3.126 \(\int \frac{x^4}{\sqrt{-3-4 x-x^2} (3+4 x+2 x^2)} \, dx\)

Optimal. Leaf size=140 \[ -\frac{1}{4} \sqrt{-x^2-4 x-3} x+\frac{5}{2} \sqrt{-x^2-4 x-3}+\frac{\tan ^{-1}\left (\frac{1-\frac{x+3}{\sqrt{-x^2-4 x-3}}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (\frac{\frac{x+3}{\sqrt{-x^2-4 x-3}}+1}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{5}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{-x^2-4 x-3}}\right )+\frac{11}{2} \sin ^{-1}(x+2) \]

[Out]

(5*Sqrt[-3 - 4*x - x^2])/2 - (x*Sqrt[-3 - 4*x - x^2])/4 + (11*ArcSin[2 + x])/2 + ArcTan[(1 - (3 + x)/Sqrt[-3 -
4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) - ArcTan[(1 + (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) - (5*ArcTan
h[x/Sqrt[-3 - 4*x - x^2]])/4

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Rubi [A]  time = 0.500627, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 14, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {6728, 619, 216, 640, 742, 1028, 986, 12, 1026, 1161, 618, 204, 1027, 206} \[ -\frac{1}{4} \sqrt{-x^2-4 x-3} x+\frac{5}{2} \sqrt{-x^2-4 x-3}+\frac{\tan ^{-1}\left (\frac{1-\frac{x+3}{\sqrt{-x^2-4 x-3}}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (\frac{\frac{x+3}{\sqrt{-x^2-4 x-3}}+1}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{5}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{-x^2-4 x-3}}\right )+\frac{11}{2} \sin ^{-1}(x+2) \]

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]

[Out]

(5*Sqrt[-3 - 4*x - x^2])/2 - (x*Sqrt[-3 - 4*x - x^2])/4 + (11*ArcSin[2 + x])/2 + ArcTan[(1 - (3 + x)/Sqrt[-3 -
4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) - ArcTan[(1 + (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) - (5*ArcTan
h[x/Sqrt[-3 - 4*x - x^2]])/4

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 1028

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol]
:> -Dist[(2*h*d - g*e)/e, Int[1/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/e, Int[(2*d + e*x)
/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0] && NeQ[2*h*d - g*e, 0]

Rule 986

Int[1/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt
[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 2]}, Dist[1/(2*q), Int[(c*d - a*f + q + (c*e - b*f)*x)/((a + b*x + c
*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Dist[1/(2*q), Int[(c*d - a*f - q + (c*e - b*f)*x)/((a + b*x + c*x^2)*Sq
rt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] &&
NeQ[c*e - b*f, 0] && NegQ[b^2 - 4*a*c]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1026

Int[(x_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*e,
Subst[Int[(1 - d*x^2)/(c*e - b*f - e*(2*c*d - b*e + 2*a*f)*x^2 + d^2*(c*e - b*f)*x^4), x], x, (1 + ((e + Sqrt[
e^2 - 4*d*f])*x)/(2*d))/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1027

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol]
:> Dist[g, Subst[Int[1/(a + (c*d - a*f)*x^2), x], x, x/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0] && EqQ[2*h*d - g*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^4}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx &=\int \left (\frac{5}{4 \sqrt{-3-4 x-x^2}}-\frac{x}{\sqrt{-3-4 x-x^2}}+\frac{x^2}{2 \sqrt{-3-4 x-x^2}}-\frac{15+8 x}{4 \sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )}\right ) \, dx\\ &=-\left (\frac{1}{4} \int \frac{15+8 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx\right )+\frac{1}{2} \int \frac{x^2}{\sqrt{-3-4 x-x^2}} \, dx+\frac{5}{4} \int \frac{1}{\sqrt{-3-4 x-x^2}} \, dx-\int \frac{x}{\sqrt{-3-4 x-x^2}} \, dx\\ &=\sqrt{-3-4 x-x^2}-\frac{1}{4} x \sqrt{-3-4 x-x^2}-\frac{1}{4} \int \frac{3+6 x}{\sqrt{-3-4 x-x^2}} \, dx+\frac{1}{2} \int \frac{-6-4 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac{5}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4}}} \, dx,x,-4-2 x\right )-\frac{3}{4} \int \frac{1}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+2 \int \frac{1}{\sqrt{-3-4 x-x^2}} \, dx\\ &=\frac{5}{2} \sqrt{-3-4 x-x^2}-\frac{1}{4} x \sqrt{-3-4 x-x^2}+\frac{5}{4} \sin ^{-1}(2+x)+\frac{1}{8} \int \frac{-6-4 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac{1}{8} \int -\frac{4 x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac{9}{4} \int \frac{1}{\sqrt{-3-4 x-x^2}} \, dx-3 \operatorname{Subst}\left (\int \frac{1}{3-3 x^2} \, dx,x,\frac{x}{\sqrt{-3-4 x-x^2}}\right )-\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4}}} \, dx,x,-4-2 x\right )\\ &=\frac{5}{2} \sqrt{-3-4 x-x^2}-\frac{1}{4} x \sqrt{-3-4 x-x^2}+\frac{13}{4} \sin ^{-1}(2+x)-\tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )+\frac{1}{2} \int \frac{x}{\sqrt{-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{3-3 x^2} \, dx,x,\frac{x}{\sqrt{-3-4 x-x^2}}\right )-\frac{9}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4}}} \, dx,x,-4-2 x\right )\\ &=\frac{5}{2} \sqrt{-3-4 x-x^2}-\frac{1}{4} x \sqrt{-3-4 x-x^2}+\frac{11}{2} \sin ^{-1}(2+x)-\frac{5}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )+4 \operatorname{Subst}\left (\int \frac{1+3 x^2}{-4-8 x^2-36 x^4} \, dx,x,\frac{1+\frac{x}{3}}{\sqrt{-3-4 x-x^2}}\right )\\ &=\frac{5}{2} \sqrt{-3-4 x-x^2}-\frac{1}{4} x \sqrt{-3-4 x-x^2}+\frac{11}{2} \sin ^{-1}(2+x)-\frac{5}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{3}-\frac{2 x}{3}+x^2} \, dx,x,\frac{1+\frac{x}{3}}{\sqrt{-3-4 x-x^2}}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{3}+\frac{2 x}{3}+x^2} \, dx,x,\frac{1+\frac{x}{3}}{\sqrt{-3-4 x-x^2}}\right )\\ &=\frac{5}{2} \sqrt{-3-4 x-x^2}-\frac{1}{4} x \sqrt{-3-4 x-x^2}+\frac{11}{2} \sin ^{-1}(2+x)-\frac{5}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-\frac{8}{9}-x^2} \, dx,x,\frac{2}{3} \left (-1+\frac{3+x}{\sqrt{-3-4 x-x^2}}\right )\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-\frac{8}{9}-x^2} \, dx,x,\frac{2}{3} \left (1+\frac{3+x}{\sqrt{-3-4 x-x^2}}\right )\right )\\ &=\frac{5}{2} \sqrt{-3-4 x-x^2}-\frac{1}{4} x \sqrt{-3-4 x-x^2}+\frac{11}{2} \sin ^{-1}(2+x)+\frac{\tan ^{-1}\left (\frac{1-\frac{3+x}{\sqrt{-3-4 x-x^2}}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{\tan ^{-1}\left (\frac{1+\frac{3+x}{\sqrt{-3-4 x-x^2}}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{5}{4} \tanh ^{-1}\left (\frac{x}{\sqrt{-3-4 x-x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.549183, size = 210, normalized size = 1.5 \[ \frac{1}{24} \left (-6 \sqrt{-x^2-4 x-3} x+60 \sqrt{-x^2-4 x-3}-\sqrt{1-2 i \sqrt{2}} \left (4 \sqrt{2}+7 i\right ) \tanh ^{-1}\left (\frac{-i \sqrt{2} x+2 x-2 i \sqrt{2}+2}{\sqrt{2+4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )-\sqrt{1+2 i \sqrt{2}} \left (4 \sqrt{2}-7 i\right ) \tanh ^{-1}\left (\frac{\left (2+i \sqrt{2}\right ) x+2 i \sqrt{2}+2}{\sqrt{2-4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )+132 \sin ^{-1}(x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]

[Out]

(60*Sqrt[-3 - 4*x - x^2] - 6*x*Sqrt[-3 - 4*x - x^2] + 132*ArcSin[2 + x] - Sqrt[1 - (2*I)*Sqrt[2]]*(7*I + 4*Sqr
t[2])*ArcTanh[(2 - (2*I)*Sqrt[2] + 2*x - I*Sqrt[2]*x)/(Sqrt[2 + (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2])] - Sqrt[1
+ (2*I)*Sqrt[2]]*(-7*I + 4*Sqrt[2])*ArcTanh[(2 + (2*I)*Sqrt[2] + (2 + I*Sqrt[2])*x)/(Sqrt[2 - (4*I)*Sqrt[2]]*
Sqrt[-3 - 4*x - x^2])])/24

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Maple [A]  time = 0.1, size = 159, normalized size = 1.1 \begin{align*} -{\frac{x}{4}\sqrt{-{x}^{2}-4\,x-3}}+{\frac{5}{2}\sqrt{-{x}^{2}-4\,x-3}}+{\frac{11\,\arcsin \left ( 2+x \right ) }{2}}+{\frac{\sqrt{4}\sqrt{3}}{24}\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12} \left ( \sqrt{2}\arctan \left ({\frac{\sqrt{2}}{6}\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12}} \right ) +5\,{\it Artanh} \left ( 3\,{\frac{x}{-3/2-x}{\frac{1}{\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12}}}} \right ) \right ){\frac{1}{\sqrt{{ \left ({{x}^{2} \left ( -{\frac{3}{2}}-x \right ) ^{-2}}-4 \right ) \left ( 1+{x \left ( -{\frac{3}{2}}-x \right ) ^{-1}} \right ) ^{-2}}}}} \left ( 1+{x \left ( -{\frac{3}{2}}-x \right ) ^{-1}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x)

[Out]

-1/4*x*(-x^2-4*x-3)^(1/2)+5/2*(-x^2-4*x-3)^(1/2)+11/2*arcsin(2+x)+1/24*3^(1/2)*4^(1/2)*(3*x^2/(-3/2-x)^2-12)^(
1/2)*(2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))+5*arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/
2)))/((x^2/(-3/2-x)^2-4)/(1+x/(-3/2-x))^2)^(1/2)/(1+x/(-3/2-x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{{\left (2 \, x^{2} + 4 \, x + 3\right )} \sqrt{-x^{2} - 4 \, x - 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/((2*x^2 + 4*x + 3)*sqrt(-x^2 - 4*x - 3)), x)

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Fricas [A]  time = 2.19684, size = 494, normalized size = 3.53 \begin{align*} -\frac{1}{4} \, \sqrt{-x^{2} - 4 \, x - 3}{\left (x - 10\right )} + \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{\sqrt{2} x + 3 \, \sqrt{2} \sqrt{-x^{2} - 4 \, x - 3}}{2 \,{\left (2 \, x + 3\right )}}\right ) + \frac{1}{8} \, \sqrt{2} \arctan \left (-\frac{\sqrt{2} x - 3 \, \sqrt{2} \sqrt{-x^{2} - 4 \, x - 3}}{2 \,{\left (2 \, x + 3\right )}}\right ) - \frac{11}{2} \, \arctan \left (\frac{\sqrt{-x^{2} - 4 \, x - 3}{\left (x + 2\right )}}{x^{2} + 4 \, x + 3}\right ) + \frac{5}{16} \, \log \left (-\frac{2 \, \sqrt{-x^{2} - 4 \, x - 3} x + 4 \, x + 3}{x^{2}}\right ) - \frac{5}{16} \, \log \left (\frac{2 \, \sqrt{-x^{2} - 4 \, x - 3} x - 4 \, x - 3}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(-x^2 - 4*x - 3)*(x - 10) + 1/8*sqrt(2)*arctan(1/2*(sqrt(2)*x + 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x
+ 3)) + 1/8*sqrt(2)*arctan(-1/2*(sqrt(2)*x - 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) - 11/2*arctan(sqrt(-x^
2 - 4*x - 3)*(x + 2)/(x^2 + 4*x + 3)) + 5/16*log(-(2*sqrt(-x^2 - 4*x - 3)*x + 4*x + 3)/x^2) - 5/16*log((2*sqrt
(-x^2 - 4*x - 3)*x - 4*x - 3)/x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{- \left (x + 1\right ) \left (x + 3\right )} \left (2 x^{2} + 4 x + 3\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(2*x**2+4*x+3)/(-x**2-4*x-3)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(x + 1)*(x + 3))*(2*x**2 + 4*x + 3)), x)

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Giac [A]  time = 1.27305, size = 254, normalized size = 1.81 \begin{align*} -\frac{1}{4} \, \sqrt{-x^{2} - 4 \, x - 3}{\left (x - 10\right )} + \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\frac{3 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) + \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\frac{\sqrt{-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) + \frac{11}{2} \, \arcsin \left (x + 2\right ) - \frac{5}{8} \, \log \left (\frac{2 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac{3 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 1\right ) + \frac{5}{8} \, \log \left (\frac{2 \,{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac{{\left (\sqrt{-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="giac")

[Out]

-1/4*sqrt(-x^2 - 4*x - 3)*(x - 10) + 1/4*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)
) + 1/4*sqrt(2)*arctan(1/2*sqrt(2)*((sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 11/2*arcsin(x + 2) - 5/8*log(2*
(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 3*(sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 + 1) + 5/8*log(2*(sqrt(-x^2 - 4*
x - 3) - 1)/(x + 2) + (sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 + 3)