Integrand size = 22, antiderivative size = 180 \[ \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {2 b}{a (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}+\frac {2 d (3 b c+a d) \sqrt {a+b x}}{3 a c (b c-a d)^2 (c+d x)^{3/2}}+\frac {2 d (3 b c-a d) (b c+3 a d) \sqrt {a+b x}}{3 a c^2 (b c-a d)^3 \sqrt {c+d x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2} c^{5/2}} \]
-2*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(3/2)/c^(5/2)+2* b/a/(-a*d+b*c)/(d*x+c)^(3/2)/(b*x+a)^(1/2)+2/3*d*(a*d+3*b*c)*(b*x+a)^(1/2) /a/c/(-a*d+b*c)^2/(d*x+c)^(3/2)+2/3*d*(-a*d+3*b*c)*(3*a*d+b*c)*(b*x+a)^(1/ 2)/a/c^2/(-a*d+b*c)^3/(d*x+c)^(1/2)
Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {2 (a+b x)^{3/2} \left (a c d^3-\frac {9 a b c d^2 (c+d x)}{a+b x}+\frac {3 a^2 d^3 (c+d x)}{a+b x}-\frac {3 b^3 c^2 (c+d x)^2}{(a+b x)^2}\right )}{3 a c^2 (-b c+a d)^3 (c+d x)^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{a^{3/2} c^{5/2}} \]
(2*(a + b*x)^(3/2)*(a*c*d^3 - (9*a*b*c*d^2*(c + d*x))/(a + b*x) + (3*a^2*d ^3*(c + d*x))/(a + b*x) - (3*b^3*c^2*(c + d*x)^2)/(a + b*x)^2))/(3*a*c^2*( -(b*c) + a*d)^3*(c + d*x)^(3/2)) - (2*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqr t[c]*Sqrt[a + b*x])])/(a^(3/2)*c^(5/2))
Time = 0.32 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {115, 27, 169, 27, 169, 27, 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.
| \(\displaystyle \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 115 |
| \(\displaystyle \frac {2 \int \frac {b c-a d+4 b d x}{2 x \sqrt {a+b x} (c+d x)^{5/2}}dx}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b c-a d+4 b d x}{x \sqrt {a+b x} (c+d x)^{5/2}}dx}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {\frac {2 d \sqrt {a+b x} (a d+3 b c)}{3 c (c+d x)^{3/2} (b c-a d)}-\frac {2 \int -\frac {3 (b c-a d)^2+2 b d (3 b c+a d) x}{2 x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {3 (b c-a d)^2+2 b d (3 b c+a d) x}{x \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} (a d+3 b c)}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {\frac {\frac {2 d \sqrt {a+b x} (3 b c-a d) (3 a d+b c)}{c \sqrt {c+d x} (b c-a d)}-\frac {2 \int -\frac {3 (b c-a d)^3}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{c (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} (a d+3 b c)}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 (b c-a d)^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{c}+\frac {2 d \sqrt {a+b x} (3 b c-a d) (3 a d+b c)}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} (a d+3 b c)}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {\frac {\frac {6 (b c-a d)^2 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{c}+\frac {2 d \sqrt {a+b x} (3 b c-a d) (3 a d+b c)}{c \sqrt {c+d x} (b c-a d)}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} (a d+3 b c)}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {2 d \sqrt {a+b x} (3 b c-a d) (3 a d+b c)}{c \sqrt {c+d x} (b c-a d)}-\frac {6 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}}{3 c (b c-a d)}+\frac {2 d \sqrt {a+b x} (a d+3 b c)}{3 c (c+d x)^{3/2} (b c-a d)}}{a (b c-a d)}+\frac {2 b}{a \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\) |
(2*b)/(a*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) + ((2*d*(3*b*c + a*d)* Sqrt[a + b*x])/(3*c*(b*c - a*d)*(c + d*x)^(3/2)) + ((2*d*(3*b*c - a*d)*(b* c + 3*a*d)*Sqrt[a + b*x])/(c*(b*c - a*d)*Sqrt[c + d*x]) - (6*(b*c - a*d)^2 *ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(3/2 )))/(3*c*(b*c - a*d)))/(a*(b*c - a*d))
3.8.90.3.1 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[(((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[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[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*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2 *n, 2*p]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n, 2*p]
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. \(1252\) vs. \(2(152)=304\).
Time = 1.72 (sec) , antiderivative size = 1253, normalized size of antiderivative = 6.96
-1/3*(-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 ^3*b*c^3*d^2+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^2*b^2*c^4*d+16*a*b^2*c*d^3*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+ 8*a^2*b*c*d^3*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+18*a*b^2*c^2*d^2*x*(a* c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-3*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^4*c^5*x+3*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^4*c^2*d^3-3*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^3*c^5+6*b^3*c^4*(a*c)^(1/2)*((b*x+a)*(d *x+c))^(1/2)+3*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^4*d^5*x^2+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^2*b^2*c^3*d^2*x+3*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^3*c^4*d*x-6*a^2*b*d^4*x^2*(a*c)^(1/2)*((b*x+a)*(d*x +c))^(1/2)+6*b^3*c^2*d^2*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*b^3*c^ 3*d*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+18*a^2*b*c^2*d^2*(a*c)^(1/2)*((b *x+a)*(d*x+c))^(1/2)-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^2*b^2*c*d^4*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^3*c^2*d^3*x^3-3*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*b*c*d^4*x^2-9*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^2*b^2*c^2*d^3*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^3*c^3*d^2*x^2...
Leaf count of result is larger than twice the leaf count of optimal. 658 vs. \(2 (152) = 304\).
Time = 0.83 (sec) , antiderivative size = 1336, normalized size of antiderivative = 7.42 \[ \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \]
[1/6*(3*(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^ 4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^ 4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + ( b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4) *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 + (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) + 4*(3*a*b^3*c^5 + 9*a^3*b*c^3*d^2 - 4*a^4*c^2*d^3 + (3*a *b^3*c^3*d^2 + 8*a^2*b^2*c^2*d^3 - 3*a^3*b*c*d^4)*x^2 + (6*a*b^3*c^4*d + 9 *a^2*b^2*c^3*d^2 + 4*a^3*b*c^2*d^3 - 3*a^4*c*d^4)*x)*sqrt(b*x + a)*sqrt(d* x + c))/(a^3*b^3*c^8 - 3*a^4*b^2*c^7*d + 3*a^5*b*c^6*d^2 - a^6*c^5*d^3 + ( a^2*b^4*c^6*d^2 - 3*a^3*b^3*c^5*d^3 + 3*a^4*b^2*c^4*d^4 - a^5*b*c^3*d^5)*x ^3 + (2*a^2*b^4*c^7*d - 5*a^3*b^3*c^6*d^2 + 3*a^4*b^2*c^5*d^3 + a^5*b*c^4* d^4 - a^6*c^3*d^5)*x^2 + (a^2*b^4*c^8 - a^3*b^3*c^7*d - 3*a^4*b^2*c^6*d^2 + 5*a^5*b*c^5*d^3 - 2*a^6*c^4*d^4)*x), 1/3*(3*(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2* b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^ 2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^ 3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)*sqrt(-a*c)*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)) + 2*(3*a*b^3*c^5 + 9*a^3*b*c^3*d^2 - 4*a^4*...
\[ \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x \left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {5}{2}} x} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (152) = 304\).
Time = 0.70 (sec) , antiderivative size = 520, normalized size of antiderivative = 2.89 \[ \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {4 \, \sqrt {b d} b^{4}}{{\left (a b^{2} c^{2} {\left | b \right |} - 2 \, a^{2} b c d {\left | b \right |} + a^{3} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} + \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (8 \, b^{6} c^{5} d^{4} {\left | b \right |} - 19 \, a b^{5} c^{4} d^{5} {\left | b \right |} + 14 \, a^{2} b^{4} c^{3} d^{6} {\left | b \right |} - 3 \, a^{3} b^{3} c^{2} d^{7} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{7} c^{9} d - 5 \, a b^{6} c^{8} d^{2} + 10 \, a^{2} b^{5} c^{7} d^{3} - 10 \, a^{3} b^{4} c^{6} d^{4} + 5 \, a^{4} b^{3} c^{5} d^{5} - a^{5} b^{2} c^{4} d^{6}} + \frac {3 \, {\left (3 \, b^{7} c^{6} d^{3} {\left | b \right |} - 10 \, a b^{6} c^{5} d^{4} {\left | b \right |} + 12 \, a^{2} b^{5} c^{4} d^{5} {\left | b \right |} - 6 \, a^{3} b^{4} c^{3} d^{6} {\left | b \right |} + a^{4} b^{3} c^{2} d^{7} {\left | b \right |}\right )}}{b^{7} c^{9} d - 5 \, a b^{6} c^{8} d^{2} + 10 \, a^{2} b^{5} c^{7} d^{3} - 10 \, a^{3} b^{4} c^{6} d^{4} + 5 \, a^{4} b^{3} c^{5} d^{5} - a^{5} b^{2} c^{4} d^{6}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {2 \, \sqrt {b d} b \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} a c^{2} {\left | b \right |}} \]
4*sqrt(b*d)*b^4/((a*b^2*c^2*abs(b) - 2*a^2*b*c*d*abs(b) + a^3*d^2*abs(b))* (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)) + 2/3*sqrt(b*x + a)*((8*b^6*c^5*d^4*abs(b) - 19*a*b^5*c^4*d^5*a bs(b) + 14*a^2*b^4*c^3*d^6*abs(b) - 3*a^3*b^3*c^2*d^7*abs(b))*(b*x + a)/(b ^7*c^9*d - 5*a*b^6*c^8*d^2 + 10*a^2*b^5*c^7*d^3 - 10*a^3*b^4*c^6*d^4 + 5*a ^4*b^3*c^5*d^5 - a^5*b^2*c^4*d^6) + 3*(3*b^7*c^6*d^3*abs(b) - 10*a*b^6*c^5 *d^4*abs(b) + 12*a^2*b^5*c^4*d^5*abs(b) - 6*a^3*b^4*c^3*d^6*abs(b) + a^4*b ^3*c^2*d^7*abs(b))/(b^7*c^9*d - 5*a*b^6*c^8*d^2 + 10*a^2*b^5*c^7*d^3 - 10* a^3*b^4*c^6*d^4 + 5*a^4*b^3*c^5*d^5 - a^5*b^2*c^4*d^6))/(b^2*c + (b*x + a) *b*d - a*b*d)^(3/2) - 2*sqrt(b*d)*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*c^2*abs(b))
Timed out. \[ \int \frac {1}{x (a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{x\,{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \]