3.372 \(\int \frac {x^2 (d+e x^2)^{3/2}}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=491 \[ \frac {\sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \left (-\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{2 c^2 \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \left (\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{2 c^2 \sqrt {\sqrt {b^2-4 a c}+b}}+\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \left (-\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \left (\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac {e x \sqrt {d+e x^2}}{2 c}+\frac {d \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 c} \]

[Out]

1/2*d*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))*e^(1/2)/c+1/2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))*(c*d-b*e+(-2*a*c*e+b
^2*e-b*c*d)/(-4*a*c+b^2)^(1/2))*e^(1/2)/c^2+1/2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))*(c*d-b*e+(2*a*c*e-b^2*e+b*c
*d)/(-4*a*c+b^2)^(1/2))*e^(1/2)/c^2+1/2*e*x*(e*x^2+d)^(1/2)/c+1/2*arctan(x*(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(1
/2)/(e*x^2+d)^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*(c*d-b*e+(-2*a*c*e+b^2*e-b*c*d)/(-4*a*c+b^2)^(1/2))*(2*c*d-e
*(b-(-4*a*c+b^2)^(1/2)))^(1/2)/c^2/(b-(-4*a*c+b^2)^(1/2))^(1/2)+1/2*arctan(x*(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^
(1/2)/(e*x^2+d)^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*(c*d-b*e+(2*a*c*e-b^2*e+b*c*d)/(-4*a*c+b^2)^(1/2))*(2*c*d-
e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)/c^2/(b+(-4*a*c+b^2)^(1/2))^(1/2)

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Rubi [A]  time = 1.80, antiderivative size = 491, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1293, 195, 217, 206, 1692, 402, 377, 205} \[ \frac {\sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \left (-\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{2 c^2 \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \left (\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{2 c^2 \sqrt {\sqrt {b^2-4 a c}+b}}+\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \left (-\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \left (\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right )}{2 c^2}+\frac {e x \sqrt {d+e x^2}}{2 c}+\frac {d \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(d + e*x^2)^(3/2))/(a + b*x^2 + c*x^4),x]

[Out]

(e*x*Sqrt[d + e*x^2])/(2*c) + (Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d - b*e - (b*c*d - b^2*e + 2*a*c*e)/
Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*
x^2])])/(2*c^2*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d - b*e + (b*c*d - b
^2*e + 2*a*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a
*c]]*Sqrt[d + e*x^2])])/(2*c^2*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (d*Sqrt[e]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])
/(2*c) + (Sqrt[e]*(c*d - b*e - (b*c*d - b^2*e + 2*a*c*e)/Sqrt[b^2 - 4*a*c])*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2
]])/(2*c^2) + (Sqrt[e]*(c*d - b*e + (b*c*d - b^2*e + 2*a*c*e)/Sqrt[b^2 - 4*a*c])*ArcTanh[(Sqrt[e]*x)/Sqrt[d +
e*x^2]])/(2*c^2)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 205

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

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])

Rule 217

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 402

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

Rule 1293

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[(
e*f^2)/c, Int[(f*x)^(m - 2)*(d + e*x^2)^(q - 1), x], x] - Dist[f^2/c, Int[((f*x)^(m - 2)*(d + e*x^2)^(q - 1)*S
imp[a*e - (c*d - b*e)*x^2, x])/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c,
0] &&  !IntegerQ[q] && GtQ[q, 0] && GtQ[m, 1] && LeQ[m, 3]

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^2 \left (d+e x^2\right )^{3/2}}{a+b x^2+c x^4} \, dx &=-\frac {\int \frac {\sqrt {d+e x^2} \left (a e-(c d-b e) x^2\right )}{a+b x^2+c x^4} \, dx}{c}+\frac {e \int \sqrt {d+e x^2} \, dx}{c}\\ &=\frac {e x \sqrt {d+e x^2}}{2 c}-\frac {\int \left (\frac {\left (-c d+b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \sqrt {d+e x^2}}{b-\sqrt {b^2-4 a c}+2 c x^2}+\frac {\left (-c d+b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \sqrt {d+e x^2}}{b+\sqrt {b^2-4 a c}+2 c x^2}\right ) \, dx}{c}+\frac {(d e) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{2 c}\\ &=\frac {e x \sqrt {d+e x^2}}{2 c}+\frac {(d e) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 c}+\frac {\left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \int \frac {\sqrt {d+e x^2}}{b+\sqrt {b^2-4 a c}+2 c x^2} \, dx}{c}-\frac {\left (-c d+b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \int \frac {\sqrt {d+e x^2}}{b-\sqrt {b^2-4 a c}+2 c x^2} \, dx}{c}\\ &=\frac {e x \sqrt {d+e x^2}}{2 c}+\frac {d \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 c}+\frac {\left (e \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{2 c^2}+\frac {\left (\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\left (b-\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}} \, dx}{2 c^2}+\frac {\left (e \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{2 c^2}+\frac {\left (\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\left (b+\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}} \, dx}{2 c^2}\\ &=\frac {e x \sqrt {d+e x^2}}{2 c}+\frac {d \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 c}+\frac {\left (e \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 c^2}+\frac {\left (\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b-\sqrt {b^2-4 a c}-\left (-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 c^2}+\frac {\left (e \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 c^2}+\frac {\left (\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b+\sqrt {b^2-4 a c}-\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 c^2}\\ &=\frac {e x \sqrt {d+e x^2}}{2 c}+\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \tan ^{-1}\left (\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{2 c^2 \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \tan ^{-1}\left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{2 c^2 \sqrt {b+\sqrt {b^2-4 a c}}}+\frac {d \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 c}+\frac {\sqrt {e} \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 c^2}+\frac {\sqrt {e} \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 c^2}\\ \end {align*}

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Mathematica [B]  time = 6.26, size = 14032, normalized size = 28.58 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2*(d + e*x^2)^(3/2))/(a + b*x^2 + c*x^4),x]

[Out]

Result too large to show

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fricas [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*(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 1.97, size = 58, normalized size = 0.12 \[ \frac {\sqrt {x^{2} e + d} x e}{2 \, c} - \frac {{\left (3 \, c d e - 2 \, b e^{2}\right )} e^{\left (-\frac {1}{2}\right )} \log \left ({\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2}\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

1/2*sqrt(x^2*e + d)*x*e/c - 1/4*(3*c*d*e - 2*b*e^2)*e^(-1/2)*log((x*e^(1/2) - sqrt(x^2*e + d))^2)/c^2

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maple [C]  time = 0.03, size = 382, normalized size = 0.78 \[ -\frac {e^{\frac {3}{2}} x^{2}}{4 c}+\frac {b \,e^{\frac {3}{2}} \ln \left (-\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{c^{2}}+\frac {d^{2} \sqrt {e}}{8 \left (-\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )^{2} c}-\frac {3 d \sqrt {e}\, \ln \left (-\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{2 c}+\frac {\sqrt {e \,x^{2}+d}\, e x}{4 c}-\frac {d \sqrt {e}}{8 c}+\frac {\sqrt {e}\, \left (a c \,d^{2} e^{2}-b^{2} d^{2} e^{2}+2 b c \,d^{3} e -c^{2} d^{4}+\left (a c \,e^{2}-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2}+2 \left (-2 a b \,e^{3}+3 a d \,e^{2} c +b^{2} d \,e^{2}-2 b c \,d^{2} e +c^{2} d^{3}\right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )+\left (-\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )^{2}\right )}{2 c^{2} \left (\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{3} c +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} b e -3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} c d +8 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) a \,e^{2}-4 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) b d e +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) c \,d^{2}+b \,d^{2} e -c \,d^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x)

[Out]

-1/4*e^(3/2)/c*x^2+1/4*e*x*(e*x^2+d)^(1/2)/c-1/8*e^(1/2)/c*d+1/2*e^(1/2)/c^2*sum(((a*c*e^2-b^2*e^2+2*b*c*d*e-c
^2*d^2)*_R^2+2*(-2*a*b*e^3+3*a*c*d*e^2+b^2*d*e^2-2*b*c*d^2*e+c^2*d^3)*_R+a*c*d^2*e^2-b^2*d^2*e^2+2*b*c*d^3*e-c
^2*d^4)/(_R^3*c+3*_R^2*b*e-3*_R^2*c*d+8*_R*a*e^2-4*_R*b*d*e+3*_R*c*d^2+b*d^2*e-c*d^3)*ln(-_R+(-e^(1/2)*x+(e*x^
2+d)^(1/2))^2),_R=RootOf(_Z^4*c+c*d^4+(4*b*e-4*c*d)*_Z^3+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^2+(4*b*d^2*e-4*c*d^3)*_
Z))+1/8*e^(1/2)/c*d^2/(-e^(1/2)*x+(e*x^2+d)^(1/2))^2+e^(3/2)/c^2*ln(-e^(1/2)*x+(e*x^2+d)^(1/2))*b-3/2*e^(1/2)/
c*ln(-e^(1/2)*x+(e*x^2+d)^(1/2))*d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c x^{4} + b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^(3/2)*x^2/(c*x^4 + b*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (e\,x^2+d\right )}^{3/2}}{c\,x^4+b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(d + e*x^2)^(3/2))/(a + b*x^2 + c*x^4),x)

[Out]

int((x^2*(d + e*x^2)^(3/2))/(a + b*x^2 + c*x^4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (d + e x^{2}\right )^{\frac {3}{2}}}{a + b x^{2} + c x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)

[Out]

Integral(x**2*(d + e*x**2)**(3/2)/(a + b*x**2 + c*x**4), x)

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