Integrand size = 20, antiderivative size = 171 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {3 (b c-5 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {(b c-5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2 (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {3 (b c-5 a d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2} \sqrt {d}} \]
3/4*(-5*a*d+b*c)*(-a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^ (1/2))/b^(7/2)/d^(1/2)+2*a*(d*x+c)^(5/2)/b/(-a*d+b*c)/(b*x+a)^(1/2)+1/2*(- 5*a*d+b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b^2/(-a*d+b*c)+3/4*(-5*a*d+b*c)*(b* x+a)^(1/2)*(d*x+c)^(1/2)/b^3
Time = 0.71 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.83 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {\frac {\sqrt {b} \sqrt {c+d x} \left (-15 a^2 d+a b (13 c-5 d x)+b^2 x (5 c+2 d x)\right )}{\sqrt {a+b x}}-\frac {6 \left (b^2 c^2-6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \left (\sqrt {a-\frac {b c}{d}}-\sqrt {a+b x}\right )}\right )}{\sqrt {d}}}{4 b^{7/2}} \]
((Sqrt[b]*Sqrt[c + d*x]*(-15*a^2*d + a*b*(13*c - 5*d*x) + b^2*x*(5*c + 2*d *x)))/Sqrt[a + b*x] - (6*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[b ]*Sqrt[c + d*x])/(Sqrt[d]*(Sqrt[a - (b*c)/d] - Sqrt[a + b*x]))])/Sqrt[d])/ (4*b^(7/2))
Time = 0.24 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 60, 60, 66, 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.
| \(\displaystyle \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(b c-5 a d) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}}dx}{b (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(b c-5 a d) \left (\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{b (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(b c-5 a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{b (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {(b c-5 a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{b (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(b c-5 a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{b (b c-a d)}+\frac {2 a (c+d x)^{5/2}}{b \sqrt {a+b x} (b c-a d)}\) |
(2*a*(c + d*x)^(5/2))/(b*(b*c - a*d)*Sqrt[a + b*x]) + ((b*c - 5*a*d)*((Sqr t[a + b*x]*(c + d*x)^(3/2))/(2*b) + (3*(b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(3/2)*Sqrt[d])))/(4*b)))/(b*(b*c - a*d))
3.8.61.3.1 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[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Leaf count of result is larger than twice the leaf count of optimal. \(454\) vs. \(2(141)=282\).
Time = 0.59 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.66
| method | result | size |
| default | \(\frac {\sqrt {d x +c}\, \left (15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{2} x -18 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x +4 b^{2} d \,x^{2} \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{2}-18 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2}-10 a b d x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+10 b^{2} c x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-30 a^{2} d \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+26 a b c \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{3}}\) | \(455\) |
1/8*(d*x+c)^(1/2)*(15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2 )+a*d+b*c)/(b*d)^(1/2))*a^2*b*d^2*x-18*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c)) ^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c*d*x+3*ln(1/2*(2*b*d*x+2*( (b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^2*x+4*b^2*d *x^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d* x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*d^2-18*ln(1/2*(2*b*d*x+2 *((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c*d+3*ln( 1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a *b^2*c^2-10*a*b*d*x*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*b^2*c*x*(b*d)^( 1/2)*((b*x+a)*(d*x+c))^(1/2)-30*a^2*d*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+ 26*a*b*c*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d )^(1/2)/(b*x+a)^(1/2)/b^3
Time = 0.29 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.54 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\left [\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b^{5} d x + a b^{4} d\right )}}, -\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{3} d^{2} x^{2} + 13 \, a b^{2} c d - 15 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b^{5} d x + a b^{4} d\right )}}\right ] \]
[1/16*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5 *a^2*b*d^2)*x)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2* c*d + a*b*d^2)*x) + 4*(2*b^3*d^2*x^2 + 13*a*b^2*c*d - 15*a^2*b*d^2 + 5*(b^ 3*c*d - a*b^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d*x + a*b^4*d), -1 /8*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5*a^ 2*b*d^2)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b* x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 2* (2*b^3*d^2*x^2 + 13*a*b^2*c*d - 15*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)* sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d*x + a*b^4*d)]
\[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x \left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.39 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.46 \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\frac {1}{4} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} d {\left | b \right |}}{b^{5}} + \frac {5 \, b^{10} c d^{2} {\left | b \right |} - 9 \, a b^{9} d^{3} {\left | b \right |}}{b^{14} d^{2}}\right )} - \frac {3 \, {\left (b^{2} c^{2} {\left | b \right |} - 6 \, a b c d {\left | b \right |} + 5 \, a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{8 \, \sqrt {b d} b^{4}} + \frac {4 \, {\left (a b^{2} c^{2} d {\left | b \right |} - 2 \, a^{2} b c d^{2} {\left | b \right |} + a^{3} d^{3} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} \sqrt {b d} b^{3}} \]
1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d*abs(b )/b^5 + (5*b^10*c*d^2*abs(b) - 9*a*b^9*d^3*abs(b))/(b^14*d^2)) - 3/8*(b^2* c^2*abs(b) - 6*a*b*c*d*abs(b) + 5*a^2*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(b*d)*b^4) + 4*(a*b^2* c^2*d*abs(b) - 2*a^2*b*c*d^2*abs(b) + a^3*d^3*abs(b))/((b^2*c - a*b*d - (s qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*sqrt(b*d) *b^3)
Timed out. \[ \int \frac {x (c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x\,{\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]