\(\int \frac {(a x+b x^3)^{3/2}}{x^4} \, dx\) [59]

Optimal result

Integrand size = 17, antiderivative size = 134 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^4} \, dx=\frac {4}{3} b \sqrt {a x+b x^3}-\frac {2 \left (a x+b x^3\right )^{3/2}}{3 x^3}+\frac {4 a^{3/4} b^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt {a x+b x^3}} \] Output:

4/3*b*(b*x^3+a*x)^(1/2)-2/3*(b*x^3+a*x)^(3/2)/x^3+4/3*a^(3/4)*b^(3/4)*x^(1 
/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJac 
obiAM(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)),1/2*2^(1/2))/(b*x^3+a*x)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^4} \, dx=-\frac {2 a \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{4},\frac {1}{4},-\frac {b x^2}{a}\right )}{3 x^2 \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

(-2*a*Sqrt[x*(a + b*x^2)]*Hypergeometric2F1[-3/2, -3/4, 1/4, -((b*x^2)/a)] 
)/(3*x^2*Sqrt[1 + (b*x^2)/a])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1926, 1927, 1917, 266, 761}

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.

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

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 1927

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 + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {2 a \sqrt {b \,x^{3}+a x}}{3 x^{2}}+\frac {2 b \sqrt {b \,x^{3}+a x}}{3}+\frac {4 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 \sqrt {b \,x^{3}+a x}}\)	\(139\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right ) \left (-b \,x^{2}+a \right )}{3 x \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {4 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 \sqrt {b \,x^{3}+a x}}\)	\(139\)
elliptic	\(-\frac {2 a \sqrt {b \,x^{3}+a x}}{3 x^{2}}+\frac {2 b \sqrt {b \,x^{3}+a x}}{3}+\frac {4 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 \sqrt {b \,x^{3}+a x}}\)	\(139\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^4} \, dx=\frac {2 \, {\left (4 \, a \sqrt {b} x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + \sqrt {b x^{3} + a x} {\left (b x^{2} - a\right )}\right )}}{3 \, x^{2}} \] Input:

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

Output:

2/3*(4*a*sqrt(b)*x^2*weierstrassPInverse(-4*a/b, 0, x) + sqrt(b*x^3 + a*x) 
*(b*x^2 - a))/x^2
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^4} \, dx=\frac {-\frac {10 \sqrt {b \,x^{2}+a}\, a}{3}+\frac {2 \sqrt {b \,x^{2}+a}\, b \,x^{2}}{3}-4 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{5}+a \,x^{3}}d x \right ) a^{2} x}{\sqrt {x}\, x} \] Input:

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

Output:

(2*( - 5*sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2 - 6*sqrt(x)*int((sqr 
t(x)*sqrt(a + b*x**2))/(a*x**3 + b*x**5),x)*a**2*x))/(3*sqrt(x)*x)