*x)*Cos[e + f*x])/f + (2*a^2*d*Sin[e + f*
x])/f^2 - (a^2*(c + d*x)*Cos[e + f*x]*Sin[e + f*x])/(2*f) + (a^2*d*Sin[e + f*x]^2)/(4*f^2)

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Rubi [A]  time = 0.10374, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3317, 3296, 2637, 3310} \[ -\frac{2 a^2 (c+d x) \cos (e+f x)}{f}-\frac{a^2 (c+d x) \sin (e+f x) \cos (e+f x)}{2 f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{1}{2} a^2 c x+\frac{a^2 d \sin ^2(e+f x)}{4 f^2}+\frac{2 a^2 d \sin (e+f x)}{f^2}+\frac{1}{4} a^2 d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*x)/2 + (a^2*d*x^2)/4 + (a^2*(c + d*x)^2)/(2*d) - (2*a^2*(c + d*x)*Cos[e + f*x])/f + (2*a^2*d*Sin[e + f*
x])/f^2 - (a^2*(c + d*x)*Cos[e + f*x]*Sin[e + f*x])/(2*f) + (a^2*d*Sin[e + f*x]^2)/(4*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]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rubi steps

\begin{align*} \int (c+d x) (a+a \sin (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)+2 a^2 (c+d x) \sin (e+f x)+a^2 (c+d x) \sin ^2(e+f x)\right ) \, dx\\ &=\frac{a^2 (c+d x)^2}{2 d}+a^2 \int (c+d x) \sin ^2(e+f x) \, dx+\left (2 a^2\right ) \int (c+d x) \sin (e+f x) \, dx\\ &=\frac{a^2 (c+d x)^2}{2 d}-\frac{2 a^2 (c+d x) \cos (e+f x)}{f}-\frac{a^2 (c+d x) \cos (e+f x) \sin (e+f x)}{2 f}+\frac{a^2 d \sin ^2(e+f x)}{4 f^2}+\frac{1}{2} a^2 \int (c+d x) \, dx+\frac{\left (2 a^2 d\right ) \int \cos (e+f x) \, dx}{f}\\ &=\frac{1}{2} a^2 c x+\frac{1}{4} a^2 d x^2+\frac{a^2 (c+d x)^2}{2 d}-\frac{2 a^2 (c+d x) \cos (e+f x)}{f}+\frac{2 a^2 d \sin (e+f x)}{f^2}-\frac{a^2 (c+d x) \cos (e+f x) \sin (e+f x)}{2 f}+\frac{a^2 d \sin ^2(e+f x)}{4 f^2}\\ \end{align*}

Mathematica [A]  time = 1.04758, size = 80, normalized size = 0.68 \[ -\frac{a^2 (6 (e+f x) (d (e-f x)-2 c f)+2 f (c+d x) \sin (2 (e+f x))+16 f (c+d x) \cos (e+f x)-16 d \sin (e+f x)+d \cos (2 (e+f x)))}{8 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(6*(e + f*x)*(-2*c*f + d*(e - f*x)) + 16*f*(c + d*x)*Cos[e + f*x] + d*Cos[2*(e + f*x)] - 16*d*Sin[e + f*
x] + 2*f*(c + d*x)*Sin[2*(e + f*x)]))/(8*f^2)

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Maple [B]  time = 0.02, size = 219, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ({\frac{{a}^{2}d}{f} \left ( \left ( fx+e \right ) \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{\frac{ \left ( fx+e \right ) ^{2}}{4}}+{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{4}} \right ) }+{a}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{\frac{{a}^{2}de}{f} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) }+2\,{\frac{{a}^{2}d \left ( \sin \left ( fx+e \right ) - \left ( fx+e \right ) \cos \left ( fx+e \right ) \right ) }{f}}-2\,{a}^{2}c\cos \left ( fx+e \right ) +2\,{\frac{{a}^{2}de\cos \left ( fx+e \right ) }{f}}+{\frac{{a}^{2}d \left ( fx+e \right ) ^{2}}{2\,f}}+{a}^{2}c \left ( fx+e \right ) -{\frac{{a}^{2}de \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))^2,x)

[Out]

1/f*(a^2/f*d*((f*x+e)*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-1/4*(f*x+e)^2+1/4*sin(f*x+e)^2)+a^2*c*(-1/2*s
in(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-a^2/f*d*e*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+2*a^2/f*d*(sin(f*x+e)
-(f*x+e)*cos(f*x+e))-2*a^2*c*cos(f*x+e)+2*a^2/f*d*e*cos(f*x+e)+1/2*a^2/f*d*(f*x+e)^2+a^2*c*(f*x+e)-a^2/f*d*e*(
f*x+e))

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Maxima [A]  time = 0.99571, size = 277, normalized size = 2.35 \begin{align*} \frac{2 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c + 8 \,{\left (f x + e\right )} a^{2} c + \frac{4 \,{\left (f x + e\right )}^{2} a^{2} d}{f} - \frac{2 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} d e}{f} - \frac{8 \,{\left (f x + e\right )} a^{2} d e}{f} - 16 \, a^{2} c \cos \left (f x + e\right ) + \frac{16 \, a^{2} d e \cos \left (f x + e\right )}{f} + \frac{{\left (2 \,{\left (f x + e\right )}^{2} - 2 \,{\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) - \cos \left (2 \, f x + 2 \, e\right )\right )} a^{2} d}{f} - \frac{16 \,{\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} a^{2} d}{f}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^2*c + 8*(f*x + e)*a^2*c + 4*(f*x + e)^2*a^2*d/f - 2*(2*f*x + 2*e - s
in(2*f*x + 2*e))*a^2*d*e/f - 8*(f*x + e)*a^2*d*e/f - 16*a^2*c*cos(f*x + e) + 16*a^2*d*e*cos(f*x + e)/f + (2*(f
*x + e)^2 - 2*(f*x + e)*sin(2*f*x + 2*e) - cos(2*f*x + 2*e))*a^2*d/f - 16*((f*x + e)*cos(f*x + e) - sin(f*x +
e))*a^2*d/f)/f

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Fricas [A]  time = 1.73688, size = 228, normalized size = 1.93 \begin{align*} \frac{3 \, a^{2} d f^{2} x^{2} + 6 \, a^{2} c f^{2} x - a^{2} d \cos \left (f x + e\right )^{2} - 8 \,{\left (a^{2} d f x + a^{2} c f\right )} \cos \left (f x + e\right ) + 2 \,{\left (4 \, a^{2} d -{\left (a^{2} d f x + a^{2} c f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, 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))^2,x, algorithm="fricas")

[Out]

1/4*(3*a^2*d*f^2*x^2 + 6*a^2*c*f^2*x - a^2*d*cos(f*x + e)^2 - 8*(a^2*d*f*x + a^2*c*f)*cos(f*x + e) + 2*(4*a^2*
d - (a^2*d*f*x + a^2*c*f)*cos(f*x + e))*sin(f*x + e))/f^2

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Sympy [A]  time = 0.844491, size = 219, normalized size = 1.86 \begin{align*} \begin{cases} \frac{a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + a^{2} c x - \frac{a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a^{2} c \cos{\left (e + f x \right )}}{f} + \frac{a^{2} d x^{2} \sin ^{2}{\left (e + f x \right )}}{4} + \frac{a^{2} d x^{2} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{a^{2} d x^{2}}{2} - \frac{a^{2} d x \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a^{2} d x \cos{\left (e + f x \right )}}{f} + \frac{2 a^{2} d \sin{\left (e + f x \right )}}{f^{2}} - \frac{a^{2} d \cos ^{2}{\left (e + f x \right )}}{4 f^{2}} & \text{for}\: f \neq 0 \\\left (a \sin{\left (e \right )} + a\right )^{2} \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))**2,x)

[Out]

Piecewise((a**2*c*x*sin(e + f*x)**2/2 + a**2*c*x*cos(e + f*x)**2/2 + a**2*c*x - a**2*c*sin(e + f*x)*cos(e + f*
x)/(2*f) - 2*a**2*c*cos(e + f*x)/f + a**2*d*x**2*sin(e + f*x)**2/4 + a**2*d*x**2*cos(e + f*x)**2/4 + a**2*d*x*
*2/2 - a**2*d*x*sin(e + f*x)*cos(e + f*x)/(2*f) - 2*a**2*d*x*cos(e + f*x)/f + 2*a**2*d*sin(e + f*x)/f**2 - a**
2*d*cos(e + f*x)**2/(4*f**2), Ne(f, 0)), ((a*sin(e) + a)**2*(c*x + d*x**2/2), True))

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Giac [A]  time = 1.10989, size = 144, normalized size = 1.22 \begin{align*} \frac{3}{4} \, a^{2} d x^{2} + \frac{3}{2} \, a^{2} c x - \frac{a^{2} d \cos \left (2 \, f x + 2 \, e\right )}{8 \, f^{2}} + \frac{2 \, a^{2} d \sin \left (f x + e\right )}{f^{2}} - \frac{2 \,{\left (a^{2} d f x + a^{2} c f\right )} \cos \left (f x + e\right )}{f^{2}} - \frac{{\left (a^{2} d f x + a^{2} c f\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, 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))^2,x, algorithm="giac")

[Out]

3/4*a^2*d*x^2 + 3/2*a^2*c*x - 1/8*a^2*d*cos(2*f*x + 2*e)/f^2 + 2*a^2*d*sin(f*x + e)/f^2 - 2*(a^2*d*f*x + a^2*c
*f)*cos(f*x + e)/f^2 - 1/4*(a^2*d*f*x + a^2*c*f)*sin(2*f*x + 2*e)/f^2