\(\int \frac {x^{9/2} (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [265]

Optimal result

Integrand size = 28, antiderivative size = 178 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x^{7/2}}{b c \sqrt {b x^2+c x^4}}+\frac {(5 b B-3 A c) \sqrt {b x^2+c x^4}}{3 b c^2 \sqrt {x}}-\frac {(5 b B-3 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{6 \sqrt [4]{b} c^{9/4} \sqrt {b x^2+c x^4}} \] Output:

-(-A*c+B*b)*x^(7/2)/b/c/(c*x^4+b*x^2)^(1/2)+1/3*(-3*A*c+5*B*b)*(c*x^4+b*x^ 
2)^(1/2)/b/c^2/x^(1/2)-1/6*(-3*A*c+5*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^(1/4)/c^(9/4)/(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 = 10.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.48 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2} \left (5 b B-3 A c+2 B c x^2+(-5 b B+3 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{3 c^2 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[(x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(x^(3/2)*(5*b*B - 3*A*c + 2*B*c*x^2 + (-5*b*B + 3*A*c)*Sqrt[1 + (c*x^2)/b] 
*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(3*c^2*Sqrt[x^2*(b + c*x 
^2)])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1943, 1429, 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.

\(\displaystyle \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1943

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[(x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

-(((b*B - A*c)*x^(7/2))/(b*c*Sqrt[b*x^2 + c*x^4])) + ((5*b*B - 3*A*c)*((2* 
Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (b^(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*c^(5/4)*Sqrt[b*x^2 + c*x^4])))/(2*b*c)
 

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 1429

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b 
*d^2*((m + 2*p - 1)/(c*(m + 4*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] && GtQ[m + 2*p - 1, 
 0] && NeQ[m + 4*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 1943

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

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.29

method	result	size
default	\(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (3 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c -5 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b +4 B \,c^{2} x^{3}-6 A \,c^{2} x +10 B b c x \right )}{6 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{3}}\)	\(230\)
risch	\(\frac {2 B \,x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{3 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (\frac {3 A \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {c \,x^{3}+b x}}-\frac {4 B b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}-3 b \left (A c -B b \right ) \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(422\)

Input:

int(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/6/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(3*A*(-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-5*B*(-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))*b+4*B*c^2 
*x^3-6*A*c^2*x+10*B*b*c*x)/c^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.57 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {{\left ({\left (5 \, B b c - 3 \, A c^{2}\right )} x^{3} + {\left (5 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (2 \, B c^{2} x^{2} + 5 \, B b c - 3 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{3 \, {\left (c^{4} x^{3} + b c^{3} x\right )}} \] Input:

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

-1/3*(((5*B*b*c - 3*A*c^2)*x^3 + (5*B*b^2 - 3*A*b*c)*x)*sqrt(c)*weierstras 
sPInverse(-4*b/c, 0, x) - (2*B*c^2*x^2 + 5*B*b*c - 3*A*c^2)*sqrt(c*x^4 + b 
*x^2)*sqrt(x))/(c^4*x^3 + b*c^3*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(x**(9/2)*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {9}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*x^(9/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Giac [F]

\[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {9}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*x^(9/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{9/2}\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int((x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-6 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a c +10 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2}+2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b c \,x^{2}+3 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) a \,b^{2} c +3 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) a b \,c^{2} x^{2}-5 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b^{4}-5 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b^{3} c \,x^{2}}{3 c^{2} \left (c \,x^{2}+b \right )} \] Input:

int(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

( - 6*sqrt(x)*sqrt(b + c*x**2)*a*c + 10*sqrt(x)*sqrt(b + c*x**2)*b**2 + 2* 
sqrt(x)*sqrt(b + c*x**2)*b*c*x**2 + 3*int((sqrt(x)*sqrt(b + c*x**2))/(b**2 
*x + 2*b*c*x**3 + c**2*x**5),x)*a*b**2*c + 3*int((sqrt(x)*sqrt(b + c*x**2) 
)/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*a*b*c**2*x**2 - 5*int((sqrt(x)*sqrt 
(b + c*x**2))/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*b**4 - 5*int((sqrt(x)*s 
qrt(b + c*x**2))/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*b**3*c*x**2)/(3*c**2 
*(b + c*x**2))