Integrand size = 28, antiderivative size = 146 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}-\frac {e \sqrt {d+e x}}{12 b (b d-a e) (a+b x)^2}+\frac {e^2 \sqrt {d+e x}}{8 b (b d-a e)^2 (a+b x)}-\frac {e^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{5/2}} \]
-1/8*e^3*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(3/2)/(-a*e+b*d )^(5/2)-1/3*(e*x+d)^(1/2)/b/(b*x+a)^3-1/12*e*(e*x+d)^(1/2)/b/(-a*e+b*d)/(b *x+a)^2+1/8*e^2*(e*x+d)^(1/2)/b/(-a*e+b*d)^2/(b*x+a)
Time = 0.54 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\sqrt {d+e x} \left (-3 a^2 e^2+2 a b e (7 d+4 e x)+b^2 \left (-8 d^2-2 d e x+3 e^2 x^2\right )\right )}{24 b (b d-a e)^2 (a+b x)^3}+\frac {e^3 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 b^{3/2} (-b d+a e)^{5/2}} \]
(Sqrt[d + e*x]*(-3*a^2*e^2 + 2*a*b*e*(7*d + 4*e*x) + b^2*(-8*d^2 - 2*d*e*x + 3*e^2*x^2)))/(24*b*(b*d - a*e)^2*(a + b*x)^3) + (e^3*ArcTan[(Sqrt[b]*Sq rt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*b^(3/2)*(-(b*d) + a*e)^(5/2))
Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1098, 27, 51, 52, 52, 73, 221}
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.
| \(\displaystyle \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1098 |
| \(\displaystyle b^4 \int \frac {\sqrt {d+e x}}{b^4 (a+b x)^4}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {\sqrt {d+e x}}{(a+b x)^4}dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {e \int \frac {1}{(a+b x)^3 \sqrt {d+e x}}dx}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {e \left (-\frac {3 e \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {e \left (-\frac {3 e \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {e \left (-\frac {3 e \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {e \left (-\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\) |
-1/3*Sqrt[d + e*x]/(b*(a + b*x)^3) + (e*(-1/2*Sqrt[d + e*x]/((b*d - a*e)*( a + b*x)^2) - (3*e*(-(Sqrt[d + e*x]/((b*d - a*e)*(a + b*x))) + (e*ArcTanh[ (Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^(3/2))))/(4 *(b*d - a*e))))/(6*b)
3.17.59.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ {a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 2.60 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.81
| method | result | size |
| pseudoelliptic | \(\frac {e^{3} \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )-\left (\left (-3 e x -4 d \right ) b +a e \right ) \sqrt {e x +d}\, \left (\frac {\left (e x -2 d \right ) b}{3}+a e \right ) \sqrt {\left (a e -b d \right ) b}}{8 \sqrt {\left (a e -b d \right ) b}\, \left (b x +a \right )^{3} \left (a e -b d \right )^{2} b}\) | \(118\) |
| derivativedivides | \(2 e^{3} \left (\frac {\frac {b \left (e x +d \right )^{\frac {5}{2}}}{16 a^{2} e^{2}-32 a b d e +16 b^{2} d^{2}}+\frac {\left (e x +d \right )^{\frac {3}{2}}}{6 a e -6 b d}-\frac {\sqrt {e x +d}}{16 b}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 b \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}}\right )\) | \(152\) |
| default | \(2 e^{3} \left (\frac {\frac {b \left (e x +d \right )^{\frac {5}{2}}}{16 a^{2} e^{2}-32 a b d e +16 b^{2} d^{2}}+\frac {\left (e x +d \right )^{\frac {3}{2}}}{6 a e -6 b d}-\frac {\sqrt {e x +d}}{16 b}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 b \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}}\right )\) | \(152\) |
1/8*(e^3*(b*x+a)^3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))-((-3*e*x-4* d)*b+a*e)*(e*x+d)^(1/2)*(1/3*(e*x-2*d)*b+a*e)*((a*e-b*d)*b)^(1/2))/((a*e-b *d)*b)^(1/2)/(b*x+a)^3/(a*e-b*d)^2/b
Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (122) = 244\).
Time = 0.43 (sec) , antiderivative size = 785, normalized size of antiderivative = 5.38 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} d^{3} - 22 \, a b^{3} d^{2} e + 17 \, a^{2} b^{2} d e^{2} - 3 \, a^{3} b e^{3} - 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (b^{4} d^{2} e - 5 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d^{3} - 3 \, a^{4} b^{4} d^{2} e + 3 \, a^{5} b^{3} d e^{2} - a^{6} b^{2} e^{3} + {\left (b^{8} d^{3} - 3 \, a b^{7} d^{2} e + 3 \, a^{2} b^{6} d e^{2} - a^{3} b^{5} e^{3}\right )} x^{3} + 3 \, {\left (a b^{7} d^{3} - 3 \, a^{2} b^{6} d^{2} e + 3 \, a^{3} b^{5} d e^{2} - a^{4} b^{4} e^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{3} - 3 \, a^{3} b^{5} d^{2} e + 3 \, a^{4} b^{4} d e^{2} - a^{5} b^{3} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (8 \, b^{4} d^{3} - 22 \, a b^{3} d^{2} e + 17 \, a^{2} b^{2} d e^{2} - 3 \, a^{3} b e^{3} - 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (b^{4} d^{2} e - 5 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d^{3} - 3 \, a^{4} b^{4} d^{2} e + 3 \, a^{5} b^{3} d e^{2} - a^{6} b^{2} e^{3} + {\left (b^{8} d^{3} - 3 \, a b^{7} d^{2} e + 3 \, a^{2} b^{6} d e^{2} - a^{3} b^{5} e^{3}\right )} x^{3} + 3 \, {\left (a b^{7} d^{3} - 3 \, a^{2} b^{6} d^{2} e + 3 \, a^{3} b^{5} d e^{2} - a^{4} b^{4} e^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{3} - 3 \, a^{3} b^{5} d^{2} e + 3 \, a^{4} b^{4} d e^{2} - a^{5} b^{3} e^{3}\right )} x\right )}}\right ] \]
[1/48*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(b^ 2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d ))/(b*x + a)) - 2*(8*b^4*d^3 - 22*a*b^3*d^2*e + 17*a^2*b^2*d*e^2 - 3*a^3*b *e^3 - 3*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^ 2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^3 - 3*a^4*b^4*d^2*e + 3*a^5*b^3*d* e^2 - a^6*b^2*e^3 + (b^8*d^3 - 3*a*b^7*d^2*e + 3*a^2*b^6*d*e^2 - a^3*b^5*e ^3)*x^3 + 3*(a*b^7*d^3 - 3*a^2*b^6*d^2*e + 3*a^3*b^5*d*e^2 - a^4*b^4*e^3)* x^2 + 3*(a^2*b^6*d^3 - 3*a^3*b^5*d^2*e + 3*a^4*b^4*d*e^2 - a^5*b^3*e^3)*x) , 1/24*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(- b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (8*b^4*d^3 - 22*a*b^3*d^2*e + 17*a^2*b^2*d*e^2 - 3*a^3*b*e^3 - 3*(b^4*d*e^ 2 - a*b^3*e^3)*x^2 + 2*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt (e*x + d))/(a^3*b^5*d^3 - 3*a^4*b^4*d^2*e + 3*a^5*b^3*d*e^2 - a^6*b^2*e^3 + (b^8*d^3 - 3*a*b^7*d^2*e + 3*a^2*b^6*d*e^2 - a^3*b^5*e^3)*x^3 + 3*(a*b^7 *d^3 - 3*a^2*b^6*d^2*e + 3*a^3*b^5*d*e^2 - a^4*b^4*e^3)*x^2 + 3*(a^2*b^6*d ^3 - 3*a^3*b^5*d^2*e + 3*a^4*b^4*d*e^2 - a^5*b^3*e^3)*x)]
Timed out. \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.28 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 8 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{3} - 3 \, \sqrt {e x + d} b^{2} d^{2} e^{3} + 8 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{4} + 6 \, \sqrt {e x + d} a b d e^{4} - 3 \, \sqrt {e x + d} a^{2} e^{5}}{24 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3}} \]
1/8*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^2 - 2*a*b^2*d *e + a^2*b*e^2)*sqrt(-b^2*d + a*b*e)) + 1/24*(3*(e*x + d)^(5/2)*b^2*e^3 - 8*(e*x + d)^(3/2)*b^2*d*e^3 - 3*sqrt(e*x + d)*b^2*d^2*e^3 + 8*(e*x + d)^(3 /2)*a*b*e^4 + 6*sqrt(e*x + d)*a*b*d*e^4 - 3*sqrt(e*x + d)*a^2*e^5)/((b^3*d ^2 - 2*a*b^2*d*e + a^2*b*e^2)*((e*x + d)*b - b*d + a*e)^3)
Time = 9.56 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {e^3\,{\left (d+e\,x\right )}^{3/2}}{3\,\left (a\,e-b\,d\right )}-\frac {e^3\,\sqrt {d+e\,x}}{8\,b}+\frac {b\,e^3\,{\left (d+e\,x\right )}^{5/2}}{8\,{\left (a\,e-b\,d\right )}^2}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2}+\frac {e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{8\,b^{3/2}\,{\left (a\,e-b\,d\right )}^{5/2}} \]
((e^3*(d + e*x)^(3/2))/(3*(a*e - b*d)) - (e^3*(d + e*x)^(1/2))/(8*b) + (b* e^3*(d + e*x)^(5/2))/(8*(a*e - b*d)^2))/((d + e*x)*(3*b^3*d^2 + 3*a^2*b*e^ 2 - 6*a*b^2*d*e) + b^3*(d + e*x)^3 - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^2 + a ^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2) + (e^3*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(8*b^(3/2)*(a*e - b*d)^(5/2))