\(\int \frac {\sqrt [4]{a x+b x^2}}{x^3} \, dx\) [151]

Optimal result

Integrand size = 17, antiderivative size = 116 \[ \int \frac {\sqrt [4]{a x+b x^2}}{x^3} \, dx=-\frac {4 \sqrt [4]{a x+b x^2}}{7 x^2}-\frac {4 b \sqrt [4]{a x+b x^2}}{21 a x}+\frac {8 b \left (\frac {b x}{a+b x}\right )^{3/4} \sqrt {a+b x} \sqrt [4]{a x+b x^2} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+b x}}\right ),2\right )}{21 a^{3/2} x} \] Output:

-4/7*(b*x^2+a*x)^(1/4)/x^2-4/21*b*(b*x^2+a*x)^(1/4)/a/x+8/21*b*(b*x/(b*x+a 
))^(3/4)*(b*x+a)^(1/2)*(b*x^2+a*x)^(1/4)*InverseJacobiAM(1/2*arcsin(a^(1/2 
)/(b*x+a)^(1/2)),2^(1/2))/a^(3/2)/x
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt [4]{a x+b x^2}}{x^3} \, dx=-\frac {4 \sqrt [4]{x (a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {1}{4},-\frac {3}{4},-\frac {b x}{a}\right )}{7 x^2 \sqrt [4]{1+\frac {b x}{a}}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {1137, 57, 61, 73, 768, 858, 807, 229}

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.

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

\(\Big \downarrow \) 1137

\(\displaystyle \frac {\sqrt [4]{a x+b x^2} \int \frac {\sqrt [4]{a+b x}}{x^{11/4}}dx}{\sqrt [4]{x} \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

\(\displaystyle \frac {\sqrt [4]{a x+b x^2} \left (\frac {1}{7} b \left (\frac {8 b^{3/2} x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{7 x^{7/4}}\right )}{\sqrt [4]{x} \sqrt [4]{a+b x}}\)

Input:

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

Output:

((a*x + b*x^2)^(1/4)*((-4*(a + b*x)^(1/4))/(7*x^(7/4)) + (b*((-4*(a + b*x) 
^(1/4))/(3*a*x^(3/4)) + (8*b^(3/2)*(1 + a/(b*x))^(3/4)*x^(3/4)*EllipticF[A 
rcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]]/2, 2])/(3*a^(3/2)*(a + b*x)^(3/4))))/7))/ 
(x^(1/4)*(a + b*x)^(1/4))
 

Defintions of rubi rules used

rule 57

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}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 1137

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( 
e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p))   Int[x^(m + p)*(b + c*x)^ 
p, x], x] /; FreeQ[{b, c, e, m}, x]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a x \right )^{\frac {1}{4}}}{x^{3}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((b*x^2+a*x)^(1/4)/x^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((b*x**2+a*x)**(1/4)/x**3,x)
 

Output:

Integral((x*(a + b*x))**(1/4)/x**3, x)
 

Maxima [F]

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

integrate((b*x^2+a*x)^(1/4)/x^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((b*x^2+a*x)^(1/4)/x^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [4]{a x+b x^2}}{x^3} \, dx=\frac {-4 \left (b x +a \right )^{\frac {1}{4}}-x^{\frac {7}{4}} \left (\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{x^{\frac {11}{4}} a +x^{\frac {15}{4}} b}d x \right ) a}{6 x^{\frac {7}{4}}} \] Input:

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

Output:

( - 4*(a + b*x)**(1/4) - x**(3/4)*int((a + b*x)**(1/4)/(x**(3/4)*a*x**2 + 
x**(3/4)*b*x**3),x)*a*x)/(6*x**(3/4)*x)