\(\int \frac {(c x)^{5/2}}{\sqrt {a+b x^2}} \, dx\) [624]

Optimal result

Integrand size = 19, antiderivative size = 273 \[ \int \frac {(c x)^{5/2}}{\sqrt {a+b x^2}} \, dx=\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c^2 \sqrt {c x} \sqrt {a+b x^2}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {6 a^{5/4} c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^2}}-\frac {3 a^{5/4} c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^2}} \] Output:

2/5*c*(c*x)^(3/2)*(b*x^2+a)^(1/2)/b-6/5*a*c^2*(c*x)^(1/2)*(b*x^2+a)^(1/2)/ 
b^(3/2)/(a^(1/2)+b^(1/2)*x)+6/5*a^(5/4)*c^(5/2)*(a^(1/2)+b^(1/2)*x)*((b*x^ 
2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*(c*x)^(1/ 
2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))/b^(7/4)/(b*x^2+a)^(1/2)-3/5*a^(5/4)*c^(5 
/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJac 
obiAM(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)),1/2*2^(1/2))/b^(7/4)/( 
b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.25 \[ \int \frac {(c x)^{5/2}}{\sqrt {a+b x^2}} \, dx=\frac {2 c (c x)^{3/2} \left (a+b x^2-a \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{5 b \sqrt {a+b x^2}} \] Input:

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

Output:

(2*c*(c*x)^(3/2)*(a + b*x^2 - a*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, 
 3/4, 7/4, -((b*x^2)/a)]))/(5*b*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {262, 266, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \int \frac {c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{5 b}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt {a} c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}-\frac {\sqrt {a} c \int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {a} c \sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{5 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt {a} c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{5 b}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{5 b}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {c^2 \sqrt {c x} \sqrt {a+b x^2}}{\sqrt {a} c+\sqrt {b} c x}}{\sqrt {b}}\right )}{5 b}\)

Input:

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

Output:

(2*c*(c*x)^(3/2)*Sqrt[a + b*x^2])/(5*b) - (6*a*c*(-((-((c^2*Sqrt[c*x]*Sqrt 
[a + b*x^2])/(Sqrt[a]*c + Sqrt[b]*c*x)) + (a^(1/4)*Sqrt[c]*(Sqrt[a]*c + Sq 
rt[b]*c*x)*Sqrt[(a*c^2 + b*c^2*x^2)/(Sqrt[a]*c + Sqrt[b]*c*x)^2]*EllipticE 
[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(b^(1/4)*Sqrt[a + 
b*x^2]))/Sqrt[b]) + (a^(1/4)*Sqrt[c]*(Sqrt[a]*c + Sqrt[b]*c*x)*Sqrt[(a*c^2 
 + b*c^2*x^2)/(Sqrt[a]*c + Sqrt[b]*c*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqr 
t[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^2])))/(5*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

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 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {c^{2} \sqrt {c x}\, \left (6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-2 b^{2} x^{4}-2 a b \,x^{2}\right )}{5 x \sqrt {b \,x^{2}+a}\, b^{2}}\)	\(210\)
risch	\(\frac {2 x^{2} \sqrt {b \,x^{2}+a}\, c^{3}}{5 b \sqrt {c x}}-\frac {3 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) c^{3} \sqrt {c x \left (b \,x^{2}+a \right )}}{5 b^{2} \sqrt {b c \,x^{3}+a c x}\, \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(217\)
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (\frac {2 c^{2} x \sqrt {b c \,x^{3}+a c x}}{5 b}-\frac {3 c^{3} a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b^{2} \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(221\)

Input:

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

Output:

-1/5*c^2/x*(c*x)^(1/2)/(b*x^2+a)^(1/2)/b^2*(6*((b*x+(-a*b)^(1/2))/(-a*b)^( 
1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-b/(-a*b)^(1 
/2)*x)^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2) 
)*a^2-3*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2 
))/(-a*b)^(1/2))^(1/2)*(-b/(-a*b)^(1/2)*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1 
/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^2-2*b^2*x^4-2*a*b*x^2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.20 \[ \int \frac {(c x)^{5/2}}{\sqrt {a+b x^2}} \, dx=\frac {2 \, {\left (\sqrt {b x^{2} + a} \sqrt {c x} b c^{2} x + 3 \, \sqrt {b c} a c^{2} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right )\right )}}{5 \, b^{2}} \] Input:

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

Output:

2/5*(sqrt(b*x^2 + a)*sqrt(c*x)*b*c^2*x + 3*sqrt(b*c)*a*c^2*weierstrassZeta 
(-4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)))/b^2
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.16 \[ \int \frac {(c x)^{5/2}}{\sqrt {a+b x^2}} \, dx=\frac {c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \] Input:

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

Output:

c**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**2*exp_polar(I 
*pi)/a)/(2*sqrt(a)*gamma(11/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*c**2*(2*sqrt(x)*sqrt(a + b*x**2)*x - 3*int((sqrt(x)*sqrt(a + b*x* 
*2))/(a + b*x**2),x)*a))/(5*b)