\(\int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx\) [286]

Optimal result

Integrand size = 21, antiderivative size = 265 \[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=-\frac {2 \sqrt {a x^2+b x^5}}{11 a x^{13/2}}+\frac {16 b \sqrt {a x^2+b x^5}}{55 a^2 x^{7/2}}+\frac {16 b^2 x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{55 \sqrt [4]{3} a^{7/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \] Output:

-2/11*(b*x^5+a*x^2)^(1/2)/a/x^(13/2)+16/55*b*(b*x^5+a*x^2)^(1/2)/a^2/x^(7/ 
2)+16/165*b^2*x^(3/2)*(a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2 
/3)*x^2)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)*InverseJacobiAM(arccos(( 
a^(1/3)+(1-3^(1/2))*b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)),1/4*6^(1/2 
)+1/4*2^(1/2))*3^(3/4)/a^(7/3)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+(1+ 
3^(1/2))*b^(1/3)*x)^2)^(1/2)/(b*x^5+a*x^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=-\frac {2 \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {1}{2},-\frac {5}{6},-\frac {b x^3}{a}\right )}{11 x^{9/2} \sqrt {x^2 \left (a+b x^3\right )}} \] Input:

Integrate[1/(x^(11/2)*Sqrt[a*x^2 + b*x^5]),x]
 

Output:

(-2*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[-11/6, 1/2, -5/6, -((b*x^3)/a)]) 
/(11*x^(9/2)*Sqrt[x^2*(a + b*x^3)])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1931, 1931, 1938, 851, 766}

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.

\(\displaystyle \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {8 b \int \frac {1}{x^{5/2} \sqrt {b x^5+a x^2}}dx}{11 a}-\frac {2 \sqrt {a x^2+b x^5}}{11 a x^{13/2}}\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {8 b \left (-\frac {2 b \int \frac {\sqrt {x}}{\sqrt {b x^5+a x^2}}dx}{5 a}-\frac {2 \sqrt {a x^2+b x^5}}{5 a x^{7/2}}\right )}{11 a}-\frac {2 \sqrt {a x^2+b x^5}}{11 a x^{13/2}}\)

\(\Big \downarrow \) 1938

\(\displaystyle -\frac {8 b \left (-\frac {2 b x \sqrt {a+b x^3} \int \frac {1}{\sqrt {x} \sqrt {b x^3+a}}dx}{5 a \sqrt {a x^2+b x^5}}-\frac {2 \sqrt {a x^2+b x^5}}{5 a x^{7/2}}\right )}{11 a}-\frac {2 \sqrt {a x^2+b x^5}}{11 a x^{13/2}}\)

\(\Big \downarrow \) 851

\(\displaystyle -\frac {8 b \left (-\frac {4 b x \sqrt {a+b x^3} \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {x}}{5 a \sqrt {a x^2+b x^5}}-\frac {2 \sqrt {a x^2+b x^5}}{5 a x^{7/2}}\right )}{11 a}-\frac {2 \sqrt {a x^2+b x^5}}{11 a x^{13/2}}\)

\(\Big \downarrow \) 766

\(\displaystyle -\frac {8 b \left (-\frac {2 b x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 \sqrt [4]{3} a^{4/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}-\frac {2 \sqrt {a x^2+b x^5}}{5 a x^{7/2}}\right )}{11 a}-\frac {2 \sqrt {a x^2+b x^5}}{11 a x^{13/2}}\)

Input:

Int[1/(x^(11/2)*Sqrt[a*x^2 + b*x^5]),x]
 

Output:

(-2*Sqrt[a*x^2 + b*x^5])/(11*a*x^(13/2)) - (8*b*((-2*Sqrt[a*x^2 + b*x^5])/ 
(5*a*x^(7/2)) - (2*b*x^(3/2)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3) 
*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF 
[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/ 
3)*x)], (2 + Sqrt[3])/4])/(5*3^(1/4)*a^(4/3)*Sqrt[(b^(1/3)*x*(a^(1/3) + b^ 
(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a*x^2 + b*x^5])))/(1 
1*a)
 

Defintions of rubi rules used

rule 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 851

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

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 [C] (verified)

Result contains complex when optimal does not.

Time = 3.22 (sec) , antiderivative size = 742, normalized size of antiderivative = 2.80

method	result	size
risch	\(-\frac {2 \left (b \,x^{3}+a \right ) \left (-8 b \,x^{3}+5 a \right )}{55 a^{2} x^{\frac {9}{2}} \sqrt {x^{2} \left (b \,x^{3}+a \right )}}+\frac {32 b^{3} \left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {\frac {\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) x}{\left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, {\left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}^{2} \sqrt {\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{b \left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, \sqrt {\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{b \left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) x}{\left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}, \sqrt {\frac {\left (\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{\left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right ) \sqrt {x}\, \sqrt {x \left (b \,x^{3}+a \right )}}{55 a^{2} \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {b x \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right ) \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}\, \sqrt {x^{2} \left (b \,x^{3}+a \right )}}\)	\(742\)
default	\(\text {Expression too large to display}\)	\(2009\)

Input:

int(1/x^(11/2)/(b*x^5+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/55*(b*x^3+a)*(-8*b*x^3+5*a)/a^2/x^(9/2)/(x^2*(b*x^3+a))^(1/2)+32/55*b^3 
/a^2*(1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*((-3/2/b*(-a*b^ 
2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*x/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^ 
(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2)*(x-1/b*(-a*b^2)^(1/3 
))^2*(1/b*(-a*b^2)^(1/3)*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^ 
(1/3))/(-1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b 
^2)^(1/3)))^(1/2)*(1/b*(-a*b^2)^(1/3)*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2 
)/b*(-a*b^2)^(1/3))/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)) 
/(x-1/b*(-a*b^2)^(1/3)))^(1/2)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a* 
b^2)^(1/3))/(-a*b^2)^(1/3)/(b*x*(x-1/b*(-a*b^2)^(1/3))*(x+1/2/b*(-a*b^2)^( 
1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2) 
/b*(-a*b^2)^(1/3)))^(1/2)*EllipticF(((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/ 
b*(-a*b^2)^(1/3))*x/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)) 
/(x-1/b*(-a*b^2)^(1/3)))^(1/2),((3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a* 
b^2)^(1/3))*(1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(1/2/b*( 
-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(3/2/b*(-a*b^2)^(1/3)-1/2*I* 
3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))/(x^2*(b*x^3+a))^(1/2)*x^(1/2)*(x*(b*x^3+ 
a))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=-\frac {2 \, {\left (16 \, \sqrt {a} b^{2} x^{7} {\rm weierstrassPInverse}\left (0, -\frac {4 \, b}{a}, \frac {1}{x}\right ) - \sqrt {b x^{5} + a x^{2}} {\left (8 \, a b x^{3} - 5 \, a^{2}\right )} \sqrt {x}\right )}}{55 \, a^{3} x^{7}} \] Input:

integrate(1/x^(11/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")
 

Output:

-2/55*(16*sqrt(a)*b^2*x^7*weierstrassPInverse(0, -4*b/a, 1/x) - sqrt(b*x^5 
 + a*x^2)*(8*a*b*x^3 - 5*a^2)*sqrt(x))/(a^3*x^7)
 

Sympy [F]

\[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=\int \frac {1}{x^{\frac {11}{2}} \sqrt {x^{2} \left (a + b x^{3}\right )}}\, dx \] Input:

integrate(1/x**(11/2)/(b*x**5+a*x**2)**(1/2),x)
 

Output:

Integral(1/(x**(11/2)*sqrt(x**2*(a + b*x**3))), x)
 

Maxima [F]

\[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=\int { \frac {1}{\sqrt {b x^{5} + a x^{2}} x^{\frac {11}{2}}} \,d x } \] Input:

integrate(1/x^(11/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(b*x^5 + a*x^2)*x^(11/2)), x)
 

Giac [F]

\[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=\int { \frac {1}{\sqrt {b x^{5} + a x^{2}} x^{\frac {11}{2}}} \,d x } \] Input:

integrate(1/x^(11/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(b*x^5 + a*x^2)*x^(11/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=\int \frac {1}{x^{11/2}\,\sqrt {b\,x^5+a\,x^2}} \,d x \] Input:

int(1/(x^(11/2)*(a*x^2 + b*x^5)^(1/2)),x)
 

Output:

int(1/(x^(11/2)*(a*x^2 + b*x^5)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^{11/2} \sqrt {a x^2+b x^5}} \, dx=\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}}{b \,x^{10}+a \,x^{7}}d x \] Input:

int(1/x^(11/2)/(b*x^5+a*x^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((sqrt(x)*sqrt(a + b*x**3))/(a*x**7 + b*x**10),x)