\(\int \frac {\sqrt {x} (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\) [254]

Optimal result

Integrand size = 28, antiderivative size = 130 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 B \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {(b B-3 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} c^{5/4} \sqrt {b x^2+c x^4}} \] Output:

2/3*B*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)-1/3*(-3*A*c+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^(5/4)/(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 = 10.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (B \left (b+c x^2\right )+(-b B+3 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{3 c \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(2*x^(3/2)*(B*(b + c*x^2) + (-(b*B) + 3*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeom 
etric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(3*c*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1945, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

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 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 = 0.59 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35

method	result	size
risch	\(\frac {2 B \,x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{3 c \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (3 A c -B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 c^{2} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(175\)
default	\(\frac {\sqrt {x}\, \left (3 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c -B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b +2 B \,c^{2} x^{3}+2 B b c x \right )}{3 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{2}}\)	\(216\)

Input:

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

Output:

2/3*B/c*x^(3/2)*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2)+1/3*(3*A*c-B*b)/c^2*(-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)*Ellipt 
icF(((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 = 51, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\left ({\left (B b - 3 \, A c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} B c \sqrt {x}\right )}}{3 \, c^{2} x} \] Input:

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

Output:

-2/3*((B*b - 3*A*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) - sqrt(c*x 
^4 + b*x^2)*B*c*sqrt(x))/(c^2*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(x)*sqrt(b + c*x**2)*b + 3*int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c* 
x**3),x)*a*c - int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c*x**3),x)*b**2)/(3*c 
)