Integrand size = 20, antiderivative size = 174 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 c (a+b x)^{5/2}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac {2 (5 b c-3 a d) (a+b x)^{3/2}}{3 d^2 (b c-a d) \sqrt {c+d x}}+\frac {b (5 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{d^3 (b c-a d)}-\frac {\sqrt {b} (5 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{7/2}} \]
-2/3*c*(b*x+a)^(5/2)/d/(-a*d+b*c)/(d*x+c)^(3/2)-(-3*a*d+5*b*c)*arctanh(d^( 1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)/d^(7/2)-2/3*(-3*a*d+5*b* c)*(b*x+a)^(3/2)/d^2/(-a*d+b*c)/(d*x+c)^(1/2)+b*(-3*a*d+5*b*c)*(b*x+a)^(1/ 2)*(d*x+c)^(1/2)/d^3/(-a*d+b*c)
Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.64 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (-2 a d (2 c+3 d x)+b \left (15 c^2+20 c d x+3 d^2 x^2\right )\right )}{3 d^3 (c+d x)^{3/2}}-\frac {\sqrt {b} (5 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{d^{7/2}} \]
(Sqrt[a + b*x]*(-2*a*d*(2*c + 3*d*x) + b*(15*c^2 + 20*c*d*x + 3*d^2*x^2))) /(3*d^3*(c + d*x)^(3/2)) - (Sqrt[b]*(5*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[ c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/d^(7/2)
Time = 0.24 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.95, 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, 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.
| \(\displaystyle \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(5 b c-3 a d) \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/2}}dx}{3 d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {(5 b c-3 a d) \left (\frac {3 b \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\right )}{3 d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(5 b c-3 a d) \left (\frac {3 b \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\right )}{3 d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {(5 b c-3 a d) \left (\frac {3 b \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\right )}{3 d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(5 b c-3 a d) \left (\frac {3 b \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{d}-\frac {2 (a+b x)^{3/2}}{d \sqrt {c+d x}}\right )}{3 d (b c-a d)}-\frac {2 c (a+b x)^{5/2}}{3 d (c+d x)^{3/2} (b c-a d)}\) |
(-2*c*(a + b*x)^(5/2))/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) + ((5*b*c - 3*a*d )*((-2*(a + b*x)^(3/2))/(d*Sqrt[c + d*x]) + (3*b*((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))))/d))/(3*d*(b*c - a*d))
3.7.42.3.1 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[(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. \(458\) vs. \(2(146)=292\).
Time = 0.56 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.64
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \left (9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,d^{3} x^{2}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c \,d^{2} x^{2}+18 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x -30 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d x +6 b \,d^{2} x^{2} \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,c^{2} d -15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{3}-12 a \,d^{2} x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+40 b c d x \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-8 a c d \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+30 b \,c^{2} \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d^{3} \left (d x +c \right )^{\frac {3}{2}}}\) | \(459\) |
1/6*(b*x+a)^(1/2)*(9*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*d^3*x^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))*b^2*c*d^2*x^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*d^2*x-30* 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+6*b*d^2*x^2*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+9*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*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^3-12*a*d^2*x*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+40*b*c*d*x*( b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-8*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*d)^(1/2)/((b*x+a)*(d *x+c))^(1/2)/d^3/(d*x+c)^(3/2)
Time = 0.37 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.48 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\left [-\frac {3 \, {\left (5 \, b c^{3} - 3 \, a c^{2} d + {\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} x^{2} + 2 \, {\left (5 \, b c^{2} d - 3 \, a c d^{2}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (3 \, b d^{2} x^{2} + 15 \, b c^{2} - 4 \, a c d + 2 \, {\left (10 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}}, \frac {3 \, {\left (5 \, b c^{3} - 3 \, a c^{2} d + {\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} x^{2} + 2 \, {\left (5 \, b c^{2} d - 3 \, a c d^{2}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (3 \, b d^{2} x^{2} + 15 \, b c^{2} - 4 \, a c d + 2 \, {\left (10 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}}\right ] \]
[-1/12*(3*(5*b*c^3 - 3*a*c^2*d + (5*b*c*d^2 - 3*a*d^3)*x^2 + 2*(5*b*c^2*d - 3*a*c*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(3*b*d^2*x^2 + 15*b*c^2 - 4*a*c*d + 2*(10*b*c *d - 3*a*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^5*x^2 + 2*c*d^4*x + c^2*d ^3), 1/6*(3*(5*b*c^3 - 3*a*c^2*d + (5*b*c*d^2 - 3*a*d^3)*x^2 + 2*(5*b*c^2* d - 3*a*c*d^2)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a )*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) + 2*(3 *b*d^2*x^2 + 15*b*c^2 - 4*a*c*d + 2*(10*b*c*d - 3*a*d^2)*x)*sqrt(b*x + a)* sqrt(d*x + c))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)]
\[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x \left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.36 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.64 \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (b^{5} c d^{4} {\left | b \right |} - a b^{4} d^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{4} c d^{5} - a b^{3} d^{6}} + \frac {4 \, {\left (5 \, b^{6} c^{2} d^{3} {\left | b \right |} - 8 \, a b^{5} c d^{4} {\left | b \right |} + 3 \, a^{2} b^{4} d^{5} {\left | b \right |}\right )}}{b^{4} c d^{5} - a b^{3} d^{6}}\right )} + \frac {3 \, {\left (5 \, b^{7} c^{3} d^{2} {\left | b \right |} - 13 \, a b^{6} c^{2} d^{3} {\left | b \right |} + 11 \, a^{2} b^{5} c d^{4} {\left | b \right |} - 3 \, a^{3} b^{4} d^{5} {\left | b \right |}\right )}}{b^{4} c d^{5} - a b^{3} d^{6}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, b c {\left | b \right |} - 3 \, a d {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{3}} \]
1/3*((b*x + a)*(3*(b^5*c*d^4*abs(b) - a*b^4*d^5*abs(b))*(b*x + a)/(b^4*c*d ^5 - a*b^3*d^6) + 4*(5*b^6*c^2*d^3*abs(b) - 8*a*b^5*c*d^4*abs(b) + 3*a^2*b ^4*d^5*abs(b))/(b^4*c*d^5 - a*b^3*d^6)) + 3*(5*b^7*c^3*d^2*abs(b) - 13*a*b ^6*c^2*d^3*abs(b) + 11*a^2*b^5*c*d^4*abs(b) - 3*a^3*b^4*d^5*abs(b))/(b^4*c *d^5 - a*b^3*d^6))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + ( 5*b*c*abs(b) - 3*a*d*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)*d^3)
Timed out. \[ \int \frac {x (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\int \frac {x\,{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \]