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

Optimal. Leaf size=212 \[ \frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}-\frac{i (c+d x)^2}{3 a^2 f}-\frac{4 i d^2 \text{Li}_2\left (-e^{i (e+f x)}\right )}{3 a^2 f^3}+\frac{2 d^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^3} \]

[Out]

((-I/3)*(c + d*x)^2)/(a^2*f) + (4*d*(c + d*x)*Log[1 + E^(I*(e + f*x))])/(3*a^2*f^2) - (((4*I)/3)*d^2*PolyLog[2
, -E^(I*(e + f*x))])/(a^2*f^3) - (d*(c + d*x)*Sec[e/2 + (f*x)/2]^2)/(3*a^2*f^2) + (2*d^2*Tan[e/2 + (f*x)/2])/(
3*a^2*f^3) + ((c + d*x)^2*Tan[e/2 + (f*x)/2])/(3*a^2*f) + ((c + d*x)^2*Sec[e/2 + (f*x)/2]^2*Tan[e/2 + (f*x)/2]
)/(6*a^2*f)

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Rubi [A]  time = 0.255528, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {3318, 4186, 3767, 8, 4184, 3719, 2190, 2279, 2391} \[ \frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{6 a^2 f}-\frac{i (c+d x)^2}{3 a^2 f}-\frac{4 i d^2 \text{Li}_2\left (-e^{i (e+f x)}\right )}{3 a^2 f^3}+\frac{2 d^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/3)*(c + d*x)^2)/(a^2*f) + (4*d*(c + d*x)*Log[1 + E^(I*(e + f*x))])/(3*a^2*f^2) - (((4*I)/3)*d^2*PolyLog[2
, -E^(I*(e + f*x))])/(a^2*f^3) - (d*(c + d*x)*Sec[e/2 + (f*x)/2]^2)/(3*a^2*f^2) + (2*d^2*Tan[e/2 + (f*x)/2])/(
3*a^2*f^3) + ((c + d*x)^2*Tan[e/2 + (f*x)/2])/(3*a^2*f) + ((c + d*x)^2*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 4186

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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]

Rule 3719

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

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{(c+d x)^2}{(a+a \cos (e+f x))^2} \, dx &=\frac{\int (c+d x)^2 \csc ^4\left (\frac{e+\pi }{2}+\frac{f x}{2}\right ) \, dx}{4 a^2}\\ &=-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{(c+d x)^2 \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)^2 \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{6 a^2}+\frac{d^2 \int \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{3 a^2 f^2}\\ &=-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x)^2 \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{\left (2 d^2\right ) \operatorname{Subst}\left (\int 1 \, dx,x,-\tan \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{3 a^2 f^3}-\frac{(2 d) \int (c+d x) \tan \left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{3 a^2 f}\\ &=-\frac{i (c+d x)^2}{3 a^2 f}-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{2 d^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^3}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x)^2 \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{(4 i d) \int \frac{e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)}{1+e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}} \, dx}{3 a^2 f}\\ &=-\frac{i (c+d x)^2}{3 a^2 f}+\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{2 d^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^3}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x)^2 \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{\left (4 d^2\right ) \int \log \left (1+e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{3 a^2 f^2}\\ &=-\frac{i (c+d x)^2}{3 a^2 f}+\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{2 d^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^3}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x)^2 \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{\left (4 i d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right )}{3 a^2 f^3}\\ &=-\frac{i (c+d x)^2}{3 a^2 f}+\frac{4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{3 a^2 f^2}-\frac{4 i d^2 \text{Li}_2\left (-e^{i (e+f x)}\right )}{3 a^2 f^3}-\frac{d (c+d x) \sec ^2\left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^2}+\frac{2 d^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f^3}+\frac{(c+d x)^2 \tan \left (\frac{e}{2}+\frac{f x}{2}\right )}{3 a^2 f}+\frac{(c+d x)^2 \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 = 1.06372, size = 212, normalized size = 1. \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right ) \left (\left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+2\right )\right ) \cos (e+f x)+2 \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+1\right )\right )\right )-2 d f (c+d x) \cos \left (\frac{1}{2} (e+f x)\right )-2 i f (c+d x) \cos ^3\left (\frac{1}{2} (e+f x)\right ) \left (f (c+d x)+4 i d \log \left (1+e^{i (e+f x)}\right )\right )-8 i d^2 \text{Li}_2\left (-e^{i (e+f x)}\right ) \cos ^3\left (\frac{1}{2} (e+f x)\right )\right )}{3 a^2 f^3 (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(e + f*x)/2]*(-2*d*f*(c + d*x)*Cos[(e + f*x)/2] - (2*I)*f*(c + d*x)*Cos[(e + f*x)/2]^3*(f*(c + d*x) + (
4*I)*d*Log[1 + E^(I*(e + f*x))]) - (8*I)*d^2*Cos[(e + f*x)/2]^3*PolyLog[2, -E^(I*(e + f*x))] + (2*(c^2*f^2 + 2
*c*d*f^2*x + d^2*(1 + f^2*x^2)) + (c^2*f^2 + 2*c*d*f^2*x + d^2*(2 + f^2*x^2))*Cos[e + f*x])*Sin[(e + f*x)/2]))
/(3*a^2*f^3*(1 + Cos[e + f*x])^2)

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Maple [B]  time = 0.349, size = 358, normalized size = 1.7 \begin{align*}{\frac{{\frac{2\,i}{3}} \left ( 2\,i{d}^{2}fx{{\rm e}^{2\,i \left ( fx+e \right ) }}+3\,{d}^{2}{f}^{2}{x}^{2}{{\rm e}^{i \left ( fx+e \right ) }}+2\,icdf{{\rm e}^{2\,i \left ( fx+e \right ) }}+2\,if{d}^{2}x{{\rm e}^{i \left ( fx+e \right ) }}+6\,cd{f}^{2}x{{\rm e}^{i \left ( fx+e \right ) }}+{f}^{2}{x}^{2}{d}^{2}+2\,ifcd{{\rm e}^{i \left ( fx+e \right ) }}+3\,{c}^{2}{f}^{2}{{\rm e}^{i \left ( fx+e \right ) }}+2\,cd{f}^{2}x+{c}^{2}{f}^{2}+2\,{d}^{2}{{\rm e}^{2\,i \left ( fx+e \right ) }}+4\,{d}^{2}{{\rm e}^{i \left ( fx+e \right ) }}+2\,{d}^{2} \right ) }{{a}^{2}{f}^{3} \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) ^{3}}}+{\frac{4\,cd\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) }{3\,{a}^{2}{f}^{2}}}-{\frac{4\,cd\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{3\,{a}^{2}{f}^{2}}}-{\frac{{\frac{2\,i}{3}}{d}^{2}{x}^{2}}{{a}^{2}f}}-{\frac{{\frac{4\,i}{3}}{d}^{2}ex}{{a}^{2}{f}^{2}}}-{\frac{{\frac{2\,i}{3}}{d}^{2}{e}^{2}}{{a}^{2}{f}^{3}}}+{\frac{4\,{d}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) x}{3\,{a}^{2}{f}^{2}}}-{\frac{{\frac{4\,i}{3}}{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{a}^{2}{f}^{3}}}+{\frac{4\,{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{3\,{a}^{2}{f}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*I*(2*I*d^2*f*x*exp(2*I*(f*x+e))+3*d^2*f^2*x^2*exp(I*(f*x+e))+2*I*c*d*f*exp(2*I*(f*x+e))+2*I*f*d^2*x*exp(I*
(f*x+e))+6*c*d*f^2*x*exp(I*(f*x+e))+f^2*x^2*d^2+2*I*f*c*d*exp(I*(f*x+e))+3*c^2*f^2*exp(I*(f*x+e))+2*c*d*f^2*x+
c^2*f^2+2*d^2*exp(2*I*(f*x+e))+4*d^2*exp(I*(f*x+e))+2*d^2)/f^3/a^2/(exp(I*(f*x+e))+1)^3+4/3/f^2*d/a^2*c*ln(exp
(I*(f*x+e))+1)-4/3/f^2*d/a^2*c*ln(exp(I*(f*x+e)))-2/3*I/f*d^2/a^2*x^2-4/3*I/f^2*d^2/a^2*e*x-2/3*I/f^3*d^2/a^2*
e^2+4/3/f^2*d^2/a^2*ln(exp(I*(f*x+e))+1)*x-4/3*I*d^2*polylog(2,-exp(I*(f*x+e)))/a^2/f^3+4/3/f^3*d^2/a^2*e*ln(e
xp(I*(f*x+e)))

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Maxima [B]  time = 2.65131, size = 1041, normalized size = 4.91 \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)^2/(a+a*cos(f*x+e))^2,x, algorithm="maxima")

[Out]

(2*c^2*f^2 + 4*d^2 + (4*d^2*f*x + 4*c*d*f + 4*(d^2*f*x + c*d*f)*cos(3*f*x + 3*e) + 12*(d^2*f*x + c*d*f)*cos(2*
f*x + 2*e) + 12*(d^2*f*x + c*d*f)*cos(f*x + e) + (4*I*d^2*f*x + 4*I*c*d*f)*sin(3*f*x + 3*e) + (12*I*d^2*f*x +
12*I*c*d*f)*sin(2*f*x + 2*e) + (12*I*d^2*f*x + 12*I*c*d*f)*sin(f*x + e))*arctan2(sin(f*x + e), cos(f*x + e) +
1) - 2*(d^2*f^2*x^2 + 2*c*d*f^2*x)*cos(3*f*x + 3*e) - (6*d^2*f^2*x^2 - 4*I*c*d*f - 4*d^2 + (12*c*d*f^2 - 4*I*d
^2*f)*x)*cos(2*f*x + 2*e) + (6*c^2*f^2 + 4*I*d^2*f*x + 4*I*c*d*f + 8*d^2)*cos(f*x + e) - (4*d^2*cos(3*f*x + 3*
e) + 12*d^2*cos(2*f*x + 2*e) + 12*d^2*cos(f*x + e) + 4*I*d^2*sin(3*f*x + 3*e) + 12*I*d^2*sin(2*f*x + 2*e) + 12
*I*d^2*sin(f*x + e) + 4*d^2)*dilog(-e^(I*f*x + I*e)) + (-2*I*d^2*f*x - 2*I*c*d*f + (-2*I*d^2*f*x - 2*I*c*d*f)*
cos(3*f*x + 3*e) + (-6*I*d^2*f*x - 6*I*c*d*f)*cos(2*f*x + 2*e) + (-6*I*d^2*f*x - 6*I*c*d*f)*cos(f*x + e) + 2*(
d^2*f*x + c*d*f)*sin(3*f*x + 3*e) + 6*(d^2*f*x + c*d*f)*sin(2*f*x + 2*e) + 6*(d^2*f*x + c*d*f)*sin(f*x + e))*l
og(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) + (-2*I*d^2*f^2*x^2 - 4*I*c*d*f^2*x)*sin(3*f*x + 3*e)
+ (-6*I*d^2*f^2*x^2 - 4*c*d*f + 4*I*d^2 - 4*(3*I*c*d*f^2 + d^2*f)*x)*sin(2*f*x + 2*e) + (6*I*c^2*f^2 - 4*d^2*
f*x - 4*c*d*f + 8*I*d^2)*sin(f*x + e))/(-3*I*a^2*f^3*cos(3*f*x + 3*e) - 9*I*a^2*f^3*cos(2*f*x + 2*e) - 9*I*a^2
*f^3*cos(f*x + e) + 3*a^2*f^3*sin(3*f*x + 3*e) + 9*a^2*f^3*sin(2*f*x + 2*e) + 9*a^2*f^3*sin(f*x + e) - 3*I*a^2
*f^3)

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Fricas [B]  time = 1.75407, size = 956, normalized size = 4.51 \begin{align*} -\frac{2 \, d^{2} f x + 2 \, c d f + 2 \,{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right ) -{\left (2 i \, d^{2} \cos \left (f x + e\right )^{2} + 4 i \, d^{2} \cos \left (f x + e\right ) + 2 i \, d^{2}\right )}{\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) -{\left (-2 i \, d^{2} \cos \left (f x + e\right )^{2} - 4 i \, d^{2} \cos \left (f x + e\right ) - 2 i \, d^{2}\right )}{\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \,{\left (d^{2} f x + c d f +{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (d^{2} f x + c d f +{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) -{\left (2 \, d^{2} f^{2} x^{2} + 4 \, c d f^{2} x + 2 \, c^{2} f^{2} + 2 \, d^{2} +{\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (a^{2} f^{3} \cos \left (f x + e\right )^{2} + 2 \, a^{2} f^{3} \cos \left (f x + e\right ) + a^{2} f^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*d^2*f*x + 2*c*d*f + 2*(d^2*f*x + c*d*f)*cos(f*x + e) - (2*I*d^2*cos(f*x + e)^2 + 4*I*d^2*cos(f*x + e)
+ 2*I*d^2)*dilog(-cos(f*x + e) + I*sin(f*x + e)) - (-2*I*d^2*cos(f*x + e)^2 - 4*I*d^2*cos(f*x + e) - 2*I*d^2)*
dilog(-cos(f*x + e) - I*sin(f*x + e)) - 2*(d^2*f*x + c*d*f + (d^2*f*x + c*d*f)*cos(f*x + e)^2 + 2*(d^2*f*x + c
*d*f)*cos(f*x + e))*log(cos(f*x + e) + I*sin(f*x + e) + 1) - 2*(d^2*f*x + c*d*f + (d^2*f*x + c*d*f)*cos(f*x +
e)^2 + 2*(d^2*f*x + c*d*f)*cos(f*x + e))*log(cos(f*x + e) - I*sin(f*x + e) + 1) - (2*d^2*f^2*x^2 + 4*c*d*f^2*x
+ 2*c^2*f^2 + 2*d^2 + (d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f^2 + 2*d^2)*cos(f*x + e))*sin(f*x + e))/(a^2*f^3*cos(
f*x + e)^2 + 2*a^2*f^3*cos(f*x + e) + a^2*f^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{2}}{\cos ^{2}{\left (e + f x \right )} + 2 \cos{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{2} x^{2}}{\cos ^{2}{\left (e + f x \right )} + 2 \cos{\left (e + f x \right )} + 1}\, dx + \int \frac{2 c d x}{\cos ^{2}{\left (e + f x \right )} + 2 \cos{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c**2/(cos(e + f*x)**2 + 2*cos(e + f*x) + 1), x) + Integral(d**2*x**2/(cos(e + f*x)**2 + 2*cos(e + f*
x) + 1), x) + Integral(2*c*d*x/(cos(e + f*x)**2 + 2*cos(e + f*x) + 1), x))/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{2}}{{\left (a \cos \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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