\(\int \frac {x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx\) [119]

Optimal result

Integrand size = 25, antiderivative size = 246 \[ \int \frac {x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {3} \sqrt {c} (c+2 d x)}{\sqrt {c^3+4 d^3 x^3}}\right )}{3 \sqrt {3} \sqrt {c} d^2}+\frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}} \] Output:

-2/9*arctan(3^(1/2)*c^(1/2)*(2*d*x+c)/(4*d^3*x^3+c^3)^(1/2))*3^(1/2)/c^(1/ 
2)/d^2+1/9*2^(1/3)*(1/2*6^(1/2)+1/2*2^(1/2))*(c+2^(2/3)*d*x)*((c^2-2^(2/3) 
*c*d*x+2*2^(1/3)*d^2*x^2)/((1+3^(1/2))*c+2^(2/3)*d*x)^2)^(1/2)*EllipticF(( 
(1-3^(1/2))*c+2^(2/3)*d*x)/((1+3^(1/2))*c+2^(2/3)*d*x),I*3^(1/2)+2*I)*3^(3 
/4)/d^2/(c*(c+2^(2/3)*d*x)/((1+3^(1/2))*c+2^(2/3)*d*x)^2)^(1/2)/(4*d^3*x^3 
+c^3)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 11.31 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.51 \[ \int \frac {x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\frac {\sqrt [6]{2} \sqrt {\frac {\sqrt [3]{2} c+2 d x}{\left (1+\sqrt [3]{-1}\right ) c}} \left (-\sqrt {\frac {\sqrt [3]{-2} c-2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}} \left (\sqrt [3]{-1} \left (2+\sqrt [3]{-2}\right ) c-2 \left (\sqrt [3]{-1}+2^{2/3}\right ) d x\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}}}{\sqrt [6]{2}}\right ),\sqrt [3]{-1}\right )+\frac {\sqrt [3]{-1} 2^{2/3} \left (1+\sqrt [3]{-1}\right ) c \sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}} \sqrt {2^{2/3}-\frac {2 \sqrt [3]{2} d x}{c}+\frac {4 d^2 x^2}{c^2}} \operatorname {EllipticPi}\left (\frac {i \sqrt [3]{2} \sqrt {3}}{2+\sqrt [3]{-2}},\arcsin \left (\frac {\sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}}}{\sqrt [6]{2}}\right ),\sqrt [3]{-1}\right )}{\sqrt {3}}\right )}{\left (2+\sqrt [3]{-2}\right ) d^2 \sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}} \sqrt {c^3+4 d^3 x^3}} \] Input:

Integrate[x/((c + d*x)*Sqrt[c^3 + 4*d^3*x^3]),x]
 

Output:

(2^(1/6)*Sqrt[(2^(1/3)*c + 2*d*x)/((1 + (-1)^(1/3))*c)]*(-(Sqrt[((-2)^(1/3 
)*c - 2*(-1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]*((-1)^(1/3)*(2 + (-2)^(1/3)) 
*c - 2*((-1)^(1/3) + 2^(2/3))*d*x)*EllipticF[ArcSin[Sqrt[(2^(1/3)*c + 2*(- 
1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]/2^(1/6)], (-1)^(1/3)]) + ((-1)^(1/3)*2 
^(2/3)*(1 + (-1)^(1/3))*c*Sqrt[(2^(1/3)*c + 2*(-1)^(2/3)*d*x)/((1 + (-1)^( 
1/3))*c)]*Sqrt[2^(2/3) - (2*2^(1/3)*d*x)/c + (4*d^2*x^2)/c^2]*EllipticPi[( 
I*2^(1/3)*Sqrt[3])/(2 + (-2)^(1/3)), ArcSin[Sqrt[(2^(1/3)*c + 2*(-1)^(2/3) 
*d*x)/((1 + (-1)^(1/3))*c)]/2^(1/6)], (-1)^(1/3)])/Sqrt[3]))/((2 + (-2)^(1 
/3))*d^2*Sqrt[(2^(1/3)*c + 2*(-1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]*Sqrt[c^ 
3 + 4*d^3*x^3])
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2564, 759, 2562, 216}

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.

\(\displaystyle \int \frac {x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx\)

\(\Big \downarrow \) 2564

\(\displaystyle \frac {\int \frac {1}{\sqrt {c^3+4 d^3 x^3}}dx}{3 d}-\frac {\int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}}dx}{3 d}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}}-\frac {\int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}}dx}{3 d}\)

\(\Big \downarrow \) 2562

\(\displaystyle \frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}}-\frac {2 c \int \frac {1}{\frac {3 c (c+2 d x)^2}{c^3+4 d^3 x^3}+1}d\frac {c+2 d x}{c \sqrt {c^3+4 d^3 x^3}}}{3 d^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}}-\frac {2 \arctan \left (\frac {\sqrt {3} \sqrt {c} (c+2 d x)}{\sqrt {c^3+4 d^3 x^3}}\right )}{3 \sqrt {3} \sqrt {c} d^2}\)

Input:

Int[x/((c + d*x)*Sqrt[c^3 + 4*d^3*x^3]),x]
 

Output:

(-2*ArcTan[(Sqrt[3]*Sqrt[c]*(c + 2*d*x))/Sqrt[c^3 + 4*d^3*x^3]])/(3*Sqrt[3 
]*Sqrt[c]*d^2) + (2^(1/3)*Sqrt[2 + Sqrt[3]]*(c + 2^(2/3)*d*x)*Sqrt[(c^2 - 
2^(2/3)*c*d*x + 2*2^(1/3)*d^2*x^2)/((1 + Sqrt[3])*c + 2^(2/3)*d*x)^2]*Elli 
pticF[ArcSin[((1 - Sqrt[3])*c + 2^(2/3)*d*x)/((1 + Sqrt[3])*c + 2^(2/3)*d* 
x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*d^2*Sqrt[(c*(c + 2^(2/3)*d*x))/((1 + Sqrt 
[3])*c + 2^(2/3)*d*x)^2]*Sqrt[c^3 + 4*d^3*x^3])
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 759

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* 
((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s 
+ r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & 
& PosQ[a]
 

rule 2562

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ 
Symbol] :> Simp[2*(e/d)   Subst[Int[1/(1 + 3*a*x^2), x], x, (1 + 2*d*(x/c)) 
/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] 
&& EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]
 

rule 2564

Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x 
_Symbol] :> Simp[(2*d*e + c*f)/(3*c*d)   Int[1/Sqrt[a + b*x^3], x], x] + Si 
mp[(d*e - c*f)/(3*c*d)   Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x], x 
] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && (EqQ[b*c^3 - 4*a* 
d^3, 0] || EqQ[b*c^3 + 8*a*d^3, 0]) && NeQ[2*d*e + c*f, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 891 vs. \(2 (197 ) = 394\).

Time = 0.38 (sec) , antiderivative size = 892, normalized size of antiderivative = 3.63

method	result	size
default	\(\text {Expression too large to display}\)	\(892\)
elliptic	\(\text {Expression too large to display}\)	\(892\)

Input:

int(x/(d*x+c)/(4*d^3*x^3+c^3)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/d*((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^ 
(1/3))*c/d)*((x-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)-1/4 
*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2)*(( 
x+1/2*2^(1/3)*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d+1/2*2^(1/3)*c/ 
d))^(1/2)*((x-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)+1/4*I 
*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2)/(4*d 
^3*x^3+c^3)^(1/2)*EllipticF(((x-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d)/( 
(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3) 
)*c/d))^(1/2),(((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I 
*3^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d+1/2*2^(1/3 
)*c/d))^(1/2))-2*c/d^2*((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/ 
3)+1/4*I*3^(1/2)*2^(1/3))*c/d)*((x-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d 
)/((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1 
/3))*c/d))^(1/2)*((x+1/2*2^(1/3)*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3)) 
*c/d+1/2*2^(1/3)*c/d))^(1/2)*((x-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d)/ 
((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3 
))*c/d))^(1/2)/(4*d^3*x^3+c^3)^(1/2)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))* 
c/d+c/d)*EllipticPi(((x-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d)/((1/4*2^( 
1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d))^ 
(1/2),((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I*3^(1/...
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.42 \[ \int \frac {x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\left [-\frac {\sqrt {3} \sqrt {-c} d^{2} \log \left (\frac {2 \, d^{6} x^{6} - 36 \, c d^{5} x^{5} - 18 \, c^{2} d^{4} x^{4} + 28 \, c^{3} d^{3} x^{3} + 18 \, c^{4} d^{2} x^{2} - c^{6} - \sqrt {3} {\left (4 \, d^{4} x^{4} - 10 \, c d^{3} x^{3} - 18 \, c^{2} d^{2} x^{2} - 8 \, c^{3} d x - c^{4}\right )} \sqrt {4 \, d^{3} x^{3} + c^{3}} \sqrt {-c}}{d^{6} x^{6} + 6 \, c d^{5} x^{5} + 15 \, c^{2} d^{4} x^{4} + 20 \, c^{3} d^{3} x^{3} + 15 \, c^{4} d^{2} x^{2} + 6 \, c^{5} d x + c^{6}}\right ) - 6 \, c \sqrt {d^{3}} {\rm weierstrassPInverse}\left (0, -\frac {c^{3}}{d^{3}}, x\right )}{18 \, c d^{4}}, \frac {\sqrt {3} \sqrt {c} d^{2} \arctan \left (\frac {\sqrt {3} \sqrt {4 \, d^{3} x^{3} + c^{3}} {\left (2 \, d^{3} x^{3} - 6 \, c d^{2} x^{2} - 6 \, c^{2} d x - c^{3}\right )} \sqrt {c}}{3 \, {\left (8 \, c d^{4} x^{4} + 4 \, c^{2} d^{3} x^{3} + 2 \, c^{4} d x + c^{5}\right )}}\right ) + 3 \, c \sqrt {d^{3}} {\rm weierstrassPInverse}\left (0, -\frac {c^{3}}{d^{3}}, x\right )}{9 \, c d^{4}}\right ] \] Input:

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

Output:

[-1/18*(sqrt(3)*sqrt(-c)*d^2*log((2*d^6*x^6 - 36*c*d^5*x^5 - 18*c^2*d^4*x^ 
4 + 28*c^3*d^3*x^3 + 18*c^4*d^2*x^2 - c^6 - sqrt(3)*(4*d^4*x^4 - 10*c*d^3* 
x^3 - 18*c^2*d^2*x^2 - 8*c^3*d*x - c^4)*sqrt(4*d^3*x^3 + c^3)*sqrt(-c))/(d 
^6*x^6 + 6*c*d^5*x^5 + 15*c^2*d^4*x^4 + 20*c^3*d^3*x^3 + 15*c^4*d^2*x^2 + 
6*c^5*d*x + c^6)) - 6*c*sqrt(d^3)*weierstrassPInverse(0, -c^3/d^3, x))/(c* 
d^4), 1/9*(sqrt(3)*sqrt(c)*d^2*arctan(1/3*sqrt(3)*sqrt(4*d^3*x^3 + c^3)*(2 
*d^3*x^3 - 6*c*d^2*x^2 - 6*c^2*d*x - c^3)*sqrt(c)/(8*c*d^4*x^4 + 4*c^2*d^3 
*x^3 + 2*c^4*d*x + c^5)) + 3*c*sqrt(d^3)*weierstrassPInverse(0, -c^3/d^3, 
x))/(c*d^4)]
 

Sympy [F]

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

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

Output:

Integral(x/((c + d*x)*sqrt(c**3 + 4*d**3*x**3)), x)
 

Maxima [F]

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

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

Output:

integrate(x/(sqrt(4*d^3*x^3 + c^3)*(d*x + c)), x)
 

Giac [F]

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

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

Output:

integrate(x/(sqrt(4*d^3*x^3 + c^3)*(d*x + c)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\int \frac {\sqrt {4 d^{3} x^{3}+c^{3}}\, x}{4 d^{4} x^{4}+4 c \,d^{3} x^{3}+c^{3} d x +c^{4}}d x \] Input:

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

Output:

int((sqrt(c**3 + 4*d**3*x**3)*x)/(c**4 + c**3*d*x + 4*c*d**3*x**3 + 4*d**4 
*x**4),x)