3.2.71 \(\int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx\) [171]

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

[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 = 0.12, 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 = {3400, 4266, 2611, 2320, 6724} \begin {gather*} -\frac {4 i x^2 \text {ArcTan}\left (e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \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 {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}} \end {gather*}

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 2320

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 2611

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 3400

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 4266

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 6724

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 ) \text {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 ) \text {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 = 0.05, size = 146, normalized size = 0.56 \begin {gather*} \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-i d^2 x^2 \text {ArcTan}\left (e^{\frac {1}{2} i (c+d x)}\right )+2 i d x \text {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )-2 i d x \text {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )-4 \text {PolyLog}\left (3,-i e^{\frac {1}{2} i (c+d x)}\right )+4 \text {PolyLog}\left (3,i e^{\frac {1}{2} i (c+d x)}\right )\right )}{d^3 \sqrt {a (1+\cos (c+d x))}} \end {gather*}

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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Maple [F]
time = 0.03, 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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

-2*(sqrt(2)*d^2*x^2*sin(1/2*d*x + 1/2*c) - 4*sqrt(2)*d*x*cos(1/2*d*x + 1/2*c) + (sqrt(2)*d^2*x^2*sin(1/2*d*x +
1/2*c) - 4*sqrt(2)*d*x*cos(1/2*d*x + 1/2*c))*cos(d*x + c) - (sqrt(2)*a*d^5*cos(d*x + c)^2 + sqrt(2)*a*d^5*sin
(d*x + c)^2 + 2*sqrt(2)*a*d^5*cos(d*x + c) + sqrt(2)*a*d^5)*integrate((x^2*cos(2*d*x + 2*c)*cos(1/2*d*x + 1/2*
c) + 2*x^2*cos(d*x + c)*cos(1/2*d*x + 1/2*c) + x^2*sin(2*d*x + 2*c)*sin(1/2*d*x + 1/2*c) + 2*x^2*sin(d*x + c)*
sin(1/2*d*x + 1/2*c) + x^2*cos(1/2*d*x + 1/2*c))/(a*d^2*cos(2*d*x + 2*c)^2 + 4*a*d^2*cos(d*x + c)^2 + a*d^2*si
n(2*d*x + 2*c)^2 + 4*a*d^2*sin(2*d*x + 2*c)*sin(d*x + c) + 4*a*d^2*sin(d*x + c)^2 + 4*a*d^2*cos(d*x + c) + a*d
^2 + 2*(2*a*d^2*cos(d*x + c) + a*d^2)*cos(2*d*x + 2*c)), x) + 4*(sqrt(2)*a*d^4*cos(d*x + c)^2 + sqrt(2)*a*d^4*
sin(d*x + c)^2 + 2*sqrt(2)*a*d^4*cos(d*x + c) + sqrt(2)*a*d^4)*integrate((x*cos(1/2*d*x + 1/2*c)*sin(2*d*x + 2
*c) + 2*x*cos(1/2*d*x + 1/2*c)*sin(d*x + c) - x*cos(2*d*x + 2*c)*sin(1/2*d*x + 1/2*c) - 2*x*cos(d*x + c)*sin(1
/2*d*x + 1/2*c) - x*sin(1/2*d*x + 1/2*c))/(a*d^2*cos(2*d*x + 2*c)^2 + 4*a*d^2*cos(d*x + c)^2 + a*d^2*sin(2*d*x
+ 2*c)^2 + 4*a*d^2*sin(2*d*x + 2*c)*sin(d*x + c) + 4*a*d^2*sin(d*x + c)^2 + 4*a*d^2*cos(d*x + c) + a*d^2 + 2*
(2*a*d^2*cos(d*x + c) + a*d^2)*cos(2*d*x + 2*c)), x) + 2*(sqrt(2)*cos(d*x + c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*
sqrt(2)*cos(d*x + c) + sqrt(2))*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) +
1) - 2*(sqrt(2)*cos(d*x + c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*log(cos(1/2*d*x +
1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - (sqrt(2)*d^2*x^2*cos(1/2*d*x + 1/2*c) + 4*s
qrt(2)*d*x*sin(1/2*d*x + 1/2*c))*sin(d*x + c))/((d^3*cos(d*x + c)^2 + d^3*sin(d*x + c)^2 + 2*d^3*cos(d*x + c)
+ d^3)*sqrt(a))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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