\(\int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx\) [266]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=-\frac {2 \sqrt {b x^2+c x^4}}{11 b x^{13/2}}+\frac {18 c \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}-\frac {30 c^2 \sqrt {b x^2+c x^4}}{77 b^3 x^{5/2}}-\frac {15 c^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 b^{13/4} \sqrt {b x^2+c x^4}} \] Output:

-2/11*(c*x^4+b*x^2)^(1/2)/b/x^(13/2)+18/77*c*(c*x^4+b*x^2)^(1/2)/b^2/x^(9/ 
2)-30/77*c^2*(c*x^4+b*x^2)^(1/2)/b^3/x^(5/2)-15/77*c^(11/4)*x*(b^(1/2)+c^( 
1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(c 
^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/b^(13/4)/(c*x^4+b*x^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.32 \[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=-\frac {2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},\frac {1}{2},-\frac {7}{4},-\frac {c x^2}{b}\right )}{11 x^{9/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[1/(x^(11/2)*Sqrt[b*x^2 + c*x^4]),x]
 

Output:

(-2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-11/4, 1/2, -7/4, -((c*x^2)/b)]) 
/(11*x^(9/2)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1430, 1430, 1430, 1431, 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.

\(\displaystyle \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx\)

\(\Big \downarrow \) 1430

\(\displaystyle -\frac {9 c \int \frac {1}{x^{7/2} \sqrt {c x^4+b x^2}}dx}{11 b}-\frac {2 \sqrt {b x^2+c x^4}}{11 b x^{13/2}}\)

\(\Big \downarrow \) 1430

\(\displaystyle -\frac {9 c \left (-\frac {5 c \int \frac {1}{x^{3/2} \sqrt {c x^4+b x^2}}dx}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )}{11 b}-\frac {2 \sqrt {b x^2+c x^4}}{11 b x^{13/2}}\)

\(\Big \downarrow \) 1430

\(\displaystyle -\frac {9 c \left (-\frac {5 c \left (-\frac {c \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 b}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )}{11 b}-\frac {2 \sqrt {b x^2+c x^4}}{11 b x^{13/2}}\)

\(\Big \downarrow \) 1431

\(\displaystyle -\frac {9 c \left (-\frac {5 c \left (-\frac {c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )}{11 b}-\frac {2 \sqrt {b x^2+c x^4}}{11 b x^{13/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {9 c \left (-\frac {5 c \left (-\frac {2 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )}{11 b}-\frac {2 \sqrt {b x^2+c x^4}}{11 b x^{13/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {9 c \left (-\frac {5 c \left (-\frac {c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^{9/2}}\right )}{11 b}-\frac {2 \sqrt {b x^2+c x^4}}{11 b x^{13/2}}\)

Input:

Int[1/(x^(11/2)*Sqrt[b*x^2 + c*x^4]),x]
 

Output:

(-2*Sqrt[b*x^2 + c*x^4])/(11*b*x^(13/2)) - (9*c*((-2*Sqrt[b*x^2 + c*x^4])/ 
(7*b*x^(9/2)) - (5*c*((-2*Sqrt[b*x^2 + c*x^4])/(3*b*x^(5/2)) - (c^(3/4)*x* 
(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[ 
2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*b^(5/4)*Sqrt[b*x^2 + c*x^4]) 
))/(7*b)))/(11*b)
 

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 1430

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( 
(m + 4*p + 3)/(b*d^2*(m + 2*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, 
 x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && LtQ[m + 2*p + 1, 0 
]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 2.73 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {15 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c^{2} x^{5}+30 c^{3} x^{6}+12 b \,c^{2} x^{4}-4 b^{2} c \,x^{2}+14 b^{3}}{77 \sqrt {c \,x^{4}+b \,x^{2}}\, x^{\frac {9}{2}} b^{3}}\)	\(147\)
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (15 c^{2} x^{4}-9 b c \,x^{2}+7 b^{2}\right )}{77 b^{3} x^{\frac {9}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {15 c^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{77 b^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(189\)

Input:

int(1/x^(11/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/77/(c*x^4+b*x^2)^(1/2)/x^(9/2)*(15*(-b*c)^(1/2)*((c*x+(-b*c)^(1/2))/(-b 
*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-c/(-b* 
c)^(1/2)*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^ 
(1/2))*c^2*x^5+30*c^3*x^6+12*b*c^2*x^4-4*b^2*c*x^2+14*b^3)/b^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.36 \[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\left (15 \, c^{\frac {5}{2}} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (15 \, c^{2} x^{4} - 9 \, b c x^{2} + 7 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{77 \, b^{3} x^{7}} \] Input:

integrate(1/x^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

-2/77*(15*c^(5/2)*x^7*weierstrassPInverse(-4*b/c, 0, x) + (15*c^2*x^4 - 9* 
b*c*x^2 + 7*b^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^3*x^7)
 

Sympy [F]

\[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=\int \frac {1}{x^{\frac {11}{2}} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:

integrate(1/x**(11/2)/(c*x**4+b*x**2)**(1/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2}} x^{\frac {11}{2}}} \,d x } \] Input:

integrate(1/x^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(c*x^4 + b*x^2)*x^(11/2)), x)
 

Giac [F]

\[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2}} x^{\frac {11}{2}}} \,d x } \] Input:

integrate(1/x^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(c*x^4 + b*x^2)*x^(11/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=\int \frac {1}{x^{11/2}\,\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:

int(1/(x^(11/2)*(b*x^2 + c*x^4)^(1/2)),x)
 

Output:

int(1/(x^(11/2)*(b*x^2 + c*x^4)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{9}+b \,x^{7}}d x \] Input:

int(1/x^(11/2)/(c*x^4+b*x^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((sqrt(x)*sqrt(b + c*x**2))/(b*x**7 + c*x**9),x)