Integrand size = 30, antiderivative size = 49 \[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {3} \sqrt {c} (c+2 d x)}{\sqrt {c^3+4 d^3 x^3}}\right )}{\sqrt {3} \sqrt {c} d} \] Output:
2/3*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
Time = 1.51 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {c^3+4 d^3 x^3}}{\sqrt {3} \sqrt {c} (c+2 d x)}\right )}{\sqrt {3} \sqrt {c} d} \] Input:
Integrate[(c - 2*d*x)/((c + d*x)*Sqrt[c^3 + 4*d^3*x^3]),x]
Output:
(-2*ArcTan[Sqrt[c^3 + 4*d^3*x^3]/(Sqrt[3]*Sqrt[c]*(c + 2*d*x))])/(Sqrt[3]* Sqrt[c]*d)
Time = 0.44 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {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.
\(\displaystyle \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx\) |
\(\Big \downarrow \) 2562 |
\(\displaystyle \frac {2 c \int \frac {1}{\frac {3 c (c+2 d x)^2}{c^3+4 d^3 x^3}+1}d\frac {c+2 d x}{c \sqrt {c^3+4 d^3 x^3}}}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 \arctan \left (\frac {\sqrt {3} \sqrt {c} (c+2 d x)}{\sqrt {c^3+4 d^3 x^3}}\right )}{\sqrt {3} \sqrt {c} d}\) |
Input:
Int[(c - 2*d*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]])/(Sqrt[3]*S qrt[c]*d)
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])
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]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.44 (sec) , antiderivative size = 889, normalized size of antiderivative = 18.14
method | result | size |
default | \(\text {Expression too large to display}\) | \(889\) |
elliptic | \(\text {Expression too large to display}\) | \(889\) |
Input:
int((-2*d*x+c)/(d*x+c)/(4*d^3*x^3+c^3)^(1/2),x,method=_RETURNVERBOSE)
Output:
-4*((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))+6*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)*((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^(1/2)*...
Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (39) = 78\).
Time = 0.18 (sec) , antiderivative size = 300, normalized size of antiderivative = 6.12 \[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\left [\frac {\sqrt {3} \sqrt {-\frac {1}{c}} \log \left (\frac {2 \, d^{6} x^{6} - 36 \, c d^{5} x^{5} - 18 \, c^{2} d^{4} x^{4} + 28 \, c^{3} d^{3} x^{3} + 18 \, c^{4} d^{2} x^{2} - c^{6} - \sqrt {3} {\left (4 \, c d^{4} x^{4} - 10 \, c^{2} d^{3} x^{3} - 18 \, c^{3} d^{2} x^{2} - 8 \, c^{4} d x - c^{5}\right )} \sqrt {4 \, d^{3} x^{3} + c^{3}} \sqrt {-\frac {1}{c}}}{d^{6} x^{6} + 6 \, c d^{5} x^{5} + 15 \, c^{2} d^{4} x^{4} + 20 \, c^{3} d^{3} x^{3} + 15 \, c^{4} d^{2} x^{2} + 6 \, c^{5} d x + c^{6}}\right )}{6 \, d}, -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {4 \, d^{3} x^{3} + c^{3}} {\left (2 \, d^{3} x^{3} - 6 \, c d^{2} x^{2} - 6 \, c^{2} d x - c^{3}\right )}}{3 \, {\left (8 \, d^{4} x^{4} + 4 \, c d^{3} x^{3} + 2 \, c^{3} d x + c^{4}\right )} \sqrt {c}}\right )}{3 \, \sqrt {c} d}\right ] \] Input:
integrate((-2*d*x+c)/(d*x+c)/(4*d^3*x^3+c^3)^(1/2),x, algorithm="fricas")
Output:
[1/6*sqrt(3)*sqrt(-1/c)*log((2*d^6*x^6 - 36*c*d^5*x^5 - 18*c^2*d^4*x^4 + 2 8*c^3*d^3*x^3 + 18*c^4*d^2*x^2 - c^6 - sqrt(3)*(4*c*d^4*x^4 - 10*c^2*d^3*x ^3 - 18*c^3*d^2*x^2 - 8*c^4*d*x - c^5)*sqrt(4*d^3*x^3 + c^3)*sqrt(-1/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))/d, -1/3*sqrt(3)*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)/((8*d^4*x^4 + 4*c*d^3*x^3 + 2*c^3*d*x + c^4)*sqrt(c)))/(sqrt(c)*d)]
\[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=- \int \left (- \frac {c}{c \sqrt {c^{3} + 4 d^{3} x^{3}} + d x \sqrt {c^{3} + 4 d^{3} x^{3}}}\right )\, dx - \int \frac {2 d x}{c \sqrt {c^{3} + 4 d^{3} x^{3}} + d x \sqrt {c^{3} + 4 d^{3} x^{3}}}\, dx \] Input:
integrate((-2*d*x+c)/(d*x+c)/(4*d**3*x**3+c**3)**(1/2),x)
Output:
-Integral(-c/(c*sqrt(c**3 + 4*d**3*x**3) + d*x*sqrt(c**3 + 4*d**3*x**3)), x) - Integral(2*d*x/(c*sqrt(c**3 + 4*d**3*x**3) + d*x*sqrt(c**3 + 4*d**3*x **3)), x)
\[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\int { -\frac {2 \, d x - c}{\sqrt {4 \, d^{3} x^{3} + c^{3}} {\left (d x + c\right )}} \,d x } \] Input:
integrate((-2*d*x+c)/(d*x+c)/(4*d^3*x^3+c^3)^(1/2),x, algorithm="maxima")
Output:
-integrate((2*d*x - c)/(sqrt(4*d^3*x^3 + c^3)*(d*x + c)), x)
\[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\int { -\frac {2 \, d x - c}{\sqrt {4 \, d^{3} x^{3} + c^{3}} {\left (d x + c\right )}} \,d x } \] Input:
integrate((-2*d*x+c)/(d*x+c)/(4*d^3*x^3+c^3)^(1/2),x, algorithm="giac")
Output:
integrate(-(2*d*x - c)/(sqrt(4*d^3*x^3 + c^3)*(d*x + c)), x)
Time = 24.73 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.94 \[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\frac {\sqrt {3}\,\ln \left (\frac {{\left (-\sqrt {c^3+4\,d^3\,x^3}+\sqrt {3}\,c^{3/2}\,1{}\mathrm {i}+\sqrt {3}\,\sqrt {c}\,d\,x\,2{}\mathrm {i}\right )}^3\,\left (\sqrt {c^3+4\,d^3\,x^3}+\sqrt {3}\,c^{3/2}\,1{}\mathrm {i}+\sqrt {3}\,\sqrt {c}\,d\,x\,2{}\mathrm {i}\right )}{{\left (c+d\,x\right )}^6}\right )\,1{}\mathrm {i}}{3\,\sqrt {c}\,d} \] Input:
int((c - 2*d*x)/((c^3 + 4*d^3*x^3)^(1/2)*(c + d*x)),x)
Output:
(3^(1/2)*log(((3^(1/2)*c^(3/2)*1i - (c^3 + 4*d^3*x^3)^(1/2) + 3^(1/2)*c^(1 /2)*d*x*2i)^3*((c^3 + 4*d^3*x^3)^(1/2) + 3^(1/2)*c^(3/2)*1i + 3^(1/2)*c^(1 /2)*d*x*2i))/(c + d*x)^6)*1i)/(3*c^(1/2)*d)
\[ \int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx=\left (\int \frac {\sqrt {4 d^{3} x^{3}+c^{3}}}{4 d^{4} x^{4}+4 c \,d^{3} x^{3}+c^{3} d x +c^{4}}d x \right ) c -2 \left (\int \frac {\sqrt {4 d^{3} x^{3}+c^{3}}\, x}{4 d^{4} x^{4}+4 c \,d^{3} x^{3}+c^{3} d x +c^{4}}d x \right ) d \] Input:
int((-2*d*x+c)/(d*x+c)/(4*d^3*x^3+c^3)^(1/2),x)
Output:
int(sqrt(c**3 + 4*d**3*x**3)/(c**4 + c**3*d*x + 4*c*d**3*x**3 + 4*d**4*x** 4),x)*c - 2*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)*d