\(\int \frac {2 b+a x^2}{x (b^2+a^2 x^2)^{3/4}} \, dx\) [1044]

Optimal result

Integrand size = 28, antiderivative size = 79 \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=\frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}-\frac {2 \arctan \left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}} \]

[Out]

2*(a^2*x^2+b^2)^(1/4)/a-2*arctan((a^2*x^2+b^2)^(1/4)/b^(1/2))/b^(1/2)-2*arctanh((a^2*x^2+b^2)^(1/4)/b^(1/2))/b
^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {457, 81, 65, 218, 212, 209} \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}}+\frac {2 \sqrt [4]{a^2 x^2+b^2}}{a} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {2 b+a x}{x \left (b^2+a^2 x\right )^{3/4}} \, dx,x,x^2\right ) \\ & = \frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}+b \text {Subst}\left (\int \frac {1}{x \left (b^2+a^2 x\right )^{3/4}} \, dx,x,x^2\right ) \\ & = \frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}+\frac {(4 b) \text {Subst}\left (\int \frac {1}{-\frac {b^2}{a^2}+\frac {x^4}{a^2}} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )}{a^2} \\ & = \frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}-2 \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )-2 \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right ) \\ & = \frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}-\frac {2 \arctan \left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=\frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}-\frac {2 \arctan \left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(-\frac {2 \left (\arctan \left (\frac {\left (a^{2} x^{2}+b^{2}\right )^{\frac {1}{4}}}{\sqrt {b}}\right ) a +\operatorname {arctanh}\left (\frac {\left (a^{2} x^{2}+b^{2}\right )^{\frac {1}{4}}}{\sqrt {b}}\right ) a -\left (a^{2} x^{2}+b^{2}\right )^{\frac {1}{4}} \sqrt {b}\right )}{a \sqrt {b}}\)	\(68\)

[In]

int((a*x^2+2*b)/x/(a^2*x^2+b^2)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-2*(arctan((a^2*x^2+b^2)^(1/4)/b^(1/2))*a+arctanh((a^2*x^2+b^2)^(1/4)/b^(1/2))*a-(a^2*x^2+b^2)^(1/4)*b^(1/2))/
a/b^(1/2)

Fricas [A] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (65) = 130\).

Time = 0.26 (sec) , antiderivative size = 264, normalized size of antiderivative = 3.34 \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=\left [-\frac {2 \, a \sqrt {b} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) - a \sqrt {b} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b^{\frac {3}{2}} + 2 \, \sqrt {a^{2} x^{2} + b^{2}} b - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {b}}{x^{2}}\right ) - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b}{a b}, \frac {2 \, a \sqrt {-b} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b}}{b}\right ) - a \sqrt {-b} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b} b - 2 \, \sqrt {a^{2} x^{2} + b^{2}} b + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {-b}}{x^{2}}\right ) + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b}{a b}\right ] \]

[In]

integrate((a*x^2+2*b)/x/(a^2*x^2+b^2)^(3/4),x, algorithm="fricas")

[Out]

[-(2*a*sqrt(b)*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b)) - a*sqrt(b)*log((a^2*x^2 + 2*b^2 - 2*(a^2*x^2 + b^2)^(1/4
)*b^(3/2) + 2*sqrt(a^2*x^2 + b^2)*b - 2*(a^2*x^2 + b^2)^(3/4)*sqrt(b))/x^2) - 2*(a^2*x^2 + b^2)^(1/4)*b)/(a*b)
, (2*a*sqrt(-b)*arctan((a^2*x^2 + b^2)^(1/4)*sqrt(-b)/b) - a*sqrt(-b)*log((a^2*x^2 + 2*b^2 - 2*(a^2*x^2 + b^2)
^(1/4)*sqrt(-b)*b - 2*sqrt(a^2*x^2 + b^2)*b + 2*(a^2*x^2 + b^2)^(3/4)*sqrt(-b))/x^2) + 2*(a^2*x^2 + b^2)^(1/4)
*b)/(a*b)]

Sympy [A] (verification not implemented)

Time = 5.91 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.44 \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=- \frac {a \left (\begin {cases} - \frac {x^{2}}{\left (b^{2}\right )^{\frac {3}{4}}} & \text {for}\: a^{2} = 0 \\- \frac {4 \sqrt [4]{a^{2} x^{2} + b^{2}}}{a^{2}} & \text {otherwise} \end {cases}\right )}{2} + b \left (\begin {cases} \frac {2 \operatorname {atan}{\left (\frac {\sqrt [4]{a^{2} x^{2} + b^{2}}}{\sqrt {- b}} \right )}}{b \sqrt {- b}} - \frac {2 \operatorname {atan}{\left (\frac {\sqrt [4]{a^{2} x^{2} + b^{2}}}{\sqrt {b}} \right )}}{b^{\frac {3}{2}}} & \text {for}\: a^{2} \neq 0 \\- \frac {\log {\left (\frac {1}{x^{2}} \right )}}{\left (b^{2}\right )^{\frac {3}{4}}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((a*x**2+2*b)/x/(a**2*x**2+b**2)**(3/4),x)

[Out]

-a*Piecewise((-x**2/(b**2)**(3/4), Eq(a**2, 0)), (-4*(a**2*x**2 + b**2)**(1/4)/a**2, True))/2 + b*Piecewise((2
*atan((a**2*x**2 + b**2)**(1/4)/sqrt(-b))/(b*sqrt(-b)) - 2*atan((a**2*x**2 + b**2)**(1/4)/sqrt(b))/b**(3/2), N
e(a**2, 0)), (-log(x**(-2))/(b**2)**(3/4), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=-b {\left (\frac {2 \, \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} - \frac {\log \left (-\frac {\sqrt {b} - {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b} + {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}\right )}{b^{\frac {3}{2}}}\right )} + \frac {2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{a} \]

[In]

integrate((a*x^2+2*b)/x/(a^2*x^2+b^2)^(3/4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.87 \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=\frac {2 \, \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {2 \, \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{a} \]

[In]

integrate((a*x^2+2*b)/x/(a^2*x^2+b^2)^(3/4),x, algorithm="giac")

[Out]

2*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(-b))/sqrt(-b) - 2*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b))/sqrt(b) + 2*(a^2*x
^2 + b^2)^(1/4)/a

Mupad [B] (verification not implemented)

Time = 6.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82 \[ \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx=\frac {2\,{\left (a^2\,x^2+b^2\right )}^{1/4}}{a}-\frac {2\,\mathrm {atanh}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2\,\mathrm {atan}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )}{\sqrt {b}} \]

[In]

int((2*b + a*x^2)/(x*(b^2 + a^2*x^2)^(3/4)),x)

[Out]

(2*(b^2 + a^2*x^2)^(1/4))/a - (2*atanh((b^2 + a^2*x^2)^(1/4)/b^(1/2)))/b^(1/2) - (2*atan((b^2 + a^2*x^2)^(1/4)
/b^(1/2)))/b^(1/2)