Integrand size = 20, antiderivative size = 159 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx=\frac {5 a^2 (8 A b-a B) \sqrt {x} \sqrt {a+b x}}{64 b}+\frac {5 a (8 A b-a B) \sqrt {x} (a+b x)^{3/2}}{96 b}+\frac {(8 A b-a B) \sqrt {x} (a+b x)^{5/2}}{24 b}+\frac {B \sqrt {x} (a+b x)^{7/2}}{4 b}+\frac {5 a^3 (8 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{3/2}} \]
5/64*a^3*(8*A*b-B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(3/2)+5/96*a *(8*A*b-B*a)*(b*x+a)^(3/2)*x^(1/2)/b+1/24*(8*A*b-B*a)*(b*x+a)^(5/2)*x^(1/2 )/b+1/4*B*(b*x+a)^(7/2)*x^(1/2)/b+5/64*a^2*(8*A*b-B*a)*x^(1/2)*(b*x+a)^(1/ 2)/b
Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (15 a^3 B+16 b^3 x^2 (4 A+3 B x)+8 a b^2 x (26 A+17 B x)+2 a^2 b (132 A+59 B x)\right )+15 a^3 (-8 A b+a B) \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{192 b^{3/2}} \]
(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(15*a^3*B + 16*b^3*x^2*(4*A + 3*B*x) + 8*a* b^2*x*(26*A + 17*B*x) + 2*a^2*b*(132*A + 59*B*x)) + 15*a^3*(-8*A*b + a*B)* Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(192*b^(3/2))
Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.82, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {90, 60, 60, 60, 65, 219}
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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(8 A b-a B) \int \frac {(a+b x)^{5/2}}{\sqrt {x}}dx}{8 b}+\frac {B \sqrt {x} (a+b x)^{7/2}}{4 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \int \frac {(a+b x)^{3/2}}{\sqrt {x}}dx+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}\right )}{8 b}+\frac {B \sqrt {x} (a+b x)^{7/2}}{4 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \int \frac {\sqrt {a+b x}}{\sqrt {x}}dx+\frac {1}{2} \sqrt {x} (a+b x)^{3/2}\right )+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}\right )}{8 b}+\frac {B \sqrt {x} (a+b x)^{7/2}}{4 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx+\sqrt {x} \sqrt {a+b x}\right )+\frac {1}{2} \sqrt {x} (a+b x)^{3/2}\right )+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}\right )}{8 b}+\frac {B \sqrt {x} (a+b x)^{7/2}}{4 b}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}+\sqrt {x} \sqrt {a+b x}\right )+\frac {1}{2} \sqrt {x} (a+b x)^{3/2}\right )+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}\right )}{8 b}+\frac {B \sqrt {x} (a+b x)^{7/2}}{4 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}}+\sqrt {x} \sqrt {a+b x}\right )+\frac {1}{2} \sqrt {x} (a+b x)^{3/2}\right )+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}\right )}{8 b}+\frac {B \sqrt {x} (a+b x)^{7/2}}{4 b}\) |
(B*Sqrt[x]*(a + b*x)^(7/2))/(4*b) + ((8*A*b - a*B)*((Sqrt[x]*(a + b*x)^(5/ 2))/3 + (5*a*((Sqrt[x]*(a + b*x)^(3/2))/2 + (3*a*(Sqrt[x]*Sqrt[a + b*x] + (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/Sqrt[b]))/4))/6))/(8*b)
3.6.4.3.1 Defintions of rubi rules used
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[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Time = 1.44 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85
| method | result | size |
| risch | \(\frac {\left (48 b^{3} B \,x^{3}+64 A \,b^{3} x^{2}+136 B a \,b^{2} x^{2}+208 a \,b^{2} A x +118 a^{2} b B x +264 a^{2} b A +15 a^{3} B \right ) \sqrt {x}\, \sqrt {b x +a}}{192 b}+\frac {5 a^{3} \left (8 A b -B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{128 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\) | \(135\) |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {x}\, \left (96 B \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}+128 A \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}+272 B a \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+416 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a x +236 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2} x +120 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b +528 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2}-15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4}+30 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{3}\right )}{384 b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}}\) | \(218\) |
1/192/b*(48*B*b^3*x^3+64*A*b^3*x^2+136*B*a*b^2*x^2+208*A*a*b^2*x+118*B*a^2 *b*x+264*A*a^2*b+15*B*a^3)*x^(1/2)*(b*x+a)^(1/2)+5/128*a^3/b^(3/2)*(8*A*b- B*a)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/( b*x+a)^(1/2)
Time = 0.24 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx=\left [-\frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (48 \, B b^{4} x^{3} + 15 \, B a^{3} b + 264 \, A a^{2} b^{2} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{2}}, \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (48 \, B b^{4} x^{3} + 15 \, B a^{3} b + 264 \, A a^{2} b^{2} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{2}}\right ] \]
[-1/384*(15*(B*a^4 - 8*A*a^3*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b )*sqrt(x) + a) - 2*(48*B*b^4*x^3 + 15*B*a^3*b + 264*A*a^2*b^2 + 8*(17*B*a* b^3 + 8*A*b^4)*x^2 + 2*(59*B*a^2*b^2 + 104*A*a*b^3)*x)*sqrt(b*x + a)*sqrt( x))/b^2, 1/192*(15*(B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt( -b)/(b*sqrt(x))) + (48*B*b^4*x^3 + 15*B*a^3*b + 264*A*a^2*b^2 + 8*(17*B*a* b^3 + 8*A*b^4)*x^2 + 2*(59*B*a^2*b^2 + 104*A*a*b^3)*x)*sqrt(b*x + a)*sqrt( x))/b^2]
Time = 33.77 (sec) , antiderivative size = 541, normalized size of antiderivative = 3.40 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx=A a^{\frac {5}{2}} \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {A a^{\frac {5}{2}} \sqrt {x}}{8 \sqrt {1 + \frac {b x}{a}}} - \frac {A a^{\frac {3}{2}} b x^{\frac {3}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} + \frac {5 A \sqrt {a} b^{2} x^{\frac {5}{2}}}{12 \sqrt {1 + \frac {b x}{a}}} + \frac {9 A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 \sqrt {b}} + 4 A a b \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) + \frac {A b^{3} x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} - \frac {11 B a^{\frac {7}{2}} \sqrt {x}}{64 b \sqrt {1 + \frac {b x}{a}}} - \frac {11 B a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 \sqrt {1 + \frac {b x}{a}}} + \frac {79 B a^{\frac {3}{2}} b x^{\frac {5}{2}}}{96 \sqrt {1 + \frac {b x}{a}}} + \frac {23 B \sqrt {a} b^{2} x^{\frac {7}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} + \frac {11 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {3}{2}}} + 2 B a^{2} \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) + \frac {B b^{3} x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]
A*a**(5/2)*sqrt(x)*sqrt(1 + b*x/a) - A*a**(5/2)*sqrt(x)/(8*sqrt(1 + b*x/a) ) - A*a**(3/2)*b*x**(3/2)/(24*sqrt(1 + b*x/a)) + 5*A*sqrt(a)*b**2*x**(5/2) /(12*sqrt(1 + b*x/a)) + 9*A*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*sqrt(b) ) + 4*A*a*b*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b* sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(8*b) + a*sqrt(x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0)), (s qrt(a)*x**(3/2)/3, True)) + A*b**3*x**(7/2)/(3*sqrt(a)*sqrt(1 + b*x/a)) - 11*B*a**(7/2)*sqrt(x)/(64*b*sqrt(1 + b*x/a)) - 11*B*a**(5/2)*x**(3/2)/(192 *sqrt(1 + b*x/a)) + 79*B*a**(3/2)*b*x**(5/2)/(96*sqrt(1 + b*x/a)) + 23*B*s qrt(a)*b**2*x**(7/2)/(24*sqrt(1 + b*x/a)) + 11*B*a**4*asinh(sqrt(b)*sqrt(x )/sqrt(a))/(64*b**(3/2)) + 2*B*a**2*Piecewise((-a**2*Piecewise((log(2*sqrt (b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x)) /sqrt(b*x), True))/(8*b) + a*sqrt(x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0)), (sqrt(a)*x**(3/2)/3, True)) + B*b**3*x**(9/2)/(4*sqr t(a)*sqrt(1 + b*x/a))
Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (121) = 242\).
Time = 0.21 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.86 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx=\frac {1}{4} \, \sqrt {b x^{2} + a x} B b^{2} x^{3} - \frac {7}{24} \, \sqrt {b x^{2} + a x} B a b x^{2} + \frac {35}{96} \, \sqrt {b x^{2} + a x} B a^{2} x + \frac {35 \, B a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {3}{2}}} + \frac {A a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} - \frac {35 \, \sqrt {b x^{2} + a x} B a^{3}}{64 \, b} + \frac {{\left (3 \, B a b^{2} + A b^{3}\right )} \sqrt {b x^{2} + a x} x^{2}}{3 \, b} - \frac {5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \sqrt {b x^{2} + a x} a x}{12 \, b^{2}} + \frac {3 \, {\left (B a^{2} b + A a b^{2}\right )} \sqrt {b x^{2} + a x} x}{2 \, b} - \frac {5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {7}{2}}} + \frac {9 \, {\left (B a^{2} b + A a b^{2}\right )} a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {5}{2}}} - \frac {{\left (B a^{3} + 3 \, A a^{2} b\right )} a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {3}{2}}} + \frac {5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \sqrt {b x^{2} + a x} a^{2}}{8 \, b^{3}} - \frac {9 \, {\left (B a^{2} b + A a b^{2}\right )} \sqrt {b x^{2} + a x} a}{4 \, b^{2}} + \frac {{\left (B a^{3} + 3 \, A a^{2} b\right )} \sqrt {b x^{2} + a x}}{b} \]
1/4*sqrt(b*x^2 + a*x)*B*b^2*x^3 - 7/24*sqrt(b*x^2 + a*x)*B*a*b*x^2 + 35/96 *sqrt(b*x^2 + a*x)*B*a^2*x + 35/128*B*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a *x)*sqrt(b))/b^(3/2) + A*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/ sqrt(b) - 35/64*sqrt(b*x^2 + a*x)*B*a^3/b + 1/3*(3*B*a*b^2 + A*b^3)*sqrt(b *x^2 + a*x)*x^2/b - 5/12*(3*B*a*b^2 + A*b^3)*sqrt(b*x^2 + a*x)*a*x/b^2 + 3 /2*(B*a^2*b + A*a*b^2)*sqrt(b*x^2 + a*x)*x/b - 5/16*(3*B*a*b^2 + A*b^3)*a^ 3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 9/8*(B*a^2*b + A* a*b^2)*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 1/2*(B*a ^3 + 3*A*a^2*b)*a*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) + 5 /8*(3*B*a*b^2 + A*b^3)*sqrt(b*x^2 + a*x)*a^2/b^3 - 9/4*(B*a^2*b + A*a*b^2) *sqrt(b*x^2 + a*x)*a/b^2 + (B*a^3 + 3*A*a^2*b)*sqrt(b*x^2 + a*x)/b
Time = 76.38 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx=\frac {{\left ({\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )} B}{b^{2}} - \frac {B a b^{2} - 8 \, A b^{3}}{b^{4}}\right )} - \frac {5 \, {\left (B a^{2} b^{2} - 8 \, A a b^{3}\right )}}{b^{4}}\right )} - \frac {15 \, {\left (B a^{3} b^{2} - 8 \, A a^{2} b^{3}\right )}}{b^{4}}\right )} \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} + \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}}\right )} b}{192 \, {\left | b \right |}} \]
1/192*((2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)*B/b^2 - (B*a*b^2 - 8*A*b^3)/ b^4) - 5*(B*a^2*b^2 - 8*A*a*b^3)/b^4) - 15*(B*a^3*b^2 - 8*A*a^2*b^3)/b^4)* sqrt((b*x + a)*b - a*b)*sqrt(b*x + a) + 15*(B*a^4 - 8*A*a^3*b)*log(abs(-sq rt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(3/2))*b/abs(b)
Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{\sqrt {x}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{\sqrt {x}} \,d x \]