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

Optimal. Leaf size=281 \[ \frac{\sqrt{c} \left (d \sqrt{b^2-4 a c}-2 a e+b d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\sqrt{c} \left (-d \sqrt{b^2-4 a c}-2 a e+b d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{\sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 1.35017, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {1251, 897, 1287, 206, 1166, 208} \[ \frac{\sqrt{c} \left (d \sqrt{b^2-4 a c}-2 a e+b d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\sqrt{c} \left (-d \sqrt{b^2-4 a c}-2 a e+b d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{\sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], 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] && IntegersQ[n,
p] && FractionQ[m]

Rule 1287

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

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 1166

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

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.800931, size = 241, normalized size = 0.86 \[ \frac{\frac{\sqrt{2} \left (\left (\sqrt{b^2-4 a c}+b\right ) \sqrt{e \sqrt{b^2-4 a c}-b e+2 c d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )+\left (\sqrt{b^2-4 a c}-b\right ) \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )\right )}{\sqrt{c} \sqrt{b^2-4 a c}}-4 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.023, size = 294, normalized size = 1.1 \begin{align*}{\frac{1}{2\,a}\sqrt{e{x}^{2}+d}}+{\frac{x}{2\,a}\sqrt{e}}-{\frac{d}{2\,a} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}-{\frac{1}{4\,a}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{8}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{6}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{4}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){{\it \_Z}}^{2}+c{d}^{4} \right ) }{\frac{{{\it \_R}}^{6}cd+ \left ( -4\,a{e}^{2}+4\,deb-3\,c{d}^{2} \right ){{\it \_R}}^{4}+d \left ( 4\,a{e}^{2}-4\,deb+3\,c{d}^{2} \right ){{\it \_R}}^{2}-c{d}^{4}}{{{\it \_R}}^{7}c+3\,{{\it \_R}}^{5}be-3\,{{\it \_R}}^{5}cd+8\,{{\it \_R}}^{3}a{e}^{2}-4\,{{\it \_R}}^{3}bde+3\,{{\it \_R}}^{3}c{d}^{2}+{\it \_R}\,b{d}^{2}e-{\it \_R}\,c{d}^{3}}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x-{\it \_R} \right ) }}-{\frac{1}{a}\sqrt{d}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a*(e*x^2+d)^(1/2)+1/2/a*e^(1/2)*x-1/2/a*d/((e*x^2+d)^(1/2)-e^(1/2)*x)-1/4/a*sum((_R^6*c*d+(-4*a*e^2+4*b*d*
e-3*c*d^2)*_R^4+d*(4*a*e^2-4*b*d*e+3*c*d^2)*_R^2-c*d^4)/(_R^7*c+3*_R^5*b*e-3*_R^5*c*d+8*_R^3*a*e^2-4*_R^3*b*d*
e+3*_R^3*c*d^2+_R*b*d^2*e-_R*c*d^3)*ln((e*x^2+d)^(1/2)-e^(1/2)*x-_R),_R=RootOf(c*_Z^8+(4*b*e-4*c*d)*_Z^6+(16*a
*e^2-8*b*d*e+6*c*d^2)*_Z^4+(4*b*d^2*e-4*c*d^3)*_Z^2+c*d^4))-1/a*d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out