Integrand size = 31, antiderivative size = 429 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (e f-d g) \sqrt {d+e x}}-\frac {8 c^2 \text {arctanh}\left (\frac {\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}{\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right )^{3/2} \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}+\frac {8 c^2 \text {arctanh}\left (\frac {\sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} \sqrt {d+e x}}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \sqrt {f+g x}}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right )^{3/2} \sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \]
4*c*e*(g*x+f)^(1/2)/(-d*g+e*f)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))/(-4*a*c+b^ 2)^(1/2)/(e*x+d)^(1/2)-4*c*e*(g*x+f)^(1/2)/(-d*g+e*f)/(-4*a*c+b^2)^(1/2)/( 2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))/(e*x+d)^(1/2)-8*c^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))/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(3/2)/(-4*a*c+b^2)^(1/ 2)/(2*c*f-g*(b-(-4*a*c+b^2)^(1/2)))^(1/2)+8*c^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))/(-4*a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(3/2)/( 2*c*f-g*(b+(-4*a*c+b^2)^(1/2)))^(1/2)
Time = 3.47 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\frac {2 e^2 \sqrt {f+g x}}{\left (c d^2+e (-b d+a e)\right ) (-e f+d g) \sqrt {d+e x}}+\frac {\sqrt {2} \left (2 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt {b^2-4 a c} d+a e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c d^2-b d e+a e^2} \sqrt {f+g x}}{\sqrt {-2 c d f+b e f+\sqrt {b^2-4 a c} e f+b d g-\sqrt {b^2-4 a c} d g-2 a e g} \sqrt {d+e x}}\right )}{\sqrt {b^2-4 a c} \left (c d^2+e (-b d+a e)\right )^{3/2} \sqrt {-2 c d f+b e f+\sqrt {b^2-4 a c} e f+b d g-\sqrt {b^2-4 a c} d g-2 a e g}}+\frac {\sqrt {2} \left (-2 c^2 d^2+b \left (-b+\sqrt {b^2-4 a c}\right ) e^2+2 c e \left (b d-\sqrt {b^2-4 a c} d+a e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c d^2-b d e+a e^2} \sqrt {f+g x}}{\sqrt {-2 c d f+b e f-\sqrt {b^2-4 a c} e f+b d g+\sqrt {b^2-4 a c} d g-2 a e g} \sqrt {d+e x}}\right )}{\sqrt {b^2-4 a c} \left (c d^2+e (-b d+a e)\right )^{3/2} \sqrt {-2 c d f+b e f-\sqrt {b^2-4 a c} e f+b d g+\sqrt {b^2-4 a c} d g-2 a e g}} \]
(2*e^2*Sqrt[f + g*x])/((c*d^2 + e*(-(b*d) + a*e))*(-(e*f) + d*g)*Sqrt[d + e*x]) + (Sqrt[2]*(2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(b*d + Sqrt[b^2 - 4*a*c]*d + a*e))*ArcTan[(Sqrt[2]*Sqrt[c*d^2 - b*d*e + a*e^2]*S qrt[f + g*x])/(Sqrt[-2*c*d*f + b*e*f + Sqrt[b^2 - 4*a*c]*e*f + b*d*g - Sqr t[b^2 - 4*a*c]*d*g - 2*a*e*g]*Sqrt[d + e*x])])/(Sqrt[b^2 - 4*a*c]*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*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^2*d^2 + b*(-b + Sqrt[b^2 - 4*a*c])*e^2 + 2*c*e*(b*d - Sqrt[b^2 - 4*a*c]*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]*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*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])
Time = 1.00 (sec) , antiderivative size = 429, 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 = {1205, 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.
\(\displaystyle \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx\) |
\(\Big \downarrow \) 1205 |
\(\displaystyle \int \left (\frac {2 c}{\sqrt {b^2-4 a c} (d+e x)^{3/2} \sqrt {f+g x} \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}-\frac {2 c}{\sqrt {b^2-4 a c} (d+e x)^{3/2} \sqrt {f+g x} \left (\sqrt {b^2-4 a c}+b+2 c x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8 c^2 \text {arctanh}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{\sqrt {b^2-4 a c} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )^{3/2} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}+\frac {8 c^2 \text {arctanh}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{\sqrt {b^2-4 a c} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )^{3/2} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}+\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \sqrt {d+e x} (e f-d g) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {4 c e \sqrt {f+g x}}{\sqrt {b^2-4 a c} \sqrt {d+e x} (e f-d g) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\) |
(4*c*e*Sqrt[f + g*x])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b - Sqrt[b^2 - 4*a*c])* e)*(e*f - d*g)*Sqrt[d + e*x]) - (4*c*e*Sqrt[f + g*x])/(Sqrt[b^2 - 4*a*c]*( 2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(e*f - d*g)*Sqrt[d + e*x]) - (8*c^2*Arc Tanh[(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])])/(Sqrt[b^2 - 4*a*c]*(2*c*d - ( b - Sqrt[b^2 - 4*a*c])*e)^(3/2)*Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g]) + (8*c^2*ArcTanh[(Sqrt[2*c*f - (b + Sqrt[b^2 - 4*a*c])*g]*Sqrt[d + e*x])/(S qrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*Sqrt[f + g*x])])/(Sqrt[b^2 - 4*a*c] *(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)^(3/2)*Sqrt[2*c*f - (b + Sqrt[b^2 - 4* a*c])*g])
3.9.53.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x _) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^ n, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
Leaf count of result is larger than twice the leaf count of optimal. \(47350\) vs. \(2(369)=738\).
Time = 0.68 (sec) , antiderivative size = 47351, normalized size of antiderivative = 110.38
Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \sqrt {f + g x} \left (a + b x + c x^{2}\right )}\, dx \]
\[ \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{\frac {3}{2}} \sqrt {g x + f}} \,d x } \]
Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {f+g x} \left (a+b x+c x^2\right )} \, dx=\int \frac {1}{\sqrt {f+g\,x}\,{\left (d+e\,x\right )}^{3/2}\,\left (c\,x^2+b\,x+a\right )} \,d x \]