Integrand size = 22, antiderivative size = 154 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 (b c-a d)^2 \sqrt {c+d x}}{4 a^3 \sqrt {a+b x}}+\frac {5 (b c-a d) (c+d x)^{3/2}}{4 a^2 x \sqrt {a+b x}}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}-\frac {15 \sqrt {c} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2}} \]
-15/4*(-a*d+b*c)^2*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))*c^ (1/2)/a^(7/2)+5/4*(-a*d+b*c)*(d*x+c)^(3/2)/a^2/x/(b*x+a)^(1/2)-1/2*(d*x+c) ^(5/2)/a/x^2/(b*x+a)^(1/2)+15/4*(-a*d+b*c)^2*(d*x+c)^(1/2)/a^3/(b*x+a)^(1/ 2)
Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (15 b^2 c^2 x^2+5 a b c x (c-5 d x)+a^2 \left (-2 c^2-9 c d x+8 d^2 x^2\right )\right )}{4 a^3 x^2 \sqrt {a+b x}}-\frac {15 \sqrt {c} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{4 a^{7/2}} \]
(Sqrt[c + d*x]*(15*b^2*c^2*x^2 + 5*a*b*c*x*(c - 5*d*x) + a^2*(-2*c^2 - 9*c *d*x + 8*d^2*x^2)))/(4*a^3*x^2*Sqrt[a + b*x]) - (15*Sqrt[c]*(b*c - a*d)^2* ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(4*a^(7/2))
Time = 0.23 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {105, 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.
| \(\displaystyle \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle -\frac {5 (b c-a d) \int \frac {(c+d x)^{3/2}}{x^2 (a+b x)^{3/2}}dx}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle -\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{x (a+b x)^{3/2}}dx}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle -\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {c \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {2 c \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {5 (b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}-\frac {2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}}\) |
-1/2*(c + d*x)^(5/2)/(a*x^2*Sqrt[a + b*x]) - (5*(b*c - a*d)*(-((c + d*x)^( 3/2)/(a*x*Sqrt[a + b*x])) - (3*(b*c - a*d)*((2*Sqrt[c + d*x])/(a*Sqrt[a + b*x]) - (2*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x]) ])/a^(3/2)))/(2*a)))/(4*a)
3.8.73.3.1 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_)^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. \(506\) vs. \(2(122)=244\).
Time = 1.71 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.29
| method | result | size |
| default | \(-\frac {\sqrt {d x +c}\, \left (15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b c \,d^{2} x^{3}-30 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} d \,x^{3}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x^{3}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} c \,d^{2} x^{2}-30 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b \,c^{2} d \,x^{2}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{3} x^{2}-16 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2} x^{2}+50 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d \,x^{2}-30 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} x^{2}+18 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c d x -10 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2} x +4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{2} \sqrt {a c}\right )}{8 a^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{2} \sqrt {a c}\, \sqrt {b x +a}}\) | \(507\) |
-1/8*(d*x+c)^(1/2)*(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*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*b^2*c^2*d*x^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)*b^3*c^3*x^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^3*c*d^2*x^2-30*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*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*b^2*c^3*x^2-16*(a*c) ^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2*x^2+50*(a*c)^(1/2)*((b*x+a)*(d*x+c) )^(1/2)*a*b*c*d*x^2-30*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2*x^2+18* (a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*c*d*x-10*(a*c)^(1/2)*((b*x+a)*(d*x +c))^(1/2)*a*b*c^2*x+4*((b*x+a)*(d*x+c))^(1/2)*a^2*c^2*(a*c)^(1/2))/a^3/(( b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1/2)/(b*x+a)^(1/2)
Time = 0.43 (sec) , antiderivative size = 481, normalized size of antiderivative = 3.12 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left ({\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {\frac {c}{a}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} - {\left (15 \, b^{2} c^{2} - 25 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{2} - {\left (5 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, \frac {15 \, {\left ({\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (15 \, b^{2} c^{2} - 25 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{2} - {\left (5 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \]
[1/16*(15*((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^3 + (a*b^2*c^2 - 2*a^2*b* c*d + a^3*d^2)*x^2)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2* d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqr t(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^2 - (15*b^2*c^2 - 25*a *b*c*d + 8*a^2*d^2)*x^2 - (5*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d* x + c))/(a^3*b*x^3 + a^4*x^2), 1/8*(15*((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2 )*x^3 + (a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(-c/a)*arctan(1/2*(2* a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a *c^2 + (b*c^2 + a*c*d)*x)) - 2*(2*a^2*c^2 - (15*b^2*c^2 - 25*a*b*c*d + 8*a ^2*d^2)*x^2 - (5*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3 *b*x^3 + a^4*x^2)]
\[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{3} \left (a + b x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 1206 vs. \(2 (122) = 244\).
Time = 1.92 (sec) , antiderivative size = 1206, normalized size of antiderivative = 7.83 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx=\text {Too large to display} \]
-15/4*(sqrt(b*d)*b^2*c^3*abs(b) - 2*sqrt(b*d)*a*b*c^2*d*abs(b) + sqrt(b*d) *a^2*c*d^2*abs(b))*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)*a^3*b) + 1/2*(7*sqrt(b*d)*b^8*c^6*abs(b) - 37*sqrt(b*d)*a*b^7*c^5*d*abs (b) + 78*sqrt(b*d)*a^2*b^6*c^4*d^2*abs(b) - 82*sqrt(b*d)*a^3*b^5*c^3*d^3*a bs(b) + 43*sqrt(b*d)*a^4*b^4*c^2*d^4*abs(b) - 9*sqrt(b*d)*a^5*b^3*c*d^5*ab s(b) - 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^6*c^5*abs(b) + 44*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^5*c^4*d*abs(b) + 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^2*b^4*c^3*d^2 *abs(b) - 52*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b *d - a*b*d))^2*a^3*b^3*c^2*d^3*abs(b) + 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^2*c*d^4*abs(b) + 21*sqr t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b ^4*c^4*abs(b) - 13*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^3*c^3*d*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^2*c^2*d^2*abs(b) - 27* sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^ 4*a^3*b*c*d^3*abs(b) - 7*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^2*c^3*abs(b) + 6*sqrt(b*d)*(sqrt(b*d)*sqrt...
Timed out. \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{x^3\,{\left (a+b\,x\right )}^{3/2}} \,d x \]