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

Optimal. Leaf size=101 \[ \frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {i (c+d x)^2}{a f}-\frac {4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right )}{a f^3} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3318, 4184, 3719, 2190, 2279, 2391} \[ \frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {i (c+d x)^2}{a f}-\frac {4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right )}{a f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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)^2}{a+a \cos (e+f x)} \, dx &=\frac {\int (c+d x)^2 \csc ^2\left (\frac {e+\pi }{2}+\frac {f x}{2}\right ) \, dx}{2 a}\\ &=\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {(2 d) \int (c+d x) \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \, dx}{a f}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a 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}{a f}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a 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}{a f^2}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a 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 )}{a f^3}\\ &=-\frac {i (c+d x)^2}{a f}+\frac {4 d (c+d x) \log \left (1+e^{i (e+f x)}\right )}{a f^2}-\frac {4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right )}{a f^3}+\frac {(c+d x)^2 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 125, normalized size = 1.24 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (f (c+d x) \left (f (c+d x) \sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right ) \left (4 d \log \left (1+e^{i (e+f x)}\right )-i f (c+d x)\right )\right )-4 i d^2 \text {Li}_2\left (-e^{i (e+f x)}\right ) \cos \left (\frac {1}{2} (e+f x)\right )\right )}{a f^3 (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 222, normalized size = 2.20 \[ \frac {{\left (2 i \, d^{2} \cos \left (f x + e\right ) + 2 i \, d^{2}\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + {\left (-2 i \, d^{2} \cos \left (f x + e\right ) - 2 i \, d^{2}\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (d^{2} f x + c d f + {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (d^{2} f x + c d f + {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \sin \left (f x + e\right )}{a f^{3} \cos \left (f x + e\right ) + a f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{2}}{a \cos \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.11, size = 197, normalized size = 1.95 \[ \frac {2 i \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}+\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a \,f^{2}}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}-\frac {2 i d^{2} x^{2}}{a f}-\frac {4 i d^{2} e x}{a \,f^{2}}-\frac {2 i d^{2} e^{2}}{a \,f^{3}}+\frac {4 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{a \,f^{2}}-\frac {4 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.06, size = 286, normalized size = 2.83 \[ \frac {2 \, c^{2} f^{2} + {\left (4 \, d^{2} f x + 4 \, c d f + 4 \, {\left (d^{2} f x + c d f\right )} \cos \left (f x + e\right ) + {\left (4 i \, d^{2} f x + 4 i \, c d f\right )} \sin \left (f x + e\right )\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) - 2 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x\right )} \cos \left (f x + e\right ) - {\left (4 \, d^{2} \cos \left (f x + e\right ) + 4 i \, d^{2} \sin \left (f x + e\right ) + 4 \, d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, f x + i \, e\right )}\right ) + {\left (-2 i \, d^{2} f x - 2 i \, c d f + {\left (-2 i \, d^{2} f x - 2 i \, c d f\right )} \cos \left (f x + e\right ) + 2 \, {\left (d^{2} f x + c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right ) + {\left (-2 i \, d^{2} f^{2} x^{2} - 4 i \, c d f^{2} x\right )} \sin \left (f x + e\right )}{-i \, a f^{3} \cos \left (f x + e\right ) + a f^{3} \sin \left (f x + e\right ) - i \, a f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^2}{a+a\,\cos \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{2}}{\cos {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} x^{2}}{\cos {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d x}{\cos {\left (e + f x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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