3.1.49 \(\int \frac {(a x+b x^3)^{3/2}}{x} \, dx\) [49]

3.1.49.1 Optimal result

Integrand size = 17, antiderivative size = 275 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x} \, dx=\frac {8 a^2 x \left (a+b x^2\right )}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}+\frac {4}{15} a x \sqrt {a x+b x^3}+\frac {2}{9} \left (a x+b x^3\right )^{3/2}-\frac {8 a^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a x+b x^3}}+\frac {4 a^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a x+b x^3}} \]

output

2/9*(b*x^3+a*x)^(3/2)+8/15*a^2*x*(b*x^2+a)/b^(1/2)/(a^(1/2)+x*b^(1/2))/(b* 
x^3+a*x)^(1/2)+4/15*a*x*(b*x^3+a*x)^(1/2)-8/15*a^(9/4)*(cos(2*arctan(b^(1/ 
4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*Ellip 
ticE(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2 
))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(3/4)/(b*x^3+a*x)^(1/ 
2)+4/15*a^(9/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arc 
tan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/ 
4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2 
))^2)^(1/2)/b^(3/4)/(b*x^3+a*x)^(1/2)
 

3.1.49.2 Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.19 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x} \, dx=\frac {2 a x \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )}{3 \sqrt {1+\frac {b x^2}{a}}} \]

input

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

output

(2*a*x*Sqrt[x*(a + b*x^2)]*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^2)/a)] 
)/(3*Sqrt[1 + (b*x^2)/a])
 

3.1.49.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {1927, 1910, 1938, 266, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 1927

\(\displaystyle \frac {2}{3} a \int \sqrt {b x^3+a x}dx+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {2}{3} a \left (\frac {2}{5} a \int \frac {x}{\sqrt {b x^3+a x}}dx+\frac {2}{5} x \sqrt {a x+b x^3}\right )+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {2}{3} a \left (\frac {2 a \sqrt {x} \sqrt {a+b x^2} \int \frac {\sqrt {x}}{\sqrt {b x^2+a}}dx}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\right )+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{3} a \left (\frac {4 a \sqrt {x} \sqrt {a+b x^2} \int \frac {x}{\sqrt {b x^2+a}}d\sqrt {x}}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\right )+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {2}{3} a \left (\frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {a} \sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\right )+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} a \left (\frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\right )+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{3} a \left (\frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\right )+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2}{3} a \left (\frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {\sqrt {x} \sqrt {a+b x^2}}{\sqrt {a}+\sqrt {b} x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\right )+\frac {2}{9} \left (a x+b x^3\right )^{3/2}\)

input

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

output

(2*(a*x + b*x^3)^(3/2))/9 + (2*a*((2*x*Sqrt[a*x + b*x^3])/5 + (4*a*Sqrt[x] 
*Sqrt[a + b*x^2]*(-((-((Sqrt[x]*Sqrt[a + b*x^2])/(Sqrt[a] + Sqrt[b]*x)) + 
(a^(1/4)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*E 
llipticE[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + b*x^ 
2]))/Sqrt[b]) + (a^(1/4)*(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])/(2*b^( 
3/4)*Sqrt[a + b*x^2])))/(5*Sqrt[a*x + b*x^3])))/3
 

3.1.49.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 834

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

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1910

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j 
 + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1))   Int[x^j*(a*x^j 
+ b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0, j, 
n] && GtQ[p, 0] && NeQ[n*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]
 

rule 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]))   Int[x^(m + j* 
p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

3.1.49.4 Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {2 b \,x^{3} \sqrt {b \,x^{3}+a x}}{9}+\frac {22 a x \sqrt {b \,x^{3}+a x}}{45}+\frac {4 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 b \sqrt {b \,x^{3}+a x}}\)	\(195\)
elliptic	\(\frac {2 b \,x^{3} \sqrt {b \,x^{3}+a x}}{9}+\frac {22 a x \sqrt {b \,x^{3}+a x}}{45}+\frac {4 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 b \sqrt {b \,x^{3}+a x}}\)	\(195\)
risch	\(\frac {2 x^{2} \left (5 b \,x^{2}+11 a \right ) \left (b \,x^{2}+a \right )}{45 \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {4 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 b \sqrt {b \,x^{3}+a x}}\)	\(196\)

input

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

output

2/9*b*x^3*(b*x^3+a*x)^(1/2)+22/45*a*x*(b*x^3+a*x)^(1/2)+4/15*a^2*(-a*b)^(1 
/2)/b*((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)*(-x/(-a*b)^(1/2)*b)^(1/2)/(b*x^3+a*x)^(1/2)*(-2*(-a*b)^ 
(1/2)/b*EllipticE(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))+( 
-a*b)^(1/2)/b*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1 
/2)))
 

3.1.49.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.21 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x} \, dx=-\frac {2 \, {\left (12 \, a^{2} \sqrt {b} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) - {\left (5 \, b^{2} x^{3} + 11 \, a b x\right )} \sqrt {b x^{3} + a x}\right )}}{45 \, b} \]

input

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

output

-2/45*(12*a^2*sqrt(b)*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a/ 
b, 0, x)) - (5*b^2*x^3 + 11*a*b*x)*sqrt(b*x^3 + a*x))/b
 

3.1.49.6 Sympy [F]

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

input

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

output

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

3.1.49.7 Maxima [F]

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

input

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

output

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

3.1.49.8 Giac [F]

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

input

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

output

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

3.1.49.9 Mupad [F(-1)]

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

input

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

output

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