\(\int \frac {\sqrt {a+b x^2}}{(c x)^{3/4}} \, dx\) [669]

Optimal result

Integrand size = 19, antiderivative size = 510 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{3/4}} \, dx=\frac {4 \sqrt [4]{c x} \sqrt {a+b x^2}}{5 c}+\frac {8 a \sqrt [4]{b} (c x)^{3/4} \sqrt {-\frac {a+b x^2}{\sqrt {a} \sqrt {b} x}} \sqrt {\frac {\left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {2}+\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a}}-\frac {2 \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )}{\sqrt [4]{b} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{5 \sqrt {2+\sqrt {2}} c \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )}-\frac {8 a \sqrt [4]{b} (c x)^{3/4} \sqrt {-\frac {a+b x^2}{\sqrt {a} \sqrt {b} x}} \sqrt {-\frac {\left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {2}+\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a}}+\frac {2 \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )}{\sqrt [4]{b} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{5 \sqrt {2+\sqrt {2}} c \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )} \] Output:

4/5*(c*x)^(1/4)*(b*x^2+a)^(1/2)/c+8/5*a*b^(1/4)*(c*x)^(3/4)*(-(b*x^2+a)/a^ 
(1/2)/b^(1/2)/x)^(1/2)*((a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^ 
(1/4)/c^(1/2)/(c*x)^(1/2))^(1/2)*EllipticF(1/2*(-a^(1/4)*c^(1/2)*(2^(1/2)+ 
2^(1/2)*b^(1/2)*x/a^(1/2)-2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/( 
c*x)^(1/2))^(1/2),(-2+2*2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)/c/(b*x^2+a)^(1/2 
)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))-8/5*a*b^(1/4)*(c*x)^(3/4)*(-(b*x^2 
+a)/a^(1/2)/b^(1/2)/x)^(1/2)*(-(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))^2/a^( 
1/4)/b^(1/4)/c^(1/2)/(c*x)^(1/2))^(1/2)*EllipticF(1/2*(a^(1/4)*c^(1/2)*(2^ 
(1/2)+2^(1/2)*b^(1/2)*x/a^(1/2)+2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^( 
1/4)/(c*x)^(1/2))^(1/2),(-2+2*2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)/c/(b*x^2+a 
)^(1/2)/(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))
 

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 = 0.11 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{3/4}} \, dx=\frac {4 x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{8},\frac {9}{8},-\frac {b x^2}{a}\right )}{(c x)^{3/4} \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

(4*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-1/2, 1/8, 9/8, -((b*x^2)/a)])/((c* 
x)^(3/4)*Sqrt[1 + (b*x^2)/a])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 537, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {248, 266, 767, 27, 2422}

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.

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

\(\Big \downarrow \) 248

\(\displaystyle \frac {4}{5} a \int \frac {1}{(c x)^{3/4} \sqrt {b x^2+a}}dx+\frac {4 \sqrt [4]{c x} \sqrt {a+b x^2}}{5 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {16 a \int \frac {1}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{5 c}+\frac {4 \sqrt [4]{c x} \sqrt {a+b x^2}}{5 c}\)

\(\Big \downarrow \) 767

\(\displaystyle \frac {16 a \left (\frac {1}{2} \int \frac {\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c} \sqrt {b x^2+a}}d\sqrt [4]{c x}+\frac {1}{2} \int \frac {\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c} \sqrt {b x^2+a}}d\sqrt [4]{c x}\right )}{5 c}+\frac {4 \sqrt [4]{c x} \sqrt {a+b x^2}}{5 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {16 a \left (\frac {\int \frac {\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{2 \sqrt [4]{a} \sqrt {c}}+\frac {\int \frac {\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{2 \sqrt [4]{a} \sqrt {c}}\right )}{5 c}+\frac {4 \sqrt [4]{c x} \sqrt {a+b x^2}}{5 c}\)

\(\Big \downarrow \) 2422

\(\displaystyle \frac {16 a \left (\frac {\sqrt [4]{b} (c x)^{3/4} \sqrt {-\frac {a c^2+b c^2 x^2}{\sqrt {a} \sqrt {b} c^2 x}} \sqrt {\frac {\left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} \sqrt {b} x c+\sqrt {2} \sqrt {a} c-2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {c x} \sqrt {c}}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )}-\frac {\sqrt [4]{b} (c x)^{3/4} \sqrt {-\frac {a c^2+b c^2 x^2}{\sqrt {a} \sqrt {b} c^2 x}} \sqrt {-\frac {\left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt {2} \sqrt {b} x c+\sqrt {2} \sqrt {a} c+2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {c x} \sqrt {c}}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )}\right )}{5 c}+\frac {4 \sqrt [4]{c x} \sqrt {a+b x^2}}{5 c}\)

Input:

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

Output:

(4*(c*x)^(1/4)*Sqrt[a + b*x^2])/(5*c) + (16*a*((b^(1/4)*(c*x)^(3/4)*Sqrt[- 
((a*c^2 + b*c^2*x^2)/(Sqrt[a]*Sqrt[b]*c^2*x))]*Sqrt[(a^(1/4)*Sqrt[c] + b^( 
1/4)*Sqrt[c*x])^2/(a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[c*x])]*EllipticF[ArcSin[Sq 
rt[-((Sqrt[2]*Sqrt[a]*c + Sqrt[2]*Sqrt[b]*c*x - 2*a^(1/4)*b^(1/4)*Sqrt[c]* 
Sqrt[c*x])/(a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[c*x]))]/2], -2*(1 - Sqrt[2])])/(2 
*Sqrt[2 + Sqrt[2]]*Sqrt[a + b*x^2]*(a^(1/4)*Sqrt[c] + b^(1/4)*Sqrt[c*x])) 
- (b^(1/4)*(c*x)^(3/4)*Sqrt[-((a*c^2 + b*c^2*x^2)/(Sqrt[a]*Sqrt[b]*c^2*x)) 
]*Sqrt[-((a^(1/4)*Sqrt[c] - b^(1/4)*Sqrt[c*x])^2/(a^(1/4)*b^(1/4)*Sqrt[c]* 
Sqrt[c*x]))]*EllipticF[ArcSin[Sqrt[(Sqrt[2]*Sqrt[a]*c + Sqrt[2]*Sqrt[b]*c* 
x + 2*a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[c*x])/(a^(1/4)*b^(1/4)*Sqrt[c]*Sqrt[c*x 
])]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*Sqrt[a + b*x^2]*(a^(1/4)*S 
qrt[c] - b^(1/4)*Sqrt[c*x]))))/(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 767

Int[1/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[1/2   Int[(1 - Rt[b/a, 4 
]*x^2)/Sqrt[a + b*x^8], x], x] + Simp[1/2   Int[(1 + Rt[b/a, 4]*x^2)/Sqrt[a 
 + b*x^8], x], x] /; FreeQ[{a, b}, x]
 

rule 2422

Int[((c_) + (d_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[(-c) 
*d*x^3*Sqrt[-(c - d*x^2)^2/(c*d*x^2)]*(Sqrt[(-d^2)*((a + b*x^8)/(b*c^2*x^4) 
)]/(Sqrt[2 + Sqrt[2]]*(c - d*x^2)*Sqrt[a + b*x^8]))*EllipticF[ArcSin[(1/2)* 
Sqrt[(Sqrt[2]*c^2 + 2*c*d*x^2 + Sqrt[2]*d^2*x^4)/(c*d*x^2)]], -2*(1 - Sqrt[ 
2])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^4 - a*d^4, 0]
 

Maple [F]

\[\int \frac {\sqrt {b \,x^{2}+a}}{\left (c x \right )^{\frac {3}{4}}}d x\]

Input:

int((b*x^2+a)^(1/2)/(c*x)^(3/4),x)
 

Output:

int((b*x^2+a)^(1/2)/(c*x)^(3/4),x)
 

Fricas [F]

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

integrate((b*x^2+a)^(1/2)/(c*x)^(3/4),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x^2 + a)*(c*x)^(1/4)/(c*x), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.72 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{3/4}} \, dx=\frac {\sqrt {a} \sqrt [4]{x} \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{8} \\ \frac {9}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {3}{4}} \Gamma \left (\frac {9}{8}\right )} \] Input:

integrate((b*x**2+a)**(1/2)/(c*x)**(3/4),x)
 

Output:

sqrt(a)*x**(1/4)*gamma(1/8)*hyper((-1/2, 1/8), (9/8,), b*x**2*exp_polar(I* 
pi)/a)/(2*c**(3/4)*gamma(9/8))
 

Maxima [F]

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

integrate((b*x^2+a)^(1/2)/(c*x)^(3/4),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^2 + a)/(c*x)^(3/4), x)
 

Giac [F]

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

integrate((b*x^2+a)^(1/2)/(c*x)^(3/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate(sqrt(b*x^2 + a)/(c*x)^(3/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{3/4}} \, dx=\int \frac {\sqrt {b\,x^2+a}}{{\left (c\,x\right )}^{3/4}} \,d x \] Input:

int((a + b*x^2)^(1/2)/(c*x)^(3/4),x)
 

Output:

int((a + b*x^2)^(1/2)/(c*x)^(3/4), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{3/4}} \, dx=\frac {\frac {4 x^{\frac {1}{4}} \sqrt {b \,x^{2}+a}}{5}+\frac {4 \left (\int \frac {\sqrt {b \,x^{2}+a}}{x^{\frac {3}{4}} a +x^{\frac {11}{4}} b}d x \right ) a}{5}}{c^{\frac {3}{4}}} \] Input:

int((b*x^2+a)^(1/2)/(c*x)^(3/4),x)
 

Output:

(4*(x**(1/4)*sqrt(a + b*x**2) + int(sqrt(a + b*x**2)/(x**(3/4)*a + x**(3/4 
)*b*x**2),x)*a))/(5*c**(3/4))