Integrand size = 35, antiderivative size = 138 \[ \int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (B d-A e) (a+b x)}{e (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-2*(A*b-B*a)*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/(-a*e +b*d)^(3/2)/b^(1/2)/((b*x+a)^2)^(1/2)-2*(-A*e+B*d)*(b*x+a)/e/(-a*e+b*d)/(e *x+d)^(1/2)/((b*x+a)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (a+b x) \left (\sqrt {b} \sqrt {-b d+a e} (-B d+A e)+(A b-a B) e \sqrt {d+e x} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{\sqrt {b} e (-b d+a e)^{3/2} \sqrt {(a+b x)^2} \sqrt {d+e x}} \]
(-2*(a + b*x)*(Sqrt[b]*Sqrt[-(b*d) + a*e]*(-(B*d) + A*e) + (A*b - a*B)*e*S qrt[d + e*x]*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/(Sqrt[b] *e*(-(b*d) + a*e)^(3/2)*Sqrt[(a + b*x)^2]*Sqrt[d + e*x])
Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1187, 27, 87, 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.
\(\displaystyle \int \frac {A+B x}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b (a+b x) (d+e x)^{3/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{(a+b x) (d+e x)^{3/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}-\frac {2 (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (\frac {2 (A b-a B) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}-\frac {2 (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a+b x) \left (-\frac {2 (A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {2 (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((-2*(B*d - A*e))/(e*(b*d - a*e)*Sqrt[d + e*x]) - (2*(A*b - a*B )*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^( 3/2))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.19.63.3.1 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] :> 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(d + e*x)^m*(f + g*x)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.07
method | result | size |
default | \(-\frac {2 \left (b x +a \right ) \left (A \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b e -B \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a e +A \sqrt {\left (a e -b d \right ) b}\, e -B \sqrt {\left (a e -b d \right ) b}\, d \right )}{\sqrt {\left (b x +a \right )^{2}}\, e \left (a e -b d \right ) \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}}\) | \(148\) |
-2*(b*x+a)*(A*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b* e-B*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*e+A*((a*e- b*d)*b)^(1/2)*e-B*((a*e-b*d)*b)^(1/2)*d)/((b*x+a)^2)^(1/2)/e/(a*e-b*d)/(e* x+d)^(1/2)/((a*e-b*d)*b)^(1/2)
Time = 0.40 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.63 \[ \int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [\frac {{\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (B b^{2} d^{2} + A a b e^{2} - {\left (B a b + A b^{2}\right )} d e\right )} \sqrt {e x + d}}{b^{3} d^{3} e - 2 \, a b^{2} d^{2} e^{2} + a^{2} b d e^{3} + {\left (b^{3} d^{2} e^{2} - 2 \, a b^{2} d e^{3} + a^{2} b e^{4}\right )} x}, -\frac {2 \, {\left ({\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (B b^{2} d^{2} + A a b e^{2} - {\left (B a b + A b^{2}\right )} d e\right )} \sqrt {e x + d}\right )}}{b^{3} d^{3} e - 2 \, a b^{2} d^{2} e^{2} + a^{2} b d e^{3} + {\left (b^{3} d^{2} e^{2} - 2 \, a b^{2} d e^{3} + a^{2} b e^{4}\right )} x}\right ] \]
[(((B*a - A*b)*e^2*x + (B*a - A*b)*d*e)*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*(B*b^2*d^ 2 + A*a*b*e^2 - (B*a*b + A*b^2)*d*e)*sqrt(e*x + d))/(b^3*d^3*e - 2*a*b^2*d ^2*e^2 + a^2*b*d*e^3 + (b^3*d^2*e^2 - 2*a*b^2*d*e^3 + a^2*b*e^4)*x), -2*(( (B*a - A*b)*e^2*x + (B*a - A*b)*d*e)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2 *d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (B*b^2*d^2 + A*a*b*e^2 - (B*a*b + A*b^2)*d*e)*sqrt(e*x + d))/(b^3*d^3*e - 2*a*b^2*d^2*e^2 + a^2*b*d*e^3 + (b^3*d^2*e^2 - 2*a*b^2*d*e^3 + a^2*b*e^4)*x)]
\[ \int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A + B x}{\left (d + e x\right )^{\frac {3}{2}} \sqrt {\left (a + b x\right )^{2}}}\, dx \]
\[ \int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {B x + A}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a \mathrm {sgn}\left (b x + a\right ) - A b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d - a e\right )}} - \frac {2 \, {\left (B d \mathrm {sgn}\left (b x + a\right ) - A e \mathrm {sgn}\left (b x + a\right )\right )}}{{\left (b d e - a e^{2}\right )} \sqrt {e x + d}} \]
-2*(B*a*sgn(b*x + a) - A*b*sgn(b*x + a))*arctan(sqrt(e*x + d)*b/sqrt(-b^2* d + a*b*e))/(sqrt(-b^2*d + a*b*e)*(b*d - a*e)) - 2*(B*d*sgn(b*x + a) - A*e *sgn(b*x + a))/((b*d*e - a*e^2)*sqrt(e*x + d))
Timed out. \[ \int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{3/2}} \,d x \]