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

Optimal. Leaf size=166 \[ \frac{e^2 \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{2 a c e+b^2 (-e)+c x^2 (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4} \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.17072, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {1247, 740, 12, 724, 206} \[ \frac{e^2 \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{2 a c e+b^2 (-e)+c x^2 (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*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] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 12

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

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

Mathematica [A]  time = 0.164521, size = 167, normalized size = 1.01 \[ -\frac{2 a c e+b^2 (-e)+c x^2 (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4} \left (e (a e-b d)+c d^2\right )}-\frac{e^2 \tanh ^{-1}\left (\frac{2 a e-b d+b e x^2-2 c d x^2}{2 \sqrt{a+b x^2+c x^4} \sqrt{e (a e-b d)+c d^2}}\right )}{2 \left (e (a e-b d)+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 454, normalized size = 2.7 \begin{align*} -2\,{\frac{c}{ \left ( e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd \right ) \left ( -4\,ac+{b}^{2} \right ) }\sqrt{c \left ({x}^{2}-1/2\,{\frac{-b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) ^{2}+\sqrt{-4\,ac+{b}^{2}} \left ({x}^{2}-1/2\,{\frac{-b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) } \left ({x}^{2}-1/2\,{\frac{-b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) ^{-1}}+2\,{\frac{ce}{ \left ( e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) }\ln \left ({ \left ( 2\,{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ({x}^{2}+{\frac{d}{e}} \right ) }+2\,\sqrt{{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}}\sqrt{c \left ({x}^{2}+{\frac{d}{e}} \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({x}^{2}+{\frac{d}{e}} \right ) }+{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}} \right ) \left ({x}^{2}+{\frac{d}{e}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}-deb+c{d}^{2}}{{e}^{2}}}}}}}+2\,{\frac{c}{ \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) \left ( -4\,ac+{b}^{2} \right ) }\sqrt{c \left ({x}^{2}+1/2\,{\frac{b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) ^{2}-\sqrt{-4\,ac+{b}^{2}} \left ({x}^{2}+1/2\,{\frac{b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) } \left ({x}^{2}+1/2\,{\frac{b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 5.24246, size = 2815, normalized size = 16.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((b^2*c - 4*a*c^2)*e^2*x^4 + (b^3 - 4*a*b*c)*e^2*x^2 + (a*b^2 - 4*a^2*c)*e^2)*sqrt(c*d^2 - b*d*e + a*e^2
)*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)) - 4*(b*c^2*d^3 - 2*(b^2*c - a*c^2)*d^2*e + (b^3 - a*b*c)
*d*e^2 - (a*b^2 - 2*a^2*c)*e^3 + (2*c^3*d^3 - 3*b*c^2*d^2*e - a*b*c*e^3 + (b^2*c + 2*a*c^2)*d*e^2)*x^2)*sqrt(c
*x^4 + b*x^2 + a))/((a*b^2*c^2 - 4*a^2*c^3)*d^4 - 2*(a*b^3*c - 4*a^2*b*c^2)*d^3*e + (a*b^4 - 2*a^2*b^2*c - 8*a
^3*c^2)*d^2*e^2 - 2*(a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^3*b^2 - 4*a^4*c)*e^4 + ((b^2*c^3 - 4*a*c^4)*d^4 - 2*(b^3*
c^2 - 4*a*b*c^3)*d^3*e + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d^2*e^2 - 2*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (a^2*b^
2*c - 4*a^3*c^2)*e^4)*x^4 + ((b^3*c^2 - 4*a*b*c^3)*d^4 - 2*(b^4*c - 4*a*b^2*c^2)*d^3*e + (b^5 - 2*a*b^3*c - 8*
a^2*b*c^2)*d^2*e^2 - 2*(a*b^4 - 4*a^2*b^2*c)*d*e^3 + (a^2*b^3 - 4*a^3*b*c)*e^4)*x^2), 1/2*(((b^2*c - 4*a*c^2)*
e^2*x^4 + (b^3 - 4*a*b*c)*e^2*x^2 + (a*b^2 - 4*a^2*c)*e^2)*sqrt(-c*d^2 + b*d*e - a*e^2)*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)) - 2*(b*c^2*d^3 - 2*(b^2*c - a*c^2)*d^2*e
+ (b^3 - a*b*c)*d*e^2 - (a*b^2 - 2*a^2*c)*e^3 + (2*c^3*d^3 - 3*b*c^2*d^2*e - a*b*c*e^3 + (b^2*c + 2*a*c^2)*d*
e^2)*x^2)*sqrt(c*x^4 + b*x^2 + a))/((a*b^2*c^2 - 4*a^2*c^3)*d^4 - 2*(a*b^3*c - 4*a^2*b*c^2)*d^3*e + (a*b^4 - 2
*a^2*b^2*c - 8*a^3*c^2)*d^2*e^2 - 2*(a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^3*b^2 - 4*a^4*c)*e^4 + ((b^2*c^3 - 4*a*c^
4)*d^4 - 2*(b^3*c^2 - 4*a*b*c^3)*d^3*e + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d^2*e^2 - 2*(a*b^3*c - 4*a^2*b*c^2)
*d*e^3 + (a^2*b^2*c - 4*a^3*c^2)*e^4)*x^4 + ((b^3*c^2 - 4*a*b*c^3)*d^4 - 2*(b^4*c - 4*a*b^2*c^2)*d^3*e + (b^5
- 2*a*b^3*c - 8*a^2*b*c^2)*d^2*e^2 - 2*(a*b^4 - 4*a^2*b^2*c)*d*e^3 + (a^2*b^3 - 4*a^3*b*c)*e^4)*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.21665, size = 613, normalized size = 3.69 \begin{align*} -\frac{\frac{{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2} + 2 \, a c^{2} d e^{2} - a b c e^{3}\right )} x^{2}}{b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}} + \frac{b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + 2 \, a c^{2} d^{2} e + b^{3} d e^{2} - a b c d e^{2} - a b^{2} e^{3} + 2 \, a^{2} c e^{3}}{b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}}}{\sqrt{c x^{4} + b x^{2} + a}} + \frac{\arctan \left (-\frac{{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} + b d e - a e^{2}}}\right ) e^{2}}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt{-c d^{2} + b d e - a e^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*c^3*d^3 - 3*b*c^2*d^2*e + b^2*c*d*e^2 + 2*a*c^2*d*e^2 - a*b*c*e^3)*x^2/(b^2*c^2*d^4 - 4*a*c^3*d^4 - 2*b^3
*c*d^3*e + 8*a*b*c^2*d^3*e + b^4*d^2*e^2 - 2*a*b^2*c*d^2*e^2 - 8*a^2*c^2*d^2*e^2 - 2*a*b^3*d*e^3 + 8*a^2*b*c*d
*e^3 + a^2*b^2*e^4 - 4*a^3*c*e^4) + (b*c^2*d^3 - 2*b^2*c*d^2*e + 2*a*c^2*d^2*e + b^3*d*e^2 - a*b*c*d*e^2 - a*b
^2*e^3 + 2*a^2*c*e^3)/(b^2*c^2*d^4 - 4*a*c^3*d^4 - 2*b^3*c*d^3*e + 8*a*b*c^2*d^3*e + b^4*d^2*e^2 - 2*a*b^2*c*d
^2*e^2 - 8*a^2*c^2*d^2*e^2 - 2*a*b^3*d*e^3 + 8*a^2*b*c*d*e^3 + a^2*b^2*e^4 - 4*a^3*c*e^4))/sqrt(c*x^4 + b*x^2
+ a) + arctan(-((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))*e^2/((c*d
^2 - b*d*e + a*e^2)*sqrt(-c*d^2 + b*d*e - a*e^2))