\(\int \frac {(a+b x^2)^{4/3}}{x^5} \, dx\) [744]

Optimal result

Integrand size = 15, antiderivative size = 133 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=-\frac {a \sqrt [3]{a+b x^2}}{4 x^4}-\frac {7 b \sqrt [3]{a+b x^2}}{12 x^2}-\frac {b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3}}-\frac {b^2 \log (x)}{9 a^{2/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{6 a^{2/3}} \] Output:

-1/4*a*(b*x^2+a)^(1/3)/x^4-7/12*b*(b*x^2+a)^(1/3)/x^2-1/9*b^2*arctan(1/3*( 
a^(1/3)+2*(b*x^2+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/a^(2/3)-1/9*b^2*ln(x)/ 
a^(2/3)+1/6*b^2*ln(a^(1/3)-(b*x^2+a)^(1/3))/a^(2/3)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=\frac {1}{36} \left (-\frac {3 \sqrt [3]{a+b x^2} \left (3 a+7 b x^2\right )}{x^4}-\frac {4 \sqrt {3} b^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{2/3}}+\frac {4 b^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )}{a^{2/3}}-\frac {2 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{a^{2/3}}\right ) \] Input:

Integrate[(a + b*x^2)^(4/3)/x^5,x]
 

Output:

((-3*(a + b*x^2)^(1/3)*(3*a + 7*b*x^2))/x^4 - (4*Sqrt[3]*b^2*ArcTan[(1 + ( 
2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3) + (4*b^2*Log[-a^(1/3) + (a 
 + b*x^2)^(1/3)])/a^(2/3) - (2*b^2*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) 
 + (a + b*x^2)^(2/3)])/a^(2/3))/36
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {243, 51, 51, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{4/3}}{x^6}dx^2\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[(a + b*x^2)^(4/3)/x^5,x]
 

Output:

(-1/2*(a + b*x^2)^(4/3)/x^4 + (2*b*(-((a + b*x^2)^(1/3)/x^2) + (b*(-((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))))/3))/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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

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 = 0.55 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(\frac {-4 b^{2} \sqrt {3}\, \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{2}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) x^{4}+4 b^{2} \ln \left (\left (b \,x^{2}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) x^{4}-2 b^{2} \ln \left (a^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}+\left (b \,x^{2}+a \right )^{\frac {2}{3}}\right ) x^{4}-21 b \,x^{2} a^{\frac {2}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}-9 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{\frac {5}{3}}}{36 x^{4} a^{\frac {2}{3}}}\)	\(136\)

Input:

int((b*x^2+a)^(4/3)/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/36*(-4*b^2*3^(1/2)*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))*3^(1/2)/a^(1/3 
))*x^4+4*b^2*ln((b*x^2+a)^(1/3)-a^(1/3))*x^4-2*b^2*ln(a^(2/3)+a^(1/3)*(b*x 
^2+a)^(1/3)+(b*x^2+a)^(2/3))*x^4-21*b*x^2*a^(2/3)*(b*x^2+a)^(1/3)-9*(b*x^2 
+a)^(1/3)*a^(5/3))/x^4/a^(2/3)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=-\frac {12 \, \sqrt {\frac {1}{3}} {\left (a^{2}\right )}^{\frac {1}{6}} a b^{2} x^{4} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (a^{2}\right )}^{\frac {1}{6}} {\left ({\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{a^{2}}\right ) + 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 4 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) + 3 \, {\left (7 \, a^{2} b x^{2} + 3 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{36 \, a^{2} x^{4}} \] Input:

integrate((b*x^2+a)^(4/3)/x^5,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

-1/36*(12*sqrt(1/3)*(a^2)^(1/6)*a*b^2*x^4*arctan(sqrt(1/3)*(a^2)^(1/6)*((a 
^2)^(1/3)*a + 2*(b*x^2 + a)^(1/3)*(a^2)^(2/3))/a^2) + 2*(a^2)^(2/3)*b^2*x^ 
4*log((b*x^2 + a)^(2/3)*a + (a^2)^(1/3)*a + (b*x^2 + a)^(1/3)*(a^2)^(2/3)) 
 - 4*(a^2)^(2/3)*b^2*x^4*log((b*x^2 + a)^(1/3)*a - (a^2)^(2/3)) + 3*(7*a^2 
*b*x^2 + 3*a^3)*(b*x^2 + a)^(1/3))/(a^2*x^4)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.32 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=- \frac {b^{\frac {4}{3}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{\frac {4}{3}} \Gamma \left (\frac {5}{3}\right )} \] Input:

integrate((b*x**2+a)**(4/3)/x**5,x)
 

Output:

-b**(4/3)*gamma(2/3)*hyper((-4/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**2) 
)/(2*x**(4/3)*gamma(5/3))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=-\frac {\sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {2}{3}}} - \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{18 \, a^{\frac {2}{3}}} + \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {2}{3}}} - \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{2} - 4 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} - 2 \, {\left (b x^{2} + a\right )} a + a^{2}\right )}} \] Input:

integrate((b*x^2+a)^(4/3)/x^5,x, algorithm="maxima")
 

Output:

-1/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3 
))/a^(2/3) - 1/18*b^2*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + 
a^(2/3))/a^(2/3) + 1/9*b^2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(2/3) - 1/12 
*(7*(b*x^2 + a)^(4/3)*b^2 - 4*(b*x^2 + a)^(1/3)*a*b^2)/((b*x^2 + a)^2 - 2* 
(b*x^2 + a)*a + a^2)
 

Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=-\frac {\frac {4 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {2}{3}}} + \frac {2 \, b^{3} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {2}{3}}} - \frac {4 \, b^{3} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {2}{3}}} + \frac {3 \, {\left (7 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{3} - 4 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{b^{2} x^{4}}}{36 \, b} \] Input:

integrate((b*x^2+a)^(4/3)/x^5,x, algorithm="giac")
 

Output:

-1/36*(4*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^ 
(1/3))/a^(2/3) + 2*b^3*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + 
 a^(2/3))/a^(2/3) - 4*b^3*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(2/3) + 
3*(7*(b*x^2 + a)^(4/3)*b^3 - 4*(b*x^2 + a)^(1/3)*a*b^3)/(b^2*x^4))/b
 

Mupad [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=\frac {b^2\,\ln \left (b^2\,{\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\,b^2\right )}{9\,a^{2/3}}-\frac {\ln \left (\frac {a^{1/3}\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{2}+b^2\,{\left (b\,x^2+a\right )}^{1/3}\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{18\,a^{2/3}}-\frac {\frac {7\,b^2\,{\left (b\,x^2+a\right )}^{4/3}}{6}-\frac {2\,a\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{3}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {b^2\,\ln \left (b^2\,{\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\,b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{2/3}} \] Input:

int((a + b*x^2)^(4/3)/x^5,x)
 

Output:

(b^2*log(b^2*(a + b*x^2)^(1/3) - a^(1/3)*b^2))/(9*a^(2/3)) - (log((a^(1/3) 
*(3^(1/2)*b^2*1i + b^2))/2 + b^2*(a + b*x^2)^(1/3))*(3^(1/2)*b^2*1i + b^2) 
)/(18*a^(2/3)) - ((7*b^2*(a + b*x^2)^(4/3))/6 - (2*a*b^2*(a + b*x^2)^(1/3) 
)/3)/(2*(a + b*x^2)^2 - 4*a*(a + b*x^2) + 2*a^2) + (b^2*log(b^2*(a + b*x^2 
)^(1/3) - a^(1/3)*b^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/(9*a 
^(2/3))
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^5} \, dx=\frac {-9 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a -21 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{2}+8 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{b \,x^{3}+a x}d x \right ) b^{2} x^{4}}{36 x^{4}} \] Input:

int((b*x^2+a)^(4/3)/x^5,x)
 

Output:

( - 9*(a + b*x**2)**(1/3)*a - 21*(a + b*x**2)**(1/3)*b*x**2 + 8*int((a + b 
*x**2)**(1/3)/(a*x + b*x**3),x)*b**2*x**4)/(36*x**4)