\(\int \frac {(b x^{2/3}+a x)^{3/2}}{x^4} \, dx\) [161]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F(-1)]
Sympy [F]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 203 \[ \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^4} \, dx=-\frac {3 a \sqrt {b x^{2/3}+a x}}{20 x^2}-\frac {3 a^2 \sqrt {b x^{2/3}+a x}}{160 b x^{5/3}}+\frac {7 a^3 \sqrt {b x^{2/3}+a x}}{320 b^2 x^{4/3}}-\frac {7 a^4 \sqrt {b x^{2/3}+a x}}{256 b^3 x}+\frac {21 a^5 \sqrt {b x^{2/3}+a x}}{512 b^4 x^{2/3}}-\frac {\left (b x^{2/3}+a x\right )^{3/2}}{2 x^3}-\frac {21 a^6 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{512 b^{9/2}} \] Output:

-3/20*a*(b*x^(2/3)+a*x)^(1/2)/x^2-3/160*a^2*(b*x^(2/3)+a*x)^(1/2)/b/x^(5/3 
)+7/320*a^3*(b*x^(2/3)+a*x)^(1/2)/b^2/x^(4/3)-7/256*a^4*(b*x^(2/3)+a*x)^(1 
/2)/b^3/x+21/512*a^5*(b*x^(2/3)+a*x)^(1/2)/b^4/x^(2/3)-1/2*(b*x^(2/3)+a*x) 
^(3/2)/x^3-21/512*a^6*arctanh(b^(1/2)*x^(1/3)/(b*x^(2/3)+a*x)^(1/2))/b^(9/ 
2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.30 \[ \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^4} \, dx=-\frac {6 a^6 \left (b+a \sqrt [3]{x}\right )^2 \sqrt {b x^{2/3}+a x} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},7,\frac {7}{2},1+\frac {a \sqrt [3]{x}}{b}\right )}{5 b^7 \sqrt [3]{x}} \] Input:

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

Output:

(-6*a^6*(b + a*x^(1/3))^2*Sqrt[b*x^(2/3) + a*x]*Hypergeometric2F1[5/2, 7, 
7/2, 1 + (a*x^(1/3))/b])/(5*b^7*x^(1/3))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {1926, 1926, 1931, 1931, 1931, 1931, 1935, 219}

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.

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1926

\(\displaystyle \frac {1}{4} a \left (\frac {1}{10} a \int \frac {1}{x^2 \sqrt {x^{2/3} b+a x}}dx-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2}\right )-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {1}{4} a \left (\frac {1}{10} a \left (-\frac {7 a \int \frac {1}{x^{5/3} \sqrt {x^{2/3} b+a x}}dx}{8 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}}\right )-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2}\right )-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {1}{4} a \left (\frac {1}{10} a \left (-\frac {7 a \left (-\frac {5 a \int \frac {1}{x^{4/3} \sqrt {x^{2/3} b+a x}}dx}{6 b}-\frac {\sqrt {a x+b x^{2/3}}}{b x^{4/3}}\right )}{8 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}}\right )-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2}\right )-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {1}{4} a \left (\frac {1}{10} a \left (-\frac {7 a \left (-\frac {5 a \left (-\frac {3 a \int \frac {1}{x \sqrt {x^{2/3} b+a x}}dx}{4 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{2 b x}\right )}{6 b}-\frac {\sqrt {a x+b x^{2/3}}}{b x^{4/3}}\right )}{8 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}}\right )-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2}\right )-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {1}{4} a \left (\frac {1}{10} a \left (-\frac {7 a \left (-\frac {5 a \left (-\frac {3 a \left (-\frac {a \int \frac {1}{x^{2/3} \sqrt {x^{2/3} b+a x}}dx}{2 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{b x^{2/3}}\right )}{4 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{2 b x}\right )}{6 b}-\frac {\sqrt {a x+b x^{2/3}}}{b x^{4/3}}\right )}{8 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}}\right )-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2}\right )-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3}\)

\(\Big \downarrow \) 1935

\(\displaystyle \frac {1}{4} a \left (\frac {1}{10} a \left (-\frac {7 a \left (-\frac {5 a \left (-\frac {3 a \left (\frac {3 a \int \frac {1}{1-\frac {b x^{2/3}}{x^{2/3} b+a x}}d\frac {\sqrt [3]{x}}{\sqrt {x^{2/3} b+a x}}}{b}-\frac {3 \sqrt {a x+b x^{2/3}}}{b x^{2/3}}\right )}{4 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{2 b x}\right )}{6 b}-\frac {\sqrt {a x+b x^{2/3}}}{b x^{4/3}}\right )}{8 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}}\right )-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2}\right )-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} a \left (\frac {1}{10} a \left (-\frac {7 a \left (-\frac {5 a \left (-\frac {3 a \left (\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{b^{3/2}}-\frac {3 \sqrt {a x+b x^{2/3}}}{b x^{2/3}}\right )}{4 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{2 b x}\right )}{6 b}-\frac {\sqrt {a x+b x^{2/3}}}{b x^{4/3}}\right )}{8 b}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}}\right )-\frac {3 \sqrt {a x+b x^{2/3}}}{5 x^2}\right )-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{2 x^3}\)

Input:

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

Output:

-1/2*(b*x^(2/3) + a*x)^(3/2)/x^3 + (a*((-3*Sqrt[b*x^(2/3) + a*x])/(5*x^2) 
+ (a*((-3*Sqrt[b*x^(2/3) + a*x])/(4*b*x^(5/3)) - (7*a*(-(Sqrt[b*x^(2/3) + 
a*x]/(b*x^(4/3))) - (5*a*((-3*Sqrt[b*x^(2/3) + a*x])/(2*b*x) - (3*a*((-3*S 
qrt[b*x^(2/3) + a*x])/(b*x^(2/3)) + (3*a*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b* 
x^(2/3) + a*x]])/b^(3/2)))/(4*b)))/(6*b)))/(8*b)))/10))/4
 

Defintions of rubi rules used

rule 219
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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1926
Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] 
 :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p 
*((n - j)/(c^n*(m + j*p + 1)))   Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), 
 x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Integer 
sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
 

rule 1931
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1)))   I 
nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] 
 &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ 
m + j*p + 1, 0]
 

rule 1935
Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp 
[-2/(n - j)   Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], 
 x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
 
Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.68

method result size
derivativedivides \(\frac {\left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} \left (105 b^{\frac {9}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {11}{2}}-595 b^{\frac {11}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {9}{2}}+1386 b^{\frac {13}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {7}{2}}-1686 b^{\frac {15}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {5}{2}}-105 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{\frac {1}{3}} a +b}}{\sqrt {b}}\right ) b^{4} a^{6} x^{2}-595 b^{\frac {17}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {3}{2}}+105 b^{\frac {19}{2}} \sqrt {x^{\frac {1}{3}} a +b}\right )}{2560 x^{3} \left (x^{\frac {1}{3}} a +b \right )^{\frac {3}{2}} b^{\frac {17}{2}}}\) \(139\)
default \(-\frac {\left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} \left (-105 b^{\frac {9}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {11}{2}}+595 b^{\frac {11}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {9}{2}}-1386 b^{\frac {13}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {7}{2}}+1686 b^{\frac {15}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {5}{2}}+595 b^{\frac {17}{2}} \left (x^{\frac {1}{3}} a +b \right )^{\frac {3}{2}}-105 b^{\frac {19}{2}} \sqrt {x^{\frac {1}{3}} a +b}+105 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{\frac {1}{3}} a +b}}{\sqrt {b}}\right ) b^{4} a^{6} x^{2}\right )}{2560 x^{3} \left (x^{\frac {1}{3}} a +b \right )^{\frac {3}{2}} b^{\frac {17}{2}}}\) \(139\)

Input:

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

Output:

1/2560*(b*x^(2/3)+a*x)^(3/2)*(105*b^(9/2)*(x^(1/3)*a+b)^(11/2)-595*b^(11/2 
)*(x^(1/3)*a+b)^(9/2)+1386*b^(13/2)*(x^(1/3)*a+b)^(7/2)-1686*b^(15/2)*(x^( 
1/3)*a+b)^(5/2)-105*arctanh((x^(1/3)*a+b)^(1/2)/b^(1/2))*b^4*a^6*x^2-595*b 
^(17/2)*(x^(1/3)*a+b)^(3/2)+105*b^(19/2)*(x^(1/3)*a+b)^(1/2))/x^3/(x^(1/3) 
*a+b)^(3/2)/b^(17/2)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^4} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.70 \[ \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^4} \, dx=\frac {\frac {105 \, a^{7} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{4}} + \frac {105 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {11}{2}} a^{7} - 595 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} a^{7} b + 1386 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} a^{7} b^{2} - 1686 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{7} b^{3} - 595 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{7} b^{4} + 105 \, \sqrt {a x^{\frac {1}{3}} + b} a^{7} b^{5}}{a^{6} b^{4} x^{2}}}{2560 \, a} \] Input:

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

Output:

1/2560*(105*a^7*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) + (105 
*(a*x^(1/3) + b)^(11/2)*a^7 - 595*(a*x^(1/3) + b)^(9/2)*a^7*b + 1386*(a*x^ 
(1/3) + b)^(7/2)*a^7*b^2 - 1686*(a*x^(1/3) + b)^(5/2)*a^7*b^3 - 595*(a*x^( 
1/3) + b)^(3/2)*a^7*b^4 + 105*sqrt(a*x^(1/3) + b)*a^7*b^5)/(a^6*b^4*x^2))/ 
a
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.79 \[ \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^4} \, dx=\frac {112 x^{\frac {5}{3}} \sqrt {x^{\frac {1}{3}} a +b}\, a^{3} b^{3}-2560 x^{\frac {2}{3}} \sqrt {x^{\frac {1}{3}} a +b}\, b^{6}+210 x^{\frac {7}{3}} \sqrt {x^{\frac {1}{3}} a +b}\, a^{5} b -96 x^{\frac {4}{3}} \sqrt {x^{\frac {1}{3}} a +b}\, a^{2} b^{4}-140 \sqrt {x^{\frac {1}{3}} a +b}\, a^{4} b^{2} x^{2}-3328 \sqrt {x^{\frac {1}{3}} a +b}\, a \,b^{5} x +105 x^{\frac {8}{3}} \sqrt {b}\, \mathrm {log}\left (\sqrt {x^{\frac {1}{3}} a +b}-\sqrt {b}\right ) a^{6}-105 x^{\frac {8}{3}} \sqrt {b}\, \mathrm {log}\left (\sqrt {x^{\frac {1}{3}} a +b}+\sqrt {b}\right ) a^{6}}{5120 x^{\frac {8}{3}} b^{5}} \] Input:

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

Output:

(112*x**(2/3)*sqrt(x**(1/3)*a + b)*a**3*b**3*x - 2560*x**(2/3)*sqrt(x**(1/ 
3)*a + b)*b**6 + 210*x**(1/3)*sqrt(x**(1/3)*a + b)*a**5*b*x**2 - 96*x**(1/ 
3)*sqrt(x**(1/3)*a + b)*a**2*b**4*x - 140*sqrt(x**(1/3)*a + b)*a**4*b**2*x 
**2 - 3328*sqrt(x**(1/3)*a + b)*a*b**5*x + 105*x**(2/3)*sqrt(b)*log(sqrt(x 
**(1/3)*a + b) - sqrt(b))*a**6*x**2 - 105*x**(2/3)*sqrt(b)*log(sqrt(x**(1/ 
3)*a + b) + sqrt(b))*a**6*x**2)/(5120*x**(2/3)*b**5*x**2)