Integrand size = 35, antiderivative size = 173 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (A b-a B) (a+b x) \sqrt {d+e x}}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{3/2}}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) \sqrt {b d-a e} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
2/3*B*(b*x+a)*(e*x+d)^(3/2)/b/e/((b*x+a)^2)^(1/2)-2*(A*b-B*a)*(b*x+a)*arct anh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*(-a*e+b*d)^(1/2)/b^(5/2)/((b*x +a)^2)^(1/2)+2*(A*b-B*a)*(b*x+a)*(e*x+d)^(1/2)/b^2/((b*x+a)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {b} \sqrt {d+e x} (3 A b e-3 a B e+b B (d+e x))+3 (-A b+a B) e \sqrt {-b d+a e} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{3 b^{5/2} e \sqrt {(a+b x)^2}} \]
(2*(a + b*x)*(Sqrt[b]*Sqrt[d + e*x]*(3*A*b*e - 3*a*B*e + b*B*(d + e*x)) + 3*(-(A*b) + a*B)*e*Sqrt[-(b*d) + a*e]*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[ -(b*d) + a*e]]))/(3*b^(5/2)*e*Sqrt[(a + b*x)^2])
Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.70, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {1187, 27, 90, 60, 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.
| \(\displaystyle \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b (a+b x) \int \frac {(A+B x) \sqrt {d+e x}}{b (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(A+B x) \sqrt {d+e x}}{a+b x}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((2*B*(d + e*x)^(3/2))/(3*b*e) + ((A*b - a*B)*((2*Sqrt[d + e*x] )/b - (2*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]) /b^(3/2)))/b))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.19.61.3.1 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 + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.77
| method | result | size |
| risch | \(\frac {2 \left (B b e x +3 A b e -3 B a e +B b d \right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{3 e \,b^{2} \left (b x +a \right )}-\frac {2 \left (A a b e -A d \,b^{2}-B e \,a^{2}+B a b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{2} \sqrt {\left (a e -b d \right ) b}\, \left (b x +a \right )}\) | \(133\) |
| default | \(\frac {2 \left (b x +a \right ) \left (-3 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a b \,e^{2}+3 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{2} d e +B \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b +3 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} e^{2}-3 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a b d e +3 A \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b e -3 B \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a e \right )}{3 \sqrt {\left (b x +a \right )^{2}}\, e \,b^{2} \sqrt {\left (a e -b d \right ) b}}\) | \(226\) |
2/3*(B*b*e*x+3*A*b*e-3*B*a*e+B*b*d)*(e*x+d)^(1/2)/e/b^2*((b*x+a)^2)^(1/2)/ (b*x+a)-2*(A*a*b*e-A*b^2*d-B*a^2*e+B*a*b*d)/b^2/((a*e-b*d)*b)^(1/2)*arctan (b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.43 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.22 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {3 \, {\left (B a - A b\right )} e \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (B b e x + B b d - 3 \, {\left (B a - A b\right )} e\right )} \sqrt {e x + d}}{3 \, b^{2} e}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} e \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (B b e x + B b d - 3 \, {\left (B a - A b\right )} e\right )} \sqrt {e x + d}\right )}}{3 \, b^{2} e}\right ] \]
[-1/3*(3*(B*a - A*b)*e*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sq rt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(B*b*e*x + B*b*d - 3*(B* a - A*b)*e)*sqrt(e*x + d))/(b^2*e), 2/3*(3*(B*a - A*b)*e*sqrt(-(b*d - a*e) /b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (B*b*e*x + B*b*d - 3*(B*a - A*b)*e)*sqrt(e*x + d))/(b^2*e)]
\[ \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {d + e x}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \]
\[ \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (B x + A\right )} \sqrt {e x + d}}{\sqrt {{\left (b x + a\right )}^{2}}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a b d \mathrm {sgn}\left (b x + a\right ) - A b^{2} d \mathrm {sgn}\left (b x + a\right ) - B a^{2} e \mathrm {sgn}\left (b x + a\right ) + A a b e \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B b^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, \sqrt {e x + d} B a b e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, \sqrt {e x + d} A b^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, b^{3} e^{3}} \]
-2*(B*a*b*d*sgn(b*x + a) - A*b^2*d*sgn(b*x + a) - B*a^2*e*sgn(b*x + a) + A *a*b*e*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^2) + 2/3*((e*x + d)^(3/2)*B*b^2*e^2*sgn(b*x + a) - 3*sqrt( e*x + d)*B*a*b*e^3*sgn(b*x + a) + 3*sqrt(e*x + d)*A*b^2*e^3*sgn(b*x + a))/ (b^3*e^3)
Timed out. \[ \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (A+B\,x\right )\,\sqrt {d+e\,x}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]