\(\int (c+d x) (a+a \sin (e+f x))^2 \, dx\) [103]

Optimal result

Integrand size = 18, antiderivative size = 118 \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=\frac {1}{2} a^2 c x+\frac {1}{4} a^2 d x^2+\frac {a^2 (c+d x)^2}{2 d}-\frac {2 a^2 (c+d x) \cos (e+f x)}{f}+\frac {2 a^2 d \sin (e+f x)}{f^2}-\frac {a^2 (c+d x) \cos (e+f x) \sin (e+f x)}{2 f}+\frac {a^2 d \sin ^2(e+f x)}{4 f^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3398, 3377, 2717, 3391} \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=-\frac {2 a^2 (c+d x) \cos (e+f x)}{f}-\frac {a^2 (c+d x) \sin (e+f x) \cos (e+f x)}{2 f}+\frac {a^2 (c+d x)^2}{2 d}+\frac {1}{2} a^2 c x+\frac {a^2 d \sin ^2(e+f x)}{4 f^2}+\frac {2 a^2 d \sin (e+f x)}{f^2}+\frac {1}{4} a^2 d x^2 \]

[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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3391

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]

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*} \text {integral}& = \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] (verified)

Time = 12.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.68 \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=-\frac {a^2 (6 (e+f x) (-2 c f+d (e-f x))+16 f (c+d x) \cos (e+f x)+d \cos (2 (e+f x))-16 d \sin (e+f x)+2 f (c+d x) \sin (2 (e+f x)))}{8 f^2} \]

[In]

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

[Out]

-1/8*(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)]))/f^2

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {\left (f \left (d x +c \right ) \sin \left (2 f x +2 e \right )+\frac {d \cos \left (2 f x +2 e \right )}{2}+8 f \left (d x +c \right ) \cos \left (f x +e \right )-8 \sin \left (f x +e \right ) d +\left (-3 d \,x^{2}-6 c x \right ) f^{2}+8 c f -\frac {d}{2}\right ) a^{2}}{4 f^{2}}\)	\(83\)
risch	\(\frac {3 a^{2} d \,x^{2}}{4}+\frac {3 a^{2} c x}{2}-\frac {2 a^{2} \left (d x +c \right ) \cos \left (f x +e \right )}{f}+\frac {2 a^{2} d \sin \left (f x +e \right )}{f^{2}}-\frac {a^{2} d \cos \left (2 f x +2 e \right )}{8 f^{2}}-\frac {a^{2} \left (d x +c \right ) \sin \left (2 f x +2 e \right )}{4 f}\)	\(92\)
parts	\(a^{2} \left (\frac {1}{2} d \,x^{2}+c x \right )+\frac {a^{2} \left (\frac {d \left (\left (f x +e \right ) \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{4}+\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{4}\right )}{f}+c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {d e \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\right )}{f}+\frac {2 a^{2} \left (\frac {d \left (\sin \left (f x +e \right )-\left (f x +e \right ) \cos \left (f x +e \right )\right )}{f}-c \cos \left (f x +e \right )+\frac {d e \cos \left (f x +e \right )}{f}\right )}{f}\)	\(185\)
derivativedivides	\(\frac {a^{2} c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {a^{2} d e \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {a^{2} d \left (\left (f x +e \right ) \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{4}+\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-2 a^{2} c \cos \left (f x +e \right )+\frac {2 a^{2} d e \cos \left (f x +e \right )}{f}+\frac {2 a^{2} d \left (\sin \left (f x +e \right )-\left (f x +e \right ) \cos \left (f x +e \right )\right )}{f}+a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}}{f}\)	\(219\)
default	\(\frac {a^{2} c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {a^{2} d e \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {a^{2} d \left (\left (f x +e \right ) \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{4}+\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-2 a^{2} c \cos \left (f x +e \right )+\frac {2 a^{2} d e \cos \left (f x +e \right )}{f}+\frac {2 a^{2} d \left (\sin \left (f x +e \right )-\left (f x +e \right ) \cos \left (f x +e \right )\right )}{f}+a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}}{f}\)	\(219\)
norman	\(\frac {\frac {a^{2} \left (c f +4 d \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f^{2}}+\frac {a^{2} d x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {4 a^{2} c}{f}+\frac {3 a^{2} d \,x^{2}}{4}-\frac {\left (4 a^{2} c f -a^{2} d \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f^{2}}+3 a^{2} c x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {3 a^{2} d \,x^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {3 a^{2} d \,x^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}-\frac {a^{2} \left (c f -4 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f^{2}}+\frac {a^{2} \left (3 c f -4 d \right ) x}{2 f}-\frac {a^{2} d x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {a^{2} \left (3 c f +4 d \right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(257\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86 \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=\frac {3 \, a^{2} d f^{2} x^{2} + 6 \, a^{2} c f^{2} x - a^{2} d \cos \left (f x + e\right )^{2} - 8 \, {\left (a^{2} d f x + a^{2} c f\right )} \cos \left (f x + e\right ) + 2 \, {\left (4 \, a^{2} d - {\left (a^{2} d f x + a^{2} c f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, f^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.86 \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=\begin {cases} \frac {a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + a^{2} c x - \frac {a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 a^{2} c \cos {\left (e + f x \right )}}{f} + \frac {a^{2} d x^{2} \sin ^{2}{\left (e + f x \right )}}{4} + \frac {a^{2} d x^{2} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {a^{2} d x^{2}}{2} - \frac {a^{2} d x \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 a^{2} d x \cos {\left (e + f x \right )}}{f} + \frac {a^{2} d \sin ^{2}{\left (e + f x \right )}}{4 f^{2}} + \frac {2 a^{2} d \sin {\left (e + f x \right )}}{f^{2}} & \text {for}\: f \neq 0 \\\left (a \sin {\left (e \right )} + a\right )^{2} \left (c x + \frac {d x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.74 \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=\frac {2 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c + 8 \, {\left (f x + e\right )} a^{2} c + \frac {4 \, {\left (f x + e\right )}^{2} a^{2} d}{f} - \frac {2 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} d e}{f} - \frac {8 \, {\left (f x + e\right )} a^{2} d e}{f} - 16 \, a^{2} c \cos \left (f x + e\right ) + \frac {16 \, a^{2} d e \cos \left (f x + e\right )}{f} + \frac {{\left (2 \, {\left (f x + e\right )}^{2} - 2 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) - \cos \left (2 \, f x + 2 \, e\right )\right )} a^{2} d}{f} - \frac {16 \, {\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} a^{2} d}{f}}{8 \, f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=\frac {3}{4} \, a^{2} d x^{2} + \frac {3}{2} \, a^{2} c x - \frac {a^{2} d \cos \left (2 \, f x + 2 \, e\right )}{8 \, f^{2}} + \frac {2 \, a^{2} d \sin \left (f x + e\right )}{f^{2}} - \frac {2 \, {\left (a^{2} d f x + a^{2} c f\right )} \cos \left (f x + e\right )}{f^{2}} - \frac {{\left (a^{2} d f x + a^{2} c f\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.08 \[ \int (c+d x) (a+a \sin (e+f x))^2 \, dx=\frac {a^2\,d\,{\sin \left (e+f\,x\right )}^2+8\,a^2\,d\,\sin \left (e+f\,x\right )+16\,a^2\,c\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a^2\,d\,f^2\,x^2-a^2\,c\,f\,\sin \left (2\,e+2\,f\,x\right )+6\,a^2\,c\,f^2\,x-a^2\,d\,f\,x\,\sin \left (2\,e+2\,f\,x\right )+8\,a^2\,d\,f\,x\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}{4\,f^2} \]

[In]

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

[Out]

(a^2*d*sin(e + f*x)^2 + 8*a^2*d*sin(e + f*x) + 16*a^2*c*f*sin(e/2 + (f*x)/2)^2 + 3*a^2*d*f^2*x^2 - a^2*c*f*sin
(2*e + 2*f*x) + 6*a^2*c*f^2*x - a^2*d*f*x*sin(2*e + 2*f*x) + 8*a^2*d*f*x*(2*sin(e/2 + (f*x)/2)^2 - 1))/(4*f^2)