3.7.18 \(\int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx\) [618]

3.7.18.1 Optimal result

Integrand size = 19, antiderivative size = 268 \[ \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx=-\frac {2 \sqrt {a+b x^2}}{a c \sqrt {c x}}+\frac {2 \sqrt {b} \sqrt {c x} \sqrt {a+b x^2}}{a c^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{a^{3/4} c^{3/2} \sqrt {a+b x^2}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{a^{3/4} c^{3/2} \sqrt {a+b x^2}} \]

output

-2*(b*x^2+a)^(1/2)/a/c/(c*x)^(1/2)+2*b^(1/2)*(c*x)^(1/2)*(b*x^2+a)^(1/2)/a 
/c^2/(a^(1/2)+x*b^(1/2))-2*b^(1/4)*(cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/ 
4)/c^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*E 
llipticE(sin(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))*( 
a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(3/4)/c^(3/2) 
/(b*x^2+a)^(1/2)+b^(1/4)*(cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2) 
))^2)^(1/2)/cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*EllipticF(s 
in(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))*(a^(1/2)+x* 
b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(3/4)/c^(3/2)/(b*x^2+a) 
^(1/2)
 

3.7.18.2 Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.20 \[ \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx=-\frac {2 x \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {b x^2}{a}\right )}{(c x)^{3/2} \sqrt {a+b x^2}} \]

input

Integrate[1/((c*x)^(3/2)*Sqrt[a + b*x^2]),x]
 

output

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

3.7.18.3 Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {264, 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.

\(\displaystyle \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx\)

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

input

Int[1/((c*x)^(3/2)*Sqrt[a + b*x^2]),x]
 

output

(-2*Sqrt[a + b*x^2])/(a*c*Sqrt[c*x]) + (2*b*(-((-((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*Ar 
cTan[(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])))/(a*c^3)
 

3.7.18.3.1 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 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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]
 

3.7.18.4 Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.73

method	result	size
default	\(\frac {2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a -\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a -2 b \,x^{2}-2 a}{\sqrt {b \,x^{2}+a}\, c \sqrt {c x}\, a}\)	\(196\)
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}}{a c \sqrt {c x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) \sqrt {c x \left (b \,x^{2}+a \right )}}{a \sqrt {b c \,x^{3}+a c x}\, c \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(212\)
elliptic	\(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {2 \left (c b \,x^{2}+a c \right )}{a \,c^{2} \sqrt {x \left (c b \,x^{2}+a c \right )}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{a c \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(223\)

input

int(1/(c*x)^(3/2)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

output

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

3.7.18.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.19 \[ \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx=-\frac {2 \, {\left (\sqrt {b c} x {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + \sqrt {b x^{2} + a} \sqrt {c x}\right )}}{a c^{2} x} \]

input

integrate(1/(c*x)^(3/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

output

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

3.7.18.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.18 \[ \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx=\frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} c^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

input

integrate(1/(c*x)**(3/2)/(b*x**2+a)**(1/2),x)
 

output

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

3.7.18.7 Maxima [F]

\[ \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} \left (c x\right )^{\frac {3}{2}}} \,d x } \]

input

integrate(1/(c*x)^(3/2)/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

output

integrate(1/(sqrt(b*x^2 + a)*(c*x)^(3/2)), x)
 

3.7.18.8 Giac [F]

\[ \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} \left (c x\right )^{\frac {3}{2}}} \,d x } \]

input

integrate(1/(c*x)^(3/2)/(b*x^2+a)^(1/2),x, algorithm="giac")
 

output

integrate(1/(sqrt(b*x^2 + a)*(c*x)^(3/2)), x)
 

3.7.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx=\int \frac {1}{{\left (c\,x\right )}^{3/2}\,\sqrt {b\,x^2+a}} \,d x \]

input

int(1/((c*x)^(3/2)*(a + b*x^2)^(1/2)),x)
 

output

int(1/((c*x)^(3/2)*(a + b*x^2)^(1/2)), x)