3.181 \(\int \frac{(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=76 \[ -\frac{2 f \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )\right )}{a d^2}+\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}+\frac{e x}{a}+\frac{f x^2}{2 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.095287, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4515, 3318, 4184, 3475} \[ -\frac{2 f \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )\right )}{a d^2}+\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}+\frac{e x}{a}+\frac{f x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4515

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/b, Int[(e + f*x)^m*Sin[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sin[c + d*x]^(n - 1)
)/(a + b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3318

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

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

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

Rubi steps

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

Mathematica [B]  time = 0.496608, size = 199, normalized size = 2.62 \[ \frac{\cos \left (\frac{d x}{2}\right ) \left (d^2 x (2 e+f x)-4 f \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+2 d^2 e x \sin \left (c+\frac{d x}{2}\right )+d^2 f x^2 \sin \left (c+\frac{d x}{2}\right )+2 d f x \cos \left (c+\frac{d x}{2}\right )-4 f \sin \left (c+\frac{d x}{2}\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-4 d e \sin \left (\frac{d x}{2}\right )-2 d f x \sin \left (\frac{d x}{2}\right )}{2 a d^2 \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.076, size = 446, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a*e/d*arctan(tan(1/2*d*x+1/2*c))+2/a*e/d/(tan(1/2*d*x+1/2*c)+1)+1/a*f/(1+tan(1/2*d*x+1/2*c)^2)/(tan(1/2*d*x+
1/2*c)+1)*x/d+1/a*f/(1+tan(1/2*d*x+1/2*c)^2)/(tan(1/2*d*x+1/2*c)+1)*x/d*tan(1/2*d*x+1/2*c)^2+1/2/a*f/(1+tan(1/
2*d*x+1/2*c)^2)/(tan(1/2*d*x+1/2*c)+1)*x^2+1/2/a*f/(1+tan(1/2*d*x+1/2*c)^2)/(tan(1/2*d*x+1/2*c)+1)*x^2*tan(1/2
*d*x+1/2*c)+1/2/a*f/(1+tan(1/2*d*x+1/2*c)^2)/(tan(1/2*d*x+1/2*c)+1)*x^2*tan(1/2*d*x+1/2*c)^2+1/2/a*f/(1+tan(1/
2*d*x+1/2*c)^2)/(tan(1/2*d*x+1/2*c)+1)*x^2*tan(1/2*d*x+1/2*c)^3-1/a*f/(1+tan(1/2*d*x+1/2*c)^2)/(tan(1/2*d*x+1/
2*c)+1)*x/d*tan(1/2*d*x+1/2*c)-1/a*f/(1+tan(1/2*d*x+1/2*c)^2)/(tan(1/2*d*x+1/2*c)+1)*x/d*tan(1/2*d*x+1/2*c)^3+
1/a*f/d^2*ln(1+tan(1/2*d*x+1/2*c)^2)-2/a*f/d^2*ln(tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

Maxima [B]  time = 1.47247, size = 369, normalized size = 4.86 \begin{align*} -\frac{4 \, c f{\left (\frac{1}{a d + \frac{a d \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac{\arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a d}\right )} - 4 \, e{\left (\frac{\arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}}\right )} - \frac{{\left ({\left (d x + c\right )}^{2} \cos \left (d x + c\right )^{2} +{\left (d x + c\right )}^{2} \sin \left (d x + c\right )^{2} + 2 \,{\left (d x + c\right )}^{2} \sin \left (d x + c\right ) +{\left (d x + c\right )}^{2} + 4 \,{\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \,{\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right )\right )} f}{a d \cos \left (d x + c\right )^{2} + a d \sin \left (d x + c\right )^{2} + 2 \, a d \sin \left (d x + c\right ) + a d}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.79171, size = 363, normalized size = 4.78 \begin{align*} \frac{d^{2} f x^{2} + 2 \, d e + 2 \,{\left (d^{2} e + d f\right )} x +{\left (d^{2} f x^{2} + 2 \, d e + 2 \,{\left (d^{2} e + d f\right )} x\right )} \cos \left (d x + c\right ) - 2 \,{\left (f \cos \left (d x + c\right ) + f \sin \left (d x + c\right ) + f\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (d^{2} f x^{2} - 2 \, d e + 2 \,{\left (d^{2} e - d f\right )} x\right )} \sin \left (d x + c\right )}{2 \,{\left (a d^{2} \cos \left (d x + c\right ) + a d^{2} \sin \left (d x + c\right ) + a d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.95445, size = 466, normalized size = 6.13 \begin{align*} \begin{cases} \frac{2 d^{2} e x \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 d^{2} e x}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{d^{2} f x^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{d^{2} f x^{2}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} - \frac{4 d e \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} - \frac{2 d f x \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 d f x}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} - \frac{4 f \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1 \right )} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} - \frac{4 f \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1 \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 f \log{\left (\tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1 \right )} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} + \frac{2 f \log{\left (\tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1 \right )}}{2 a d^{2} \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d^{2}} & \text{for}\: d \neq 0 \\\frac{\left (e x + \frac{f x^{2}}{2}\right ) \sin{\left (c \right )}}{a \sin{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*d**2*e*x*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 2*d**2*e*x/(2*a*d**2*tan(c/2 +
d*x/2) + 2*a*d**2) + d**2*f*x**2*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + d**2*f*x**2/(2*a*d
**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*d*e*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*d*f*x*tan
(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 2*d*f*x/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*f*lo
g(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*f*log(tan(c/2 + d*x/2) + 1
)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 2*f*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2
+ d*x/2) + 2*a*d**2) + 2*f*log(tan(c/2 + d*x/2)**2 + 1)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2), Ne(d, 0)), ((
e*x + f*x**2/2)*sin(c)/(a*sin(c) + a), True))

________________________________________________________________________________________

Giac [B]  time = 1.52989, size = 1042, normalized size = 13.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d^2*f*x^2*tan(1/2*d*x)*tan(1/2*c) - d^2*f*x^2*tan(1/2*d*x) - d^2*f*x^2*tan(1/2*c) + 2*d^2*x*e*tan(1/2*d*x
)*tan(1/2*c) - d^2*f*x^2 - 2*d^2*x*e*tan(1/2*d*x) - 2*d^2*x*e*tan(1/2*c) + 2*d*f*x*tan(1/2*d*x)*tan(1/2*c) - 2
*d^2*x*e + 2*d*f*x*tan(1/2*d*x) + 2*d*f*x*tan(1/2*c) + 2*d*e*tan(1/2*d*x)*tan(1/2*c) - 2*f*log(2*(tan(1/2*c)^2
+ 1)/(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x
)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(
1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1))*tan(1/2*d*x)*tan(1/2*c) - 2*d*f*x + 2*d*e*tan(1/2*d*x) + 2*f*lo
g(2*(tan(1/2*c)^2 + 1)/(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/2*d*x)^3*tan(1/2*c
)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(
1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1))*tan(1/2*d*x) + 2*d*e*tan(1/2*c) + 2*f*log(2*(t
an(1/2*c)^2 + 1)/(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/2*d*x)^3*tan(1/2*c)^2 +
tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*
x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1))*tan(1/2*c) - 2*d*e + 2*f*log(2*(tan(1/2*c)^2 + 1)/(t
an(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x)^4 + 2*
tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(1/2*c)^2
+ 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)))/(a*d^2*tan(1/2*d*x)*tan(1/2*c) - a*d^2*tan(1/2*d*x) - a*d^2*tan(1/2*c)
- a*d^2)