Integrand size = 21, antiderivative size = 110 \[ \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx=\frac {3 e (2 c d-b e) \sqrt {b x+c x^2}}{4 c^2}+\frac {e (d+e x) \sqrt {b x+c x^2}}{2 c}+\frac {\left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{5/2}} \]
1/4*(3*b^2*e^2-8*b*c*d*e+8*c^2*d^2)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c ^(5/2)+3/4*e*(-b*e+2*c*d)*(c*x^2+b*x)^(1/2)/c^2+1/2*e*(e*x+d)*(c*x^2+b*x)^ (1/2)/c
Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} e x (b+c x) (8 c d-3 b e+2 c e x)+\left (-8 c^2 d^2+8 b c d e-3 b^2 e^2\right ) \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{4 c^{5/2} \sqrt {x (b+c x)}} \]
(Sqrt[c]*e*x*(b + c*x)*(8*c*d - 3*b*e + 2*c*e*x) + (-8*c^2*d^2 + 8*b*c*d*e - 3*b^2*e^2)*Sqrt[x]*Sqrt[b + c*x]*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x] ])/(4*c^(5/2)*Sqrt[x*(b + c*x)])
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1166, 27, 1160, 1091, 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.
\(\displaystyle \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1166 |
\(\displaystyle \frac {\int \frac {d (4 c d-b e)+3 e (2 c d-b e) x}{2 \sqrt {c x^2+b x}}dx}{2 c}+\frac {e \sqrt {b x+c x^2} (d+e x)}{2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {d (4 c d-b e)+3 e (2 c d-b e) x}{\sqrt {c x^2+b x}}dx}{4 c}+\frac {e \sqrt {b x+c x^2} (d+e x)}{2 c}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\frac {\left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right ) \int \frac {1}{\sqrt {c x^2+b x}}dx}{2 c}+\frac {3 e \sqrt {b x+c x^2} (2 c d-b e)}{c}}{4 c}+\frac {e \sqrt {b x+c x^2} (d+e x)}{2 c}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {\frac {\left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right ) \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{c}+\frac {3 e \sqrt {b x+c x^2} (2 c d-b e)}{c}}{4 c}+\frac {e \sqrt {b x+c x^2} (d+e x)}{2 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{c^{3/2}}+\frac {3 e \sqrt {b x+c x^2} (2 c d-b e)}{c}}{4 c}+\frac {e \sqrt {b x+c x^2} (d+e x)}{2 c}\) |
(e*(d + e*x)*Sqrt[b*x + c*x^2])/(2*c) + ((3*e*(2*c*d - b*e)*Sqrt[b*x + c*x ^2])/c + ((8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(3/2))/(4*c)
3.4.16.3.1 Defintions of rubi rules used
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[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{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[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
Time = 2.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \left (b^{2} e^{2}-\frac {8}{3} b c d e +\frac {8}{3} c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{4}-\frac {3 \left (\frac {2 \left (-e x -4 d \right ) c^{\frac {3}{2}}}{3}+\sqrt {c}\, b e \right ) \sqrt {x \left (c x +b \right )}\, e}{4}}{c^{\frac {5}{2}}}\) | \(79\) |
risch | \(-\frac {\left (-2 c e x +3 b e -8 c d \right ) e x \left (c x +b \right )}{4 c^{2} \sqrt {x \left (c x +b \right )}}+\frac {\left (3 b^{2} e^{2}-8 b c d e +8 c^{2} d^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}\) | \(89\) |
default | \(\frac {d^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}+e^{2} \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+2 d e \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )\) | \(157\) |
3/4/c^(5/2)*((b^2*e^2-8/3*b*c*d*e+8/3*c^2*d^2)*arctanh((x*(c*x+b))^(1/2)/x /c^(1/2))-(2/3*(-e*x-4*d)*c^(3/2)+c^(1/2)*b*e)*(x*(c*x+b))^(1/2)*e)
Time = 0.44 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.73 \[ \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx=\left [\frac {{\left (8 \, c^{2} d^{2} - 8 \, b c d e + 3 \, b^{2} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{2} e^{2} x + 8 \, c^{2} d e - 3 \, b c e^{2}\right )} \sqrt {c x^{2} + b x}}{8 \, c^{3}}, -\frac {{\left (8 \, c^{2} d^{2} - 8 \, b c d e + 3 \, b^{2} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, c^{2} e^{2} x + 8 \, c^{2} d e - 3 \, b c e^{2}\right )} \sqrt {c x^{2} + b x}}{4 \, c^{3}}\right ] \]
[1/8*((8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c *x^2 + b*x)*sqrt(c)) + 2*(2*c^2*e^2*x + 8*c^2*d*e - 3*b*c*e^2)*sqrt(c*x^2 + b*x))/c^3, -1/4*((8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*sqrt(-c)*arctan(sqr t(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*c^2*e^2*x + 8*c^2*d*e - 3*b*c*e^2)*sqr t(c*x^2 + b*x))/c^3]
Time = 0.54 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.68 \[ \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \sqrt {b x + c x^{2}} \left (\frac {e^{2} x}{2 c} + \frac {- \frac {3 b e^{2}}{4 c} + 2 d e}{c}\right ) + \left (- \frac {b \left (- \frac {3 b e^{2}}{4 c} + 2 d e\right )}{2 c} + d^{2}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (d^{2} \sqrt {b x} + \frac {2 d e \left (b x\right )^{\frac {3}{2}}}{3 b} + \frac {e^{2} \left (b x\right )^{\frac {5}{2}}}{5 b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\begin {cases} d^{2} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{3}}{3 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(b*x + c*x**2)*(e**2*x/(2*c) + (-3*b*e**2/(4*c) + 2*d*e)/c) + (-b*(-3*b*e**2/(4*c) + 2*d*e)/(2*c) + d**2)*Piecewise((log(b + 2*sqrt(c )*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log( b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)), Ne(c, 0)), (2*(d**2*sqrt(b* x) + 2*d*e*(b*x)**(3/2)/(3*b) + e**2*(b*x)**(5/2)/(5*b**2))/b, Ne(b, 0)), (zoo*Piecewise((d**2*x, Eq(e, 0)), ((d + e*x)**3/(3*e), True)), True))
Time = 0.20 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.40 \[ \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} e^{2} x}{2 \, c} + \frac {d^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{\sqrt {c}} - \frac {b d e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} + \frac {3 \, b^{2} e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} + \frac {2 \, \sqrt {c x^{2} + b x} d e}{c} - \frac {3 \, \sqrt {c x^{2} + b x} b e^{2}}{4 \, c^{2}} \]
1/2*sqrt(c*x^2 + b*x)*e^2*x/c + d^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq rt(c))/sqrt(c) - b*d*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2 ) + 3/8*b^2*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 2*s qrt(c*x^2 + b*x)*d*e/c - 3/4*sqrt(c*x^2 + b*x)*b*e^2/c^2
Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, e^{2} x}{c} + \frac {8 \, c d e - 3 \, b e^{2}}{c^{2}}\right )} - \frac {{\left (8 \, c^{2} d^{2} - 8 \, b c d e + 3 \, b^{2} e^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {5}{2}}} \]
1/4*sqrt(c*x^2 + b*x)*(2*e^2*x/c + (8*c*d*e - 3*b*e^2)/c^2) - 1/8*(8*c^2*d ^2 - 8*b*c*d*e + 3*b^2*e^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt (c) + b))/c^(5/2)
Timed out. \[ \int \frac {(d+e x)^2}{\sqrt {b x+c x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{\sqrt {c\,x^2+b\,x}} \,d x \]