3.335 \(\int \frac{1}{x (d+e x^2) \sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=138 \[ -\frac{e \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 d \sqrt{a e^2-b d e+c d^2}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{a} d} \]

[Out]

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

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Rubi [A]  time = 0.194722, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1251, 960, 724, 206} \[ -\frac{e \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 d \sqrt{a e^2-b d e+c d^2}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

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

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 960

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, 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] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 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 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{1}{x \left (d+e x^2\right ) \sqrt{a+b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x (d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{1}{d x \sqrt{a+b x+c x^2}}-\frac{e}{d (d+e x) \sqrt{a+b x+c x^2}}\right ) \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{2 d}-\frac{e \operatorname{Subst}\left (\int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{\operatorname{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{e \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 )}{d}\\ &=-\frac{\tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{a} d}-\frac{e \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 \sqrt{c d^2-b d e+a e^2}}\\ \end{align*}

Mathematica [A]  time = 0.140591, size = 134, normalized size = 0.97 \[ -\frac{\frac{e \tanh ^{-1}\left (\frac{-2 a e+b \left (d-e x^2\right )+2 c d x^2}{2 \sqrt{a+b x^2+c x^4} \sqrt{e (a e-b d)+c d^2}}\right )}{\sqrt{e (a e-b d)+c d^2}}+\frac{\tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{\sqrt{a}}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

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

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Maple [A]  time = 0.012, size = 207, normalized size = 1.5 \begin{align*} -{\frac{1}{2\,d}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{1}{2\,d}\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}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{4} + b x^{2} + a}{\left (e x^{2} + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.69411, size = 2381, normalized size = 17.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (d + e x^{2}\right ) \sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError