\(\int \frac {1}{\sqrt [4]{c x} (a+b x^2)^{3/2}} \, dx\) [690]

Optimal result

Integrand size = 19, antiderivative size = 512 \[ \int \frac {1}{\sqrt [4]{c x} \left (a+b x^2\right )^{3/2}} \, dx=\frac {(c x)^{3/4}}{a c \sqrt {a+b x^2}}-\frac {(c x)^{3/4} \sqrt {-\frac {a+b x^2}{\sqrt {a} \sqrt {b} x}} \sqrt {\frac {\left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {2}+\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a}}-\frac {2 \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )}{\sqrt [4]{b} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} a^{3/4} \sqrt {c} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}+\sqrt [4]{b} \sqrt {c x}\right )}-\frac {(c x)^{3/4} \sqrt {-\frac {a+b x^2}{\sqrt {a} \sqrt {b} x}} \sqrt {-\frac {\left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )^2}{\sqrt [4]{a} \sqrt [4]{b} \sqrt {c} \sqrt {c x}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {2}+\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a}}+\frac {2 \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )}{\sqrt [4]{b} \sqrt {c x}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} a^{3/4} \sqrt {c} \sqrt {a+b x^2} \left (\sqrt [4]{a} \sqrt {c}-\sqrt [4]{b} \sqrt {c x}\right )} \] Output:

(c*x)^(3/4)/a/c/(b*x^2+a)^(1/2)-1/2*(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)/a^(3/4)/c^(1/2)/(b*x^2+a)^(1 
/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))-1/2*(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)/a^(3/4)/c^(1/2)/(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 = 10.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.12 \[ \int \frac {1}{\sqrt [4]{c x} \left (a+b x^2\right )^{3/2}} \, dx=\frac {4 x \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {3}{2},\frac {11}{8},-\frac {b x^2}{a}\right )}{3 a \sqrt [4]{c x} \sqrt {a+b x^2}} \] Input:

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

Output:

(4*x*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[3/8, 3/2, 11/8, -((b*x^2)/a)])/ 
(3*a*(c*x)^(1/4)*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {253, 266, 838, 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.

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

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 838

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2422

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

Input:

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

Output:

(c*x)^(3/4)/(a*c*Sqrt[a + b*x^2]) + (-1/2*(a^(1/4)*Sqrt[c]*(c*x)^(3/4)*Sqr 
t[-((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])]) 
/(Sqrt[2 + Sqrt[2]]*Sqrt[a + b*x^2]*(a^(1/4)*Sqrt[c] + b^(1/4)*Sqrt[c*x])) 
 - (a^(1/4)*Sqrt[c]*(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]*S 
qrt[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])))/(a*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 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -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 838

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

\[\int \frac {1}{\left (c x \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(b*x^2 + a)*(c*x)^(3/4)/(b^2*c*x^5 + 2*a*b*c*x^3 + a^2*c*x), 
x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.09 \[ \int \frac {1}{\sqrt [4]{c x} \left (a+b x^2\right )^{3/2}} \, dx=\frac {x^{\frac {3}{4}} \Gamma \left (\frac {3}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{8}, \frac {3}{2} \\ \frac {11}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \sqrt [4]{c} \Gamma \left (\frac {11}{8}\right )} \] Input:

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

Output:

x**(3/4)*gamma(3/8)*hyper((3/8, 3/2), (11/8,), b*x**2*exp_polar(I*pi)/a)/( 
2*a**(3/2)*c**(1/4)*gamma(11/8))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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