3.4.25 \(\int \frac {(a+b x^2+c x^4)^{3/2}}{x (d+e x^2)} \, dx\) [325]

Optimal. Leaf size=350 \[ \frac {a \sqrt {a+b x^2+c x^4}}{2 d}-\frac {\left (4 c d^2-e (5 b d-4 a e)-2 c d e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 d e^2}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 d}+\frac {a b \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} d}+\frac {\left (8 c^2 d^3+b e^2 (3 b d-4 a e)-12 c d e (b d-a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 \sqrt {c} d e^3}-\frac {\left (c d^2-b d e+a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x^2}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x^2+c x^4}}\right )}{2 d e^3} \]

[Out]

-1/2*a^(3/2)*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d-1/2*(a*e^2-b*d*e+c*d^2)^(3/2)*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))/d/e^3+1/4*a*b*arctanh(1/2*(2*c
*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d/c^(1/2)+1/16*(8*c^2*d^3+b*e^2*(-4*a*e+3*b*d)-12*c*d*e*(-a*e+b*d))*arc
tanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d/e^3/c^(1/2)+1/2*a*(c*x^4+b*x^2+a)^(1/2)/d-1/8*(4*c*d^2-e
*(-4*a*e+5*b*d)-2*c*d*e*x^2)*(c*x^4+b*x^2+a)^(1/2)/d/e^2

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Rubi [A]
time = 0.36, antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1265, 909, 748, 857, 635, 212, 738, 828} \begin {gather*} -\frac {a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 d}+\frac {\left (-12 c d e (b d-a e)+b e^2 (3 b d-4 a e)+8 c^2 d^3\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 \sqrt {c} d e^3}-\frac {\sqrt {a+b x^2+c x^4} \left (-e (5 b d-4 a e)+4 c d^2-2 c d e x^2\right )}{8 d e^2}-\frac {\left (a e^2-b d e+c d^2\right )^{3/2} \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 d e^3}+\frac {a \sqrt {a+b x^2+c x^4}}{2 d}+\frac {a b \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a + b*x^2 + c*x^4])/(2*d) - ((4*c*d^2 - e*(5*b*d - 4*a*e) - 2*c*d*e*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*d
*e^2) - (a^(3/2)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*d) + (a*b*ArcTanh[(b + 2*c*x^2
)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*Sqrt[c]*d) + ((8*c^2*d^3 + b*e^2*(3*b*d - 4*a*e) - 12*c*d*e*(b*d -
a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*Sqrt[c]*d*e^3) - ((c*d^2 - b*d*e + a*e^2
)^(3/2)*ArcTanh[(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
*d*e^3)

Rule 212

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

Rule 635

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 738

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 748

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

Rule 828

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 857

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 909

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

Rule 1265

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 {\left (a+b x^2+c x^4\right )^{3/2}}{x \left (d+e x^2\right )} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x (d+e x)} \, dx,x,x^2\right )\\ &=-\frac {\text {Subst}\left (\int \frac {(-b d+a e-c d x) \sqrt {a+b x+c x^2}}{d+e x} \, dx,x,x^2\right )}{2 d}+\frac {a \text {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x} \, dx,x,x^2\right )}{2 d}\\ &=\frac {a \sqrt {a+b x^2+c x^4}}{2 d}-\frac {\left (4 c d^2-e (5 b d-4 a e)-2 c d e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 d e^2}-\frac {a \text {Subst}\left (\int \frac {-2 a-b x}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4 d}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} c \left (4 b c d^3-5 b^2 d^2 e-4 a c d^2 e+12 a b d e^2-8 a^2 e^3\right )+\frac {1}{2} c \left (8 c^2 d^3+b e^2 (3 b d-4 a e)-12 c d e (b d-a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{8 c d e^2}\\ &=\frac {a \sqrt {a+b x^2+c x^4}}{2 d}-\frac {\left (4 c d^2-e (5 b d-4 a e)-2 c d e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 d e^2}+\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{2 d}+\frac {(a b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4 d}-\frac {\left (c d^2-b d e+a e^2\right )^2 \text {Subst}\left (\int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{2 d e^3}+\frac {\left (8 c^2 d^3+b e^2 (3 b d-4 a e)-12 c d e (b d-a e)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 d e^3}\\ &=\frac {a \sqrt {a+b x^2+c x^4}}{2 d}-\frac {\left (4 c d^2-e (5 b d-4 a e)-2 c d e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 d e^2}-\frac {a^2 \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )}{d}+\frac {(a b) \text {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 )}{2 d}+\frac {\left (c d^2-b d e+a e^2\right )^2 \text {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 )}{d e^3}+\frac {\left (8 c^2 d^3+b e^2 (3 b d-4 a e)-12 c d e (b d-a e)\right ) \text {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 d e^3}\\ &=\frac {a \sqrt {a+b x^2+c x^4}}{2 d}-\frac {\left (4 c d^2-e (5 b d-4 a e)-2 c d e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 d e^2}-\frac {a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 d}+\frac {a b \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} d}+\frac {\left (8 c^2 d^3+b e^2 (3 b d-4 a e)-12 c d 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 \sqrt {c} d e^3}-\frac {\left (c d^2-b d e+a e^2\right )^{3/2} \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 d e^3}\\ \end {align*}

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Mathematica [A]
time = 0.94, size = 258, normalized size = 0.74 \begin {gather*} \frac {\left (-4 c d+5 b e+2 c e x^2\right ) \sqrt {a+b x^2+c x^4}}{8 e^2}-\frac {\sqrt {-c d^2+b d e-a e^2} \left (c d^2+e (-b d+a e)\right ) \tan ^{-1}\left (\frac {\sqrt {c} \left (d+e x^2\right )-e \sqrt {a+b x^2+c x^4}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{d e^3}+\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{d}-\frac {\left (8 c^2 d^2+3 b^2 e^2+12 c e (-b d+a e)\right ) \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{16 \sqrt {c} e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*c*d + 5*b*e + 2*c*e*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*e^2) - (Sqrt[-(c*d^2) + b*d*e - a*e^2]*(c*d^2 + e*(-
(b*d) + a*e))*ArcTan[(Sqrt[c]*(d + e*x^2) - e*Sqrt[a + b*x^2 + c*x^4])/Sqrt[-(c*d^2) + e*(b*d - a*e)]])/(d*e^3
) + (a^(3/2)*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/d - ((8*c^2*d^2 + 3*b^2*e^2 + 12*c*e*(-
(b*d) + a*e))*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(16*Sqrt[c]*e^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1612\) vs. \(2(302)=604\).
time = 0.14, size = 1613, normalized size = 4.61

method	result	size
elliptic	\(\frac {c \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 e}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 e}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 e \sqrt {c}}+\frac {3 a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 e}-\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}\, d}{2 e^{2}}-\frac {3 b \sqrt {c}\, d \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 e^{2}}+\frac {c^{\frac {3}{2}} d^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 e^{3}}+\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -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 ) a^{2}}{2 d \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -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 ) a b}{e \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}+\frac {d \ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -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 ) a c}{e^{2} \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}+\frac {d \ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -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 ) b^{2}}{2 e^{2} \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}-\frac {d^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -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 ) b c}{e^{3} \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}+\frac {d^{3} \ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -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 ) c^{2}}{2 e^{4} \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 d}\)	\(1270\)
default	\(\text {Expression too large to display}\)	\(1613\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e/d*(3/4*c^(1/2)/e^3*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*b*d^2+1/6/e*c*x^4*(c*x^4+b*x^2+a)^(1/2)+
7/24/e*b*x^2*(c*x^4+b*x^2+a)^(1/2)+1/16/e/c*b^2*(c*x^4+b*x^2+a)^(1/2)-1/2*c^(3/2)/e^4*ln((1/2*b+c*x^2)/c^(1/2)
+(c*x^4+b*x^2+a)^(1/2))*d^3-1/4/e^2*x^2*c*(c*x^4+b*x^2+a)^(1/2)*d-3/16/c^(1/2)/e^2*ln((1/2*b+c*x^2)/c^(1/2)+(c
*x^4+b*x^2+a)^(1/2))*b^2*d-5/8/e^2*b*(c*x^4+b*x^2+a)^(1/2)*d+1/2/e^3*c*(c*x^4+b*x^2+a)^(1/2)*d^2+3/8/c^(1/2)/e
*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*a*b+2/3/e*a*(c*x^4+b*x^2+a)^(1/2)-1/32/c^(3/2)/e*ln((1/2*b+c*
x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*b^3-3/4*c^(1/2)/e^2*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*a*d-1/
2/e/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^
2)/e^2)^(1/2)*(c*(x^2+d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*a^2+1/e^2/((a*
e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^
(1/2)*(c*(x^2+d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*d*a*b-c/e^3/((a*e^2-b*
d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*
(c*(x^2+d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*a*d^2-1/2/e^3/((a*e^2-b*d*e+
c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(
x^2+d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*b^2*d^2+c/e^4/((a*e^2-b*d*e+c*d^
2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+
d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*d^3*b-1/2*c^2/e^5/((a*e^2-b*d*e+c*d^
2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+
d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*d^4)+1/d*(1/6*c*x^4*(c*x^4+b*x^2+a)^
(1/2)+7/24*b*x^2*(c*x^4+b*x^2+a)^(1/2)+1/16/c*b^2*(c*x^4+b*x^2+a)^(1/2)-1/32/c^(3/2)*b^3*ln((1/2*b+c*x^2)/c^(1
/2)+(c*x^4+b*x^2+a)^(1/2))+3/8*a*b*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))/c^(1/2)+2/3*a*(c*x^4+b*x^2+
a)^(1/2)-1/2*a^(3/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2))

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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