Integrand size = 20, antiderivative size = 174 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 c (a+b x)^{5/2}}{d (b c-a d) \sqrt {c+d x}}-\frac {3 (5 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {(5 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2 (b c-a d)}+\frac {3 (b c-a d) (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{7/2}} \]
3/4*(-a*d+b*c)*(-a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^ (1/2))/d^(7/2)/b^(1/2)-2*c*(b*x+a)^(5/2)/d/(-a*d+b*c)/(d*x+c)^(1/2)+1/2*(- a*d+5*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^2/(-a*d+b*c)-3/4*(-a*d+5*b*c)*(b* x+a)^(1/2)*(d*x+c)^(1/2)/d^3
Time = 0.63 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.83 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {\frac {\sqrt {d} \sqrt {a+b x} \left (a d (13 c+5 d x)+b \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )}{\sqrt {c+d x}}-\frac {6 \left (5 b^2 c^2-6 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \left (\sqrt {a-\frac {b c}{d}}-\sqrt {a+b x}\right )}\right )}{\sqrt {b}}}{4 d^{7/2}} \]
((Sqrt[d]*Sqrt[a + b*x]*(a*d*(13*c + 5*d*x) + b*(-15*c^2 - 5*c*d*x + 2*d^2 *x^2)))/Sqrt[c + d*x] - (6*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt [b]*Sqrt[c + d*x])/(Sqrt[d]*(Sqrt[a - (b*c)/d] - Sqrt[a + b*x]))])/Sqrt[b] )/(4*d^(7/2))
Time = 0.24 (sec) , antiderivative size = 172, 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.
| \(\displaystyle \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(5 b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(5 b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(5 b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {(5 b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(5 b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{d \sqrt {c+d x} (b c-a d)}\) |
(-2*c*(a + b*x)^(5/2))/(d*(b*c - a*d)*Sqrt[c + d*x]) + ((5*b*c - a*d)*(((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(d*(b*c - a*d))
3.7.35.3.1 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. \(2(144)=288\).
Time = 0.55 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.61
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \left (3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{3} x -18 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x +15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d x +4 b \,d^{2} x^{2} \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} c \,d^{2}-18 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,c^{2} d +15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{3}+10 a \,d^{2} x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-10 b c d x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+26 a c d \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-30 b \,c^{2} \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {d x +c}\, d^{3}}\) | \(455\) |
1/8*(b*x+a)^(1/2)*(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^2*d^3*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*c*d^2*x+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))*b^2*c^2*d*x+4*b*d^2 *x^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+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^2*c*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*b*c^2*d+15*l n(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^2*c^3+10*a*d^2*x*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*b*c*d*x*(b*d)^( 1/2)*((b*x+a)*(d*x+c))^(1/2)+26*a*c*d*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)- 30*b*c^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d )^(1/2)/(d*x+c)^(1/2)/d^3
Time = 0.31 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.49 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + {\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b d^{5} x + b c d^{4}\right )}}, -\frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + {\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{3} x^{2} - 15 \, b^{2} c^{2} d + 13 \, a b c d^{2} - 5 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b d^{5} x + b c d^{4}\right )}}\right ] \]
[1/16*(3*(5*b^2*c^3 - 6*a*b*c^2*d + a^2*c*d^2 + (5*b^2*c^2*d - 6*a*b*c*d^2 + a^2*d^3)*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^2*d^3*x^2 - 15*b^2*c^2*d + 13*a*b*c*d^2 - 5*(b^ 2*c*d^2 - a*b*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^5*x + b*c*d^4), -1 /8*(3*(5*b^2*c^3 - 6*a*b*c^2*d + a^2*c*d^2 + (5*b^2*c^2*d - 6*a*b*c*d^2 + a^2*d^3)*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^2*d^3*x^2 - 15*b^2*c^2*d + 13*a*b*c*d^2 - 5*(b^2*c*d^2 - a*b*d^3)*x)* sqrt(b*x + a)*sqrt(d*x + c))/(b*d^5*x + b*c*d^4)]
\[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\int \frac {x \left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.33 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b x + a\right )} {\left | b \right |}}{b d} - \frac {5 \, b^{2} c d^{3} {\left | b \right |} - a b d^{4} {\left | b \right |}}{b^{2} d^{5}}\right )} - \frac {3 \, {\left (5 \, b^{3} c^{2} d^{2} {\left | b \right |} - 6 \, a b^{2} c d^{3} {\left | b \right |} + a^{2} b d^{4} {\left | b \right |}\right )}}{b^{2} d^{5}}\right )}}{4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {3 \, {\left (5 \, b^{2} c^{2} {\left | b \right |} - 6 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} b d^{3}} \]
1/4*sqrt(b*x + a)*((b*x + a)*(2*(b*x + a)*abs(b)/(b*d) - (5*b^2*c*d^3*abs( b) - a*b*d^4*abs(b))/(b^2*d^5)) - 3*(5*b^3*c^2*d^2*abs(b) - 6*a*b^2*c*d^3* abs(b) + a^2*b*d^4*abs(b))/(b^2*d^5))/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) - 3/4*(5*b^2*c^2*abs(b) - 6*a*b*c*d*abs(b) + a^2*d^2*abs(b))*log(abs(-sqrt (b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d ^3)
Timed out. \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{3/2}} \, dx=\int \frac {x\,{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{3/2}} \,d x \]