\(\int \frac {A+B x^2}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx\) [260]

Optimal result

Integrand size = 28, antiderivative size = 204 \[ \int \frac {A+B x^2}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {2 (11 b B-9 A c) \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac {10 c (11 b B-9 A c) \sqrt {b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac {5 c^{7/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 b^{13/4} \sqrt {b x^2+c x^4}} \] Output:

-2/11*A*(c*x^4+b*x^2)^(1/2)/b/x^(13/2)-2/77*(-9*A*c+11*B*b)*(c*x^4+b*x^2)^ 
(1/2)/b^2/x^(9/2)+10/231*c*(-9*A*c+11*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^(5/2) 
+5/231*c^(7/4)*(-9*A*c+11*B*b)*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 = 10.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.41 \[ \int \frac {A+B x^2}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=-\frac {2 \left (7 A \left (b+c x^2\right )+(11 b B-9 A c) x^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{2},-\frac {3}{4},-\frac {c x^2}{b}\right )\right )}{77 b x^{9/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(-2*(7*A*(b + c*x^2) + (11*b*B - 9*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeome 
tric2F1[-7/4, 1/2, -3/4, -((c*x^2)/b)]))/(77*b*x^(9/2)*Sqrt[x^2*(b + c*x^2 
)])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1944, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

(-2*A*Sqrt[b*x^2 + c*x^4])/(11*b*x^(13/2)) + ((11*b*B - 9*A*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]
 

rule 1944

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

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.06

method	result	size
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (45 x^{4} A \,c^{2}-55 x^{4} B b c -27 A b c \,x^{2}+33 x^{2} B \,b^{2}+21 b^{2} A \right )}{231 b^{3} x^{\frac {9}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {5 c \left (9 A c -11 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 b^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(216\)
default	\(-\frac {45 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c^{2} x^{5}-55 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c \,x^{5}+90 A \,c^{3} x^{6}-110 B b \,c^{2} x^{6}+36 A b \,c^{2} x^{4}-44 x^{4} B \,b^{2} c -12 A \,b^{2} c \,x^{2}+66 x^{2} B \,b^{3}+42 A \,b^{3}}{231 \sqrt {c \,x^{4}+b \,x^{2}}\, x^{\frac {9}{2}} b^{3}}\)	\(274\)

Input:

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

Output:

-2/231*(c*x^2+b)*(45*A*c^2*x^4-55*B*b*c*x^4-27*A*b*c*x^2+33*B*b^2*x^2+21*A 
*b^2)/b^3/x^(9/2)/(x^2*(c*x^2+b))^(1/2)-5/231*c*(9*A*c-11*B*b)/b^3*(-b*c)^ 
(1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/2)) 
*c/(-b*c)^(1/2))^(1/2)*(-c/(-b*c)^(1/2)*x)^(1/2)/(c*x^3+b*x)^(1/2)*Ellipti 
cF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^(1/2)/(x^2*( 
c*x^2+b))^(1/2)*(x*(c*x^2+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.47 \[ \int \frac {A+B x^2}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=\frac {2 \, {\left (5 \, {\left (11 \, B b c - 9 \, A c^{2}\right )} \sqrt {c} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (5 \, {\left (11 \, B b c - 9 \, A c^{2}\right )} x^{4} - 21 \, A b^{2} - 3 \, {\left (11 \, B b^{2} - 9 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, b^{3} x^{7}} \] Input:

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

Output:

2/231*(5*(11*B*b*c - 9*A*c^2)*sqrt(c)*x^7*weierstrassPInverse(-4*b/c, 0, x 
) + (5*(11*B*b*c - 9*A*c^2)*x^4 - 21*A*b^2 - 3*(11*B*b^2 - 9*A*b*c)*x^2)*s 
qrt(c*x^4 + b*x^2)*sqrt(x))/(b^3*x^7)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x^2}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, b +9 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{9}+b \,x^{7}}d x \right ) a c \,x^{5}-11 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{9}+b \,x^{7}}d x \right ) b^{2} x^{5}}{9 \sqrt {x}\, c \,x^{5}} \] Input:

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

Output:

( - 2*sqrt(b + c*x**2)*b + 9*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x** 
7 + c*x**9),x)*a*c*x**5 - 11*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b*x** 
7 + c*x**9),x)*b**2*x**5)/(9*sqrt(x)*c*x**5)