Integrand size = 31, antiderivative size = 417 \[ \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\frac {2 e^{3/2} \text {arctanh}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{c \sqrt {g}}-\frac {2 \left (e (2 c d-b e)+\frac {2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}{\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}\right )}{c \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}-\frac {2 \left (e (2 c d-b e)-\frac {2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}\right )}{c \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \]
2*e^(3/2)*arctanh(g^(1/2)*(e*x+d)^(1/2)/e^(1/2)/(g*x+f)^(1/2))/c/g^(1/2)-2 *arctanh((e*x+d)^(1/2)*(2*c*f-g*(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(g*x+f)^(1/2 )/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(1/2))*(e*(-b*e+2*c*d)+(2*c^2*d^2+b^2*e ^2-2*c*e*(a*e+b*d))/(-4*a*c+b^2)^(1/2))/c/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))) ^(1/2)/(2*c*f-g*(b-(-4*a*c+b^2)^(1/2)))^(1/2)-2*arctanh((e*x+d)^(1/2)*(2*c *f-g*(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(g*x+f)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^ (1/2)))^(1/2))*(e*(-b*e+2*c*d)+(-2*c^2*d^2-b^2*e^2+2*c*e*(a*e+b*d))/(-4*a* c+b^2)^(1/2))/c/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(2*c*f-g*(b+(-4*a*c +b^2)^(1/2)))^(1/2)
Time = 3.56 (sec) , antiderivative size = 473, normalized size of antiderivative = 1.13 \[ \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\frac {\frac {\sqrt {2} \left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \sqrt {c d^2+e (-b d+a e)} \arctan \left (\frac {\sqrt {2} \sqrt {c d^2-b d e+a e^2} \sqrt {f+g x}}{\sqrt {-2 c d f+b e f+\sqrt {b^2-4 a c} e f+b d g-\sqrt {b^2-4 a c} d g-2 a e g} \sqrt {d+e x}}\right )}{\sqrt {b^2-4 a c} \sqrt {-2 c d f+b e f+\sqrt {b^2-4 a c} e f+b d g-\sqrt {b^2-4 a c} d g-2 a e g}}+\frac {\sqrt {2} \left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \sqrt {c d^2+e (-b d+a e)} \arctan \left (\frac {\sqrt {2} \sqrt {c d^2-b d e+a e^2} \sqrt {f+g x}}{\sqrt {-2 c d f+b e f-\sqrt {b^2-4 a c} e f+b d g+\sqrt {b^2-4 a c} d g-2 a e g} \sqrt {d+e x}}\right )}{\sqrt {b^2-4 a c} \sqrt {-2 c d f+b e f-\sqrt {b^2-4 a c} e f+b d g+\sqrt {b^2-4 a c} d g-2 a e g}}+\frac {2 e^{3/2} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{\sqrt {g}}}{c} \]
((Sqrt[2]*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*Sqrt[c*d^2 + e*(-(b*d) + a* e)]*ArcTan[(Sqrt[2]*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[f + g*x])/(Sqrt[-2*c* d*f + b*e*f + Sqrt[b^2 - 4*a*c]*e*f + b*d*g - Sqrt[b^2 - 4*a*c]*d*g - 2*a* e*g]*Sqrt[d + e*x])])/(Sqrt[b^2 - 4*a*c]*Sqrt[-2*c*d*f + b*e*f + Sqrt[b^2 - 4*a*c]*e*f + b*d*g - Sqrt[b^2 - 4*a*c]*d*g - 2*a*e*g]) + (Sqrt[2]*(-2*c* d + (b + Sqrt[b^2 - 4*a*c])*e)*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*ArcTan[(Sqrt [2]*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[f + g*x])/(Sqrt[-2*c*d*f + b*e*f - Sq rt[b^2 - 4*a*c]*e*f + b*d*g + Sqrt[b^2 - 4*a*c]*d*g - 2*a*e*g]*Sqrt[d + e* x])])/(Sqrt[b^2 - 4*a*c]*Sqrt[-2*c*d*f + b*e*f - Sqrt[b^2 - 4*a*c]*e*f + b *d*g + Sqrt[b^2 - 4*a*c]*d*g - 2*a*e*g]) + (2*e^(3/2)*ArcTanh[(Sqrt[e]*Sqr t[f + g*x])/(Sqrt[g]*Sqrt[d + e*x])])/Sqrt[g])/c
Time = 1.62 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1204, 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.
\(\displaystyle \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx\) |
\(\Big \downarrow \) 1204 |
\(\displaystyle \int \left (\frac {-a e^2+e x (2 c d-b e)+c d^2}{c \sqrt {d+e x} \sqrt {f+g x} \left (a+b x+c x^2\right )}+\frac {e^2}{c \sqrt {d+e x} \sqrt {f+g x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}+e (2 c d-b e)\right ) \text {arctanh}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{c \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}-\frac {2 \left (e (2 c d-b e)-\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{c \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}+\frac {2 e^{3/2} \text {arctanh}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{c \sqrt {g}}\) |
(2*e^(3/2)*ArcTanh[(Sqrt[g]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[f + g*x])])/(c*Sq rt[g]) - (2*(e*(2*c*d - b*e) + (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))/S qrt[b^2 - 4*a*c])*ArcTanh[(Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])/(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/(c*Sqrt[ 2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g] ) - (2*(e*(2*c*d - b*e) - (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))/Sqrt[b ^2 - 4*a*c])*ArcTanh[(Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x ])/(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/(c*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g])
3.9.50.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[m + 1/2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(11685\) vs. \(2(361)=722\).
Time = 0.65 (sec) , antiderivative size = 11686, normalized size of antiderivative = 28.02
Timed out. \[ \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}}}{\sqrt {f + g x} \left (a + b x + c x^{2}\right )}\, dx \]
\[ \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )} \sqrt {g x + f}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(d+e x)^{3/2}}{\sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\int \frac {{\left (d+e\,x\right )}^{3/2}}{\sqrt {f+g\,x}\,\left (c\,x^2+b\,x+a\right )} \,d x \]