\(\int \frac {x^{9/2} (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\) [250]

Optimal result

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

10/231*b*(-11*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)-2/77*(-11*A*c+9*B 
*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2+2/11*B*x^(7/2)*(c*x^4+b*x^2)^(1/2)/c-5 
/231*b^(7/4)*(-11*A*c+9*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))/c^(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 = 10.12 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.60 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (\left (b+c x^2\right ) \left (45 b^2 B+3 c^2 x^2 \left (11 A+7 B x^2\right )-b c \left (55 A+27 B x^2\right )\right )+5 b^2 (-9 b B+11 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{231 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(2*x^(3/2)*((b + c*x^2)*(45*b^2*B + 3*c^2*x^2*(11*A + 7*B*x^2) - b*c*(55*A 
 + 27*B*x^2)) + 5*b^2*(-9*b*B + 11*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric 
2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(231*c^3*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.64 (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 = {1945, 1429, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

(2*B*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(11*c) - ((9*b*B - 11*A*c)*((2*x^(3/2)*S 
qrt[b*x^2 + c*x^4])/(7*c) - (5*b*((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])))/(7*c)))/(11*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 1945

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {2 \left (-21 B \,c^{2} x^{4}-33 A \,c^{2} x^{2}+27 x^{2} B b c +55 A b c -45 B \,b^{2}\right ) x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{231 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {5 b^{2} \left (11 A c -9 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 c^{4} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(215\)
default	\(\frac {\sqrt {x}\, \left (42 B \,c^{4} x^{7}+55 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c +66 A \,c^{4} x^{5}-45 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{3}-12 B b \,c^{3} x^{5}-44 A b \,c^{3} x^{3}+36 B \,b^{2} c^{2} x^{3}-110 A \,b^{2} c^{2} x +90 B \,b^{3} c x \right )}{231 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{4}}\)	\(274\)

Input:

int(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/231*(-21*B*c^2*x^4-33*A*c^2*x^2+27*B*b*c*x^2+55*A*b*c-45*B*b^2)/c^3*x^( 
3/2)*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2)+5/231*b^2*(11*A*c-9*B*b)/c^4*(-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)*Elliptic 
F(((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 = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.49 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\left (5 \, {\left (9 \, B b^{3} - 11 \, A b^{2} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (21 \, B c^{3} x^{4} + 45 \, B b^{2} c - 55 \, A b c^{2} - 3 \, {\left (9 \, B b c^{2} - 11 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, c^{4} x} \] Input:

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

-2/231*(5*(9*B*b^3 - 11*A*b^2*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, 
x) - (21*B*c^3*x^4 + 45*B*b^2*c - 55*A*b*c^2 - 3*(9*B*b*c^2 - 11*A*c^3)*x^ 
2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^4*x)
 

Sympy [F(-1)]

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

integrate(x**(9/2)*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2),x)
 

Output:

int((x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {-110 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a b c +66 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{2}+90 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3}-54 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{2}+42 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}+55 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) a \,b^{2} c -45 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{4}}{231 c^{3}} \] Input:

int(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)
 

Output:

( - 110*sqrt(x)*sqrt(b + c*x**2)*a*b*c + 66*sqrt(x)*sqrt(b + c*x**2)*a*c** 
2*x**2 + 90*sqrt(x)*sqrt(b + c*x**2)*b**3 - 54*sqrt(x)*sqrt(b + c*x**2)*b* 
*2*c*x**2 + 42*sqrt(x)*sqrt(b + c*x**2)*b*c**2*x**4 + 55*int((sqrt(x)*sqrt 
(b + c*x**2))/(b*x + c*x**3),x)*a*b**2*c - 45*int((sqrt(x)*sqrt(b + c*x**2 
))/(b*x + c*x**3),x)*b**4)/(231*c**3)