Integrand size = 21, antiderivative size = 101 \[ \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (d+e x) (b d+(2 c d-b e) x)}{b^2 \sqrt {b x+c x^2}}+\frac {2 e (2 c d-b e) \sqrt {b x+c x^2}}{b^2 c}+\frac {2 e^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]
2*e^2*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(3/2)-2*(e*x+d)*(b*d+(-b*e+2* c*d)*x)/b^2/(c*x^2+b*x)^(1/2)+2*e*(-b*e+2*c*d)*(c*x^2+b*x)^(1/2)/b^2/c
Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (\sqrt {c} \left (2 c^2 d^2 x+b^2 e^2 x+b c d (d-2 e x)\right )+b^2 e^2 \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{b^2 c^{3/2} \sqrt {x (b+c x)}} \]
(-2*(Sqrt[c]*(2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x)) + b^2*e^2*Sqrt[ x]*Sqrt[b + c*x]*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]))/(b^2*c^(3/2)*Sq rt[x*(b + c*x)])
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1164, 25, 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.
\(\displaystyle \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1164 |
\(\displaystyle -\frac {2 \int -\frac {e (b d+(2 c d-b e) x)}{\sqrt {c x^2+b x}}dx}{b^2}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \int \frac {e (b d+(2 c d-b e) x)}{\sqrt {c x^2+b x}}dx}{b^2}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 e \int \frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}dx}{b^2}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {2 e \left (\frac {b^2 e \int \frac {1}{\sqrt {c x^2+b x}}dx}{2 c}+\frac {\sqrt {b x+c x^2} (2 c d-b e)}{c}\right )}{b^2}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {2 e \left (\frac {b^2 e \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{c}+\frac {\sqrt {b x+c x^2} (2 c d-b e)}{c}\right )}{b^2}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 e \left (\frac {b^2 e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}+\frac {\sqrt {b x+c x^2} (2 c d-b e)}{c}\right )}{b^2}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}\) |
(-2*(d + e*x)*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (2*e*(((2 *c*d - b*e)*Sqrt[b*x + c*x^2])/c + (b^2*e*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c *x^2]])/c^(3/2)))/b^2
3.4.23.3.1 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[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* c)) Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int QuadraticQ[a, b, c, d, e, m, p, x]
Time = 2.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81
method | result | size |
pseudoelliptic | \(\frac {-2 b d \left (-2 e x +d \right ) c^{\frac {3}{2}}-4 c^{\frac {5}{2}} d^{2} x +2 \left (\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) \sqrt {x \left (c x +b \right )}-\sqrt {c}\, x \right ) e^{2} b^{2}}{c^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}\, b^{2}}\) | \(82\) |
risch | \(-\frac {2 d^{2} \left (c x +b \right )}{b^{2} \sqrt {x \left (c x +b \right )}}+\frac {\frac {e^{2} b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}+\frac {2 \left (-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \sqrt {c \left (\frac {b}{c}+x \right )^{2}-b \left (\frac {b}{c}+x \right )}}{c^{2} b \left (\frac {b}{c}+x \right )}}{b}\) | \(125\) |
default | \(-\frac {2 d^{2} \left (2 c x +b \right )}{b^{2} \sqrt {c \,x^{2}+b x}}+e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )+2 d e \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )\) | \(169\) |
2*(-b*d*(-2*e*x+d)*c^(3/2)-2*c^(5/2)*d^2*x+(arctanh((x*(c*x+b))^(1/2)/x/c^ (1/2))*(x*(c*x+b))^(1/2)-c^(1/2)*x)*e^2*b^2)/(x*(c*x+b))^(1/2)/c^(3/2)/b^2
Time = 0.27 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.40 \[ \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {{\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (b c^{2} d^{2} + {\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + b^{2} c e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{b^{2} c^{3} x^{2} + b^{3} c^{2} x}, -\frac {2 \, {\left ({\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (b c^{2} d^{2} + {\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + b^{2} c e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{b^{2} c^{3} x^{2} + b^{3} c^{2} x}\right ] \]
[((b^2*c*e^2*x^2 + b^3*e^2*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)* sqrt(c)) - 2*(b*c^2*d^2 + (2*c^3*d^2 - 2*b*c^2*d*e + b^2*c*e^2)*x)*sqrt(c* x^2 + b*x))/(b^2*c^3*x^2 + b^3*c^2*x), -2*((b^2*c*e^2*x^2 + b^3*e^2*x)*sqr t(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (b*c^2*d^2 + (2*c^3*d^2 - 2*b*c^2*d*e + b^2*c*e^2)*x)*sqrt(c*x^2 + b*x))/(b^2*c^3*x^2 + b^3*c^2*x)]
\[ \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {4 \, c d^{2} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {4 \, d e x}{\sqrt {c x^{2} + b x} b} - \frac {2 \, e^{2} x}{\sqrt {c x^{2} + b x} c} + \frac {e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} - \frac {2 \, d^{2}}{\sqrt {c x^{2} + b x} b} \]
-4*c*d^2*x/(sqrt(c*x^2 + b*x)*b^2) + 4*d*e*x/(sqrt(c*x^2 + b*x)*b) - 2*e^2 *x/(sqrt(c*x^2 + b*x)*c) + e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c) )/c^(3/2) - 2*d^2/(sqrt(c*x^2 + b*x)*b)
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {e^{2} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}} - \frac {2 \, {\left (\frac {d^{2}}{b} + \frac {{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} x}{b^{2} c}\right )}}{\sqrt {c x^{2} + b x}} \]
-e^2*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(3/2) - 2*( d^2/b + (2*c^2*d^2 - 2*b*c*d*e + b^2*e^2)*x/(b^2*c))/sqrt(c*x^2 + b*x)
Time = 9.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {e^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{c^{3/2}}-\frac {d^2\,\left (2\,b+4\,c\,x\right )}{b^2\,\sqrt {c\,x^2+b\,x}}-\frac {2\,e^2\,x}{c\,\sqrt {c\,x^2+b\,x}}+\frac {4\,d\,e\,x}{b\,\sqrt {x\,\left (b+c\,x\right )}} \]