\(\int \frac {(a+c x^4)^{3/2}}{x^2} \, dx\) [349]

Optimal result

Integrand size = 15, antiderivative size = 252 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^2} \, dx=-\frac {a \sqrt {a+c x^4}}{x}+\frac {1}{5} c x^3 \sqrt {a+c x^4}+\frac {12 a \sqrt {c} x \sqrt {a+c x^4}}{5 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {12 a^{5/4} \sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+c x^4}}+\frac {6 a^{5/4} \sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+c x^4}} \] Output:

-a*(c*x^4+a)^(1/2)/x+1/5*c*x^3*(c*x^4+a)^(1/2)+12*a*c^(1/2)*x*(c*x^4+a)^(1 
/2)/(5*a^(1/2)+5*c^(1/2)*x^2)-12/5*a^(5/4)*c^(1/4)*(a^(1/2)+c^(1/2)*x^2)*( 
(c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x/ 
a^(1/4))),1/2*2^(1/2))/(c*x^4+a)^(1/2)+6/5*a^(5/4)*c^(1/4)*(a^(1/2)+c^(1/2 
)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan( 
c^(1/4)*x/a^(1/4)),1/2*2^(1/2))/(c*x^4+a)^(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 = 50, normalized size of antiderivative = 0.20 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^2} \, dx=-\frac {a \sqrt {a+c x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},-\frac {c x^4}{a}\right )}{x \sqrt {1+\frac {c x^4}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {809, 811, 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.

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

\(\Big \downarrow \) 809

\(\displaystyle 6 c \int x^2 \sqrt {c x^4+a}dx-\frac {\left (a+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 811

\(\displaystyle 6 c \left (\frac {2}{5} a \int \frac {x^2}{\sqrt {c x^4+a}}dx+\frac {1}{5} x^3 \sqrt {a+c x^4}\right )-\frac {\left (a+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 834

\(\displaystyle 6 c \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+a}}dx}{\sqrt {c}}\right )+\frac {1}{5} x^3 \sqrt {a+c x^4}\right )-\frac {\left (a+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 27

\(\displaystyle 6 c \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}\right )+\frac {1}{5} x^3 \sqrt {a+c x^4}\right )-\frac {\left (a+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

-((a + c*x^4)^(3/2)/x) + 6*c*((x^3*Sqrt[a + c*x^4])/5 + (2*a*(-((-((x*Sqrt 
[a + c*x^4])/(Sqrt[a] + Sqrt[c]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*S 
qrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/ 
a^(1/4)], 1/2])/(c^(1/4)*Sqrt[a + c*x^4]))/Sqrt[c]) + (a^(1/4)*(Sqrt[a] + 
Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTa 
n[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[a + c*x^4])))/5)
 

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 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 809

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ 
[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntB 
inomialQ[a, b, c, n, m, p, x]
 

rule 811

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

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 [C] (verified)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.48

method	result	size
risch	\(-\frac {\sqrt {c \,x^{4}+a}\, \left (-c \,x^{4}+5 a \right )}{5 x}+\frac {12 i a^{\frac {3}{2}} \sqrt {c}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(122\)
default	\(-\frac {a \sqrt {c \,x^{4}+a}}{x}+\frac {c \,x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {12 i a^{\frac {3}{2}} \sqrt {c}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(128\)
elliptic	\(-\frac {a \sqrt {c \,x^{4}+a}}{x}+\frac {c \,x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {12 i a^{\frac {3}{2}} \sqrt {c}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(128\)

Input:

int((c*x^4+a)^(3/2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

-1/5*(c*x^4+a)^(1/2)*(-c*x^4+5*a)/x+12/5*I*a^(3/2)*c^(1/2)/(I/a^(1/2)*c^(1 
/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2) 
/(c*x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/ 
a^(1/2)*c^(1/2))^(1/2),I))
 

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.16 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^2} \, dx=\frac {a^{\frac {3}{2}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.16 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^2} \, dx=\frac {{\left (c\,x^4+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {5}{4};\ -\frac {1}{4};\ -\frac {a}{c\,x^4}\right )}{5\,x\,{\left (\frac {a}{c\,x^4}+1\right )}^{3/2}} \] Input:

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

Output:

((a + c*x^4)^(3/2)*hypergeom([-3/2, -5/4], -1/4, -a/(c*x^4)))/(5*x*(a/(c*x 
^4) + 1)^(3/2))
 

Reduce [F]

\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^2} \, dx=\frac {7 \sqrt {c \,x^{4}+a}\, a +\sqrt {c \,x^{4}+a}\, c \,x^{4}+12 \left (\int \frac {\sqrt {c \,x^{4}+a}}{c \,x^{6}+a \,x^{2}}d x \right ) a^{2} x}{5 x} \] Input:

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

Output:

(7*sqrt(a + c*x**4)*a + sqrt(a + c*x**4)*c*x**4 + 12*int(sqrt(a + c*x**4)/ 
(a*x**2 + c*x**6),x)*a**2*x)/(5*x)