\(\int \frac {(a+b x^2)^{4/3}}{x} \, dx\) [742]

Optimal result

Integrand size = 15, antiderivative size = 117 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=\frac {3}{2} a \sqrt [3]{a+b x^2}+\frac {3}{8} \left (a+b x^2\right )^{4/3}-\frac {1}{2} \sqrt {3} a^{4/3} \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )-\frac {1}{2} a^{4/3} \log (x)+\frac {3}{4} a^{4/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \] Output:

3/2*a*(b*x^2+a)^(1/3)+3/8*(b*x^2+a)^(4/3)-1/2*3^(1/2)*a^(4/3)*arctan(1/3*( 
a^(1/3)+2*(b*x^2+a)^(1/3))*3^(1/2)/a^(1/3))-1/2*a^(4/3)*ln(x)+3/4*a^(4/3)* 
ln(a^(1/3)-(b*x^2+a)^(1/3))
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=\frac {1}{8} \left (3 \sqrt [3]{a+b x^2} \left (5 a+b x^2\right )-4 \sqrt {3} a^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 a^{4/3} \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )-2 a^{4/3} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )\right ) \] Input:

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

Output:

(3*(a + b*x^2)^(1/3)*(5*a + b*x^2) - 4*Sqrt[3]*a^(4/3)*ArcTan[(1 + (2*(a + 
 b*x^2)^(1/3))/a^(1/3))/Sqrt[3]] + 4*a^(4/3)*Log[-a^(1/3) + (a + b*x^2)^(1 
/3)] - 2*a^(4/3)*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/ 
3)])/8
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {243, 60, 60, 69, 16, 1082, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{4/3}}{x^2}dx^2\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{2} \left (a \int \frac {\sqrt [3]{b x^2+a}}{x^2}dx^2+\frac {3}{4} \left (a+b x^2\right )^{4/3}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{2} \left (a \left (a \int \frac {1}{x^2 \left (b x^2+a\right )^{2/3}}dx^2+3 \sqrt [3]{a+b x^2}\right )+\frac {3}{4} \left (a+b x^2\right )^{4/3}\right )\)

\(\Big \downarrow \) 69

\(\displaystyle \frac {1}{2} \left (a \left (a \left (-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 a^{2/3}}-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )+\frac {3}{4} \left (a+b x^2\right )^{4/3}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} \left (a \left (a \left (-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )+\frac {3}{4} \left (a+b x^2\right )^{4/3}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{2} \left (a \left (a \left (\frac {3 \int \frac {1}{-x^4-3}d\left (\frac {2 \sqrt [3]{b x^2+a}}{\sqrt [3]{a}}+1\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )+\frac {3}{4} \left (a+b x^2\right )^{4/3}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (a \left (a \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )+\frac {3}{4} \left (a+b x^2\right )^{4/3}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

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

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (b \,x^{2}+5 a \right )}{8}+\frac {\left (-\sqrt {3}\, \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{2}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right )+\ln \left (\left (b \,x^{2}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )-\frac {\ln \left (a^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}+\left (b \,x^{2}+a \right )^{\frac {2}{3}}\right )}{2}\right ) a^{\frac {4}{3}}}{2}\)	\(102\)

Input:

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

Output:

3/8*(b*x^2+a)^(1/3)*(b*x^2+5*a)+1/2*(-3^(1/2)*arctan(1/3*(a^(1/3)+2*(b*x^2 
+a)^(1/3))*3^(1/2)/a^(1/3))+ln((b*x^2+a)^(1/3)-a^(1/3))-1/2*ln(a^(2/3)+a^( 
1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3)))*a^(4/3)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=-\frac {1}{2} \, \sqrt {3} a^{\frac {4}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) - \frac {1}{4} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{8} \, {\left (b x^{2} + 5 \, a\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \] Input:

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

Output:

-1/2*sqrt(3)*a^(4/3)*arctan(1/3*(2*sqrt(3)*(b*x^2 + a)^(1/3)*a^(2/3) + sqr 
t(3)*a)/a) - 1/4*a^(4/3)*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) 
 + a^(2/3)) + 1/2*a^(4/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3/8*(b*x^2 + 
5*a)*(b*x^2 + a)^(1/3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=- \frac {b^{\frac {4}{3}} x^{\frac {8}{3}} \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {4}{3} \\ - \frac {1}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (- \frac {1}{3}\right )} \] Input:

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

Output:

-b**(4/3)*x**(8/3)*gamma(-4/3)*hyper((-4/3, -4/3), (-1/3,), a*exp_polar(I* 
pi)/(b*x**2))/(2*gamma(-1/3))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=-\frac {1}{2} \, \sqrt {3} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{4} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{8} \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a \] Input:

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

Output:

-1/2*sqrt(3)*a^(4/3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^ 
(1/3)) - 1/4*a^(4/3)*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a 
^(2/3)) + 1/2*a^(4/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3/8*(b*x^2 + a)^( 
4/3) + 3/2*(b*x^2 + a)^(1/3)*a
 

Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=-\frac {1}{2} \, \sqrt {3} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{4} \, a^{\frac {4}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {4}{3}} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) + \frac {3}{8} \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a \] Input:

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

Output:

-1/2*sqrt(3)*a^(4/3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^ 
(1/3)) - 1/4*a^(4/3)*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a 
^(2/3)) + 1/2*a^(4/3)*log(abs((b*x^2 + a)^(1/3) - a^(1/3))) + 3/8*(b*x^2 + 
 a)^(4/3) + 3/2*(b*x^2 + a)^(1/3)*a
 

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=\frac {3\,a\,{\left (b\,x^2+a\right )}^{1/3}}{2}+\frac {3\,{\left (b\,x^2+a\right )}^{4/3}}{8}+\frac {a^{4/3}\,\ln \left (\frac {9\,a^2\,{\left (b\,x^2+a\right )}^{1/3}}{2}-\frac {9\,a^{7/3}}{2}\right )}{2}-\frac {a^{4/3}\,\ln \left (\frac {9\,a^{7/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2}+\frac {9\,a^2\,{\left (b\,x^2+a\right )}^{1/3}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2}+a^{4/3}\,\ln \left (9\,a^{7/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\frac {9\,a^2\,{\left (b\,x^2+a\right )}^{1/3}}{2}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \] Input:

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

Output:

(3*a*(a + b*x^2)^(1/3))/2 + (3*(a + b*x^2)^(4/3))/8 + (a^(4/3)*log((9*a^2* 
(a + b*x^2)^(1/3))/2 - (9*a^(7/3))/2))/2 - (a^(4/3)*log((9*a^(7/3)*((3^(1/ 
2)*1i)/2 + 1/2))/2 + (9*a^2*(a + b*x^2)^(1/3))/2)*((3^(1/2)*1i)/2 + 1/2))/ 
2 + a^(4/3)*log(9*a^(7/3)*((3^(1/2)*1i)/4 - 1/4) - (9*a^2*(a + b*x^2)^(1/3 
))/2)*((3^(1/2)*1i)/4 - 1/4)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{x} \, dx=\frac {15 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a}{8}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{2}}{8}+\left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{b \,x^{3}+a x}d x \right ) a^{2} \] Input:

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

Output:

(15*(a + b*x**2)**(1/3)*a + 3*(a + b*x**2)**(1/3)*b*x**2 + 8*int((a + b*x* 
*2)**(1/3)/(a*x + b*x**3),x)*a**2)/8