3.135 \(\int \frac{c+d x}{(a+a \cos (e+f x))^2} \, dx\)

Optimal. Leaf size=123 \[ \frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}+\frac{2 d \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 a^2 f^2} \]

[Out]

(2*d*Log[Cos[e/2 + (f*x)/2]])/(3*a^2*f^2) - (d*Sec[e/2 + (f*x)/2]^2)/(6*a^2*f^2) + ((c + d*x)*Tan[e/2 + (f*x)/
2])/(3*a^2*f) + ((c + d*x)*Sec[e/2 + (f*x)/2]^2*Tan[e/2 + (f*x)/2])/(6*a^2*f)

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Rubi [A]  time = 0.0946936, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3318, 4185, 4184, 3475} \[ \frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}+\frac{2 d \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 a^2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*Log[Cos[e/2 + (f*x)/2]])/(3*a^2*f^2) - (d*Sec[e/2 + (f*x)/2]^2)/(6*a^2*f^2) + ((c + d*x)*Tan[e/2 + (f*x)/
2])/(3*a^2*f) + ((c + d*x)*Sec[e/2 + (f*x)/2]^2*Tan[e/2 + (f*x)/2])/(6*a^2*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 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

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))^2} \, dx &=\frac{\int (c+d x) \csc ^4\left (\frac{e+\pi }{2}+\frac{f x}{2}\right ) \, dx}{4 a^2}\\ &=-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}+\frac{(c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}+\frac{\int (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{6 a^2}\\ &=-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}+\frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}-\frac{d \int \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{3 a^2 f}\\ &=\frac{2 d \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 a^2 f^2}-\frac{d \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f^2}+\frac{(c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}\\ \end{align*}

Mathematica [A]  time = 0.507986, size = 113, normalized size = 0.92 \[ \frac{\cos \left (\frac{1}{2} (e+f x)\right ) \left (f (c+d x) \left (3 \sin \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{3}{2} (e+f x)\right )\right )+2 d \cos \left (\frac{3}{2} (e+f x)\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+2 d \cos \left (\frac{1}{2} (e+f x)\right ) \left (3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )-1\right )\right )}{3 a^2 f^2 (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(e + f*x)/2]*(2*d*Cos[(3*(e + f*x))/2]*Log[Cos[(e + f*x)/2]] + 2*d*Cos[(e + f*x)/2]*(-1 + 3*Log[Cos[(e +
f*x)/2]]) + f*(c + d*x)*(3*Sin[(e + f*x)/2] + Sin[(3*(e + f*x))/2])))/(3*a^2*f^2*(1 + Cos[e + f*x])^2)

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Maple [A]  time = 0.11, size = 123, normalized size = 1. \begin{align*}{\frac{c}{6\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+{\frac{c}{2\,{a}^{2}f}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-{\frac{d}{6\,{a}^{2}{f}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2}}+{\frac{dx}{2\,{a}^{2}f}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+{\frac{dx}{6\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{d}{3\,{a}^{2}{f}^{2}}\ln \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a^2*c/f*tan(1/2*f*x+1/2*e)^3+1/2/a^2*c/f*tan(1/2*f*x+1/2*e)-1/6/a^2*d/f^2*tan(1/2*f*x+1/2*e)^2+1/2/a^2/f*x
*d*tan(1/2*f*x+1/2*e)+1/6/a^2/f*x*d*tan(1/2*f*x+1/2*e)^3-1/3/a^2*d/f^2*ln(1+tan(1/2*f*x+1/2*e)^2)

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Maxima [B]  time = 1.2705, size = 1030, normalized size = 8.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*(2*(3*(f*x + e)*sin(f*x + e) + cos(2*f*x + 2*e) + cos(f*x + e))*cos(3*f*x + 3*e) + 2*(9*(f*x + e)*sin(
f*x + e) + 6*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + 6*cos(2*f*x + 2*e)^2 + 6*cos(f*x + e)^2 - (2*(3*cos(2*f*x +
2*e) + 3*cos(f*x + e) + 1)*cos(3*f*x + 3*e) + cos(3*f*x + 3*e)^2 + 6*(3*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + 9
*cos(2*f*x + 2*e)^2 + 9*cos(f*x + e)^2 + 6*(sin(2*f*x + 2*e) + sin(f*x + e))*sin(3*f*x + 3*e) + sin(3*f*x + 3*
e)^2 + 9*sin(2*f*x + 2*e)^2 + 18*sin(2*f*x + 2*e)*sin(f*x + e) + 9*sin(f*x + e)^2 + 6*cos(f*x + e) + 1)*log(co
s(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) - 2*(f*x + 3*(f*x + e)*cos(f*x + e) + e - sin(2*f*x + 2*e)
- sin(f*x + e))*sin(3*f*x + 3*e) - 6*(f*x + 3*(f*x + e)*cos(f*x + e) + e - 2*sin(f*x + e))*sin(2*f*x + 2*e) +
6*sin(2*f*x + 2*e)^2 + 6*sin(f*x + e)^2 + 2*cos(f*x + e))*d/(a^2*f*cos(3*f*x + 3*e)^2 + 9*a^2*f*cos(2*f*x + 2
*e)^2 + 9*a^2*f*cos(f*x + e)^2 + a^2*f*sin(3*f*x + 3*e)^2 + 9*a^2*f*sin(2*f*x + 2*e)^2 + 18*a^2*f*sin(2*f*x +
2*e)*sin(f*x + e) + 9*a^2*f*sin(f*x + e)^2 + 6*a^2*f*cos(f*x + e) + a^2*f + 2*(3*a^2*f*cos(2*f*x + 2*e) + 3*a^
2*f*cos(f*x + e) + a^2*f)*cos(3*f*x + 3*e) + 6*(3*a^2*f*cos(f*x + e) + a^2*f)*cos(2*f*x + 2*e) + 6*(a^2*f*sin(
2*f*x + 2*e) + a^2*f*sin(f*x + e))*sin(3*f*x + 3*e)) - c*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(
cos(f*x + e) + 1)^3)/a^2 + d*e*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2*
f))/f

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Fricas [A]  time = 1.67137, size = 298, normalized size = 2.42 \begin{align*} -\frac{d \cos \left (f x + e\right ) -{\left (d \cos \left (f x + e\right )^{2} + 2 \, 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 (2 \, d f x + 2 \, c f +{\left (d f x + c f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) + d}{3 \,{\left (a^{2} f^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} f^{2} \cos \left (f x + e\right ) + a^{2} f^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+a*cos(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.43222, size = 146, normalized size = 1.19 \begin{align*} \begin{cases} \frac{c \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{6 a^{2} f} + \frac{c \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{2 a^{2} f} + \frac{d x \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{6 a^{2} f} + \frac{d x \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{2 a^{2} f} - \frac{d \log{\left (\tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 1 \right )}}{3 a^{2} f^{2}} - \frac{d \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{6 a^{2} f^{2}} & \text{for}\: f \neq 0 \\\frac{c x + \frac{d x^{2}}{2}}{\left (a \cos{\left (e \right )} + a\right )^{2}} & \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))**2,x)

[Out]

Piecewise((c*tan(e/2 + f*x/2)**3/(6*a**2*f) + c*tan(e/2 + f*x/2)/(2*a**2*f) + d*x*tan(e/2 + f*x/2)**3/(6*a**2*
f) + d*x*tan(e/2 + f*x/2)/(2*a**2*f) - d*log(tan(e/2 + f*x/2)**2 + 1)/(3*a**2*f**2) - d*tan(e/2 + f*x/2)**2/(6
*a**2*f**2), Ne(f, 0)), ((c*x + d*x**2/2)/(a*cos(e) + a)**2, True))

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Giac [B]  time = 1.50304, size = 1022, normalized size = 8.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+a*cos(f*x+e))^2,x, algorithm="giac")

[Out]

-1/6*(3*d*f*x*tan(1/2*f*x)^3*tan(1/2*e)^2 + 3*d*f*x*tan(1/2*f*x)^2*tan(1/2*e)^3 - 2*d*log(4*(tan(1/2*e)^2 + 1)
/(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*f*x)^3*tan(1/2*e)^3 + 3*c*f*tan(1/2*f*x)^3*tan(1/2*e)^2 + 3*c*f*tan(1/2
*f*x)^2*tan(1/2*e)^3 + d*tan(1/2*f*x)^3*tan(1/2*e)^3 + d*f*x*tan(1/2*f*x)^3 - 3*d*f*x*tan(1/2*f*x)^2*tan(1/2*e
) - 3*d*f*x*tan(1/2*f*x)*tan(1/2*e)^2 + 6*d*log(4*(tan(1/2*e)^2 + 1)/(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*f*x
)^2*tan(1/2*e)^2 + d*f*x*tan(1/2*e)^3 + c*f*tan(1/2*f*x)^3 - 3*c*f*tan(1/2*f*x)^2*tan(1/2*e) + d*tan(1/2*f*x)^
3*tan(1/2*e) - 3*c*f*tan(1/2*f*x)*tan(1/2*e)^2 - d*tan(1/2*f*x)^2*tan(1/2*e)^2 + c*f*tan(1/2*e)^3 + d*tan(1/2*
f*x)*tan(1/2*e)^3 + 3*d*f*x*tan(1/2*f*x) + 3*d*f*x*tan(1/2*e) - 6*d*log(4*(tan(1/2*e)^2 + 1)/(tan(1/2*f*x)^4*t
an(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*f*x)*tan(1/2*e) + 3*c*f*tan(1/2*f*x) - d*tan(1/2*f*x)^2 + 3*c*f*tan(1/2*e) + d*tan(1/2*f*
x)*tan(1/2*e) - d*tan(1/2*e)^2 + 2*d*log(4*(tan(1/2*e)^2 + 1)/(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)) - d)/(a^2*f^2*tan(
1/2*f*x)^3*tan(1/2*e)^3 - 3*a^2*f^2*tan(1/2*f*x)^2*tan(1/2*e)^2 + 3*a^2*f^2*tan(1/2*f*x)*tan(1/2*e) - a^2*f^2)