Integrand size = 22, antiderivative size = 164 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx=\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}-2 a^{3/2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{3/2}} \]
-1/4*(-3*a^2*d^2-6*a*b*c*d+b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/ (d*x+c)^(1/2))/d^(3/2)/b^(1/2)-2*a^(3/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^( 1/2)/(d*x+c)^(1/2))*c^(1/2)+1/2*(b*x+a)^(3/2)*(d*x+c)^(1/2)+1/4*(3*a*d+b*c )*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d
Time = 0.52 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx=\frac {1}{4} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} (5 a d+b (c+2 d x))}{d}-8 a^{3/2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\frac {\left (-b^2 c^2+6 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {b} d^{3/2}}\right ) \]
((Sqrt[a + b*x]*Sqrt[c + d*x]*(5*a*d + b*(c + 2*d*x)))/d - 8*a^(3/2)*Sqrt[ c]*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])] + ((-(b^2*c^2) + 6*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(Sqrt[b]*d^(3/2)))/4
Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {112, 27, 171, 27, 175, 66, 104, 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.
| \(\displaystyle \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx\) |
| \(\Big \downarrow \) 112 |
| \(\displaystyle \frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}-\frac {1}{2} \int -\frac {\sqrt {a+b x} (4 a c+(b c+3 a d) x)}{2 x \sqrt {c+d x}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt {a+b x} (4 a c+(b c+3 a d) x)}{x \sqrt {c+d x}}dx+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {8 a^2 c d-\left (b^2 c^2-6 a b d c-3 a^2 d^2\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{d}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{d}\right )+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {\int \frac {8 a^2 c d-\left (b^2 c^2-6 a b d c-3 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{d}\right )+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{4} \left (\frac {8 a^2 c d \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-\left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{d}\right )+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{4} \left (\frac {8 a^2 c d \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx-2 \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 d}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{d}\right )+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{4} \left (\frac {16 a^2 c d \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}-2 \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 d}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{d}\right )+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{4} \left (\frac {-16 a^{3/2} \sqrt {c} d \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {2 \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}}}{2 d}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{d}\right )+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}\) |
((a + b*x)^(3/2)*Sqrt[c + d*x])/2 + (((b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/d + (-16*a^(3/2)*Sqrt[c]*d*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a] *Sqrt[c + d*x])] - (2*(b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*ArcTanh[(Sqrt[d]*S qrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*Sqrt[d]))/(2*d))/4
3.7.1.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^m*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + p + 1))), x] - Simp[1/(f*(m + n + p + 1)) Int[(a + b*x)^(m - 1)*(c + d*x) ^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a *f) + b*n*(d*e - c*f))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (IntegersQ[2*m, 2*n, 2*p ] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
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. \(332\) vs. \(2(126)=252\).
Time = 0.54 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.03
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b d x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{2} d^{2}+6 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a b c d -\ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{2} c^{2}-8 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} c d +10 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a d +2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c \right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, d \sqrt {b d}\, \sqrt {a c}}\) | \(333\) |
1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(4*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c ))^(1/2)*b*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))*(a*c)^(1/2)*a^2*d^2+6*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*c)^(1/2)*a*b*c*d-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*c)^(1/ 2)*b^2*c^2-8*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^( 1/2)+2*a*c)/x)*a^2*c*d+10*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)* a*d+2*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*c)/((b*x+a)*(d*x+c ))^(1/2)/d/(b*d)^(1/2)/(a*c)^(1/2)
Time = 1.08 (sec) , antiderivative size = 979, normalized size of antiderivative = 5.97 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx=\left [\frac {8 \, \sqrt {a c} a b d^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b d^{2}}, \frac {4 \, \sqrt {a c} a b d^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b d^{2}}, \frac {16 \, \sqrt {-a c} a b d^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b d^{2}}, \frac {8 \, \sqrt {-a c} a b d^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b d^{2}}\right ] \]
[1/16*(8*sqrt(a*c)*a*b*d^2*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2 )*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*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^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2), 1 /8*(4*sqrt(a*c)*a*b*d^2*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x ^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*( a*b*c^2 + a^2*c*d)*x)/x^2) + (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*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^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2), 1/16*(16*sqrt(-a*c)*a*b *d^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - (b^2*c^2 - 6*a*b*c *d - 3*a^2*d^2)*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^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*s qrt(d*x + c))/(b*d^2), 1/8*(8*sqrt(-a*c)*a*b*d^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) + (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*sqrt(-b*d)*...
\[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}} \sqrt {c + d x}}{x}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx=\text {Exception raised: ValueError} \]
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
Exception generated. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}\,\sqrt {c+d\,x}}{x} \,d x \]