Integrand size = 35, antiderivative size = 230 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) (b d-a e)^{3/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
2/3*(A*b-B*a)*(b*x+a)*(e*x+d)^(3/2)/b^2/((b*x+a)^2)^(1/2)+2/5*B*(b*x+a)*(e *x+d)^(5/2)/b/e/((b*x+a)^2)^(1/2)-2*(A*b-B*a)*(-a*e+b*d)^(3/2)*(b*x+a)*arc tanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)/((b*x+a)^2)^(1/2)+2*( A*b-B*a)*(-a*e+b*d)*(b*x+a)*(e*x+d)^(1/2)/b^3/((b*x+a)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.64 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\frac {\sqrt {b} \sqrt {d+e x} \left (15 a^2 B e^2-5 a b e (4 B d+3 A e+B e x)+b^2 \left (3 B (d+e x)^2+5 A e (4 d+e x)\right )\right )}{e}+15 (A b-a B) (-b d+a e)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{15 b^{7/2} \sqrt {(a+b x)^2}} \]
(2*(a + b*x)*((Sqrt[b]*Sqrt[d + e*x]*(15*a^2*B*e^2 - 5*a*b*e*(4*B*d + 3*A* e + B*e*x) + b^2*(3*B*(d + e*x)^2 + 5*A*e*(4*d + e*x))))/e + 15*(A*b - a*B )*(-(b*d) + a*e)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]) )/(15*b^(7/2)*Sqrt[(a + b*x)^2])
Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.65, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1187, 27, 90, 60, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b (a+b x) \int \frac {(A+B x) (d+e x)^{3/2}}{b (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((2*B*(d + e*x)^(5/2))/(5*b*e) + ((A*b - a*B)*((2*(d + e*x)^(3/ 2))/(3*b) + ((b*d - a*e)*((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))/b))/Sqrt[a^2 + 2* a*b*x + b^2*x^2]
3.19.60.3.1 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 = 225, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(-\frac {2 \left (-3 B \,b^{2} e^{2} x^{2}-5 A \,b^{2} e^{2} x +5 B a b \,e^{2} x -6 B \,b^{2} d e x +15 A a b \,e^{2}-20 A \,b^{2} d e -15 a^{2} B \,e^{2}+20 B a b d e -3 B \,b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 e \,b^{3} \left (b x +a \right )}+\frac {2 \left (A \,a^{2} b \,e^{2}-2 A \,b^{2} d e a +A \,d^{2} b^{3}-B \,e^{2} a^{3}+2 B \,a^{2} b d e -B \,b^{2} d^{2} a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{3} \sqrt {\left (a e -b d \right ) b}\, \left (b x +a \right )}\) | \(225\) |
| default | \(\frac {2 \left (b x +a \right ) \left (3 B \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2}+5 A \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2} e +15 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b \,e^{3}-30 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} d \,e^{2}+15 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} d^{2} e -5 B \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a b e -15 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} e^{3}+30 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b d \,e^{2}-15 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} d^{2} e -15 A \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2}+15 A \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{2} d e +15 B \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}-15 B \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a b d e \right )}{15 \sqrt {\left (b x +a \right )^{2}}\, e \,b^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(414\) |
-2/15/e*(-3*B*b^2*e^2*x^2-5*A*b^2*e^2*x+5*B*a*b*e^2*x-6*B*b^2*d*e*x+15*A*a *b*e^2-20*A*b^2*d*e-15*B*a^2*e^2+20*B*a*b*d*e-3*B*b^2*d^2)*(e*x+d)^(1/2)/b ^3*((b*x+a)^2)^(1/2)/(b*x+a)+2*(A*a^2*b*e^2-2*A*a*b^2*d*e+A*b^3*d^2-B*a^3* e^2+2*B*a^2*b*d*e-B*a*b^2*d^2)/b^3/((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 = 0.35 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.62 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3} e}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3} e}\right ] \]
[-1/15*(15*((B*a*b - A*b^2)*d*e - (B*a^2 - A*a*b)*e^2)*sqrt((b*d - a*e)/b) *log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(3*B*b^2*e^2*x^2 + 3*B*b^2*d^2 - 20*(B*a*b - A*b^2)*d*e + 15*(B*a^ 2 - A*a*b)*e^2 + (6*B*b^2*d*e - 5*(B*a*b - A*b^2)*e^2)*x)*sqrt(e*x + d))/( b^3*e), 2/15*(15*((B*a*b - A*b^2)*d*e - (B*a^2 - A*a*b)*e^2)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (3*B*b ^2*e^2*x^2 + 3*B*b^2*d^2 - 20*(B*a*b - A*b^2)*d*e + 15*(B*a^2 - A*a*b)*e^2 + (6*B*b^2*d*e - 5*(B*a*b - A*b^2)*e^2)*x)*sqrt(e*x + d))/(b^3*e)]
\[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{\frac {3}{2}}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \]
\[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{\sqrt {{\left (b x + a\right )}^{2}}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, B a^{2} b d e \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + B a^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{3} e^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{4} e^{5} \mathrm {sgn}\left (b x + a\right ) - 15 \, \sqrt {e x + d} B a b^{3} d e^{5} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {e x + d} A b^{4} d e^{5} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {e x + d} B a^{2} b^{2} e^{6} \mathrm {sgn}\left (b x + a\right ) - 15 \, \sqrt {e x + d} A a b^{3} e^{6} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, b^{5} e^{5}} \]
-2*(B*a*b^2*d^2*sgn(b*x + a) - A*b^3*d^2*sgn(b*x + a) - 2*B*a^2*b*d*e*sgn( b*x + a) + 2*A*a*b^2*d*e*sgn(b*x + a) + B*a^3*e^2*sgn(b*x + a) - A*a^2*b*e ^2*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^3) + 2/15*(3*(e*x + d)^(5/2)*B*b^4*e^4*sgn(b*x + a) - 5*(e*x + d)^(3/2)*B*a*b^3*e^5*sgn(b*x + a) + 5*(e*x + d)^(3/2)*A*b^4*e^5*sgn(b*x + a) - 15*sqrt(e*x + d)*B*a*b^3*d*e^5*sgn(b*x + a) + 15*sqrt(e*x + d)*A*b^4 *d*e^5*sgn(b*x + a) + 15*sqrt(e*x + d)*B*a^2*b^2*e^6*sgn(b*x + a) - 15*sqr t(e*x + d)*A*a*b^3*e^6*sgn(b*x + a))/(b^5*e^5)
Timed out. \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]