Integrand size = 35, antiderivative size = 60 \[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=-\frac {x \sqrt {c+b x^2+a x^3}}{c+a x^3}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {c+b x^2+a x^3}}\right )}{\sqrt {b}} \] Output:
-x*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)-arctanh(b^(1/2)*x/(a*x^3+b*x^2+c)^(1/2) )/b^(1/2)
Time = 0.47 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=-\frac {x \sqrt {c+x^2 (b+a x)}}{c+a x^3}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {c+x^2 (b+a x)}}\right )}{\sqrt {b}} \] Input:
Integrate[((-2*c + a*x^3)*Sqrt[c + b*x^2 + a*x^3])/(c + a*x^3)^2,x]
Output:
-((x*Sqrt[c + x^2*(b + a*x)])/(c + a*x^3)) - ArcTanh[(Sqrt[b]*x)/Sqrt[c + x^2*(b + a*x)]]/Sqrt[b]
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.
\(\displaystyle \int \frac {\left (a x^3-2 c\right ) \sqrt {a x^3+b x^2+c}}{\left (a x^3+c\right )^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\sqrt {a x^3+b x^2+c}}{a x^3+c}-\frac {3 c \sqrt {a x^3+b x^2+c}}{\left (a x^3+c\right )^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {\sqrt {a x^3+b x^2+c}}{a x^3+c}-\frac {3 c \sqrt {a x^3+b x^2+c}}{\left (a x^3+c\right )^2}\right )dx\) |
Input:
Int[((-2*c + a*x^3)*Sqrt[c + b*x^2 + a*x^3])/(c + a*x^3)^2,x]
Output:
$Aborted
Time = 3.48 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92
method | result | size |
default | \(-\frac {x \sqrt {a \,x^{3}+b \,x^{2}+c}}{a \,x^{3}+c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a \,x^{3}+b \,x^{2}+c}}{x \sqrt {b}}\right )}{\sqrt {b}}\) | \(55\) |
pseudoelliptic | \(-\frac {x \sqrt {a \,x^{3}+b \,x^{2}+c}}{a \,x^{3}+c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a \,x^{3}+b \,x^{2}+c}}{x \sqrt {b}}\right )}{\sqrt {b}}\) | \(55\) |
elliptic | \(\text {Expression too large to display}\) | \(5094\) |
Input:
int((a*x^3-2*c)*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)^2,x,method=_RETURNVERBOSE)
Output:
-x*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)-1/b^(1/2)*arctanh((a*x^3+b*x^2+c)^(1/2) /x/b^(1/2))
Time = 0.11 (sec) , antiderivative size = 244, normalized size of antiderivative = 4.07 \[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=\left [-\frac {4 \, \sqrt {a x^{3} + b x^{2} + c} b x - {\left (a x^{3} + c\right )} \sqrt {b} \log \left (\frac {a^{2} x^{6} + 8 \, a b x^{5} + 8 \, b^{2} x^{4} + 2 \, a c x^{3} + 8 \, b c x^{2} - 4 \, {\left (a x^{4} + 2 \, b x^{3} + c x\right )} \sqrt {a x^{3} + b x^{2} + c} \sqrt {b} + c^{2}}{a^{2} x^{6} + 2 \, a c x^{3} + c^{2}}\right )}{4 \, {\left (a b x^{3} + b c\right )}}, -\frac {2 \, \sqrt {a x^{3} + b x^{2} + c} b x - {\left (a x^{3} + c\right )} \sqrt {-b} \arctan \left (\frac {{\left (a x^{3} + 2 \, b x^{2} + c\right )} \sqrt {a x^{3} + b x^{2} + c} \sqrt {-b}}{2 \, {\left (a b x^{4} + b^{2} x^{3} + b c x\right )}}\right )}{2 \, {\left (a b x^{3} + b c\right )}}\right ] \] Input:
integrate((a*x^3-2*c)*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)^2,x, algorithm="fric as")
Output:
[-1/4*(4*sqrt(a*x^3 + b*x^2 + c)*b*x - (a*x^3 + c)*sqrt(b)*log((a^2*x^6 + 8*a*b*x^5 + 8*b^2*x^4 + 2*a*c*x^3 + 8*b*c*x^2 - 4*(a*x^4 + 2*b*x^3 + c*x)* sqrt(a*x^3 + b*x^2 + c)*sqrt(b) + c^2)/(a^2*x^6 + 2*a*c*x^3 + c^2)))/(a*b* x^3 + b*c), -1/2*(2*sqrt(a*x^3 + b*x^2 + c)*b*x - (a*x^3 + c)*sqrt(-b)*arc tan(1/2*(a*x^3 + 2*b*x^2 + c)*sqrt(a*x^3 + b*x^2 + c)*sqrt(-b)/(a*b*x^4 + b^2*x^3 + b*c*x)))/(a*b*x^3 + b*c)]
Timed out. \[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((a*x**3-2*c)*(a*x**3+b*x**2+c)**(1/2)/(a*x**3+c)**2,x)
Output:
Timed out
\[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=\int { \frac {\sqrt {a x^{3} + b x^{2} + c} {\left (a x^{3} - 2 \, c\right )}}{{\left (a x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate((a*x^3-2*c)*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)^2,x, algorithm="maxi ma")
Output:
integrate(sqrt(a*x^3 + b*x^2 + c)*(a*x^3 - 2*c)/(a*x^3 + c)^2, x)
\[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=\int { \frac {\sqrt {a x^{3} + b x^{2} + c} {\left (a x^{3} - 2 \, c\right )}}{{\left (a x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate((a*x^3-2*c)*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)^2,x, algorithm="giac ")
Output:
integrate(sqrt(a*x^3 + b*x^2 + c)*(a*x^3 - 2*c)/(a*x^3 + c)^2, x)
Time = 9.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32 \[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=\frac {\ln \left (\frac {c+a\,x^3+2\,b\,x^2-2\,\sqrt {b}\,x\,\sqrt {a\,x^3+b\,x^2+c}}{c^2+a\,c\,x^3}\right )}{2\,\sqrt {b}}-\frac {x\,\sqrt {a\,x^3+b\,x^2+c}}{a\,x^3+c} \] Input:
int(-((2*c - a*x^3)*(c + a*x^3 + b*x^2)^(1/2))/(c + a*x^3)^2,x)
Output:
log((c + a*x^3 + 2*b*x^2 - 2*b^(1/2)*x*(c + a*x^3 + b*x^2)^(1/2))/(c^2 + a *c*x^3))/(2*b^(1/2)) - (x*(c + a*x^3 + b*x^2)^(1/2))/(c + a*x^3)
\[ \int \frac {\left (-2 c+a x^3\right ) \sqrt {c+b x^2+a x^3}}{\left (c+a x^3\right )^2} \, dx=\int \frac {\left (a \,x^{3}-2 c \right ) \sqrt {a \,x^{3}+b \,x^{2}+c}}{\left (a \,x^{3}+c \right )^{2}}d x \] Input:
int((a*x^3-2*c)*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)^2,x)
Output:
int((a*x^3-2*c)*(a*x^3+b*x^2+c)^(1/2)/(a*x^3+c)^2,x)