Integrand size = 30, antiderivative size = 119 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x)}{(b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 \sqrt {b} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-2*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*b^(1/2)/(-a*e+b *d)^(3/2)/((b*x+a)^2)^(1/2)+2*(b*x+a)/(-a*e+b*d)/(e*x+d)^(1/2)/((b*x+a)^2) ^(1/2)
Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (a+b x) \left (\sqrt {-b d+a e}+\sqrt {b} \sqrt {d+e x} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{(-b d+a e)^{3/2} \sqrt {(a+b x)^2} \sqrt {d+e x}} \]
(-2*(a + b*x)*(Sqrt[-(b*d) + a*e] + Sqrt[b]*Sqrt[d + e*x]*ArcTan[(Sqrt[b]* Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/((-(b*d) + a*e)^(3/2)*Sqrt[(a + b*x)^ 2]*Sqrt[d + e*x])
Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1102, 27, 61, 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.
\(\displaystyle \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {b (a+b x) \int \frac {1}{b (a+b x) (d+e x)^{3/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(a+b x) \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a+b x) \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sq rt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/Sqrt[a^2 + 2*a*b*x + b^ 2*x^2]
3.18.8.3.1 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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
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 = 2.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76
method | result | size |
default | \(-\frac {2 \left (b x +a \right ) \left (b \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {e x +d}+\sqrt {\left (a e -b d \right ) b}\right )}{\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right ) \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}}\) | \(90\) |
-2*(b*x+a)*(b*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*(e*x+d)^(1/2)+(( a*e-b*d)*b)^(1/2))/((b*x+a)^2)^(1/2)/(a*e-b*d)/(e*x+d)^(1/2)/((a*e-b*d)*b) ^(1/2)
Time = 0.27 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.80 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {{\left (e x + d\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, \sqrt {e x + d}}{b d^{2} - a d e + {\left (b d e - a e^{2}\right )} x}, -\frac {2 \, {\left ({\left (e x + d\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - \sqrt {e x + d}\right )}}{b d^{2} - a d e + {\left (b d e - a e^{2}\right )} x}\right ] \]
[-((e*x + d)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)* sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*sqrt(e*x + d))/(b*d^2 - a*d*e + (b*d*e - a*e^2)*x), -2*((e*x + d)*sqrt(-b/(b*d - a*e))*arctan(-(b* d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - sqrt(e*x + d) )/(b*d^2 - a*d*e + (b*d*e - a*e^2)*x)]
\[ \int \frac {1}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \sqrt {\left (a + b x\right )^{2}}}\, dx \]
\[ \int \frac {1}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {1}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=2 \, {\left (\frac {b \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d - a e\right )}} + \frac {1}{{\left (b d - a e\right )} \sqrt {e x + d}}\right )} \mathrm {sgn}\left (b x + a\right ) \]
2*(b*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*(b *d - a*e)) + 1/((b*d - a*e)*sqrt(e*x + d)))*sgn(b*x + a)
Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{3/2}} \,d x \]