\(\int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [625]

Optimal result

Integrand size = 26, antiderivative size = 75 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{2 b^2}-\frac {a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1125, 654, 622, 31} \[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{2 b^2}-\frac {a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[x^3/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/(2*b^2) - (a*(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 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{2 b^2}-\frac {a \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 {\sqrt {a^2+2 a b x^2+b^2 x^4}}{2 b^2}-\frac {\left (a \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 {\sqrt {a^2+2 a b x^2+b^2 x^4}}{2 b^2}-\frac {a \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 [A] (verified)

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.51 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\frac {b x^2 \left (-\sqrt {a^2} \left (a+b x^2\right )+a \sqrt {\left (a+b x^2\right )^2}\right )}{a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}}+2 a \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )}{2 b^2} \]

[In]

Integrate[x^3/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

((b*x^2*(-(Sqrt[a^2]*(a + b*x^2)) + a*Sqrt[(a + b*x^2)^2]))/(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2]) +
2*a*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])])/(2*b^2)

Maple [C] (warning: unable to verify)

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

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.41

method	result	size
pseudoelliptic	\(-\frac {\left (-b \,x^{2}+\ln \left (b \,x^{2}+a \right ) a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(31\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (-b \,x^{2}+\ln \left (b \,x^{2}+a \right ) a \right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{2}}\)	\(41\)
risch	\(\frac {x^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 \left (b \,x^{2}+a \right ) b}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{2}}\)	\(64\)

[In]

int(x^3/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.29 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {b x^{2} - a \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

integrate(x^3/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/2*(b*x^2 - a*log(b*x^2 + a))/b^2

Sympy [F]

\[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {x^{3}}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx \]

[In]

integrate(x**3/((b*x**2+a)**2)**(1/2),x)

[Out]

Integral(x**3/sqrt((a + b*x**2)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.31 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {x^{2}}{2 \, b} - \frac {a \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

integrate(x^3/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*x^2/b - 1/2*a*log(b*x^2 + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.44 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {1}{2} \, {\left (\frac {x^{2}}{b} - \frac {a \log \left ({\left | b x^{2} + a \right |}\right )}{b^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^3/((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,b^2}-\frac {a\,b\,\ln \left (a\,b+\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {b^2}+b^2\,x^2\right )}{2\,{\left (b^2\right )}^{3/2}} \]

[In]

int(x^3/((a + b*x^2)^2)^(1/2),x)

[Out]

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