Integrand size = 30, antiderivative size = 172 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 e \sqrt {d+e x}}{4 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{2 (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-3/4*e^2*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/(-a*e+b*d )^(5/2)/b^(1/2)/((b*x+a)^2)^(1/2)+3/4*e*(e*x+d)^(1/2)/(-a*e+b*d)^2/((b*x+a )^2)^(1/2)-1/2*(e*x+d)^(1/2)/(-a*e+b*d)/(b*x+a)/((b*x+a)^2)^(1/2)
Time = 0.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} \sqrt {-b d+a e} \sqrt {d+e x} (-2 b d+5 a e+3 b e x)+3 e^2 (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 \sqrt {b} (-b d+a e)^{5/2} (a+b x) \sqrt {(a+b x)^2}} \]
(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x]*(-2*b*d + 5*a*e + 3*b*e*x) + 3*e ^2*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*Sqrt [b]*(-(b*d) + a*e)^(5/2)*(a + b*x)*Sqrt[(a + b*x)^2])
Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1102, 27, 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.
| \(\displaystyle \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2} \sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {1}{b^3 (a+b x)^3 \sqrt {d+e x}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^3 \sqrt {d+e x}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {3 e \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {3 e \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {3 e \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-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))))/Sqrt[a^2 + 2* a*b*x + b^2*x^2]
3.18.16.3.1 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 + 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_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.76 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.18
| method | result | size |
| default | \(\frac {\left (3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{2} e^{2} x^{2}+6 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a b \,e^{2} x +3 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b e x +3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} e^{2}+5 \sqrt {e x +d}\, a e \sqrt {\left (a e -b d \right ) b}-2 \sqrt {e x +d}\, d b \sqrt {\left (a e -b d \right ) b}\right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(203\) |
1/4*(3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^2*e^2*x^2+6*arctan(b* (e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b*e^2*x+3*(e*x+d)^(1/2)*((a*e-b*d)*b) ^(1/2)*b*e*x+3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*e^2+5*(e*x+ d)^(1/2)*a*e*((a*e-b*d)*b)^(1/2)-2*(e*x+d)^(1/2)*d*b*((a*e-b*d)*b)^(1/2))* (b*x+a)/((a*e-b*d)*b)^(1/2)/(a*e-b*d)^2/((b*x+a)^2)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (119) = 238\).
Time = 0.32 (sec) , antiderivative size = 549, normalized size of antiderivative = 3.19 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, b^{3} d^{2} - 7 \, a b^{2} d e + 5 \, a^{2} b e^{2} - 3 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} + {\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \, {\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, b^{3} d^{2} - 7 \, a b^{2} d e + 5 \, a^{2} b e^{2} - 3 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} + {\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \, {\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}}\right ] \]
[1/8*(3*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*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*(2* b^3*d^2 - 7*a*b^2*d*e + 5*a^2*b*e^2 - 3*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4*d^3 - 3*a^3*b^3*d^2*e + 3*a^4*b^2*d*e^2 - a^5*b*e^3 + (b^6* d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*e^2 - a^3*b^3*e^3)*x^2 + 2*(a*b^5*d^3 - 3*a^2*b^4*d^2*e + 3*a^3*b^3*d*e^2 - a^4*b^2*e^3)*x), 1/4*(3*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*s qrt(e*x + d)/(b*e*x + b*d)) - (2*b^3*d^2 - 7*a*b^2*d*e + 5*a^2*b*e^2 - 3*( b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4*d^3 - 3*a^3*b^3*d^2*e + 3* a^4*b^2*d*e^2 - a^5*b*e^3 + (b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*e^2 - a ^3*b^3*e^3)*x^2 + 2*(a*b^5*d^3 - 3*a^2*b^4*d^2*e + 3*a^3*b^3*d*e^2 - a^4*b ^2*e^3)*x)]
\[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {d + e x} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} \sqrt {e x + d}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, e^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (e x + d\right )}^{\frac {3}{2}} b e^{2} - 5 \, \sqrt {e x + d} b d e^{2} + 5 \, \sqrt {e x + d} a e^{3}}{4 \, {\left (b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} \]
3/4*e^2*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^2*d^2*sgn(b*x + a ) - 2*a*b*d*e*sgn(b*x + a) + a^2*e^2*sgn(b*x + a))*sqrt(-b^2*d + a*b*e)) + 1/4*(3*(e*x + d)^(3/2)*b*e^2 - 5*sqrt(e*x + d)*b*d*e^2 + 5*sqrt(e*x + d)* a*e^3)/((b^2*d^2*sgn(b*x + a) - 2*a*b*d*e*sgn(b*x + a) + a^2*e^2*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^2)
Timed out. \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]