Integrand size = 22, antiderivative size = 149 \[ \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx=\frac {(3 b c-5 a d) (a+b x)^{3/2}}{3 a c^2 (c+d x)^{3/2}}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}}+\frac {(3 b c-5 a d) \sqrt {a+b x}}{c^3 \sqrt {c+d x}}-\frac {\sqrt {a} (3 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{7/2}} \]
1/3*(-5*a*d+3*b*c)*(b*x+a)^(3/2)/a/c^2/(d*x+c)^(3/2)-(b*x+a)^(5/2)/a/c/x/( d*x+c)^(3/2)-(-5*a*d+3*b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^ (1/2))*a^(1/2)/c^(7/2)+(-5*a*d+3*b*c)*(b*x+a)^(1/2)/c^3/(d*x+c)^(1/2)
Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (2 b c x (3 c+2 d x)-a \left (3 c^2+20 c d x+15 d^2 x^2\right )\right )}{3 c^3 x (c+d x)^{3/2}}+\frac {\sqrt {a} (-3 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{c^{7/2}} \]
(Sqrt[a + b*x]*(2*b*c*x*(3*c + 2*d*x) - a*(3*c^2 + 20*c*d*x + 15*d^2*x^2)) )/(3*c^3*x*(c + d*x)^(3/2)) + (Sqrt[a]*(-3*b*c + 5*a*d)*ArcTanh[(Sqrt[a]*S qrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/c^(7/2)
Time = 0.23 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {107, 105, 105, 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.
| \(\displaystyle \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 107 |
| \(\displaystyle \frac {(3 b c-5 a d) \int \frac {(a+b x)^{3/2}}{x (c+d x)^{5/2}}dx}{2 a c}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle \frac {(3 b c-5 a d) \left (\frac {a \int \frac {\sqrt {a+b x}}{x (c+d x)^{3/2}}dx}{c}+\frac {2 (a+b x)^{3/2}}{3 c (c+d x)^{3/2}}\right )}{2 a c}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle \frac {(3 b c-5 a d) \left (\frac {a \left (\frac {a \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{c}+\frac {2 \sqrt {a+b x}}{c \sqrt {c+d x}}\right )}{c}+\frac {2 (a+b x)^{3/2}}{3 c (c+d x)^{3/2}}\right )}{2 a c}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {(3 b c-5 a d) \left (\frac {a \left (\frac {2 a \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{c}+\frac {2 \sqrt {a+b x}}{c \sqrt {c+d x}}\right )}{c}+\frac {2 (a+b x)^{3/2}}{3 c (c+d x)^{3/2}}\right )}{2 a c}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(3 b c-5 a d) \left (\frac {a \left (\frac {2 \sqrt {a+b x}}{c \sqrt {c+d x}}-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}\right )}{c}+\frac {2 (a+b x)^{3/2}}{3 c (c+d x)^{3/2}}\right )}{2 a c}-\frac {(a+b x)^{5/2}}{a c x (c+d x)^{3/2}}\) |
-((a + b*x)^(5/2)/(a*c*x*(c + d*x)^(3/2))) + ((3*b*c - 5*a*d)*((2*(a + b*x )^(3/2))/(3*c*(c + d*x)^(3/2)) + (a*((2*Sqrt[a + b*x])/(c*Sqrt[c + d*x]) - (2*Sqrt[a]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(3 /2)))/c))/(2*a*c)
3.7.45.3.1 Defintions of rubi rules used
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 + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
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. \(2(123)=246\).
Time = 1.57 (sec) , antiderivative size = 459, normalized size of antiderivative = 3.08
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \left (15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} d^{3} x^{3}-9 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b c \,d^{2} x^{3}+30 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} c \,d^{2} x^{2}-18 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b \,c^{2} d \,x^{2}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} c^{2} d x -9 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b \,c^{3} x -30 a \,d^{2} x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+8 b c d \,x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-40 a c d x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+12 b \,c^{2} x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-6 a \,c^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{6 c^{3} \sqrt {a c}\, x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}}}\) | \(459\) |
1/6*(b*x+a)^(1/2)/c^3*(15*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*d^3*x^3-9*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*b*c*d^2*x^3+30*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^2*x^2-18*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*b*c^2*d*x^2+15*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^2*d*x-9*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*b*c^3*x-30*a*d^2*x^2*(a*c)^ (1/2)*((b*x+a)*(d*x+c))^(1/2)+8*b*c*d*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1 /2)-40*a*c*d*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*b*c^2*x*(a*c)^(1/2)* ((b*x+a)*(d*x+c))^(1/2)-6*a*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/(a*c) ^(1/2)/x/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)
Time = 0.66 (sec) , antiderivative size = 463, normalized size of antiderivative = 3.11 \[ \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (3 \, b c d^{2} - 5 \, a d^{3}\right )} x^{3} + 2 \, {\left (3 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2} + {\left (3 \, b c^{3} - 5 \, a c^{2} d\right )} x\right )} \sqrt {\frac {a}{c}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (3 \, a c^{2} - {\left (4 \, b c d - 15 \, a d^{2}\right )} x^{2} - 2 \, {\left (3 \, b c^{2} - 10 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (c^{3} d^{2} x^{3} + 2 \, c^{4} d x^{2} + c^{5} x\right )}}, \frac {3 \, {\left ({\left (3 \, b c d^{2} - 5 \, a d^{3}\right )} x^{3} + 2 \, {\left (3 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2} + {\left (3 \, b c^{3} - 5 \, a c^{2} d\right )} x\right )} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) - 2 \, {\left (3 \, a c^{2} - {\left (4 \, b c d - 15 \, a d^{2}\right )} x^{2} - 2 \, {\left (3 \, b c^{2} - 10 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (c^{3} d^{2} x^{3} + 2 \, c^{4} d x^{2} + c^{5} x\right )}}\right ] \]
[-1/12*(3*((3*b*c*d^2 - 5*a*d^3)*x^3 + 2*(3*b*c^2*d - 5*a*c*d^2)*x^2 + (3* b*c^3 - 5*a*c^2*d)*x)*sqrt(a/c)*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^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*s qrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(3*a*c^2 - (4*b*c*d - 15*a*d^ 2)*x^2 - 2*(3*b*c^2 - 10*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^3*d^2*x ^3 + 2*c^4*d*x^2 + c^5*x), 1/6*(3*((3*b*c*d^2 - 5*a*d^3)*x^3 + 2*(3*b*c^2* d - 5*a*c*d^2)*x^2 + (3*b*c^3 - 5*a*c^2*d)*x)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2* c + (a*b*c + a^2*d)*x)) - 2*(3*a*c^2 - (4*b*c*d - 15*a*d^2)*x^2 - 2*(3*b*c ^2 - 10*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^3*d^2*x^3 + 2*c^4*d*x^2 + c^5*x)]
\[ \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}}}{x^{2} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (123) = 246\).
Time = 1.06 (sec) , antiderivative size = 612, normalized size of antiderivative = 4.11 \[ \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx=\frac {2 \, \sqrt {b x + a} {\left (\frac {2 \, {\left (b^{5} c^{5} d^{2} {\left | b \right |} - 4 \, a b^{4} c^{4} d^{3} {\left | b \right |} + 3 \, a^{2} b^{3} c^{3} d^{4} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{3} c^{7} d - a b^{2} c^{6} d^{2}} + \frac {3 \, {\left (b^{6} c^{6} d {\left | b \right |} - 4 \, a b^{5} c^{5} d^{2} {\left | b \right |} + 5 \, a^{2} b^{4} c^{4} d^{3} {\left | b \right |} - 2 \, a^{3} b^{3} c^{3} d^{4} {\left | b \right |}\right )}}{b^{3} c^{7} d - a b^{2} c^{6} d^{2}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {{\left (3 \, \sqrt {b d} a b^{3} c - 5 \, \sqrt {b d} a^{2} b^{2} d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c^{3} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} a b^{5} c^{2} - 2 \, \sqrt {b d} a^{2} b^{4} c d + \sqrt {b d} a^{3} b^{3} d^{2} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} c - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} c^{3} {\left | b \right |}} \]
2/3*sqrt(b*x + a)*(2*(b^5*c^5*d^2*abs(b) - 4*a*b^4*c^4*d^3*abs(b) + 3*a^2* b^3*c^3*d^4*abs(b))*(b*x + a)/(b^3*c^7*d - a*b^2*c^6*d^2) + 3*(b^6*c^6*d*a bs(b) - 4*a*b^5*c^5*d^2*abs(b) + 5*a^2*b^4*c^4*d^3*abs(b) - 2*a^3*b^3*c^3* d^4*abs(b))/(b^3*c^7*d - a*b^2*c^6*d^2))/(b^2*c + (b*x + a)*b*d - a*b*d)^( 3/2) - (3*sqrt(b*d)*a*b^3*c - 5*sqrt(b*d)*a^2*b^2*d)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2) /(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c^3*abs(b)) - 2*(sqrt(b*d)*a*b^5*c^ 2 - 2*sqrt(b*d)*a^2*b^4*c*d + sqrt(b*d)*a^3*b^3*d^2 - sqrt(b*d)*(sqrt(b*d) *sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c - sqrt(b*d )*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^ 2*d)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*c^3*abs(b))
Timed out. \[ \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}}{x^2\,{\left (c+d\,x\right )}^{5/2}} \,d x \]