Integrand size = 44, antiderivative size = 200 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx=-\frac {(2 c e f-4 c d g+b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (2 c d-b e)}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (2 c d-b e) (d+e x)^2}-\frac {(2 c e f-4 c d g+b e g) \arctan \left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{2 \sqrt {c} e^2} \]
-2*(-d*g+e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(3/2)/e^2/(-b*e+2*c*d)/(e*x +d)^2-1/2*(b*e*g-4*c*d*g+2*c*e*f)*arctan(1/2*e*(2*c*x+b)/c^(1/2)/(d*(-b*e+ c*d)-b*e^2*x-c*e^2*x^2)^(1/2))/e^2/c^(1/2)-(b*e*g-4*c*d*g+2*c*e*f)*(d*(-b* e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/e^2/(-b*e+2*c*d)
Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx=\frac {\sqrt {(d+e x) (-b e+c (d-e x))} \left (\frac {-2 e f+3 d g+e g x}{d+e x}+\frac {(2 c e f-4 c d g+b e g) \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {c} \sqrt {d+e x}}\right )}{\sqrt {c} \sqrt {d+e x} \sqrt {-b e+c (d-e x)}}\right )}{e^2} \]
(Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))]*((-2*e*f + 3*d*g + e*g*x)/(d + e*x ) + ((2*c*e*f - 4*c*d*g + b*e*g)*ArcTan[Sqrt[c*d - b*e - c*e*x]/(Sqrt[c]*S qrt[d + e*x])])/(Sqrt[c]*Sqrt[d + e*x]*Sqrt[-(b*e) + c*(d - e*x)])))/e^2
Time = 0.43 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1216, 1211, 27, 1160, 1092, 217}
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.
| \(\displaystyle \int \frac {(f+g x) \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 1216 |
| \(\displaystyle \int \frac {(f+g x) \left (\frac {c d^2-b d e}{d}-c e x\right )^2}{\left (-b d e-b e^2 x+c d^2-c e^2 x^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 1211 |
| \(\displaystyle -\frac {\int \frac {c e^2 (c e f-2 c d g+b e g+c e g x)}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{c e^3}-\frac {2 (e f-d g) (-b e+c d-c e x)}{e^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {c e f-2 c d g+b e g+c e g x}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{e}-\frac {2 (e f-d g) (-b e+c d-c e x)}{e^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle -\frac {\frac {1}{2} (b e g-4 c d g+2 c e f) \int \frac {1}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx-\frac {g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e}}{e}-\frac {2 (e f-d g) (-b e+c d-c e x)}{e^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle -\frac {(b e g-4 c d g+2 c e f) \int \frac {1}{-\frac {(b+2 c x)^2 e^4}{-c x^2 e^2-b x e^2+d (c d-b e)}-4 c e^2}d\left (-\frac {e^2 (b+2 c x)}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}\right )-\frac {g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e}}{e}-\frac {2 (e f-d g) (-b e+c d-c e x)}{e^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {\frac {\arctan \left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right ) (b e g-4 c d g+2 c e f)}{2 \sqrt {c} e}-\frac {g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e}}{e}-\frac {2 (e f-d g) (-b e+c d-c e x)}{e^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\) |
(-2*(e*f - d*g)*(c*d - b*e - c*e*x))/(e^2*Sqrt[d*(c*d - b*e) - b*e^2*x - c *e^2*x^2]) - (-((g*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/e) + ((2*c*e *f - 4*c*d*g + b*e*g)*ArcTan[(e*(b + 2*c*x))/(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])])/(2*Sqrt[c]*e))/e
3.22.76.3.1 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[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[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[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*(2*c*d - b*e)^(m - 2)*(c*( e*f + d*g) - b*e*g)^n*((d + e*x)/(c^(m + n - 1)*e^(n - 1)*Sqrt[a + b*x + c* x^2])), x] + Simp[1/(c^(m + n - 1)*e^(n - 2)) Int[ExpandToSum[((2*c*d - b *e)^(m - 1)*(c*(e*f + d*g) - b*e*g)^n - c^(m + n - 1)*e^n*(d + e*x)^(m - 1) *(f + g*x)^n)/(c*d - b*e - c*e*x), x]/Sqrt[a + b*x + c*x^2], x], x] /; Free Q[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*Sqrt[(a_.) + (b_.)* (x_) + (c_.)*(x_)^2], x_Symbol] :> Int[((f + g*x)^n*(a + b*x + c*x^2)^(m + 1/2))/(a/d + c*(x/e))^m, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[c* d^2 - b*d*e + a*e^2, 0] && ILtQ[m, 0] && IntegerQ[n]
Time = 0.76 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.86
| method | result | size |
| default | \(\frac {g \left (\sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}+\frac {\left (-b \,e^{2}+2 c d e \right ) \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 c d e}{2 c \,e^{2}}\right )}{\sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {c \,e^{2}}}\right )}{e^{2}}+\frac {\left (-d g +e f \right ) \left (-\frac {2 \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )^{2}}-\frac {2 c \,e^{2} \left (\sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}+\frac {\left (-b \,e^{2}+2 c d e \right ) \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 c d e}{2 c \,e^{2}}\right )}{\sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 c d e \right ) \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {c \,e^{2}}}\right )}{-b \,e^{2}+2 c d e}\right )}{e^{3}}\) | \(371\) |
g/e^2*((-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)+1/2*(-b*e^2+2*c*d *e)/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/ (-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)))+(-d*g+e*f)/e^3*(-2/(-b *e^2+2*c*d*e)/(x+d/e)^2*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(3/2)- 2*c*e^2/(-b*e^2+2*c*d*e)*((-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2 )+1/2*(-b*e^2+2*c*d*e)/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e ^2+2*c*d*e)/c/e^2)/(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2))))
Time = 1.01 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.00 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx=\left [-\frac {{\left (2 \, c d e f - {\left (4 \, c d^{2} - b d e\right )} g + {\left (2 \, c e^{2} f - {\left (4 \, c d e - b e^{2}\right )} g\right )} x\right )} \sqrt {-c} \log \left (8 \, c^{2} e^{2} x^{2} + 8 \, b c e^{2} x - 4 \, c^{2} d^{2} + 4 \, b c d e + b^{2} e^{2} + 4 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {-c}\right ) - 4 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (c e g x - 2 \, c e f + 3 \, c d g\right )}}{4 \, {\left (c e^{3} x + c d e^{2}\right )}}, \frac {{\left (2 \, c d e f - {\left (4 \, c d^{2} - b d e\right )} g + {\left (2 \, c e^{2} f - {\left (4 \, c d e - b e^{2}\right )} g\right )} x\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {c}}{2 \, {\left (c^{2} e^{2} x^{2} + b c e^{2} x - c^{2} d^{2} + b c d e\right )}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (c e g x - 2 \, c e f + 3 \, c d g\right )}}{2 \, {\left (c e^{3} x + c d e^{2}\right )}}\right ] \]
[-1/4*((2*c*d*e*f - (4*c*d^2 - b*d*e)*g + (2*c*e^2*f - (4*c*d*e - b*e^2)*g )*x)*sqrt(-c)*log(8*c^2*e^2*x^2 + 8*b*c*e^2*x - 4*c^2*d^2 + 4*b*c*d*e + b^ 2*e^2 + 4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt( -c)) - 4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(c*e*g*x - 2*c*e*f + 3 *c*d*g))/(c*e^3*x + c*d*e^2), 1/2*((2*c*d*e*f - (4*c*d^2 - b*d*e)*g + (2*c *e^2*f - (4*c*d*e - b*e^2)*g)*x)*sqrt(c)*arctan(1/2*sqrt(-c*e^2*x^2 - b*e^ 2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(c)/(c^2*e^2*x^2 + b*c*e^2*x - c^ 2*d^2 + b*c*d*e)) + 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(c*e*g*x - 2*c*e*f + 3*c*d*g))/(c*e^3*x + c*d*e^2)]
\[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx=\int \frac {\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )} \left (f + g x\right )}{\left (d + e x\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-2*c*d>0)', see `assume?` for more deta
Time = 0.38 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.47 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx={\left (\frac {{\left (2 \, c e f \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 4 \, c d g \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + b e g \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} \arctan \left (\frac {\sqrt {-c + \frac {2 \, c d}{e x + d} - \frac {b e}{e x + d}}}{\sqrt {c}}\right )}{\sqrt {c} e^{3}} - \frac {2 \, {\left (\sqrt {-c + \frac {2 \, c d}{e x + d} - \frac {b e}{e x + d}} e f \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - \sqrt {-c + \frac {2 \, c d}{e x + d} - \frac {b e}{e x + d}} d g \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )}}{e^{3}} + \frac {2 \, c \sqrt {-c + \frac {2 \, c d}{e x + d} - \frac {b e}{e x + d}} d g \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - b \sqrt {-c + \frac {2 \, c d}{e x + d} - \frac {b e}{e x + d}} e g \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{{\left (\frac {2 \, c d}{e x + d} - \frac {b e}{e x + d}\right )} e^{3}}\right )} {\left | e \right |} \]
((2*c*e*f*sgn(1/(e*x + d))*sgn(e) - 4*c*d*g*sgn(1/(e*x + d))*sgn(e) + b*e* g*sgn(1/(e*x + d))*sgn(e))*arctan(sqrt(-c + 2*c*d/(e*x + d) - b*e/(e*x + d ))/sqrt(c))/(sqrt(c)*e^3) - 2*(sqrt(-c + 2*c*d/(e*x + d) - b*e/(e*x + d))* e*f*sgn(1/(e*x + d))*sgn(e) - sqrt(-c + 2*c*d/(e*x + d) - b*e/(e*x + d))*d *g*sgn(1/(e*x + d))*sgn(e))/e^3 + (2*c*sqrt(-c + 2*c*d/(e*x + d) - b*e/(e* x + d))*d*g*sgn(1/(e*x + d))*sgn(e) - b*sqrt(-c + 2*c*d/(e*x + d) - b*e/(e *x + d))*e*g*sgn(1/(e*x + d))*sgn(e))/((2*c*d/(e*x + d) - b*e/(e*x + d))*e ^3))*abs(e)
Timed out. \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx=\int \frac {\left (f+g\,x\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}{{\left (d+e\,x\right )}^2} \,d x \]