Integrand size = 15, antiderivative size = 143 \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\frac {3 a^4 \sqrt {x} \sqrt {a+b x}}{128 b^3}-\frac {a^3 x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a^2 x^{5/2} \sqrt {a+b x}}{80 b}+\frac {3}{40} a x^{7/2} \sqrt {a+b x}+\frac {1}{5} x^{7/2} (a+b x)^{3/2}-\frac {3 a^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{7/2}} \]
1/5*x^(7/2)*(b*x+a)^(3/2)-3/128*a^5*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2)) /b^(7/2)-1/64*a^3*x^(3/2)*(b*x+a)^(1/2)/b^2+1/80*a^2*x^(5/2)*(b*x+a)^(1/2) /b+3/40*a*x^(7/2)*(b*x+a)^(1/2)+3/128*a^4*x^(1/2)*(b*x+a)^(1/2)/b^3
Time = 0.43 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (15 a^4-10 a^3 b x+8 a^2 b^2 x^2+176 a b^3 x^3+128 b^4 x^4\right )+30 a^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{640 b^{7/2}} \]
(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(15*a^4 - 10*a^3*b*x + 8*a^2*b^2*x^2 + 176* a*b^3*x^3 + 128*b^4*x^4) + 30*a^5*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqr t[a + b*x])])/(640*b^(7/2))
Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {60, 60, 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.
| \(\displaystyle \int x^{5/2} (a+b x)^{3/2} \, dx\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{10} a \int x^{5/2} \sqrt {a+b x}dx+\frac {1}{5} x^{7/2} (a+b x)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{10} a \left (\frac {1}{8} a \int \frac {x^{5/2}}{\sqrt {a+b x}}dx+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )+\frac {1}{5} x^{7/2} (a+b x)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{10} a \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \int \frac {x^{3/2}}{\sqrt {a+b x}}dx}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )+\frac {1}{5} x^{7/2} (a+b x)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{10} a \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \int \frac {\sqrt {x}}{\sqrt {a+b x}}dx}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )+\frac {1}{5} x^{7/2} (a+b x)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{10} a \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx}{2 b}\right )}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )+\frac {1}{5} x^{7/2} (a+b x)^{3/2}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {3}{10} a \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{b}\right )}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )+\frac {1}{5} x^{7/2} (a+b x)^{3/2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{10} a \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}\right )}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )+\frac {1}{5} x^{7/2} (a+b x)^{3/2}\) |
(x^(7/2)*(a + b*x)^(3/2))/5 + (3*a*((x^(7/2)*Sqrt[a + b*x])/4 + (a*((x^(5/ 2)*Sqrt[a + b*x])/(3*b) - (5*a*((x^(3/2)*Sqrt[a + b*x])/(2*b) - (3*a*((Sqr t[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/b^(3/ 2)))/(4*b)))/(6*b)))/8))/10
3.6.21.3.1 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_)^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 = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(\frac {\left (128 b^{4} x^{4}+176 a \,b^{3} x^{3}+8 a^{2} b^{2} x^{2}-10 a^{3} b x +15 a^{4}\right ) \sqrt {x}\, \sqrt {b x +a}}{640 b^{3}}-\frac {3 a^{5} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{256 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +a}}\) | \(109\) |
| default | \(\frac {x^{\frac {5}{2}} \left (b x +a \right )^{\frac {5}{2}}}{5 b}-\frac {a \left (\frac {x^{\frac {3}{2}} \left (b x +a \right )^{\frac {5}{2}}}{4 b}-\frac {3 a \left (\frac {\sqrt {x}\, \left (b x +a \right )^{\frac {5}{2}}}{3 b}-\frac {a \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 a \left (\sqrt {x}\, \sqrt {b x +a}+\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b x +a}\, \sqrt {x}\, \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )}{2 b}\) | \(144\) |
1/640*(128*b^4*x^4+176*a*b^3*x^3+8*a^2*b^2*x^2-10*a^3*b*x+15*a^4)*x^(1/2)* (b*x+a)^(1/2)/b^3-3/256*a^5/b^(7/2)*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 = 0.23 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.29 \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\left [\frac {15 \, a^{5} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (128 \, b^{5} x^{4} + 176 \, a b^{4} x^{3} + 8 \, a^{2} b^{3} x^{2} - 10 \, a^{3} b^{2} x + 15 \, a^{4} b\right )} \sqrt {b x + a} \sqrt {x}}{1280 \, b^{4}}, \frac {15 \, a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (128 \, b^{5} x^{4} + 176 \, a b^{4} x^{3} + 8 \, a^{2} b^{3} x^{2} - 10 \, a^{3} b^{2} x + 15 \, a^{4} b\right )} \sqrt {b x + a} \sqrt {x}}{640 \, b^{4}}\right ] \]
[1/1280*(15*a^5*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(128*b^5*x^4 + 176*a*b^4*x^3 + 8*a^2*b^3*x^2 - 10*a^3*b^2*x + 15*a^4*b) *sqrt(b*x + a)*sqrt(x))/b^4, 1/640*(15*a^5*sqrt(-b)*arctan(sqrt(b*x + a)*s qrt(-b)/(b*sqrt(x))) + (128*b^5*x^4 + 176*a*b^4*x^3 + 8*a^2*b^3*x^2 - 10*a ^3*b^2*x + 15*a^4*b)*sqrt(b*x + a)*sqrt(x))/b^4]
Time = 80.46 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.24 \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\frac {3 a^{\frac {9}{2}} \sqrt {x}}{128 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {a^{\frac {7}{2}} x^{\frac {3}{2}}}{128 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {a^{\frac {5}{2}} x^{\frac {5}{2}}}{320 b \sqrt {1 + \frac {b x}{a}}} + \frac {23 a^{\frac {3}{2}} x^{\frac {7}{2}}}{80 \sqrt {1 + \frac {b x}{a}}} + \frac {19 \sqrt {a} b x^{\frac {9}{2}}}{40 \sqrt {1 + \frac {b x}{a}}} - \frac {3 a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{128 b^{\frac {7}{2}}} + \frac {b^{2} x^{\frac {11}{2}}}{5 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]
3*a**(9/2)*sqrt(x)/(128*b**3*sqrt(1 + b*x/a)) + a**(7/2)*x**(3/2)/(128*b** 2*sqrt(1 + b*x/a)) - a**(5/2)*x**(5/2)/(320*b*sqrt(1 + b*x/a)) + 23*a**(3/ 2)*x**(7/2)/(80*sqrt(1 + b*x/a)) + 19*sqrt(a)*b*x**(9/2)/(40*sqrt(1 + b*x/ a)) - 3*a**5*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(128*b**(7/2)) + b**2*x**(11/2 )/(5*sqrt(a)*sqrt(1 + b*x/a))
Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (103) = 206\).
Time = 0.31 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.48 \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\frac {3 \, a^{5} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{256 \, b^{\frac {7}{2}}} + \frac {\frac {15 \, \sqrt {b x + a} a^{5} b^{4}}{\sqrt {x}} - \frac {70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} b^{3}}{x^{\frac {3}{2}}} - \frac {128 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} b^{2}}{x^{\frac {5}{2}}} + \frac {70 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{5} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{5}}{x^{\frac {9}{2}}}}{640 \, {\left (b^{8} - \frac {5 \, {\left (b x + a\right )} b^{7}}{x} + \frac {10 \, {\left (b x + a\right )}^{2} b^{6}}{x^{2}} - \frac {10 \, {\left (b x + a\right )}^{3} b^{5}}{x^{3}} + \frac {5 \, {\left (b x + a\right )}^{4} b^{4}}{x^{4}} - \frac {{\left (b x + a\right )}^{5} b^{3}}{x^{5}}\right )}} \]
3/256*a^5*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/ sqrt(x)))/b^(7/2) + 1/640*(15*sqrt(b*x + a)*a^5*b^4/sqrt(x) - 70*(b*x + a) ^(3/2)*a^5*b^3/x^(3/2) - 128*(b*x + a)^(5/2)*a^5*b^2/x^(5/2) + 70*(b*x + a )^(7/2)*a^5*b/x^(7/2) - 15*(b*x + a)^(9/2)*a^5/x^(9/2))/(b^8 - 5*(b*x + a) *b^7/x + 10*(b*x + a)^2*b^6/x^2 - 10*(b*x + a)^3*b^5/x^3 + 5*(b*x + a)^4*b ^4/x^4 - (b*x + a)^5*b^3/x^5)
Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (103) = 206\).
Time = 236.80 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.49 \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\frac {3 \, {\left (\frac {315 \, a^{5} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {7}{2}}} + {\left (2 \, {\left (4 \, {\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {8 \, {\left (b x + a\right )}}{b^{4}} - \frac {41 \, a}{b^{4}}\right )} + \frac {171 \, a^{2}}{b^{4}}\right )} - \frac {745 \, a^{3}}{b^{4}}\right )} {\left (b x + a\right )} + \frac {965 \, a^{4}}{b^{4}}\right )} \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a}\right )} {\left | b \right |} + \frac {80 \, {\left (\frac {15 \, a^{3} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}} + \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} - \frac {13 \, a}{b^{2}}\right )} + \frac {33 \, a^{2}}{b^{2}}\right )}\right )} a^{2} {\left | b \right |}}{b^{2}} - \frac {20 \, {\left (\frac {105 \, a^{4} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {5}{2}}} - {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{3}} - \frac {25 \, a}{b^{3}}\right )} + \frac {163 \, a^{2}}{b^{3}}\right )} - \frac {279 \, a^{3}}{b^{3}}\right )} \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a}\right )} a {\left | b \right |}}{b}}{1920 \, b} \]
1/1920*(3*(315*a^5*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b )))/b^(7/2) + (2*(4*(b*x + a)*(2*(b*x + a)*(8*(b*x + a)/b^4 - 41*a/b^4) + 171*a^2/b^4) - 745*a^3/b^4)*(b*x + a) + 965*a^4/b^4)*sqrt((b*x + a)*b - a* b)*sqrt(b*x + a))*abs(b) + 80*(15*a^3*log(abs(-sqrt(b*x + a)*sqrt(b) + sqr t((b*x + a)*b - a*b)))/b^(3/2) + sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2* (b*x + a)*(4*(b*x + a)/b^2 - 13*a/b^2) + 33*a^2/b^2))*a^2*abs(b)/b^2 - 20* (105*a^4*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(5/2 ) - (2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 - 25*a/b^3) + 163*a^2/b^3) - 279*a^3/b^3)*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a))*a*abs(b)/b)/b
Timed out. \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\int x^{5/2}\,{\left (a+b\,x\right )}^{3/2} \,d x \]
Time = 0.00 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.80 \[ \int x^{5/2} (a+b x)^{3/2} \, dx=\frac {15 \sqrt {x}\, \sqrt {b x +a}\, a^{4} b -10 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b^{2} x +8 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{3} x^{2}+176 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{4} x^{3}+128 \sqrt {x}\, \sqrt {b x +a}\, b^{5} x^{4}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{5}}{640 b^{4}} \]