Integrand size = 20, antiderivative size = 171 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {(b c-a d) (b c+5 a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(b c-a d)^2 (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{3/2}} \]
-1/8*(-a*d+b*c)^2*(5*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^(3/2)-1/12*(5*a*d+b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b^2/ d+1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/b/d-1/8*(-a*d+b*c)*(5*a*d+b*c)*(b*x+a)^( 1/2)*(d*x+c)^(1/2)/b^3/d
Time = 0.32 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.80 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-2 a b d (11 c+5 d x)+b^2 \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )}{24 b^3 d}-\frac {(b c-a d)^2 (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{3/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^2*d^2 - 2*a*b*d*(11*c + 5*d*x) + b^2*(3 *c^2 + 14*c*d*x + 8*d^2*x^2)))/(24*b^3*d) - ((b*c - a*d)^2*(b*c + 5*a*d)*A rcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(7/2)*d^(3/2 ))
Time = 0.23 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {90, 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.
\(\displaystyle \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}}dx}{6 b d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\) |
(Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b*d) - ((b*c + 5*a*d)*((Sqrt[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)))/(6*b*d)
3.8.7.3.1 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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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. \(394\) vs. \(2(139)=278\).
Time = 1.64 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.31
method | result | size |
default | \(-\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (-16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}-27 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x -28 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+44 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{48 b^{3} d \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}\) | \(395\) |
-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-16*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2) *(b*d)^(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^3-27*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^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^2*c^2*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))*b^3*c^3+20* (b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*d^2*x-28*(b*d)^(1/2)*((b*x+a)*(d*x +c))^(1/2)*b^2*c*d*x-30*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+44*(b* d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d-6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^( 1/2)*b^2*c^2)/b^3/d/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)
Time = 0.25 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.41 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\left [\frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{2}}, \frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{2}}\right ] \]
[1/96*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(b*d)*l og(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*(8*b ^3*d^3*x^2 + 3*b^3*c^2*d - 22*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(7*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^2), 1/48*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*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*(8*b^3*d^3*x^2 + 3*b^3*c^2*d - 22*a*b^2* c*d^2 + 15*a^2*b*d^3 + 2*(7*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt (d*x + c))/(b^4*d^2)]
\[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\int \frac {x \left (c + d x\right )^{\frac {3}{2}}}{\sqrt {a + b x}}\, dx \]
Exception generated. \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (139) = 278\).
Time = 0.34 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.02 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} d {\left | b \right |}}{b^{2}} + \frac {6 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} c {\left | b \right |}}{b^{3}}}{24 \, b} \]
1/24*((sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*( b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2* a*b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^ 2*b*c*d^2 - 5*a^3*d^3)*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^2))*d*abs(b)/b^2 + 6*(sqrt(b^2*c + (b *x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*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))*c*abs(b)/b^3)/b
Timed out. \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\int \frac {x\,{\left (c+d\,x\right )}^{3/2}}{\sqrt {a+b\,x}} \,d x \]