3.4.35 \(\int \frac {1}{x (d+e x^2) \sqrt {a+b x^2+c x^4}} \, dx\) [335]

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

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 138, normalized size of antiderivative = 1.00, 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 = {1265, 974, 738, 212} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])]/(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 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 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 974

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

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Mathematica [A]
time = 0.48, size = 144, normalized size = 1.04 \begin {gather*} \frac {-\frac {e \sqrt {-c d^2+b d e-a e^2} \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 )}{c d^2+e (-b d+a e)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a}}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 207, normalized size = 1.50

method	result	size
default	\(\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{2 d \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 d \sqrt {a}}\)	\(207\)
elliptic	\(\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{2 d \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 d \sqrt {a}}\)	\(207\)

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,method=_RETURNVERBOSE)

[Out]

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))-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)

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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(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)*(x^2*e + d)*x), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (121) = 242\).
time = 0.53, size = 1121, normalized size = 8.12 \begin {gather*} \left [\frac {\sqrt {c d^{2} - b d e + a e^{2}} a e \log \left (-\frac {8 \, c^{2} d^{2} x^{4} + 8 \, b c d^{2} x^{2} + {\left (b^{2} + 4 \, a c\right )} d^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {c d^{2} - b d e + a e^{2}} + {\left ({\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 8 \, a^{2}\right )} e^{2} - 2 \, {\left (4 \, b c d x^{4} + {\left (3 \, b^{2} + 4 \, a c\right )} d x^{2} + 4 \, a b d\right )} e}{x^{4} e^{2} + 2 \, d x^{2} e + d^{2}}\right ) + {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right )}{4 \, {\left (a c d^{3} - a b d^{2} e + a^{2} d e^{2}\right )}}, -\frac {2 \, \sqrt {-c d^{2} + b d e - a e^{2}} a \arctan \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {-c d^{2} + b d e - a e^{2}}}{2 \, {\left (c^{2} d^{2} x^{4} + b c d^{2} x^{2} + a c d^{2} + {\left (a c x^{4} + a b x^{2} + a^{2}\right )} e^{2} - {\left (b c d x^{4} + b^{2} d x^{2} + a b d\right )} e\right )}}\right ) e - {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right )}{4 \, {\left (a c d^{3} - a b d^{2} e + a^{2} d e^{2}\right )}}, \frac {\sqrt {c d^{2} - b d e + a e^{2}} a e \log \left (-\frac {8 \, c^{2} d^{2} x^{4} + 8 \, b c d^{2} x^{2} + {\left (b^{2} + 4 \, a c\right )} d^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {c d^{2} - b d e + a e^{2}} + {\left ({\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 8 \, a^{2}\right )} e^{2} - 2 \, {\left (4 \, b c d x^{4} + {\left (3 \, b^{2} + 4 \, a c\right )} d x^{2} + 4 \, a b d\right )} e}{x^{4} e^{2} + 2 \, d x^{2} e + d^{2}}\right ) + 2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right )}{4 \, {\left (a c d^{3} - a b d^{2} e + a^{2} d e^{2}\right )}}, -\frac {\sqrt {-c d^{2} + b d e - a e^{2}} a \arctan \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c d x^{2} + b d - {\left (b x^{2} + 2 \, a\right )} e\right )} \sqrt {-c d^{2} + b d e - a e^{2}}}{2 \, {\left (c^{2} d^{2} x^{4} + b c d^{2} x^{2} + a c d^{2} + {\left (a c x^{4} + a b x^{2} + a^{2}\right )} e^{2} - {\left (b c d x^{4} + b^{2} d x^{2} + a b d\right )} e\right )}}\right ) e - {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right )}{2 \, {\left (a c d^{3} - a b d^{2} e + a^{2} d e^{2}\right )}}\right ] \end {gather*}

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*x^4 + 8*b*c*d^2*x^2 + (b^2 + 4*a*c)*d^2 - 4*sqrt(c*x^4 +
b*x^2 + a)*(2*c*d*x^2 + b*d - (b*x^2 + 2*a)*e)*sqrt(c*d^2 - b*d*e + a*e^2) + ((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 +
8*a^2)*e^2 - 2*(4*b*c*d*x^4 + (3*b^2 + 4*a*c)*d*x^2 + 4*a*b*d)*e)/(x^4*e^2 + 2*d*x^2*e + 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*arctan(-1/2*sqrt(c*x^4 +
b*x^2 + a)*(2*c*d*x^2 + b*d - (b*x^2 + 2*a)*e)*sqrt(-c*d^2 + b*d*e - a*e^2)/(c^2*d^2*x^4 + b*c*d^2*x^2 + a*c*d
^2 + (a*c*x^4 + a*b*x^2 + a^2)*e^2 - (b*c*d*x^4 + b^2*d*x^2 + a*b*d)*e))*e - (c*d^2 - b*d*e + a*e^2)*sqrt(a)*l
og(-((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*x^4 + 8*b*c*d^2*x^2 + (b^2 + 4*a
*c)*d^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*d*x^2 + b*d - (b*x^2 + 2*a)*e)*sqrt(c*d^2 - b*d*e + a*e^2) + ((b^2 +
4*a*c)*x^4 + 8*a*b*x^2 + 8*a^2)*e^2 - 2*(4*b*c*d*x^4 + (3*b^2 + 4*a*c)*d*x^2 + 4*a*b*d)*e)/(x^4*e^2 + 2*d*x^2*
e + 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*arctan(-1/2*sqr
t(c*x^4 + b*x^2 + a)*(2*c*d*x^2 + b*d - (b*x^2 + 2*a)*e)*sqrt(-c*d^2 + b*d*e - a*e^2)/(c^2*d^2*x^4 + b*c*d^2*x
^2 + a*c*d^2 + (a*c*x^4 + a*b*x^2 + a^2)*e^2 - (b*c*d*x^4 + b^2*d*x^2 + a*b*d)*e))*e - (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)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (d + e x^{2}\right ) \sqrt {a + b x^{2} + c x^{4}}}\, dx \end {gather*}

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 = 0.00, 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(1/x/(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),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 = -0.01 \begin {gather*} \int \frac {1}{x\,\left (e\,x^2+d\right )\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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