Integrand size = 29, antiderivative size = 187 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=-\frac {a^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {3 a^2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b^2 B \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}-\frac {A (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 a x^4}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \] Output:
-1/3*a^3*B*((b*x+a)^2)^(1/2)/x^3/(b*x+a)-3/2*a^2*b*B*((b*x+a)^2)^(1/2)/x^2 /(b*x+a)-3*a*b^2*B*((b*x+a)^2)^(1/2)/x/(b*x+a)-1/4*A*(b*x+a)^3*((b*x+a)^2) ^(1/2)/a/x^4+b^3*B*((b*x+a)^2)^(1/2)*ln(x)/(b*x+a)
Time = 1.05 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.49 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=\frac {\sqrt {a^2} \left (12 A b^3 x^3+18 a b^2 x^2 (A+2 B x)+6 a^2 b x (2 A+3 B x)+a^3 (3 A+4 B x)\right )-\sqrt {(a+b x)^2} \left (3 A b^3 x^3+a^3 (3 A+4 B x)+a^2 b x (9 A+14 B x)+a b^2 x^2 (9 A+22 B x)\right )-24 a b^3 B x^4 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )-24 \sqrt {a^2} b^3 B x^4 \log (x)+12 \sqrt {a^2} b^3 B x^4 \log \left (a \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )\right )+12 \sqrt {a^2} b^3 B x^4 \log \left (a \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right )}{24 a x^4} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^5,x]
Output:
(Sqrt[a^2]*(12*A*b^3*x^3 + 18*a*b^2*x^2*(A + 2*B*x) + 6*a^2*b*x*(2*A + 3*B *x) + a^3*(3*A + 4*B*x)) - Sqrt[(a + b*x)^2]*(3*A*b^3*x^3 + a^3*(3*A + 4*B *x) + a^2*b*x*(9*A + 14*B*x) + a*b^2*x^2*(9*A + 22*B*x)) - 24*a*b^3*B*x^4* ArcTanh[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])] - 24*Sqrt[a^2]*b^3*B*x^4*Lo g[x] + 12*Sqrt[a^2]*b^3*B*x^4*Log[a*(Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2])] + 12*Sqrt[a^2]*b^3*B*x^4*Log[a*(Sqrt[a^2] + b*x - Sqrt[(a + b*x)^2])])/(2 4*a*x^4)
Time = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.46, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1187, 27, 87, 49, 2009}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^5} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^5}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^5}dx}{a+b x}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (B \int \frac {(a+b x)^3}{x^4}dx-\frac {A (a+b x)^4}{4 a x^4}\right )}{a+b x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (B \int \left (\frac {a^3}{x^4}+\frac {3 b a^2}{x^3}+\frac {3 b^2 a}{x^2}+\frac {b^3}{x}\right )dx-\frac {A (a+b x)^4}{4 a x^4}\right )}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (B \left (-\frac {a^3}{3 x^3}-\frac {3 a^2 b}{2 x^2}-\frac {3 a b^2}{x}+b^3 \log (x)\right )-\frac {A (a+b x)^4}{4 a x^4}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^5,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/4*(A*(a + b*x)^4)/(a*x^4) + B*(-1/3*a^3 /x^3 - (3*a^2*b)/(2*x^2) - (3*a*b^2)/x + b^3*Log[x])))/(a + b*x)
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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(d + e*x)^m*(f + g*x)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 1.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.50
method | result | size |
default | \(-\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (-12 B \,b^{3} \ln \left (x \right ) x^{4}+12 A \,b^{3} x^{3}+36 B a \,b^{2} x^{3}+18 A a \,b^{2} x^{2}+18 B \,a^{2} b \,x^{2}+12 A \,a^{2} b x +4 B \,a^{3} x +3 a^{3} A \right )}{12 \left (b x +a \right )^{3} x^{4}}\) | \(94\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-A \,b^{3}-3 B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{2} A a \,b^{2}-\frac {3}{2} B \,a^{2} b \right ) x^{2}+\left (-A \,a^{2} b -\frac {1}{3} B \,a^{3}\right ) x -\frac {a^{3} A}{4}\right )}{\left (b x +a \right ) x^{4}}+\frac {b^{3} B \sqrt {\left (b x +a \right )^{2}}\, \ln \left (x \right )}{b x +a}\) | \(105\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/12*((b*x+a)^2)^(3/2)*(-12*B*b^3*ln(x)*x^4+12*A*b^3*x^3+36*B*a*b^2*x^3+1 8*A*a*b^2*x^2+18*B*a^2*b*x^2+12*A*a^2*b*x+4*B*a^3*x+3*a^3*A)/(b*x+a)^3/x^4
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.40 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=\frac {12 \, B b^{3} x^{4} \log \left (x\right ) - 3 \, A a^{3} - 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x, algorithm="fricas")
Output:
1/12*(12*B*b^3*x^4*log(x) - 3*A*a^3 - 12*(3*B*a*b^2 + A*b^3)*x^3 - 18*(B*a ^2*b + A*a*b^2)*x^2 - 4*(B*a^3 + 3*A*a^2*b)*x)/x^4
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{5}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**5,x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**5, x)
Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (126) = 252\).
Time = 0.04 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.03 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} B b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{4} x}{2 \, a^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{3}}{2 \, a} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{3}}{6 \, a^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{4}}{4 \, a^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2}}{2 \, a^{2} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{4 \, a^{3} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{4 \, a^{4} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{3 \, a^{2} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{4 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{4 \, a^{2} x^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x, algorithm="maxima")
Output:
(-1)^(2*b^2*x + 2*a*b)*B*b^3*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2) *B*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a ^2)*B*b^4*x/a^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^3/a - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^3/a^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A* b^4/a^4 - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^2/(a^2*x) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/(a^3*x) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2 )*B*b/(a^3*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^2) - 1/ 3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a ^2)^(5/2)*A*b/(a^3*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^4)
Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.65 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=B b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {3 \, A a^{3} \mathrm {sgn}\left (b x + a\right ) + 12 \, {\left (3 \, B a b^{2} \mathrm {sgn}\left (b x + a\right ) + A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 18 \, {\left (B a^{2} b \mathrm {sgn}\left (b x + a\right ) + A a b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 4 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} x}{12 \, x^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x, algorithm="giac")
Output:
B*b^3*log(abs(x))*sgn(b*x + a) - 1/12*(3*A*a^3*sgn(b*x + a) + 12*(3*B*a*b^ 2*sgn(b*x + a) + A*b^3*sgn(b*x + a))*x^3 + 18*(B*a^2*b*sgn(b*x + a) + A*a* b^2*sgn(b*x + a))*x^2 + 4*(B*a^3*sgn(b*x + a) + 3*A*a^2*b*sgn(b*x + a))*x) /x^4
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^5} \,d x \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^5,x)
Output:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^5, x)
Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.26 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx=\frac {12 \,\mathrm {log}\left (x \right ) b^{4} x^{4}-3 a^{4}-16 a^{3} b x -36 a^{2} b^{2} x^{2}-48 a \,b^{3} x^{3}}{12 x^{4}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^5,x)
Output:
(12*log(x)*b**4*x**4 - 3*a**4 - 16*a**3*b*x - 36*a**2*b**2*x**2 - 48*a*b** 3*x**3)/(12*x**4)