\(\int \sqrt {c x} \sqrt {a+b x^2} \, dx\) [594]

Optimal result

Integrand size = 19, antiderivative size = 269 \[ \int \sqrt {c x} \sqrt {a+b x^2} \, dx=\frac {2 (c x)^{3/2} \sqrt {a+b x^2}}{5 c}+\frac {4 a \sqrt {c x} \sqrt {a+b x^2}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {4 a^{5/4} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^2}}+\frac {2 a^{5/4} \sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^2}} \] Output:

2/5*(c*x)^(3/2)*(b*x^2+a)^(1/2)/c+4/5*a*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b^(1/2 
)/(a^(1/2)+b^(1/2)*x)-4/5*a^(5/4)*c^(1/2)*(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^(3/4)/(b*x^2+a)^(1/2)+2/5*a^(5/4)*c^(1/2)*(a 
^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM( 
2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)),1/2*2^(1/2))/b^(3/4)/(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 = 7.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.21 \[ \int \sqrt {c x} \sqrt {a+b x^2} \, dx=\frac {2 x \sqrt {c x} \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )}{3 \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {248, 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.

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

\(\Big \downarrow \) 248

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

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

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 248

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) 
  Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ 
p, 0] && 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 = 0.42 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {2 \sqrt {c x}\, \left (2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}+b^{2} x^{4}+a b \,x^{2}\right )}{5 \sqrt {b \,x^{2}+a}\, b x}\)	\(205\)
risch	\(\frac {2 x^{2} \sqrt {b \,x^{2}+a}\, c}{5 \sqrt {c x}}+\frac {2 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) c \sqrt {c x \left (b \,x^{2}+a \right )}}{5 b \sqrt {b c \,x^{3}+a c x}\, \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(210\)
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (\frac {2 x \sqrt {b c \,x^{3}+a c x}}{5}+\frac {2 a c \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(213\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.18 \[ \int \sqrt {c x} \sqrt {a+b x^2} \, dx=\frac {2 \, {\left (\sqrt {b x^{2} + a} \sqrt {c x} b x - 2 \, \sqrt {b c} a {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right )\right )}}{5 \, b} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.17 \[ \int \sqrt {c x} \sqrt {a+b x^2} \, dx=\frac {\sqrt {a} \sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

sqrt(a)*sqrt(c)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**2*exp_ 
polar(I*pi)/a)/(2*gamma(7/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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