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

Optimal. Leaf size=269 \[ \frac {\sqrt {a+b x^2+c x^4} \left (-2 c e (5 b d-4 a e)+b^2 e^2-2 c e x^2 (2 c d-b e)+8 c^2 d^2\right )}{16 c e^3}-\frac {(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{3/2} e^4}+\frac {\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^4}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 e} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1247, 734, 814, 843, 621, 206, 724} \[ \frac {\sqrt {a+b x^2+c x^4} \left (-2 c e (5 b d-4 a e)+b^2 e^2-2 c e x^2 (2 c d-b e)+8 c^2 d^2\right )}{16 c e^3}-\frac {(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{3/2} e^4}+\frac {\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^4}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 4*a*e) - 2*c*e*(2*c*d - b*e)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*c*e^3) +
(a + b*x^2 + c*x^4)^(3/2)/(6*e) - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 - 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b + 2
*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(3/2)*e^4) + ((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*e^4)

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 734

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 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 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(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]

Rubi steps

\begin {align*} \int \frac {x \left (a+b x^2+c x^4\right )^{3/2}}{d+e x^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx,x,x^2\right )\\ &=\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 e}-\frac {\operatorname {Subst}\left (\int \frac {(b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{d+e x} \, dx,x,x^2\right )}{4 e}\\ &=\frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c e^3}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 e}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (4 c e (b d-2 a e)^2-d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )-\frac {1}{2} (2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 c e^3}\\ &=\frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c e^3}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 e}+\frac {\left (c d^2-b d e+a e^2\right )^2 \operatorname {Subst}\left (\int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{2 e^4}-\frac {\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{32 c e^4}\\ &=\frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c e^3}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 e}-\frac {\left (c d^2-b d e+a e^2\right )^2 \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^4}-\frac {\left ((2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right )\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 )}{16 c e^4}\\ &=\frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c e^3}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 e}-\frac {(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{3/2} e^4}+\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 e^4}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 255, normalized size = 0.95 \[ \frac {2 \sqrt {c} \left (e \sqrt {a+b x^2+c x^4} \left (2 c e \left (16 a e-15 b d+7 b e x^2\right )+3 b^2 e^2+4 c^2 \left (6 d^2-3 d e x^2+2 e^2 x^4\right )\right )-24 c \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {2 a e-b d+b e x^2-2 c d x^2}{2 \sqrt {a+b x^2+c x^4} \sqrt {e (a e-b d)+c d^2}}\right )\right )-3 (2 c d-b e) \left (4 c e (3 a e-2 b d)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{96 c^{3/2} e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 + 4*c*e*(-2*b*d + 3*a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x
^2 + c*x^4])] + 2*Sqrt[c]*(e*Sqrt[a + b*x^2 + c*x^4]*(3*b^2*e^2 + 2*c*e*(-15*b*d + 16*a*e + 7*b*e*x^2) + 4*c^2
*(6*d^2 - 3*d*e*x^2 + 2*e^2*x^4)) - 24*c*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x^2
+ b*e*x^2)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + b*x^2 + c*x^4])]))/(96*c^(3/2)*e^4)

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

[Out]

Timed out

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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*(c*x^4+b*x^2+a)^(3/2)/(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 = 1411, normalized size = 5.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*(c*x^4+b*x^2+a)^(3/2)/(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(a*e^2-b*d*e>0)', see `assume?`
for more details)Is a*e^2-b*d*e                            +c*d^2 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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