\(\int \frac {x (d+e x^2)}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [76]

Optimal result

Integrand size = 31, antiderivative size = 83 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {e \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 b^2}+\frac {(b d-a e) \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1261, 654, 622, 31} \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\left (a+b x^2\right ) (b d-a e) \log \left (a+b x^2\right )}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {e \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 b^2} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 622

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1261

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(276\) vs. \(2(83)=166\).

Time = 0.55 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.33 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\left (2 a+b x^2\right ) \left (b e x^2 \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )-(-b d+a e) \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right ) \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+(-b d+a e) \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right ) \log \left (b^2 \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )\right )}{2 b^2 \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )} \]

[In]

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

[Out]

-1/2*((2*a + b*x^2)*(b*e*x^2*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])) - (-(b*d) + a*e)*(-a^2 -
a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2])*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + (-(b*d) + a*e)*(-a^2 -
a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2])*Log[b^2*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])]))/(b^2*(Sqrt[a^2]
- Sqrt[(a + b*x^2)^2])*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54

method	result	size
pseudoelliptic	\(-\frac {\left (-e \,x^{2} b +\ln \left (b \,x^{2}+a \right ) a e -\ln \left (b \,x^{2}+a \right ) b d \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(45\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (-e \,x^{2} b +\ln \left (b \,x^{2}+a \right ) a e -\ln \left (b \,x^{2}+a \right ) b d \right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{2}}\)	\(55\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, e \,x^{2}}{2 \left (b \,x^{2}+a \right ) b}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -b d \right ) \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{2}}\)	\(72\)

[In]

int(x*(e*x^2+d)/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(-e*x^2*b+ln(b*x^2+a)*a*e-ln(b*x^2+a)*b*d)*csgn(b*x^2+a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.35 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {b e x^{2} + {\left (b d - a e\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.91 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.81 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\begin {cases} \frac {\left (\frac {a}{b} + x^{2}\right ) \left (- \frac {a e}{b} + d\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{\sqrt {b^{2} \left (\frac {a}{b} + x^{2}\right )^{2}}} + \frac {e \sqrt {a^{2} + 2 a b x^{2} + b^{2} x^{4}}}{b^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {2 d \sqrt {a^{2} + 2 a b x^{2}} + \frac {e \left (- a^{2} \sqrt {a^{2} + 2 a b x^{2}} + \frac {\left (a^{2} + 2 a b x^{2}\right )^{\frac {3}{2}}}{3}\right )}{a b}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {d x^{2} + \frac {e x^{4}}{2}}{\sqrt {a^{2}}} & \text {otherwise} \end {cases}}{2} \]

[In]

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

[Out]

Piecewise(((a/b + x**2)*(-a*e/b + d)*log(a/b + x**2)/sqrt(b**2*(a/b + x**2)**2) + e*sqrt(a**2 + 2*a*b*x**2 + b
**2*x**4)/b**2, Ne(b**2, 0)), ((2*d*sqrt(a**2 + 2*a*b*x**2) + e*(-a**2*sqrt(a**2 + 2*a*b*x**2) + (a**2 + 2*a*b
*x**2)**(3/2)/3)/(a*b))/(2*a*b), Ne(a*b, 0)), ((d*x**2 + e*x**4/2)/sqrt(a**2), True))/2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.37 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {e x^{2}}{2 \, b} + \frac {{\left (b d - a e\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.48 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {1}{2} \, {\left (\frac {e x^{2}}{b} + \frac {{\left (b d - a e\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{b^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

1/2*(e*x^2/b + (b*d - a*e)*log(abs(b*x^2 + a))/b^2)*sgn(b*x^2 + a)

Mupad [B] (verification not implemented)

Time = 8.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.24 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {e\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,b^2}-\frac {a\,b\,e\,\ln \left (a\,b+\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {b^2}+b^2\,x^2\right )}{2\,{\left (b^2\right )}^{3/2}}+\frac {b^2\,d\,\ln \left (b^2\,x^2+a\,b\right )\,\mathrm {sign}\left (2\,b^2\,x^2+2\,a\,b\right )}{2\,{\left (b^2\right )}^{3/2}} \]

[In]

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

[Out]

(e*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(2*b^2) - (a*b*e*log(a*b + ((a + b*x^2)^2)^(1/2)*(b^2)^(1/2) + b^2*x^2))
/(2*(b^2)^(3/2)) + (b^2*d*log(a*b + b^2*x^2)*sign(2*a*b + 2*b^2*x^2))/(2*(b^2)^(3/2))