3.140 \(\int \frac{c+d x}{a-a \cos (e+f x)} \, dx\)

Optimal. Leaf size=50 \[ \frac{2 d \log \left (\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{a f^2}-\frac{(c+d x) \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f} \]

[Out]

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

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Rubi [A]  time = 0.0652033, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3318, 4184, 3475} \[ \frac{2 d \log \left (\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{a f^2}-\frac{(c+d x) \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a - a*Cos[e + f*x]),x]

[Out]

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

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

Mathematica [A]  time = 0.262909, size = 57, normalized size = 1.14 \[ \frac{f (c+d x) \sin (e+f x)-4 d \sin ^2\left (\frac{1}{2} (e+f x)\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{a f^2 (\cos (e+f x)-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d*Log[Sin[(e + f*x)/2]]*Sin[(e + f*x)/2]^2 + f*(c + d*x)*Sin[e + f*x])/(a*f^2*(-1 + Cos[e + f*x]))

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Maple [A]  time = 0.063, size = 85, normalized size = 1.7 \begin{align*} -{\frac{c}{af} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-1}}-{\frac{dx}{af} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-1}}-{\frac{d}{a{f}^{2}}\ln \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) }+2\,{\frac{d\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{a{f}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16745, size = 216, normalized size = 4.32 \begin{align*} \frac{\frac{{\left ({\left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right ) - 2 \,{\left (f x + e\right )} \sin \left (f x + e\right )\right )} d}{a f \cos \left (f x + e\right )^{2} + a f \sin \left (f x + e\right )^{2} - 2 \, a f \cos \left (f x + e\right ) + a f} - \frac{c{\left (\cos \left (f x + e\right ) + 1\right )}}{a \sin \left (f x + e\right )} + \frac{d e{\left (\cos \left (f x + e\right ) + 1\right )}}{a f \sin \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.63022, size = 151, normalized size = 3.02 \begin{align*} -\frac{d f x - d \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) + c f +{\left (d f x + c f\right )} \cos \left (f x + e\right )}{a f^{2} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.896187, size = 90, normalized size = 1.8 \begin{align*} \begin{cases} - \frac{c}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}} - \frac{d x}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}} - \frac{d \log{\left (\tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 1 \right )}}{a f^{2}} + \frac{2 d \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} \right )}}{a f^{2}} & \text{for}\: f \neq 0 \\\frac{c x + \frac{d x^{2}}{2}}{- a \cos{\left (e \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.16138, size = 309, normalized size = 6.18 \begin{align*} \frac{d f x \tan \left (\frac{1}{2} \, f x\right ) \tan \left (\frac{1}{2} \, e\right ) + c f \tan \left (\frac{1}{2} \, f x\right ) \tan \left (\frac{1}{2} \, e\right ) - d f x + d \log \left (\frac{4 \,{\left (\tan \left (\frac{1}{2} \, e\right )^{2} + 1\right )}}{\tan \left (\frac{1}{2} \, f x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, f x\right )^{3} \tan \left (\frac{1}{2} \, e\right ) + \tan \left (\frac{1}{2} \, f x\right )^{2} \tan \left (\frac{1}{2} \, e\right )^{2} + \tan \left (\frac{1}{2} \, f x\right )^{2} + 2 \, \tan \left (\frac{1}{2} \, f x\right ) \tan \left (\frac{1}{2} \, e\right ) + \tan \left (\frac{1}{2} \, e\right )^{2}}\right ) \tan \left (\frac{1}{2} \, f x\right ) + d \log \left (\frac{4 \,{\left (\tan \left (\frac{1}{2} \, e\right )^{2} + 1\right )}}{\tan \left (\frac{1}{2} \, f x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, f x\right )^{3} \tan \left (\frac{1}{2} \, e\right ) + \tan \left (\frac{1}{2} \, f x\right )^{2} \tan \left (\frac{1}{2} \, e\right )^{2} + \tan \left (\frac{1}{2} \, f x\right )^{2} + 2 \, \tan \left (\frac{1}{2} \, f x\right ) \tan \left (\frac{1}{2} \, e\right ) + \tan \left (\frac{1}{2} \, e\right )^{2}}\right ) \tan \left (\frac{1}{2} \, e\right ) - c f}{a f^{2} \tan \left (\frac{1}{2} \, f x\right ) + a f^{2} \tan \left (\frac{1}{2} \, e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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