3.186 \(\int \frac{x^2}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=329 \[ -\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}+\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}-\frac{2 i \text{Li}_3\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \sqrt{a^2-b^2}}+\frac{2 i \text{Li}_3\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{d^3 \sqrt{a^2-b^2}}-\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \sqrt{a^2-b^2}}+\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \sqrt{a^2-b^2}} \]

[Out]

((-I)*x^2*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) + (I*x^2*Log[1 + (b*E^(I*(c
+ d*x)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) - (2*x*PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b
^2]))])/(Sqrt[a^2 - b^2]*d^2) + (2*x*PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))])/(Sqrt[a^2 - b^2
]*d^2) - ((2*I)*PolyLog[3, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))])/(Sqrt[a^2 - b^2]*d^3) + ((2*I)*PolyL
og[3, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))])/(Sqrt[a^2 - b^2]*d^3)

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Rubi [A]  time = 0.662918, antiderivative size = 329, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3321, 2264, 2190, 2531, 2282, 6589} \[ -\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}+\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}-\frac{2 i \text{Li}_3\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \sqrt{a^2-b^2}}+\frac{2 i \text{Li}_3\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{d^3 \sqrt{a^2-b^2}}-\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \sqrt{a^2-b^2}}+\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*Cos[c + d*x]),x]

[Out]

((-I)*x^2*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) + (I*x^2*Log[1 + (b*E^(I*(c
+ d*x)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) - (2*x*PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b
^2]))])/(Sqrt[a^2 - b^2]*d^2) + (2*x*PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))])/(Sqrt[a^2 - b^2
]*d^2) - ((2*I)*PolyLog[3, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))])/(Sqrt[a^2 - b^2]*d^3) + ((2*I)*PolyL
og[3, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))])/(Sqrt[a^2 - b^2]*d^3)

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{x^2}{a+b \cos (c+d x)} \, dx &=2 \int \frac{e^{i (c+d x)} x^2}{b+2 a e^{i (c+d x)}+b e^{2 i (c+d x)}} \, dx\\ &=\frac{(2 b) \int \frac{e^{i (c+d x)} x^2}{2 a-2 \sqrt{a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2}}-\frac{(2 b) \int \frac{e^{i (c+d x)} x^2}{2 a+2 \sqrt{a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2}}\\ &=-\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{(2 i) \int x \log \left (1+\frac{2 b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} d}-\frac{(2 i) \int x \log \left (1+\frac{2 b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} d}\\ &=-\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}-\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}+\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}+\frac{2 \int \text{Li}_2\left (-\frac{2 b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} d^2}-\frac{2 \int \text{Li}_2\left (-\frac{2 b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} d^2}\\ &=-\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}-\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}+\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{b x}{-a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt{a^2-b^2} d^3}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt{a^2-b^2} d^3}\\ &=-\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i x^2 \log \left (1+\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}-\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}+\frac{2 x \text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}-\frac{2 i \text{Li}_3\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^3}+\frac{2 i \text{Li}_3\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^3}\\ \end{align*}

Mathematica [A]  time = 0.701405, size = 379, normalized size = 1.15 \[ \frac{e^{i c} \left (-i \left (d^2 x^2 \log \left (1+\frac{b e^{i (2 c+d x)}}{a e^{i c}-\sqrt{e^{2 i c} \left (a^2-b^2\right )}}\right )-d^2 x^2 \log \left (1+\frac{b e^{i (2 c+d x)}}{\sqrt{e^{2 i c} \left (a^2-b^2\right )}+a e^{i c}}\right )+2 i d x \text{Li}_2\left (-\frac{b e^{i (2 c+d x)}}{e^{i c} a+\sqrt{\left (a^2-b^2\right ) e^{2 i c}}}\right )+2 \text{Li}_3\left (-\frac{b e^{i (2 c+d x)}}{a e^{i c}-\sqrt{\left (a^2-b^2\right ) e^{2 i c}}}\right )-2 \text{Li}_3\left (-\frac{b e^{i (2 c+d x)}}{e^{i c} a+\sqrt{\left (a^2-b^2\right ) e^{2 i c}}}\right )\right )-2 d x \text{Li}_2\left (-\frac{b e^{i (2 c+d x)}}{a e^{i c}-\sqrt{\left (a^2-b^2\right ) e^{2 i c}}}\right )\right )}{d^3 \sqrt{e^{2 i c} \left (a^2-b^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*Cos[c + d*x]),x]

[Out]

(E^(I*c)*(-2*d*x*PolyLog[2, -((b*E^(I*(2*c + d*x)))/(a*E^(I*c) - Sqrt[(a^2 - b^2)*E^((2*I)*c)]))] - I*(d^2*x^2
*Log[1 + (b*E^(I*(2*c + d*x)))/(a*E^(I*c) - Sqrt[(a^2 - b^2)*E^((2*I)*c)])] - d^2*x^2*Log[1 + (b*E^(I*(2*c + d
*x)))/(a*E^(I*c) + Sqrt[(a^2 - b^2)*E^((2*I)*c)])] + (2*I)*d*x*PolyLog[2, -((b*E^(I*(2*c + d*x)))/(a*E^(I*c) +
Sqrt[(a^2 - b^2)*E^((2*I)*c)]))] + 2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(a*E^(I*c) - Sqrt[(a^2 - b^2)*E^((2*I
)*c)]))] - 2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(a*E^(I*c) + Sqrt[(a^2 - b^2)*E^((2*I)*c)]))])))/(d^3*Sqrt[(a^
2 - b^2)*E^((2*I)*c)])

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Maple [F]  time = 0.44, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{a+b\cos \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.51322, size = 3218, normalized size = 9.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*b*d*x*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(d*x + c) + 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*
b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) - 4*b*d*x*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(d*x +
c) + 2*I*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) + 4*b*d*x*
sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c)
)*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) - 4*b*d*x*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(d*x + c) - 2*I*a*sin
(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*I*b*c^2*sqrt((a^2 -
b^2)/b^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) + 2*I*b*c^2*sqrt((a^2 -
b^2)/b^2)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) - 2*I*b*c^2*sqrt((a^2
- b^2)/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) - 2*a) + 2*I*b*c^2*sqrt((a^
2 - b^2)/b^2)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) - 2*a) + 2*(I*b*d^2*x^2 -
I*b*c^2)*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(d*x + c) + 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d
*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b) + 2*(-I*b*d^2*x^2 + I*b*c^2)*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(d
*x + c) + 2*I*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b) + 2*(-I*b
*d^2*x^2 + I*b*c^2)*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) -
I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b) + 2*(I*b*d^2*x^2 - I*b*c^2)*sqrt((a^2 - b^2)/b^2)*log(1/2*(
2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b)
+ 4*I*b*sqrt((a^2 - b^2)/b^2)*polylog(3, -1/2*(2*a*cos(d*x + c) + 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b
*sin(d*x + c))*sqrt((a^2 - b^2)/b^2))/b) - 4*I*b*sqrt((a^2 - b^2)/b^2)*polylog(3, -1/2*(2*a*cos(d*x + c) + 2*I
*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2))/b) - 4*I*b*sqrt((a^2 - b^2)/b^2
)*polylog(3, -1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b
^2)/b^2))/b) + 4*I*b*sqrt((a^2 - b^2)/b^2)*polylog(3, -1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) - 2*(b*cos(d
*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2))/b))/((a^2 - b^2)*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a + b*cos(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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