\(\int \sqrt {x} \sqrt {b x^2+c x^4} \, dx\) [231]

Optimal result

Integrand size = 21, antiderivative size = 146 \[ \int \sqrt {x} \sqrt {b x^2+c x^4} \, dx=\frac {4 b \sqrt {b x^2+c x^4}}{21 c \sqrt {x}}+\frac {2}{7} x^{3/2} \sqrt {b x^2+c x^4}-\frac {2 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 c^{5/4} \sqrt {b x^2+c x^4}} \] Output:

4/21*b*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)+2/7*x^(3/2)*(c*x^4+b*x^2)^(1/2)-2/21* 
b^(7/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*Inve 
rseJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/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 = 9.82 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.59 \[ \int \sqrt {x} \sqrt {b x^2+c x^4} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}}-b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{7 c \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[Sqrt[x]*Sqrt[b*x^2 + c*x^4],x]
 

Output:

(2*Sqrt[x^2*(b + c*x^2)]*((b + c*x^2)*Sqrt[1 + (c*x^2)/b] - b*Hypergeometr 
ic2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)]))/(7*c*Sqrt[x]*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1426, 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.

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[Sqrt[x]*Sqrt[b*x^2 + c*x^4],x]
 

Output:

(2*x^(3/2)*Sqrt[b*x^2 + c*x^4])/7 + (2*b*((2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqr 
t[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)*S 
qrt[b*x^2 + c*x^4])))/7
 

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 1426

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

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]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99

method	result	size
default	\(-\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (b^{2} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-3 c^{3} x^{5}-5 b \,c^{2} x^{3}-2 b^{2} c x \right )}{21 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c^{2}}\)	\(145\)
risch	\(\frac {2 \left (3 c \,x^{2}+2 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{21 c \sqrt {x}}-\frac {2 b^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 c^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(180\)

Input:

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

Output:

-2/21*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)*(b^2*(-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))-3*c^3*x^5-5*b*c^2*x^3-2*b^2*c*x)/c^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.40 \[ \int \sqrt {x} \sqrt {b x^2+c x^4} \, dx=-\frac {2 \, {\left (2 \, b^{2} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} {\left (3 \, c^{2} x^{2} + 2 \, b c\right )} \sqrt {x}\right )}}{21 \, c^{2} x} \] Input:

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

Output:

-2/21*(2*b^2*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) - sqrt(c*x^4 + b* 
x^2)*(3*c^2*x^2 + 2*b*c)*sqrt(x))/(c^2*x)
 

Sympy [F]

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

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

Output:

Integral(sqrt(x)*sqrt(x**2*(b + c*x**2)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {x} \sqrt {b x^2+c x^4} \, dx=\frac {\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b}{21}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, c \,x^{2}}{7}-\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{2}}{21}}{c} \] Input:

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

Output:

(2*(2*sqrt(x)*sqrt(b + c*x**2)*b + 3*sqrt(x)*sqrt(b + c*x**2)*c*x**2 - int 
((sqrt(x)*sqrt(b + c*x**2))/(b*x + c*x**3),x)*b**2))/(21*c)