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

Optimal. Leaf size=49 \[ \frac {(c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {2 d \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{a f^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3318, 4184, 3475} \[ \frac {(c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {2 d \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{a f^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + a*Cos[e + f*x]),x]

[Out]

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

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 3475

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

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]

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

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Mathematica [A]  time = 0.08, size = 70, normalized size = 1.43 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (f (c+d x) \sin \left (\frac {1}{2} (e+f x)\right )+2 d \cos \left (\frac {1}{2} (e+f x)\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{a f^2 (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 58, normalized size = 1.18 \[ \frac {{\left (d \cos \left (f x + e\right ) + d\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left (d f x + c f\right )} \sin \left (f x + e\right )}{a f^{2} \cos \left (f x + e\right ) + a f^{2}} \]

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

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giac [B]  time = 0.61, size = 234, normalized size = 4.78 \[ -\frac {d f x \tan \left (\frac {1}{2} \, f x\right ) + d f x \tan \left (\frac {1}{2} \, e\right ) - d \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, f x\right )^{4} \tan \left (\frac {1}{2} \, e\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, f x\right )^{3} \tan \left (\frac {1}{2} \, e\right ) + \tan \left (\frac {1}{2} \, f x\right )^{2} \tan \left (\frac {1}{2} \, e\right )^{2} + \tan \left (\frac {1}{2} \, f x\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, f x\right ) \tan \left (\frac {1}{2} \, e\right ) + 1\right )}}{\tan \left (\frac {1}{2} \, e\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, f x\right ) \tan \left (\frac {1}{2} \, e\right ) + c f \tan \left (\frac {1}{2} \, f x\right ) + c f \tan \left (\frac {1}{2} \, e\right ) + d \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, f x\right )^{4} \tan \left (\frac {1}{2} \, e\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, f x\right )^{3} \tan \left (\frac {1}{2} \, e\right ) + \tan \left (\frac {1}{2} \, f x\right )^{2} \tan \left (\frac {1}{2} \, e\right )^{2} + \tan \left (\frac {1}{2} \, f x\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, f x\right ) \tan \left (\frac {1}{2} \, e\right ) + 1\right )}}{\tan \left (\frac {1}{2} \, e\right )^{2} + 1}\right )}{a f^{2} \tan \left (\frac {1}{2} \, f x\right ) \tan \left (\frac {1}{2} \, e\right ) - a f^{2}} \]

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

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maple [A]  time = 0.09, size = 60, normalized size = 1.22 \[ \frac {c \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {d x \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {d \ln \left (1+\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{a \,f^{2}} \]

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

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maxima [B]  time = 0.43, size = 160, normalized size = 3.27 \[ \frac {\frac {{\left ({\left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right ) + 2 \, {\left (f x + e\right )} \sin \left (f x + e\right )\right )} d}{a f \cos \left (f x + e\right )^{2} + a f \sin \left (f x + e\right )^{2} + 2 \, a f \cos \left (f x + e\right ) + a f} + \frac {c \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}} - \frac {d e \sin \left (f x + e\right )}{a f {\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \]

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*si
n(f*x + e)/(a*(cos(f*x + e) + 1)) - d*e*sin(f*x + e)/(a*f*(cos(f*x + e) + 1)))/f

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mupad [B]  time = 0.66, size = 65, normalized size = 1.33 \[ \frac {2\,d\,\ln \left ({\mathrm {e}}^{e\,1{}\mathrm {i}}\,{\mathrm {e}}^{f\,x\,1{}\mathrm {i}}+1\right )}{a\,f^2}+\frac {\left (c+d\,x\right )\,2{}\mathrm {i}}{a\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}-\frac {d\,x\,2{}\mathrm {i}}{a\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + a*cos(e + f*x)),x)

[Out]

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

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sympy [A]  time = 0.65, size = 70, normalized size = 1.43 \[ \begin {cases} \frac {c \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f} + \frac {d x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f} - \frac {d \log {\left (\tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 1 \right )}}{a f^{2}} & \text {for}\: f \neq 0 \\\frac {c x + \frac {d x^{2}}{2}}{a \cos {\relax (e )} + a} & \text {otherwise} \end {cases} \]

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

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