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

Optimal. Leaf size=208 \[ \frac {\left (-4 c e (b d-a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} e^3}-\frac {d \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^3}-\frac {\sqrt {a+b x^2+c x^4} \left (-b e+4 c d-2 c e x^2\right )}{8 c e^2} \]

[Out]

1/16*(8*c^2*d^2-b^2*e^2-4*c*e*(-a*e+b*d))*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(3/2)/e^3-1
/2*d*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x^2)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2))*(a*e^2-b*d*e+c*
d^2)^(1/2)/e^3-1/8*(-2*c*e*x^2-b*e+4*c*d)*(c*x^4+b*x^2+a)^(1/2)/c/e^2

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Rubi [A]  time = 0.31, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1251, 814, 843, 621, 206, 724} \[ \frac {\left (-4 c e (b d-a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} e^3}-\frac {d \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^3}-\frac {\sqrt {a+b x^2+c x^4} \left (-b e+4 c d-2 c e x^2\right )}{8 c e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*c*d - b*e - 2*c*e*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c*e^2) + ((8*c^2*d^2 - b^2*e^2 - 4*c*e*(b*d - a*e))*Ar
cTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(3/2)*e^3) - (d*Sqrt[c*d^2 - b*d*e + a*e^2]*Ar
cTanh[(b*d - 2*a*e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x^2 + c*x^4])])/(2*e^3)

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 621

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

Rule 724

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

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {a+b x^2+c x^4}}{d+e x^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x \sqrt {a+b x+c x^2}}{d+e x} \, dx,x,x^2\right )\\ &=-\frac {\left (4 c d-b e-2 c e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c e^2}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2} d \left (4 b c d-b^2 e-4 a c e\right )-\frac {1}{2} \left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{8 c e^2}\\ &=-\frac {\left (4 c d-b e-2 c e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c e^2}-\frac {\left (d \left (c d^2-b d e+a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{2 e^3}+\frac {\left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 c e^3}\\ &=-\frac {\left (4 c d-b e-2 c e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c e^2}+\frac {\left (d \left (c d^2-b d e+a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x^2}{\sqrt {a+b x^2+c x^4}}\right )}{e^3}+\frac {\left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^2}{\sqrt {a+b x^2+c x^4}}\right )}{8 c e^3}\\ &=-\frac {\left (4 c d-b e-2 c e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c e^2}+\frac {\left (8 c^2 d^2-b^2 e^2-4 c e (b d-a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} e^3}-\frac {d \sqrt {c d^2-b d e+a e^2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x^2}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x^2+c x^4}}\right )}{2 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 205, normalized size = 0.99 \[ \frac {\left (4 c e (a e-b d)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )+2 \sqrt {c} \left (4 c d \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {2 a e-b d+b e x^2-2 c d x^2}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )+e \sqrt {a+b x^2+c x^4} \left (b e-4 c d+2 c e x^2\right )\right )}{16 c^{3/2} e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*c^2*d^2 - b^2*e^2 + 4*c*e*(-(b*d) + a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])] + 2*S
qrt[c]*(e*(-4*c*d + b*e + 2*c*e*x^2)*Sqrt[a + b*x^2 + c*x^4] + 4*c*d*Sqrt[c*d^2 - b*d*e + a*e^2]*ArcTanh[(-(b*
d) + 2*a*e - 2*c*d*x^2 + b*e*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x^2 + c*x^4])]))/(16*c^(3/2)*e^3)

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fricas [A]  time = 46.63, size = 1231, normalized size = 5.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x, algorithm="fricas")

[Out]

[1/32*(8*sqrt(c*d^2 - b*d*e + a*e^2)*c^2*d*log(-((8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^4 - 8*a*b*d*e +
8*a^2*e^2 + (b^2 + 4*a*c)*d^2 + 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x^2 - 4*sqrt(c*x^4 + b*x^2 +
a)*sqrt(c*d^2 - b*d*e + a*e^2)*((2*c*d - b*e)*x^2 + b*d - 2*a*e))/(e^2*x^4 + 2*d*e*x^2 + d^2)) + (8*c^2*d^2 -
4*b*c*d*e - (b^2 - 4*a*c)*e^2)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 +
b)*sqrt(c) - 4*a*c) + 4*(2*c^2*e^2*x^2 - 4*c^2*d*e + b*c*e^2)*sqrt(c*x^4 + b*x^2 + a))/(c^2*e^3), -1/32*(16*s
qrt(-c*d^2 + b*d*e - a*e^2)*c^2*d*arctan(-1/2*sqrt(c*x^4 + b*x^2 + a)*sqrt(-c*d^2 + b*d*e - a*e^2)*((2*c*d - b
*e)*x^2 + b*d - 2*a*e)/((c^2*d^2 - b*c*d*e + a*c*e^2)*x^4 + a*c*d^2 - a*b*d*e + a^2*e^2 + (b*c*d^2 - b^2*d*e +
a*b*e^2)*x^2)) - (8*c^2*d^2 - 4*b*c*d*e - (b^2 - 4*a*c)*e^2)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqr
t(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) - 4*(2*c^2*e^2*x^2 - 4*c^2*d*e + b*c*e^2)*sqrt(c*x^4 + b*x
^2 + a))/(c^2*e^3), 1/16*(4*sqrt(c*d^2 - b*d*e + a*e^2)*c^2*d*log(-((8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2
)*x^4 - 8*a*b*d*e + 8*a^2*e^2 + (b^2 + 4*a*c)*d^2 + 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x^2 - 4*sq
rt(c*x^4 + b*x^2 + a)*sqrt(c*d^2 - b*d*e + a*e^2)*((2*c*d - b*e)*x^2 + b*d - 2*a*e))/(e^2*x^4 + 2*d*e*x^2 + d^
2)) - (8*c^2*d^2 - 4*b*c*d*e - (b^2 - 4*a*c)*e^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sq
rt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(2*c^2*e^2*x^2 - 4*c^2*d*e + b*c*e^2)*sqrt(c*x^4 + b*x^2 + a))/(c^2*e^3)
, -1/16*(8*sqrt(-c*d^2 + b*d*e - a*e^2)*c^2*d*arctan(-1/2*sqrt(c*x^4 + b*x^2 + a)*sqrt(-c*d^2 + b*d*e - a*e^2)
*((2*c*d - b*e)*x^2 + b*d - 2*a*e)/((c^2*d^2 - b*c*d*e + a*c*e^2)*x^4 + a*c*d^2 - a*b*d*e + a^2*e^2 + (b*c*d^2
- b^2*d*e + a*b*e^2)*x^2)) + (8*c^2*d^2 - 4*b*c*d*e - (b^2 - 4*a*c)*e^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x
^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) - 2*(2*c^2*e^2*x^2 - 4*c^2*d*e + b*c*e^2)*sqrt(c*x^4
+ b*x^2 + a))/(c^2*e^3)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.01, size = 887, normalized size = 4.26 \[ \frac {a d \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x^{2}+\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, e^{2}}-\frac {b \,d^{2} \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x^{2}+\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, e^{3}}+\frac {c \,d^{3} \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x^{2}+\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, e^{4}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{4 e}+\frac {a \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 \sqrt {c}\, e}-\frac {b^{2} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {3}{2}} e}-\frac {b d \ln \left (\frac {\left (x^{2}+\frac {d}{e}\right ) c +\frac {b e -2 c d}{2 e}}{\sqrt {c}}+\sqrt {\left (x^{2}+\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\right )}{4 \sqrt {c}\, e^{2}}+\frac {\sqrt {c}\, d^{2} \ln \left (\frac {\left (x^{2}+\frac {d}{e}\right ) c +\frac {b e -2 c d}{2 e}}{\sqrt {c}}+\sqrt {\left (x^{2}+\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\right )}{2 e^{3}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{8 c e}-\frac {\sqrt {\left (x^{2}+\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, d}{2 e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x)

[Out]

1/4/e*(c*x^4+b*x^2+a)^(1/2)*x^2+1/8/e/c*(c*x^4+b*x^2+a)^(1/2)*b+1/4/e/c^(1/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+
b*x^2+a)^(1/2))*a-1/16/e/c^(3/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*b^2-1/2*d/e^2*((x^2+d/e)^2*c+
(b*e-2*c*d)*(x^2+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)-1/4*d/e^2*ln(((x^2+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((
x^2+d/e)^2*c+(b*e-2*c*d)*(x^2+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)*b+1/2*d^2/e^3*ln(((x^2+d/e)*c+1/2
*(b*e-2*c*d)/e)/c^(1/2)+((x^2+d/e)^2*c+(b*e-2*c*d)*(x^2+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)+1/2*d/e
^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x^2+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2
)/e^2)^(1/2)*((x^2+d/e)^2*c+(b*e-2*c*d)*(x^2+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*a-1/2*d^2/e^3/(
(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x^2+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^
2)^(1/2)*((x^2+d/e)^2*c+(b*e-2*c*d)*(x^2+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*b+1/2*d^3/e^4/((a*e
^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x^2+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(
1/2)*((x^2+d/e)^2*c+(b*e-2*c*d)*(x^2+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*c

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^4+b*x^2+a)^(1/2)/(e*x^2+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*x^2 + c*x^4)^(1/2))/(d + e*x^2),x)

[Out]

int((x^3*(a + b*x^2 + c*x^4)^(1/2))/(d + e*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c*x**4+b*x**2+a)**(1/2)/(e*x**2+d),x)

[Out]

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

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