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

Optimal. Leaf size=262 \[ -\frac {16 \text {Li}_3\left (-i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3 \sqrt {a \cos (c+d x)+a}}+\frac {16 \text {Li}_3\left (i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3 \sqrt {a \cos (c+d x)+a}}+\frac {8 i x \text {Li}_2\left (-i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^2 \sqrt {a \cos (c+d x)+a}}-\frac {8 i x \text {Li}_2\left (i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^2 \sqrt {a \cos (c+d x)+a}}-\frac {4 i x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \tan ^{-1}\left (e^{\frac {1}{2} i (c+d x)}\right )}{d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-4*I*x^2*arctan(exp(1/2*I*(d*x+c)))*cos(1/2*d*x+1/2*c)/d/(a+a*cos(d*x+c))^(1/2)+8*I*x*cos(1/2*d*x+1/2*c)*polyl
og(2,-I*exp(1/2*I*(d*x+c)))/d^2/(a+a*cos(d*x+c))^(1/2)-8*I*x*cos(1/2*d*x+1/2*c)*polylog(2,I*exp(1/2*I*(d*x+c))
)/d^2/(a+a*cos(d*x+c))^(1/2)-16*cos(1/2*d*x+1/2*c)*polylog(3,-I*exp(1/2*I*(d*x+c)))/d^3/(a+a*cos(d*x+c))^(1/2)
+16*cos(1/2*d*x+1/2*c)*polylog(3,I*exp(1/2*I*(d*x+c)))/d^3/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3319, 4181, 2531, 2282, 6589} \[ \frac {8 i x \text {Li}_2\left (-i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^2 \sqrt {a \cos (c+d x)+a}}-\frac {8 i x \text {Li}_2\left (i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^2 \sqrt {a \cos (c+d x)+a}}-\frac {16 \text {Li}_3\left (-i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3 \sqrt {a \cos (c+d x)+a}}+\frac {16 \text {Li}_3\left (i e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3 \sqrt {a \cos (c+d x)+a}}-\frac {4 i x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \tan ^{-1}\left (e^{\frac {1}{2} i (c+d x)}\right )}{d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

((-4*I)*x^2*ArcTan[E^((I/2)*(c + d*x))]*Cos[c/2 + (d*x)/2])/(d*Sqrt[a + a*Cos[c + d*x]]) + ((8*I)*x*Cos[c/2 +
(d*x)/2]*PolyLog[2, (-I)*E^((I/2)*(c + d*x))])/(d^2*Sqrt[a + a*Cos[c + d*x]]) - ((8*I)*x*Cos[c/2 + (d*x)/2]*Po
lyLog[2, I*E^((I/2)*(c + d*x))])/(d^2*Sqrt[a + a*Cos[c + d*x]]) - (16*Cos[c/2 + (d*x)/2]*PolyLog[3, (-I)*E^((I
/2)*(c + d*x))])/(d^3*Sqrt[a + a*Cos[c + d*x]]) + (16*Cos[c/2 + (d*x)/2]*PolyLog[3, I*E^((I/2)*(c + d*x))])/(d
^3*Sqrt[a + a*Cos[c + d*x]])

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 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 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 4181

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

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}{\sqrt {a+a \cos (c+d x)}} \, dx &=\frac {\sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right ) \int x^2 \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{\sqrt {a+a \cos (c+d x)}}\\ &=-\frac {4 i x^2 \tan ^{-1}\left (e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cos (c+d x)}}-\frac {\left (4 \sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x \log \left (1-i e^{i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{d \sqrt {a+a \cos (c+d x)}}+\frac {\left (4 \sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x \log \left (1+i e^{i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{d \sqrt {a+a \cos (c+d x)}}\\ &=-\frac {4 i x^2 \tan ^{-1}\left (e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cos (c+d x)}}+\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_2\left (-i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_2\left (i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {\left (8 i \sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \text {Li}_2\left (-i e^{i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{d^2 \sqrt {a+a \cos (c+d x)}}+\frac {\left (8 i \sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \text {Li}_2\left (i e^{i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{d^2 \sqrt {a+a \cos (c+d x)}}\\ &=-\frac {4 i x^2 \tan ^{-1}\left (e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cos (c+d x)}}+\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_2\left (-i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_2\left (i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {\left (16 \sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{d^3 \sqrt {a+a \cos (c+d x)}}+\frac {\left (16 \sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{d^3 \sqrt {a+a \cos (c+d x)}}\\ &=-\frac {4 i x^2 \tan ^{-1}\left (e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cos (c+d x)}}+\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_2\left (-i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_2\left (i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {16 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_3\left (-i e^{\frac {1}{2} i (c+d x)}\right )}{d^3 \sqrt {a+a \cos (c+d x)}}+\frac {16 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Li}_3\left (i e^{\frac {1}{2} i (c+d x)}\right )}{d^3 \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 146, normalized size = 0.56 \[ \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-i d^2 x^2 \tan ^{-1}\left (e^{\frac {1}{2} i (c+d x)}\right )+2 i d x \text {Li}_2\left (-i e^{\frac {1}{2} i (c+d x)}\right )-2 i d x \text {Li}_2\left (i e^{\frac {1}{2} i (c+d x)}\right )-4 \text {Li}_3\left (-i e^{\frac {1}{2} i (c+d x)}\right )+4 \text {Li}_3\left (i e^{\frac {1}{2} i (c+d x)}\right )\right )}{d^3 \sqrt {a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(4*Cos[(c + d*x)/2]*((-I)*d^2*x^2*ArcTan[E^((I/2)*(c + d*x))] + (2*I)*d*x*PolyLog[2, (-I)*E^((I/2)*(c + d*x))]
- (2*I)*d*x*PolyLog[2, I*E^((I/2)*(c + d*x))] - 4*PolyLog[3, (-I)*E^((I/2)*(c + d*x))] + 4*PolyLog[3, I*E^((I
/2)*(c + d*x))]))/(d^3*Sqrt[a*(1 + Cos[c + d*x])])

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\sqrt {a \cos \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(x^2/sqrt(a*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a +a \cos \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+a*cos(d*x+c))^(1/2),x)

[Out]

int(x^2/(a+a*cos(d*x+c))^(1/2),x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + a*cos(c + d*x))^(1/2),x)

[Out]

int(x^2/(a + a*cos(c + d*x))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+a*cos(d*x+c))**(1/2),x)

[Out]

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

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