\(\int \frac {(b+a x^2)^{3/4}}{x} \, dx\) [853]

Optimal result

Integrand size = 15, antiderivative size = 65 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=\frac {2}{3} \left (b+a x^2\right )^{3/4}+b^{3/4} \arctan \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )-b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 52, 65, 304, 209, 212} \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=b^{3/4} \arctan \left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )-b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )+\frac {2}{3} \left (a x^2+b\right )^{3/4} \]

[In]

Int[(b + a*x^2)^(3/4)/x,x]

[Out]

(2*(b + a*x^2)^(3/4))/3 + b^(3/4)*ArcTan[(b + a*x^2)^(1/4)/b^(1/4)] - b^(3/4)*ArcTanh[(b + a*x^2)^(1/4)/b^(1/4
)]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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 272

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

Rule 304

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=\frac {2}{3} \left (b+a x^2\right )^{3/4}+b^{3/4} \arctan \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )-b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right ) \]

[In]

Integrate[(b + a*x^2)^(3/4)/x,x]

[Out]

(2*(b + a*x^2)^(3/4))/3 + b^(3/4)*ArcTan[(b + a*x^2)^(1/4)/b^(1/4)] - b^(3/4)*ArcTanh[(b + a*x^2)^(1/4)/b^(1/4
)]

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {2 \left (a \,x^{2}+b \right )^{\frac {3}{4}}}{3}+b^{\frac {3}{4}} \arctan \left (\frac {\left (a \,x^{2}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )-\frac {\ln \left (\frac {\left (a \,x^{2}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{2}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right ) b^{\frac {3}{4}}}{2}\)	\(68\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.94 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=-\frac {1}{2} \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{2} + b\right )}^{\frac {1}{4}} b^{2} + {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{2} i \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{2} + b\right )}^{\frac {1}{4}} b^{2} + i \, {\left (b^{3}\right )}^{\frac {3}{4}}\right ) - \frac {1}{2} i \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{2} + b\right )}^{\frac {1}{4}} b^{2} - i \, {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{2} \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{2} + b\right )}^{\frac {1}{4}} b^{2} - {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {2}{3} \, {\left (a x^{2} + b\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=- \frac {a^{\frac {3}{4}} x^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

-a**(3/4)*x**(3/2)*gamma(-3/4)*hyper((-3/4, -3/4), (1/4,), b*exp_polar(I*pi)/(a*x**2))/(2*gamma(1/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=\frac {1}{2} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{2} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}}\right )} + \frac {2}{3} \, {\left (a x^{2} + b\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (49) = 98\).

Time = 0.28 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.85 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=-\frac {1}{2} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) - \frac {1}{2} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (a x^{2} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{2} + b} + \sqrt {-b}\right ) - \frac {1}{4} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (a x^{2} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{2} + b} + \sqrt {-b}\right ) + \frac {2}{3} \, {\left (a x^{2} + b\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.42 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x} \, dx=b^{3/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^2+b\right )}^{1/4}}{b^{1/4}}\right )-b^{3/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^2+b\right )}^{1/4}}{b^{1/4}}\right )+\frac {2\,{\left (a\,x^2+b\right )}^{3/4}}{3} \]

[In]

int((b + a*x^2)^(3/4)/x,x)

[Out]

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