\(\int x^{5/2} \sqrt {b x^2+c x^4} \, dx\) [229]

Optimal result

Integrand size = 21, antiderivative size = 176 \[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=-\frac {20 b^2 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {10 b^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}} \] Output:

-20/231*b^2*(c*x^4+b*x^2)^(1/2)/c^2/x^(1/2)+4/77*b*x^(3/2)*(c*x^4+b*x^2)^( 
1/2)/c+2/11*x^(7/2)*(c*x^4+b*x^2)^(1/2)+10/231*b^(11/4)*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^(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 = 10.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.58 \[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {1+\frac {c x^2}{b}} \left (-5 b^2+2 b c x^2+7 c^2 x^4\right )+5 b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{77 c^2 \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[x^(5/2)*Sqrt[b*x^2 + c*x^4],x]
 

Output:

(2*Sqrt[x^2*(b + c*x^2)]*(Sqrt[1 + (c*x^2)/b]*(-5*b^2 + 2*b*c*x^2 + 7*c^2* 
x^4) + 5*b^2*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)]))/(77*c^2*Sqr 
t[x]*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1426, 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.

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[x^(5/2)*Sqrt[b*x^2 + c*x^4],x]
 

Output:

(2*x^(7/2)*Sqrt[b*x^2 + c*x^4])/11 + (2*b*((2*x^(3/2)*Sqrt[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
 

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 = 0.36 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (21 c^{4} x^{7}+5 b^{3} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+27 b \,c^{3} x^{5}-4 b^{2} c^{2} x^{3}-10 b^{3} c x \right )}{231 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c^{3}}\)	\(157\)
risch	\(-\frac {2 \left (-21 c^{2} x^{4}-6 b c \,x^{2}+10 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{231 c^{2} \sqrt {x}}+\frac {10 b^{3} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 c^{3} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(191\)

Input:

int(x^(5/2)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/231*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)*(21*c^4*x^7+5*b^3*(-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))+27*b*c^3*x^5-4*b^2*c^2*x^3-10*b^3*c*x)/c^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.39 \[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=\frac {2 \, {\left (10 \, b^{3} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (21 \, c^{3} x^{4} + 6 \, b c^{2} x^{2} - 10 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, c^{3} x} \] Input:

integrate(x^(5/2)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

2/231*(10*b^3*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) + (21*c^3*x^4 + 
6*b*c^2*x^2 - 10*b^2*c)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^3*x)
 

Sympy [F]

\[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=\int x^{\frac {5}{2}} \sqrt {x^{2} \left (b + c x^{2}\right )}\, dx \] Input:

integrate(x**(5/2)*(c*x**4+b*x**2)**(1/2),x)
 

Output:

Integral(x**(5/2)*sqrt(x**2*(b + c*x**2)), x)
 

Maxima [F]

\[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=\int { \sqrt {c x^{4} + b x^{2}} x^{\frac {5}{2}} \,d x } \] Input:

integrate(x^(5/2)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*x^(5/2), x)
 

Giac [F]

\[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=\int { \sqrt {c x^{4} + b x^{2}} x^{\frac {5}{2}} \,d x } \] Input:

integrate(x^(5/2)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^4 + b*x^2)*x^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=\int x^{5/2}\,\sqrt {c\,x^4+b\,x^2} \,d x \] Input:

int(x^(5/2)*(b*x^2 + c*x^4)^(1/2),x)
 

Output:

int(x^(5/2)*(b*x^2 + c*x^4)^(1/2), x)
 

Reduce [F]

\[ \int x^{5/2} \sqrt {b x^2+c x^4} \, dx=\frac {-\frac {20 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2}}{231}+\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b c \,x^{2}}{77}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, c^{2} x^{4}}{11}+\frac {10 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{3}}{231}}{c^{2}} \] Input:

int(x^(5/2)*(c*x^4+b*x^2)^(1/2),x)
 

Output:

(2*( - 10*sqrt(x)*sqrt(b + c*x**2)*b**2 + 6*sqrt(x)*sqrt(b + c*x**2)*b*c*x 
**2 + 21*sqrt(x)*sqrt(b + c*x**2)*c**2*x**4 + 5*int((sqrt(x)*sqrt(b + c*x* 
*2))/(b*x + c*x**3),x)*b**3))/(231*c**2)