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

Optimal. Leaf size=102 \[ \frac {4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {i (c+d x)^2}{a f}-\frac {4 i d^2 \text {Li}_2\left (e^{i (e+f x)}\right )}{a f^3} \]

[Out]

-I*(d*x+c)^2/a/f-(d*x+c)^2*cot(1/2*e+1/2*f*x)/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

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Rubi [A]  time = 0.20, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3318, 4184, 3717, 2190, 2279, 2391} \[ \frac {4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {i (c+d x)^2}{a f}-\frac {4 i d^2 \text {Li}_2\left (e^{i (e+f x)}\right )}{a f^3} \]

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) - ((c + d*x)^2*Cot[e/2 + (f*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)

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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}{2}+\frac {f x}{2}\right ) \, dx}{2 a}\\ &=-\frac {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {(2 d) \int (c+d x) \cot \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 \cot \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 {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\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 {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {4 d (c+d x) \log \left (1-e^{i (e+f x)}\right )}{a f^2}+\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 {(c+d x)^2 \cot \left (\frac {e}{2}+\frac {f x}{2}\right )}{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}\\ \end {align*}

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Mathematica [B]  time = 5.53, size = 292, normalized size = 2.86 \[ \frac {2 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (-2 c d f \sin \left (\frac {1}{2} (e+f x)\right ) \left (f x \cos \left (\frac {e}{2}\right )-2 \sin \left (\frac {e}{2}\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+f^2 (c+d x)^2 \sin \left (\frac {f x}{2}\right )+d^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (f^2 x^2 \cos \left (\frac {e}{2}\right ) \left (-e^{i \tan ^{-1}\left (\tan \left (\frac {e}{2}\right )\right )}\right ) \sqrt {\sec ^2\left (\frac {e}{2}\right )}-4 \sin \left (\frac {e}{2}\right ) \left (i \text {Li}_2\left (e^{i \left (f x+2 \tan ^{-1}\left (\tan \left (\frac {e}{2}\right )\right )\right )}\right )-\frac {1}{2} i f x \left (\pi -2 \tan ^{-1}\left (\tan \left (\frac {e}{2}\right )\right )\right )-\left (2 \tan ^{-1}\left (\tan \left (\frac {e}{2}\right )\right )+f x\right ) \log \left (1-e^{i \left (2 \tan ^{-1}\left (\tan \left (\frac {e}{2}\right )\right )+f x\right )}\right )+2 \tan ^{-1}\left (\tan \left (\frac {e}{2}\right )\right ) \log \left (\sin \left (\tan ^{-1}\left (\tan \left (\frac {e}{2}\right )\right )+\frac {f x}{2}\right )\right )-\pi \log \left (1+e^{-i f x}\right )+\pi \log \left (\cos \left (\frac {f x}{2}\right )\right )\right )\right )\right )}{f^3 (a-a \cos (e+f x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Csc[e/2]*Sin[(e + f*x)/2]*(f^2*(c + d*x)^2*Sin[(f*x)/2] - 2*c*d*f*(f*x*Cos[e/2] - 2*Log[Sin[(e + f*x)/2]]*S
in[e/2])*Sin[(e + f*x)/2] + d^2*(-(E^(I*ArcTan[Tan[e/2]])*f^2*x^2*Cos[e/2]*Sqrt[Sec[e/2]^2]) - 4*((-1/2*I)*f*x
*(Pi - 2*ArcTan[Tan[e/2]]) - Pi*Log[1 + E^((-I)*f*x)] - (f*x + 2*ArcTan[Tan[e/2]])*Log[1 - E^(I*(f*x + 2*ArcTa
n[Tan[e/2]]))] + Pi*Log[Cos[(f*x)/2]] + 2*ArcTan[Tan[e/2]]*Log[Sin[(f*x)/2 + ArcTan[Tan[e/2]]]] + I*PolyLog[2,
E^(I*(f*x + 2*ArcTan[Tan[e/2]]))])*Sin[e/2])*Sin[(e + f*x)/2]))/(f^3*(a - a*Cos[e + f*x]))

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fricas [B]  time = 0.65, size = 283, normalized size = 2.77 \[ -\frac {d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2} + 2 i \, d^{2} {\rm Li}_2\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 2 i \, d^{2} {\rm Li}_2\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) + 2 \, {\left (d^{2} e - c d f\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2} i \, \sin \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) + 2 \, {\left (d^{2} e - c d f\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) - \frac {1}{2} i \, \sin \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) - 2 \, {\left (d^{2} f x + d^{2} e\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 2 \, {\left (d^{2} f x + d^{2} e\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} \cos \left (f x + e\right )}{a f^{3} \sin \left (f x + e\right )} \]

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]

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

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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 = 0.12, size = 247, normalized size = 2.42 \[ -\frac {2 i \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}+\frac {4 d c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{a \,f^{2}}-\frac {2 i d^{2} x^{2}}{a f}-\frac {4 i d^{2} e x}{a \,f^{2}}-\frac {2 i d^{2} e^{2}}{a \,f^{3}}+\frac {4 d^{2} \ln \left (1-{\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{2}}+\frac {4 d^{2} \ln \left (1-{\mathrm e}^{i \left (f x +e \right )}\right ) e}{a \,f^{3}}-\frac {4 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{a \,f^{3}} \]

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

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maxima [B]  time = 0.68, size = 314, normalized size = 3.08 \[ -\frac {2 \, c^{2} f^{2} - 4 \, {\left (c d f \cos \left (f x + e\right ) + i \, c d f \sin \left (f x + e\right ) - c d f\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) + {\left (4 \, d^{2} f x \cos \left (f x + e\right ) + 4 i \, d^{2} f x \sin \left (f x + e\right ) - 4 \, d^{2} f x\right )} \arctan \left (\sin \left (f x + e\right ), -\cos \left (f x + e\right ) + 1\right ) + 2 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x\right )} \cos \left (f x + e\right ) + {\left (4 \, d^{2} \cos \left (f x + e\right ) + 4 i \, d^{2} \sin \left (f x + e\right ) - 4 \, d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, f x + i \, e\right )}\right ) - {\left (2 i \, d^{2} f x + 2 i \, c d f + {\left (-2 i \, d^{2} f x - 2 i \, c d f\right )} \cos \left (f x + e\right ) + 2 \, {\left (d^{2} f x + c d f\right )} \sin \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right ) - {\left (-2 i \, d^{2} f^{2} x^{2} - 4 i \, c d f^{2} x\right )} \sin \left (f x + e\right )}{-i \, a f^{3} \cos \left (f x + e\right ) + a f^{3} \sin \left (f x + e\right ) + i \, a f^{3}} \]

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*(c*d*f*cos(f*x + e) + I*c*d*f*sin(f*x + e) - c*d*f)*arctan2(sin(f*x + e), cos(f*x + e) - 1) +
(4*d^2*f*x*cos(f*x + e) + 4*I*d^2*f*x*sin(f*x + e) - 4*d^2*f*x)*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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