Integrand size = 26, antiderivative size = 621 \[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx=\frac {2 \sqrt {c} f \arctan \left (\frac {2 \sqrt {c} \sqrt {-a+b x-c x^2}}{b-\sqrt {b^2-4 a c}-2 c x}\right )}{e^2+c f^2}+\frac {2 (2 c d+b e) f \arctan \left (\frac {\sqrt {b^2-4 a c} f-\frac {4 c d \sqrt {-a+b x-c x^2}}{b-\sqrt {b^2-4 a c}-2 c x}-\frac {2 \left (b-\sqrt {b^2-4 a c}\right ) e \sqrt {-a+b x-c x^2}}{b-\sqrt {b^2-4 a c}-2 c x}}{\sqrt {4 b d e+4 a e^2-b^2 f^2+4 c \left (d^2+a f^2\right )}}\right )}{\left (e^2+c f^2\right ) \sqrt {4 b d e+4 a e^2-b^2 f^2+4 c \left (d^2+a f^2\right )}}-\frac {e \log \left (\frac {c \left (b^2-4 a c-b \sqrt {b^2-4 a c}+2 c \sqrt {b^2-4 a c} x\right )}{\left (b-\sqrt {b^2-4 a c}-2 c x\right )^2}\right )}{e^2+c f^2}+\frac {e \log \left (\frac {c^2 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d+2 c^3 \sqrt {b^2-4 a c} d x+b^2 c^2 e x-4 a c^3 e x-b c^2 \sqrt {b^2-4 a c} e x+2 c^3 \sqrt {b^2-4 a c} e x^2+b^2 c^2 f \sqrt {-a+b x-c x^2}-4 a c^3 f \sqrt {-a+b x-c x^2}-b c^2 \sqrt {b^2-4 a c} f \sqrt {-a+b x-c x^2}+2 c^3 \sqrt {b^2-4 a c} f x \sqrt {-a+b x-c x^2}}{\left (b-\sqrt {b^2-4 a c}-2 c x\right )^2}\right )}{e^2+c f^2} \] Output:
2*c^(1/2)*f*arctan(2*c^(1/2)*(-c*x^2+b*x-a)^(1/2)/(b-(-4*a*c+b^2)^(1/2)-2* c*x))/(c*f^2+e^2)+2*(b*e+2*c*d)*f*arctan((f*(-4*a*c+b^2)^(1/2)-4*c*d*(-c*x ^2+b*x-a)^(1/2)/(b-(-4*a*c+b^2)^(1/2)-2*c*x)-2*(b-(-4*a*c+b^2)^(1/2))*e*(- c*x^2+b*x-a)^(1/2)/(b-(-4*a*c+b^2)^(1/2)-2*c*x))/(4*b*d*e+4*a*e^2-b^2*f^2+ 4*c*(a*f^2+d^2))^(1/2))/(c*f^2+e^2)/(4*b*d*e+4*a*e^2-b^2*f^2+4*c*(a*f^2+d^ 2))^(1/2)-e*ln(c*(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2)+2*(-4*a*c+b^2)^(1/2)*c*x) /(b-(-4*a*c+b^2)^(1/2)-2*c*x)^2)/(c*f^2+e^2)+e*ln((c^2*(b^2-4*a*c-b*(-4*a* c+b^2)^(1/2))*d+2*c^3*(-4*a*c+b^2)^(1/2)*d*x+b^2*c^2*e*x-4*a*c^3*e*x-b*c^2 *(-4*a*c+b^2)^(1/2)*e*x+2*c^3*(-4*a*c+b^2)^(1/2)*e*x^2+b^2*c^2*f*(-c*x^2+b *x-a)^(1/2)-4*a*c^3*f*(-c*x^2+b*x-a)^(1/2)-b*c^2*(-4*a*c+b^2)^(1/2)*f*(-c* x^2+b*x-a)^(1/2)+2*c^3*(-4*a*c+b^2)^(1/2)*f*x*(-c*x^2+b*x-a)^(1/2))/(b-(-4 *a*c+b^2)^(1/2)-2*c*x)^2)/(c*f^2+e^2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 2.47 (sec) , antiderivative size = 1737, normalized size of antiderivative = 2.80 \[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx =\text {Too large to display} \] Input:
Integrate[(d + e*x + f*Sqrt[-a + b*x - c*x^2])^(-1),x]
Output:
(-2*Sqrt[c]*f*ArcTan[(Sqrt[c]*x)/(Sqrt[-a] - Sqrt[-a + x*(b - c*x)])] + 2* e*Log[x] - e*Log[2*a - b*x + 2*Sqrt[-a]*Sqrt[-a + x*(b - c*x)]] - Sqrt[-a] *(c*d + b*e)*(e^2 + c*f^2)*RootSum[c^2*d^2 + 2*b*c*d*e + b^2*e^2 + a*c^2*f ^2 + 2*b*c*d*f*#1 + 2*b^2*e*f*#1 - 4*a*c*e*f*#1 + 2*c*d^2*#1^2 + 2*b*d*e*# 1^2 + 4*a*e^2*#1^2 + b^2*f^2*#1^2 - 2*a*c*f^2*#1^2 + 2*b*d*f*#1^3 + 4*a*e* f*#1^3 + d^2*#1^4 + a*f^2*#1^4 & , (-Log[x] + Log[Sqrt[-a] - Sqrt[-a + b*x - c*x^2] + x*#1])/(b*c*d*f + b^2*e*f - 2*a*c*e*f + 2*c*d^2*#1 + 2*b*d*e*# 1 + 4*a*e^2*#1 + b^2*f^2*#1 - 2*a*c*f^2*#1 + 3*b*d*f*#1^2 + 6*a*e*f*#1^2 + 2*d^2*#1^3 + 2*a*f^2*#1^3) & ] + Sqrt[-a]*d*(e^2 + c*f^2)*RootSum[c^2*d^2 + 2*b*c*d*e + b^2*e^2 + a*c^2*f^2 + 2*b*c*d*f*#1 + 2*b^2*e*f*#1 - 4*a*c*e *f*#1 + 2*c*d^2*#1^2 + 2*b*d*e*#1^2 + 4*a*e^2*#1^2 + b^2*f^2*#1^2 - 2*a*c* f^2*#1^2 + 2*b*d*f*#1^3 + 4*a*e*f*#1^3 + d^2*#1^4 + a*f^2*#1^4 & , (-(Log[ x]*#1^2) + Log[Sqrt[-a] - Sqrt[-a + b*x - c*x^2] + x*#1]*#1^2)/(b*c*d*f + b^2*e*f - 2*a*c*e*f + 2*c*d^2*#1 + 2*b*d*e*#1 + 4*a*e^2*#1 + b^2*f^2*#1 - 2*a*c*f^2*#1 + 3*b*d*f*#1^2 + 6*a*e*f*#1^2 + 2*d^2*#1^3 + 2*a*f^2*#1^3) & ] + RootSum[c^2*d^2 + 2*b*c*d*e + b^2*e^2 + a*c^2*f^2 + 2*b*c*d*f*#1 + 2*b ^2*e*f*#1 - 4*a*c*e*f*#1 + 2*c*d^2*#1^2 + 2*b*d*e*#1^2 + 4*a*e^2*#1^2 + b^ 2*f^2*#1^2 - 2*a*c*f^2*#1^2 + 2*b*d*f*#1^3 + 4*a*e*f*#1^3 + d^2*#1^4 + a*f ^2*#1^4 & , (-(c^2*d^2*f*Log[x]) - 2*b*c*d*e*f*Log[x] - b^2*e^2*f*Log[x] + a*c*e^2*f*Log[x] + c^2*d^2*f*Log[Sqrt[-a] - Sqrt[-a + b*x - c*x^2] + x...
Time = 1.88 (sec) , antiderivative size = 397, normalized size of antiderivative = 0.64, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {1}{f \sqrt {-a+b x-c x^2}+d+e x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {f \sqrt {-a+b x-c x^2}}{-a f^2-x \left (2 d e-b f^2\right )-x^2 \left (c f^2+e^2\right )-d^2}+\frac {d+e x}{a f^2+x \left (2 d e-b f^2\right )+x^2 \left (c f^2+e^2\right )+d^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f (b e+2 c d) \arctan \left (\frac {2 a e-x (b e+2 c d)+b d}{\sqrt {-a+b x-c x^2} \sqrt {4 c \left (a f^2+d^2\right )+4 a e^2-b^2 f^2+4 b d e}}\right )}{\left (c f^2+e^2\right ) \sqrt {4 c \left (a f^2+d^2\right )+4 a e^2-b^2 f^2+4 b d e}}+\frac {f (b e+2 c d) \arctan \left (\frac {-b f^2+2 x \left (c f^2+e^2\right )+2 d e}{f \sqrt {4 c \left (a f^2+d^2\right )+4 a e^2-b^2 f^2+4 b d e}}\right )}{\left (c f^2+e^2\right ) \sqrt {4 c \left (a f^2+d^2\right )+4 a e^2-b^2 f^2+4 b d e}}-\frac {\sqrt {c} f \arctan \left (\frac {b-2 c x}{2 \sqrt {c} \sqrt {-a+b x-c x^2}}\right )}{c f^2+e^2}+\frac {e \text {arctanh}\left (\frac {d+e x}{f \sqrt {-a+b x-c x^2}}\right )}{c f^2+e^2}+\frac {e \log \left (a f^2+x \left (2 d e-b f^2\right )+x^2 \left (c f^2+e^2\right )+d^2\right )}{2 \left (c f^2+e^2\right )}\) |
Input:
Int[(d + e*x + f*Sqrt[-a + b*x - c*x^2])^(-1),x]
Output:
((2*c*d + b*e)*f*ArcTan[(2*d*e - b*f^2 + 2*(e^2 + c*f^2)*x)/(f*Sqrt[4*b*d* e + 4*a*e^2 - b^2*f^2 + 4*c*(d^2 + a*f^2)])])/((e^2 + c*f^2)*Sqrt[4*b*d*e + 4*a*e^2 - b^2*f^2 + 4*c*(d^2 + a*f^2)]) - (Sqrt[c]*f*ArcTan[(b - 2*c*x)/ (2*Sqrt[c]*Sqrt[-a + b*x - c*x^2])])/(e^2 + c*f^2) + ((2*c*d + b*e)*f*ArcT an[(b*d + 2*a*e - (2*c*d + b*e)*x)/(Sqrt[4*b*d*e + 4*a*e^2 - b^2*f^2 + 4*c *(d^2 + a*f^2)]*Sqrt[-a + b*x - c*x^2])])/((e^2 + c*f^2)*Sqrt[4*b*d*e + 4* a*e^2 - b^2*f^2 + 4*c*(d^2 + a*f^2)]) + (e*ArcTanh[(d + e*x)/(f*Sqrt[-a + b*x - c*x^2])])/(e^2 + c*f^2) + (e*Log[d^2 + a*f^2 + (2*d*e - b*f^2)*x + ( e^2 + c*f^2)*x^2])/(2*(e^2 + c*f^2))
Leaf count of result is larger than twice the leaf count of optimal. \(4814\) vs. \(2(569)=1138\).
Time = 0.15 (sec) , antiderivative size = 4815, normalized size of antiderivative = 7.75
Input:
int(1/(d+e*x+f*(-c*x^2+b*x-a)^(1/2)),x,method=_RETURNVERBOSE)
Output:
f*(2*(c*f^2+e^2)/(-f^2*(4*a*c*f^2-b^2*f^2+4*a*e^2+4*b*d*e+4*c*d^2))^(1/2)/ (2*c*f^2+2*e^2)*(1/2*(-4*(x+(-b*f^2+2*d*e+(-f^2*(4*a*c*f^2-b^2*f^2+4*a*e^2 +4*b*d*e+4*c*d^2))^(1/2))/(2*c*f^2+2*e^2))^2*c+4*(b*e^2+2*c*d*e+(f^2*(-4*a *c*f^2+b^2*f^2-4*a*e^2-4*b*d*e-4*c*d^2))^(1/2)*c)/(c*f^2+e^2)*(x+(-b*f^2+2 *d*e+(-f^2*(4*a*c*f^2-b^2*f^2+4*a*e^2+4*b*d*e+4*c*d^2))^(1/2))/(2*c*f^2+2* e^2))-2*(2*a*c*e^2*f^2-b^2*e^2*f^2-2*b*c*d*e*f^2-2*c^2*d^2*f^2+2*e^4*a+2*d *e^3*b+2*d^2*e^2*c+(f^2*(-4*a*c*f^2+b^2*f^2-4*a*e^2-4*b*d*e-4*c*d^2))^(1/2 )*b*e^2+2*(f^2*(-4*a*c*f^2+b^2*f^2-4*a*e^2-4*b*d*e-4*c*d^2))^(1/2)*c*d*e)/ (c*f^2+e^2)^2)^(1/2)+1/2*(b*e^2+2*c*d*e+(f^2*(-4*a*c*f^2+b^2*f^2-4*a*e^2-4 *b*d*e-4*c*d^2))^(1/2)*c)/(c*f^2+e^2)/c^(1/2)*arctan(2*c^(1/2)*(x+(-b*f^2+ 2*d*e+(-f^2*(4*a*c*f^2-b^2*f^2+4*a*e^2+4*b*d*e+4*c*d^2))^(1/2))/(2*c*f^2+2 *e^2)-1/2*(b*e^2+2*c*d*e+(f^2*(-4*a*c*f^2+b^2*f^2-4*a*e^2-4*b*d*e-4*c*d^2) )^(1/2)*c)/(c*f^2+e^2)/c)/(-4*(x+(-b*f^2+2*d*e+(-f^2*(4*a*c*f^2-b^2*f^2+4* a*e^2+4*b*d*e+4*c*d^2))^(1/2))/(2*c*f^2+2*e^2))^2*c+4*(b*e^2+2*c*d*e+(f^2* (-4*a*c*f^2+b^2*f^2-4*a*e^2-4*b*d*e-4*c*d^2))^(1/2)*c)/(c*f^2+e^2)*(x+(-b* f^2+2*d*e+(-f^2*(4*a*c*f^2-b^2*f^2+4*a*e^2+4*b*d*e+4*c*d^2))^(1/2))/(2*c*f ^2+2*e^2))-2*(2*a*c*e^2*f^2-b^2*e^2*f^2-2*b*c*d*e*f^2-2*c^2*d^2*f^2+2*e^4* a+2*d*e^3*b+2*d^2*e^2*c+(f^2*(-4*a*c*f^2+b^2*f^2-4*a*e^2-4*b*d*e-4*c*d^2)) ^(1/2)*b*e^2+2*(f^2*(-4*a*c*f^2+b^2*f^2-4*a*e^2-4*b*d*e-4*c*d^2))^(1/2)*c* d*e)/(c*f^2+e^2)^2)^(1/2))+(2*a*c*e^2*f^2-b^2*e^2*f^2-2*b*c*d*e*f^2-2*c...
Timed out. \[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx=\text {Timed out} \] Input:
integrate(1/(d+e*x+f*(-c*x^2+b*x-a)^(1/2)),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx=\int \frac {1}{d + e x + f \sqrt {- a + b x - c x^{2}}}\, dx \] Input:
integrate(1/(d+e*x+f*(-c*x**2+b*x-a)**(1/2)),x)
Output:
Integral(1/(d + e*x + f*sqrt(-a + b*x - c*x**2)), x)
\[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx=\int { \frac {1}{e x + \sqrt {-c x^{2} + b x - a} f + d} \,d x } \] Input:
integrate(1/(d+e*x+f*(-c*x^2+b*x-a)^(1/2)),x, algorithm="maxima")
Output:
integrate(1/(e*x + sqrt(-c*x^2 + b*x - a)*f + d), x)
Exception generated. \[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(d+e*x+f*(-c*x^2+b*x-a)^(1/2)),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx=\int \frac {1}{d+f\,\sqrt {-c\,x^2+b\,x-a}+e\,x} \,d x \] Input:
int(1/(d + f*(b*x - a - c*x^2)^(1/2) + e*x),x)
Output:
int(1/(d + f*(b*x - a - c*x^2)^(1/2) + e*x), x)
\[ \int \frac {1}{d+e x+f \sqrt {-a+b x-c x^2}} \, dx=\int \frac {1}{d +e x +f \sqrt {-c \,x^{2}+b x -a}}d x \] Input:
int(1/(d+e*x+f*(-c*x^2+b*x-a)^(1/2)),x)
Output:
int(1/(d+e*x+f*(-c*x^2+b*x-a)^(1/2)),x)