Integrand size = 29, antiderivative size = 102 \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\frac {b \sqrt {d+e x+f x^2}}{4 f}+\frac {b x \sqrt {d+e x+f x^2}}{2 e}+\frac {\left (8 a f-b \left (e+\frac {4 d f}{e}\right )\right ) \text {arctanh}\left (\frac {e+2 f x}{2 \sqrt {f} \sqrt {d+e x+f x^2}}\right )}{8 f^{3/2}} \] Output:
1/4*b*(f*x^2+e*x+d)^(1/2)/f+1/2*b*x*(f*x^2+e*x+d)^(1/2)/e+1/8*(8*a*f-b*(e+ 4*d*f/e))*arctanh(1/2*(2*f*x+e)/f^(1/2)/(f*x^2+e*x+d)^(1/2))/f^(3/2)
Time = 0.96 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.85 \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\frac {b \sqrt {f} (e+2 f x) \sqrt {d+x (e+f x)}+\left (-8 a e f+b \left (e^2+4 d f\right )\right ) \text {arctanh}\left (\frac {\sqrt {f} x}{\sqrt {d}-\sqrt {d+x (e+f x)}}\right )}{4 e f^{3/2}} \] Input:
Integrate[(a + b*x + (b*f*x^2)/e)/Sqrt[d + e*x + f*x^2],x]
Output:
(b*Sqrt[f]*(e + 2*f*x)*Sqrt[d + x*(e + f*x)] + (-8*a*e*f + b*(e^2 + 4*d*f) )*ArcTanh[(Sqrt[f]*x)/(Sqrt[d] - Sqrt[d + x*(e + f*x)])])/(4*e*f^(3/2))
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2192, 27, 1160, 1092, 219}
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.
\(\displaystyle \int \frac {a+\frac {b f x^2}{e}+b x}{\sqrt {d+e x+f x^2}} \, dx\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {\int \frac {f \left (2 \left (2 a-\frac {b d}{e}\right )+b x\right )}{2 \sqrt {f x^2+e x+d}}dx}{2 f}+\frac {b x \sqrt {d+e x+f x^2}}{2 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {2 \left (2 a-\frac {b d}{e}\right )+b x}{\sqrt {f x^2+e x+d}}dx+\frac {b x \sqrt {d+e x+f x^2}}{2 e}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{4} \left (\frac {b \sqrt {d+e x+f x^2}}{f}-\frac {\left (-8 a f+\frac {4 b d f}{e}+b e\right ) \int \frac {1}{\sqrt {f x^2+e x+d}}dx}{2 f}\right )+\frac {b x \sqrt {d+e x+f x^2}}{2 e}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{4} \left (\frac {b \sqrt {d+e x+f x^2}}{f}-\frac {\left (-8 a f+\frac {4 b d f}{e}+b e\right ) \int \frac {1}{4 f-\frac {(e+2 f x)^2}{f x^2+e x+d}}d\frac {e+2 f x}{\sqrt {f x^2+e x+d}}}{f}\right )+\frac {b x \sqrt {d+e x+f x^2}}{2 e}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {b \sqrt {d+e x+f x^2}}{f}-\frac {\left (-8 a f+\frac {4 b d f}{e}+b e\right ) \text {arctanh}\left (\frac {e+2 f x}{2 \sqrt {f} \sqrt {d+e x+f x^2}}\right )}{2 f^{3/2}}\right )+\frac {b x \sqrt {d+e x+f x^2}}{2 e}\) |
Input:
Int[(a + b*x + (b*f*x^2)/e)/Sqrt[d + e*x + f*x^2],x]
Output:
(b*x*Sqrt[d + e*x + f*x^2])/(2*e) + ((b*Sqrt[d + e*x + f*x^2])/f - ((b*e - 8*a*f + (4*b*d*f)/e)*ArcTanh[(e + 2*f*x)/(2*Sqrt[f]*Sqrt[d + e*x + f*x^2] )])/(2*f^(3/2)))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Time = 1.61 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {b \left (2 f x +e \right ) \sqrt {f \,x^{2}+e x +d}}{4 f e}+\frac {\left (8 a e f -4 b d f -b \,e^{2}\right ) \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{8 f^{\frac {3}{2}} e}\) | \(79\) |
default | \(\frac {\frac {a e \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{\sqrt {f}}+b e \left (\frac {\sqrt {f \,x^{2}+e x +d}}{f}-\frac {e \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{2 f^{\frac {3}{2}}}\right )+f b \left (\frac {x \sqrt {f \,x^{2}+e x +d}}{2 f}-\frac {3 e \left (\frac {\sqrt {f \,x^{2}+e x +d}}{f}-\frac {e \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{2 f^{\frac {3}{2}}}\right )}{4 f}-\frac {d \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{2 f^{\frac {3}{2}}}\right )}{e}\) | \(195\) |
Input:
int((a+b*x+b*f*x^2/e)/(f*x^2+e*x+d)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4*b*(2*f*x+e)*(f*x^2+e*x+d)^(1/2)/f/e+1/8*(8*a*e*f-4*b*d*f-b*e^2)/f^(3/2 )*ln((f*x+1/2*e)/f^(1/2)+(f*x^2+e*x+d)^(1/2))/e
Time = 0.09 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.01 \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\left [-\frac {{\left (b e^{2} + 4 \, {\left (b d - 2 \, a e\right )} f\right )} \sqrt {f} \log \left (-8 \, f^{2} x^{2} - 8 \, e f x - e^{2} - 4 \, \sqrt {f x^{2} + e x + d} {\left (2 \, f x + e\right )} \sqrt {f} - 4 \, d f\right ) - 4 \, {\left (2 \, b f^{2} x + b e f\right )} \sqrt {f x^{2} + e x + d}}{16 \, e f^{2}}, \frac {{\left (b e^{2} + 4 \, {\left (b d - 2 \, a e\right )} f\right )} \sqrt {-f} \arctan \left (\frac {\sqrt {f x^{2} + e x + d} {\left (2 \, f x + e\right )} \sqrt {-f}}{2 \, {\left (f^{2} x^{2} + e f x + d f\right )}}\right ) + 2 \, {\left (2 \, b f^{2} x + b e f\right )} \sqrt {f x^{2} + e x + d}}{8 \, e f^{2}}\right ] \] Input:
integrate((a+b*x+b*f*x^2/e)/(f*x^2+e*x+d)^(1/2),x, algorithm="fricas")
Output:
[-1/16*((b*e^2 + 4*(b*d - 2*a*e)*f)*sqrt(f)*log(-8*f^2*x^2 - 8*e*f*x - e^2 - 4*sqrt(f*x^2 + e*x + d)*(2*f*x + e)*sqrt(f) - 4*d*f) - 4*(2*b*f^2*x + b *e*f)*sqrt(f*x^2 + e*x + d))/(e*f^2), 1/8*((b*e^2 + 4*(b*d - 2*a*e)*f)*sqr t(-f)*arctan(1/2*sqrt(f*x^2 + e*x + d)*(2*f*x + e)*sqrt(-f)/(f^2*x^2 + e*f *x + d*f)) + 2*(2*b*f^2*x + b*e*f)*sqrt(f*x^2 + e*x + d))/(e*f^2)]
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (85) = 170\).
Time = 0.47 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.10 \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\begin {cases} \left (\frac {b}{4 f} + \frac {b x}{2 e}\right ) \sqrt {d + e x + f x^{2}} + \left (a - \frac {b d}{2 e} - \frac {b e}{8 f}\right ) \left (\begin {cases} \frac {\log {\left (e + 2 \sqrt {f} \sqrt {d + e x + f x^{2}} + 2 f x \right )}}{\sqrt {f}} & \text {for}\: d - \frac {e^{2}}{4 f} \neq 0 \\\frac {\left (\frac {e}{2 f} + x\right ) \log {\left (\frac {e}{2 f} + x \right )}}{\sqrt {f \left (\frac {e}{2 f} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: f \neq 0 \\\frac {2 a \sqrt {d + e x} + \frac {2 b \left (- d \sqrt {d + e x} + \frac {\left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{e} + \frac {2 b f \left (d^{2} \sqrt {d + e x} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{3}}}{e} & \text {for}\: e \neq 0 \\\frac {a x + \tilde {\infty } b f x^{3} + \frac {b x^{2}}{2}}{\sqrt {d}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*x+b*f*x**2/e)/(f*x**2+e*x+d)**(1/2),x)
Output:
Piecewise(((b/(4*f) + b*x/(2*e))*sqrt(d + e*x + f*x**2) + (a - b*d/(2*e) - b*e/(8*f))*Piecewise((log(e + 2*sqrt(f)*sqrt(d + e*x + f*x**2) + 2*f*x)/s qrt(f), Ne(d - e**2/(4*f), 0)), ((e/(2*f) + x)*log(e/(2*f) + x)/sqrt(f*(e/ (2*f) + x)**2), True)), Ne(f, 0)), ((2*a*sqrt(d + e*x) + 2*b*(-d*sqrt(d + e*x) + (d + e*x)**(3/2)/3)/e + 2*b*f*(d**2*sqrt(d + e*x) - 2*d*(d + e*x)** (3/2)/3 + (d + e*x)**(5/2)/5)/e**3)/e, Ne(e, 0)), ((a*x + zoo*b*f*x**3 + b *x**2/2)/sqrt(d), True))
Exception generated. \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*x+b*f*x^2/e)/(f*x^2+e*x+d)^(1/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` for more deta
Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\frac {1}{4} \, \sqrt {f x^{2} + e x + d} {\left (\frac {2 \, b x}{e} + \frac {b}{f}\right )} + \frac {{\left (b e^{2} + 4 \, b d f - 8 \, a e f\right )} \log \left ({\left | 2 \, {\left (\sqrt {f} x - \sqrt {f x^{2} + e x + d}\right )} \sqrt {f} + e \right |}\right )}{8 \, e f^{\frac {3}{2}}} \] Input:
integrate((a+b*x+b*f*x^2/e)/(f*x^2+e*x+d)^(1/2),x, algorithm="giac")
Output:
1/4*sqrt(f*x^2 + e*x + d)*(2*b*x/e + b/f) + 1/8*(b*e^2 + 4*b*d*f - 8*a*e*f )*log(abs(2*(sqrt(f)*x - sqrt(f*x^2 + e*x + d))*sqrt(f) + e))/(e*f^(3/2))
Timed out. \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\int \frac {a+b\,x+\frac {b\,f\,x^2}{e}}{\sqrt {f\,x^2+e\,x+d}} \,d x \] Input:
int((a + b*x + (b*f*x^2)/e)/(d + e*x + f*x^2)^(1/2),x)
Output:
int((a + b*x + (b*f*x^2)/e)/(d + e*x + f*x^2)^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.70 \[ \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx=\frac {2 \sqrt {f \,x^{2}+e x +d}\, b e f +4 \sqrt {f \,x^{2}+e x +d}\, b \,f^{2} x +8 \sqrt {f}\, \mathrm {log}\left (\frac {2 \sqrt {f}\, \sqrt {f \,x^{2}+e x +d}+e +2 f x}{\sqrt {4 d f -e^{2}}}\right ) a e f -4 \sqrt {f}\, \mathrm {log}\left (\frac {2 \sqrt {f}\, \sqrt {f \,x^{2}+e x +d}+e +2 f x}{\sqrt {4 d f -e^{2}}}\right ) b d f -\sqrt {f}\, \mathrm {log}\left (\frac {2 \sqrt {f}\, \sqrt {f \,x^{2}+e x +d}+e +2 f x}{\sqrt {4 d f -e^{2}}}\right ) b \,e^{2}}{8 e \,f^{2}} \] Input:
int((a+b*x+b*f*x^2/e)/(f*x^2+e*x+d)^(1/2),x)
Output:
(2*sqrt(d + e*x + f*x**2)*b*e*f + 4*sqrt(d + e*x + f*x**2)*b*f**2*x + 8*sq rt(f)*log((2*sqrt(f)*sqrt(d + e*x + f*x**2) + e + 2*f*x)/sqrt(4*d*f - e**2 ))*a*e*f - 4*sqrt(f)*log((2*sqrt(f)*sqrt(d + e*x + f*x**2) + e + 2*f*x)/sq rt(4*d*f - e**2))*b*d*f - sqrt(f)*log((2*sqrt(f)*sqrt(d + e*x + f*x**2) + e + 2*f*x)/sqrt(4*d*f - e**2))*b*e**2)/(8*e*f**2)