Integrand size = 20, antiderivative size = 117 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\frac {b (2 A b+3 a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 (2 A b+3 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}}+\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]
-2/3*A*(b*x+a)^(5/2)/a/x^(3/2)+(2*A*b+3*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+ a)^(1/2))*b^(1/2)-2/3*(2*A*b+3*B*a)*(b*x+a)^(3/2)/a/x^(1/2)+b*(2*A*b+3*B*a )*x^(1/2)*(b*x+a)^(1/2)/a
Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (-2 a A-8 A b x-6 a B x+3 b B x^2\right )}{3 x^{3/2}}+2 \sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right ) \]
(Sqrt[a + b*x]*(-2*a*A - 8*A*b*x - 6*a*B*x + 3*b*B*x^2))/(3*x^(3/2)) + 2*S qrt[b]*(2*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x] )]
Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 57, 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.
\(\displaystyle \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(3 a B+2 A b) \int \frac {(a+b x)^{3/2}}{x^{3/2}}dx}{3 a}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {(3 a B+2 A b) \left (3 b \int \frac {\sqrt {a+b x}}{\sqrt {x}}dx-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )}{3 a}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(3 a B+2 A b) \left (3 b \left (\frac {1}{2} a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx+\sqrt {x} \sqrt {a+b x}\right )-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )}{3 a}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {(3 a B+2 A b) \left (3 b \left (a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}+\sqrt {x} \sqrt {a+b x}\right )-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )}{3 a}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(3 a B+2 A b) \left (3 b \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}}+\sqrt {x} \sqrt {a+b x}\right )-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )}{3 a}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}}\) |
(-2*A*(a + b*x)^(5/2))/(3*a*x^(3/2)) + ((2*A*b + 3*a*B)*((-2*(a + b*x)^(3/ 2))/Sqrt[x] + 3*b*(Sqrt[x]*Sqrt[a + b*x] + (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sq rt[a + b*x]])/Sqrt[b])))/(3*a)
3.5.95.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, 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[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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.50 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (-3 b B \,x^{2}+8 A b x +6 B a x +2 A a \right )}{3 x^{\frac {3}{2}}}+\frac {\sqrt {b}\, \left (2 A b +3 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{2 \sqrt {x}\, \sqrt {b x +a}}\) | \(93\) |
default | \(\frac {\sqrt {b x +a}\, \left (6 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b^{2} x^{2}+9 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a b \,x^{2}+6 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}-16 A \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}-12 B a x \sqrt {x \left (b x +a \right )}\, \sqrt {b}-4 A a \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{6 x^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\) | \(162\) |
-1/3*(b*x+a)^(1/2)*(-3*B*b*x^2+8*A*b*x+6*B*a*x+2*A*a)/x^(3/2)+1/2*b^(1/2)* (2*A*b+3*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 = 163, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\left [\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {b} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \]
[1/6*(3*(3*B*a + 2*A*b)*sqrt(b)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sq rt(x) + a) + 2*(3*B*b*x^2 - 2*A*a - 2*(3*B*a + 4*A*b)*x)*sqrt(b*x + a)*sqr t(x))/x^2, -1/3*(3*(3*B*a + 2*A*b)*sqrt(-b)*x^2*arctan(sqrt(b*x + a)*sqrt( -b)/(b*sqrt(x))) - (3*B*b*x^2 - 2*A*a - 2*(3*B*a + 4*A*b)*x)*sqrt(b*x + a) *sqrt(x))/x^2]
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (109) = 218\).
Time = 3.83 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.89 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=- \frac {2 A \sqrt {a} b}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {2 A a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 A b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} + 2 A b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 A b^{2} \sqrt {x}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}} - \frac {2 B a^{\frac {3}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + B \sqrt {a} b \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {2 B \sqrt {a} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 3 B a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} \]
-2*A*sqrt(a)*b/(sqrt(x)*sqrt(1 + b*x/a)) - 2*A*a*sqrt(b)*sqrt(a/(b*x) + 1) /(3*x) - 2*A*b**(3/2)*sqrt(a/(b*x) + 1)/3 + 2*A*b**(3/2)*asinh(sqrt(b)*sqr t(x)/sqrt(a)) - 2*A*b**2*sqrt(x)/(sqrt(a)*sqrt(1 + b*x/a)) - 2*B*a**(3/2)/ (sqrt(x)*sqrt(1 + b*x/a)) + B*sqrt(a)*b*sqrt(x)*sqrt(1 + b*x/a) - 2*B*sqrt (a)*b*sqrt(x)/sqrt(1 + b*x/a) + 3*B*a*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a ))
Time = 0.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.25 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\frac {3}{2} \, B a \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + A b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {3 \, \sqrt {b x^{2} + a x} B a}{x} - \frac {7 \, \sqrt {b x^{2} + a x} A b}{3 \, x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{x^{2}} - \frac {\sqrt {b x^{2} + a x} A a}{3 \, x^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{3 \, x^{3}} \]
3/2*B*a*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + A*b^(3/2)*l og(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 3*sqrt(b*x^2 + a*x)*B*a/x - 7/3*sqrt(b*x^2 + a*x)*A*b/x + (b*x^2 + a*x)^(3/2)*B/x^2 - 1/3*sqrt(b*x^2 + a*x)*A*a/x^2 - 1/3*(b*x^2 + a*x)^(3/2)*A/x^3
Time = 75.34 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=-\frac {{\left (\frac {3 \, {\left (3 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \frac {{\left ({\left (3 \, {\left (b x + a\right )} B b^{2} - \frac {4 \, {\left (3 \, B a^{2} b^{3} + 2 \, A a b^{4}\right )}}{a b}\right )} {\left (b x + a\right )} + \frac {3 \, {\left (3 \, B a^{3} b^{3} + 2 \, A a^{2} b^{4}\right )}}{a b}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}}\right )} b}{3 \, {\left | b \right |}} \]
-1/3*(3*(3*B*a*b + 2*A*b^2)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a )*b - a*b)))/sqrt(b) - ((3*(b*x + a)*B*b^2 - 4*(3*B*a^2*b^3 + 2*A*a*b^4)/( a*b))*(b*x + a) + 3*(3*B*a^3*b^3 + 2*A*a^2*b^4)/(a*b))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(3/2))*b/abs(b)
Timed out. \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{x^{5/2}} \,d x \]