Integrand size = 31, antiderivative size = 286 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 a^3 (A b-a B) \sqrt {x} (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 (A b-a B) x^{3/2} (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a (A b-a B) x^{5/2} (a+b x)}{5 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{7/2} (a+b x)}{7 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{9/2} (a+b x)}{9 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{7/2} (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
2/3*a^2*(A*b-B*a)*x^(3/2)*(b*x+a)/b^4/((b*x+a)^2)^(1/2)-2/5*a*(A*b-B*a)*x^ (5/2)*(b*x+a)/b^3/((b*x+a)^2)^(1/2)+2/7*(A*b-B*a)*x^(7/2)*(b*x+a)/b^2/((b* x+a)^2)^(1/2)+2/9*B*x^(9/2)*(b*x+a)/b/((b*x+a)^2)^(1/2)+2*a^(7/2)*(A*b-B*a )*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(11/2)/((b*x+a)^2)^(1/2)-2*a^3 *(A*b-B*a)*(b*x+a)*x^(1/2)/b^5/((b*x+a)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.49 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {b} \sqrt {x} \left (315 a^4 B-105 a^3 b (3 A+B x)+21 a^2 b^2 x (5 A+3 B x)-9 a b^3 x^2 (7 A+5 B x)+5 b^4 x^3 (9 A+7 B x)\right )-315 a^{7/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{315 b^{11/2} \sqrt {(a+b x)^2}} \]
(2*(a + b*x)*(Sqrt[b]*Sqrt[x]*(315*a^4*B - 105*a^3*b*(3*A + B*x) + 21*a^2* b^2*x*(5*A + 3*B*x) - 9*a*b^3*x^2*(7*A + 5*B*x) + 5*b^4*x^3*(9*A + 7*B*x)) - 315*a^(7/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(315*b^( 11/2)*Sqrt[(a + b*x)^2])
Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.52, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {1187, 27, 90, 60, 60, 60, 60, 73, 218}
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.
| \(\displaystyle \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b (a+b x) \int \frac {x^{7/2} (A+B x)}{b (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {x^{7/2} (A+B x)}{a+b x}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \int \frac {x^{7/2}}{a+b x}dx}{b}+\frac {2 B x^{9/2}}{9 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \int \frac {x^{5/2}}{a+b x}dx}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \int \frac {x^{3/2}}{a+b x}dx}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((2*B*x^(9/2))/(9*b) + ((A*b - a*B)*((2*x^(7/2))/(7*b) - (a*((2 *x^(5/2))/(5*b) - (a*((2*x^(3/2))/(3*b) - (a*((2*Sqrt[x])/b - (2*Sqrt[a]*A rcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/b))/b))/b))/b))/Sqrt[a^2 + 2*a *b*x + b^2*x^2]
3.9.9.3.1 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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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 = 0.14 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.55
| method | result | size |
| risch | \(-\frac {2 \left (-35 b^{4} B \,x^{4}-45 A \,b^{4} x^{3}+45 B a \,b^{3} x^{3}+63 A a \,b^{3} x^{2}-63 B \,a^{2} b^{2} x^{2}-105 A \,a^{2} b^{2} x +105 B \,a^{3} b x +315 A \,a^{3} b -315 B \,a^{4}\right ) \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{315 b^{5} \left (b x +a \right )}+\frac {2 a^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{5} \sqrt {b a}\, \left (b x +a \right )}\) | \(156\) |
| default | \(\frac {2 \left (b x +a \right ) \left (35 B \,x^{\frac {9}{2}} \sqrt {b a}\, b^{4}+45 A \,x^{\frac {7}{2}} \sqrt {b a}\, b^{4}-45 B \,x^{\frac {7}{2}} \sqrt {b a}\, a \,b^{3}-63 A \,x^{\frac {5}{2}} \sqrt {b a}\, a \,b^{3}+63 B \,x^{\frac {5}{2}} \sqrt {b a}\, a^{2} b^{2}+105 A \,x^{\frac {3}{2}} \sqrt {b a}\, a^{2} b^{2}-105 B \,x^{\frac {3}{2}} \sqrt {b a}\, a^{3} b -315 A \sqrt {x}\, \sqrt {b a}\, a^{3} b +315 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{4} b +315 B \sqrt {x}\, \sqrt {b a}\, a^{4}-315 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{5}\right )}{315 \sqrt {\left (b x +a \right )^{2}}\, b^{5} \sqrt {b a}}\) | \(197\) |
-2/315*(-35*B*b^4*x^4-45*A*b^4*x^3+45*B*a*b^3*x^3+63*A*a*b^3*x^2-63*B*a^2* b^2*x^2-105*A*a^2*b^2*x+105*B*a^3*b*x+315*A*a^3*b-315*B*a^4)*x^(1/2)/b^5*( (b*x+a)^2)^(1/2)/(b*x+a)+2*a^4*(A*b-B*a)/b^5/(b*a)^(1/2)*arctan(b*x^(1/2)/ (b*a)^(1/2))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.30 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.97 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{315 \, b^{5}}, -\frac {2 \, {\left (315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}\right )}}{315 \, b^{5}}\right ] \]
[-1/315*(315*(B*a^4 - A*a^3*b)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b ) - a)/(b*x + a)) - 2*(35*B*b^4*x^4 + 315*B*a^4 - 315*A*a^3*b - 45*(B*a*b^ 3 - A*b^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2)* x)*sqrt(x))/b^5, -2/315*(315*(B*a^4 - A*a^3*b)*sqrt(a/b)*arctan(b*sqrt(x)* sqrt(a/b)/a) - (35*B*b^4*x^4 + 315*B*a^4 - 315*A*a^3*b - 45*(B*a*b^3 - A*b ^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2)*x)*sqrt (x))/b^5]
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Timed out} \]
Time = 0.32 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.90 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {10 \, {\left (7 \, B b^{4} x^{2} + 9 \, B a b^{3} x\right )} x^{\frac {7}{2}} - 2 \, {\left (5 \, {\left (11 \, B a b^{3} - 9 \, A b^{4}\right )} x^{2} + 9 \, {\left (9 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 6 \, {\left (3 \, {\left (11 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{2} + 7 \, {\left (9 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + 21 \, {\left (3 \, {\left (11 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 5 \, {\left (9 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {x}}{315 \, {\left (b^{5} x + a b^{4}\right )}} - \frac {2 \, {\left (B a^{5} - A a^{4} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} - \frac {{\left (11 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {x}}{3 \, b^{5}} \]
1/315*(10*(7*B*b^4*x^2 + 9*B*a*b^3*x)*x^(7/2) - 2*(5*(11*B*a*b^3 - 9*A*b^4 )*x^2 + 9*(9*B*a^2*b^2 - 7*A*a*b^3)*x)*x^(5/2) + 6*(3*(11*B*a^2*b^2 - 9*A* a*b^3)*x^2 + 7*(9*B*a^3*b - 7*A*a^2*b^2)*x)*x^(3/2) + 21*(3*(11*B*a^3*b - 9*A*a^2*b^2)*x^2 + 5*(9*B*a^4 - 7*A*a^3*b)*x)*sqrt(x))/(b^5*x + a*b^4) - 2 *(B*a^5 - A*a^4*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/3*((11* B*a^3*b - 9*A*a^2*b^2)*x^(3/2) - 6*(B*a^4 - A*a^3*b)*sqrt(x))/b^5
Time = 0.26 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.72 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a^{5} \mathrm {sgn}\left (b x + a\right ) - A a^{4} b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} + \frac {2 \, {\left (35 \, B b^{8} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) - 45 \, B a b^{7} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + 45 \, A b^{8} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + 63 \, B a^{2} b^{6} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) - 63 \, A a b^{7} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) - 105 \, B a^{3} b^{5} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 105 \, A a^{2} b^{6} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 315 \, B a^{4} b^{4} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - 315 \, A a^{3} b^{5} \sqrt {x} \mathrm {sgn}\left (b x + a\right )\right )}}{315 \, b^{9}} \]
-2*(B*a^5*sgn(b*x + a) - A*a^4*b*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b)) /(sqrt(a*b)*b^5) + 2/315*(35*B*b^8*x^(9/2)*sgn(b*x + a) - 45*B*a*b^7*x^(7/ 2)*sgn(b*x + a) + 45*A*b^8*x^(7/2)*sgn(b*x + a) + 63*B*a^2*b^6*x^(5/2)*sgn (b*x + a) - 63*A*a*b^7*x^(5/2)*sgn(b*x + a) - 105*B*a^3*b^5*x^(3/2)*sgn(b* x + a) + 105*A*a^2*b^6*x^(3/2)*sgn(b*x + a) + 315*B*a^4*b^4*sqrt(x)*sgn(b* x + a) - 315*A*a^3*b^5*sqrt(x)*sgn(b*x + a))/b^9
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]