\(\int \frac {1}{(a x+b x^3)^{3/2}} \, dx\) [79]

Optimal result

Integrand size = 13, antiderivative size = 273 \[ \int \frac {1}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {1}{a \sqrt {a x+b x^3}}+\frac {3 \sqrt {b} x \left (a+b x^2\right )}{a^2 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {3 \sqrt {a x+b x^3}}{a^2 x}-\frac {3 \sqrt [4]{b} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{7/4} \sqrt {a x+b x^3}}+\frac {3 \sqrt [4]{b} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 a^{7/4} \sqrt {a x+b x^3}} \] Output:

1/a/(b*x^3+a*x)^(1/2)+3*b^(1/2)*x*(b*x^2+a)/a^2/(a^(1/2)+b^(1/2)*x)/(b*x^3 
+a*x)^(1/2)-3*(b*x^3+a*x)^(1/2)/a^2/x-3*b^(1/4)*x^(1/2)*(a^(1/2)+b^(1/2)*x 
)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x 
^(1/2)/a^(1/4))),1/2*2^(1/2))/a^(7/4)/(b*x^3+a*x)^(1/2)+3/2*b^(1/4)*x^(1/2 
)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacob 
iAM(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)),1/2*2^(1/2))/a^(7/4)/(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 = 51, normalized size of antiderivative = 0.19 \[ \int \frac {1}{\left (a x+b x^3\right )^{3/2}} \, dx=-\frac {2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},-\frac {b x^2}{a}\right )}{a \sqrt {x \left (a+b x^2\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {1912, 1931, 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.

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

\(\Big \downarrow \) 1912

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1938

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

1/(a*Sqrt[a*x + b*x^3]) + (3*((-2*Sqrt[a*x + b*x^3])/(a*x) + (2*b*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]*El 
lipticE[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])))/(a*Sqrt[a*x + b*x^3])))/(2*a)
 

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 1912

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

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 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]
 

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.75

method	result	size
default	\(-\frac {2 \left (b \,x^{2}+a \right )}{a^{2} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {b \,x^{2}}{a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {3 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{2 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(206\)
elliptic	\(-\frac {2 \left (b \,x^{2}+a \right )}{a^{2} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {b \,x^{2}}{a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {3 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{2 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(206\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )}{a^{2} \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {b^{2} \left (\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{b^{2} \sqrt {b \,x^{3}+a x}}-\frac {a \left (\frac {x^{2}}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}-\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{2 a b \sqrt {b \,x^{3}+a x}}\right )}{b}\right )}{a^{2}}\)	\(379\)

Input:

int(1/(b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.26 \[ \int \frac {1}{\left (a x+b x^3\right )^{3/2}} \, dx=-\frac {3 \, {\left (b x^{3} + a x\right )} \sqrt {b} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + \sqrt {b x^{3} + a x} {\left (3 \, b x^{2} + 2 \, a\right )}}{a^{2} b x^{3} + a^{3} x} \] Input:

integrate(1/(b*x^3+a*x)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(1/(b*x**3+a*x)**(3/2),x)
 

Output:

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

Maxima [F]

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

integrate(1/(b*x^3+a*x)^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/(b*x^3+a*x)^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.15 \[ \int \frac {1}{\left (a x+b x^3\right )^{3/2}} \, dx=-\frac {2\,x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{2};\ \frac {3}{4};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^3+a\,x\right )}^{3/2}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{6}+2 a b \,x^{4}+a^{2} x^{2}}d x \] Input:

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

Output:

int((sqrt(x)*sqrt(a + b*x**2))/(a**2*x**2 + 2*a*b*x**4 + b**2*x**6),x)