Integrand size = 33, antiderivative size = 260 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {3 b e^2}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{3 (b d-a e)^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b e}{(b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^3 (a+b x)}{(b d-a e)^4 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 b e^3 (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 b e^3 (a+b x) \log (d+e x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
-3*b*e^2/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)-1/3*b/(-a*e+b*d)^2/(b*x+a)^2/((b*x +a)^2)^(1/2)+b*e/(-a*e+b*d)^3/(b*x+a)/((b*x+a)^2)^(1/2)-e^3*(b*x+a)/(-a*e+ b*d)^4/(e*x+d)/((b*x+a)^2)^(1/2)-4*b*e^3*(b*x+a)*ln(b*x+a)/(-a*e+b*d)^5/(( b*x+a)^2)^(1/2)+4*b*e^3*(b*x+a)*ln(e*x+d)/(-a*e+b*d)^5/((b*x+a)^2)^(1/2)
Time = 1.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.55 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-b (b d-a e)^3+3 b e (b d-a e)^2 (a+b x)-9 b e^2 (b d-a e) (a+b x)^2+\frac {3 e^3 (-b d+a e) (a+b x)^3}{d+e x}-12 b e^3 (a+b x)^3 \log (a+b x)+12 b e^3 (a+b x)^3 \log (d+e x)}{3 (b d-a e)^5 \left ((a+b x)^2\right )^{3/2}} \] Input:
Integrate[(a + b*x)/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
Output:
(-(b*(b*d - a*e)^3) + 3*b*e*(b*d - a*e)^2*(a + b*x) - 9*b*e^2*(b*d - a*e)* (a + b*x)^2 + (3*e^3*(-(b*d) + a*e)*(a + b*x)^3)/(d + e*x) - 12*b*e^3*(a + b*x)^3*Log[a + b*x] + 12*b*e^3*(a + b*x)^3*Log[d + e*x])/(3*(b*d - a*e)^5 *((a + b*x)^2)^(3/2))
Time = 0.60 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 27, 54, 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.
| \(\displaystyle \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {1}{b^5 (a+b x)^4 (d+e x)^2}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^4 (d+e x)^2}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {4 b e^4}{(b d-a e)^5 (d+e x)}+\frac {e^4}{(b d-a e)^4 (d+e x)^2}-\frac {4 b^2 e^3}{(b d-a e)^5 (a+b x)}+\frac {3 b^2 e^2}{(b d-a e)^4 (a+b x)^2}-\frac {2 b^2 e}{(b d-a e)^3 (a+b x)^3}+\frac {b^2}{(b d-a e)^2 (a+b x)^4}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {e^3}{(d+e x) (b d-a e)^4}-\frac {4 b e^3 \log (a+b x)}{(b d-a e)^5}+\frac {4 b e^3 \log (d+e x)}{(b d-a e)^5}-\frac {3 b e^2}{(a+b x) (b d-a e)^4}+\frac {b e}{(a+b x)^2 (b d-a e)^3}-\frac {b}{3 (a+b x)^3 (b d-a e)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(a + b*x)/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
Output:
((a + b*x)*(-1/3*b/((b*d - a*e)^2*(a + b*x)^3) + (b*e)/((b*d - a*e)^3*(a + b*x)^2) - (3*b*e^2)/((b*d - a*e)^4*(a + b*x)) - e^3/((b*d - a*e)^4*(d + e *x)) - (4*b*e^3*Log[a + b*x])/(b*d - a*e)^5 + (4*b*e^3*Log[d + e*x])/(b*d - a*e)^5))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
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[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(482\) vs. \(2(192)=384\).
Time = 1.54 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.86
| method | result | size |
| default | \(\frac {\left (-3 a^{4} e^{4}-36 \ln \left (e x +d \right ) a \,b^{3} e^{4} x^{3}-12 \ln \left (e x +d \right ) b^{4} d \,e^{3} x^{3}-36 \ln \left (e x +d \right ) a^{2} b^{2} e^{4} x^{2}-12 \ln \left (e x +d \right ) a^{3} b \,e^{4} x -10 a^{3} b d \,e^{3}-6 a \,b^{3} d^{3} e +18 a^{2} b^{2} d^{2} e^{2}+36 \ln \left (b x +a \right ) x^{3} a \,b^{3} e^{4}+12 \ln \left (b x +a \right ) x^{3} b^{4} d \,e^{3}+36 \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{4}+b^{4} d^{4}-12 \ln \left (e x +d \right ) a^{3} b d \,e^{3}+36 \ln \left (b x +a \right ) x \,a^{2} b^{2} d \,e^{3}+36 \ln \left (b x +a \right ) x^{2} a \,b^{3} d \,e^{3}+12 \ln \left (b x +a \right ) b^{4} e^{4} x^{4}-12 \ln \left (e x +d \right ) b^{4} e^{4} x^{4}-36 \ln \left (e x +d \right ) a \,b^{3} d \,e^{3} x^{2}-36 \ln \left (e x +d \right ) a^{2} b^{2} d \,e^{3} x +6 x^{2} b^{4} d^{2} e^{2}-22 x \,a^{3} b \,e^{4}-2 x \,b^{4} d^{3} e +12 \ln \left (b x +a \right ) x \,a^{3} b \,e^{4}+12 \ln \left (b x +a \right ) a^{3} b d \,e^{3}+24 x^{2} a \,b^{3} d \,e^{3}+6 x \,a^{2} b^{2} d \,e^{3}+18 x a \,b^{3} d^{2} e^{2}-12 x^{3} a \,b^{3} e^{4}+12 x^{3} b^{4} d \,e^{3}-30 x^{2} a^{2} b^{2} e^{4}\right ) \left (b x +a \right )^{2}}{3 \left (e x +d \right ) \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(483\) |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {4 b^{3} e^{3} x^{3}}{a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}-\frac {2 b^{2} \left (5 a e +b d \right ) e^{2} x^{2}}{a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}-\frac {2 \left (11 e^{2} a^{2}+8 a b d e -b^{2} d^{2}\right ) b e x}{3 \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}-\frac {3 e^{3} a^{3}+13 a^{2} b d \,e^{2}-5 a \,b^{2} d^{2} e +b^{3} d^{3}}{3 \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\right )}{\left (b x +a \right )^{4} \left (e x +d \right )}-\frac {4 \sqrt {\left (b x +a \right )^{2}}\, b \,e^{3} \ln \left (e x +d \right )}{\left (b x +a \right ) \left (e^{5} a^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}+\frac {4 \sqrt {\left (b x +a \right )^{2}}\, b \,e^{3} \ln \left (-b x -a \right )}{\left (b x +a \right ) \left (e^{5} a^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}\) | \(518\) |
Input:
int((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*(-3*a^4*e^4-36*ln(e*x+d)*a*b^3*e^4*x^3-12*ln(e*x+d)*b^4*d*e^3*x^3-36*l n(e*x+d)*a^2*b^2*e^4*x^2-12*ln(e*x+d)*a^3*b*e^4*x-10*a^3*b*d*e^3-6*a*b^3*d ^3*e+18*a^2*b^2*d^2*e^2+36*ln(b*x+a)*x^3*a*b^3*e^4+12*ln(b*x+a)*x^3*b^4*d* e^3+36*ln(b*x+a)*x^2*a^2*b^2*e^4+b^4*d^4-12*ln(e*x+d)*a^3*b*d*e^3+36*ln(b* x+a)*x*a^2*b^2*d*e^3+36*ln(b*x+a)*x^2*a*b^3*d*e^3+12*ln(b*x+a)*b^4*e^4*x^4 -12*ln(e*x+d)*b^4*e^4*x^4-36*ln(e*x+d)*a*b^3*d*e^3*x^2-36*ln(e*x+d)*a^2*b^ 2*d*e^3*x+6*x^2*b^4*d^2*e^2-22*x*a^3*b*e^4-2*x*b^4*d^3*e+12*ln(b*x+a)*x*a^ 3*b*e^4+12*ln(b*x+a)*a^3*b*d*e^3+24*x^2*a*b^3*d*e^3+6*x*a^2*b^2*d*e^3+18*x *a*b^3*d^2*e^2-12*x^3*a*b^3*e^4+12*x^3*b^4*d*e^3-30*x^2*a^2*b^2*e^4)*(b*x+ a)^2/(e*x+d)/(a*e-b*d)^5/((b*x+a)^2)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 751 vs. \(2 (192) = 384\).
Time = 0.09 (sec) , antiderivative size = 751, normalized size of antiderivative = 2.89 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {b^{4} d^{4} - 6 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 10 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} + 4 \, a b^{3} d e^{3} - 5 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (b^{4} d^{3} e - 9 \, a b^{3} d^{2} e^{2} - 3 \, a^{2} b^{2} d e^{3} + 11 \, a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} e^{4} x^{4} + a^{3} b d e^{3} + {\left (b^{4} d e^{3} + 3 \, a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + {\left (3 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right ) - 12 \, {\left (b^{4} e^{4} x^{4} + a^{3} b d e^{3} + {\left (b^{4} d e^{3} + 3 \, a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + {\left (3 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (a^{3} b^{5} d^{6} - 5 \, a^{4} b^{4} d^{5} e + 10 \, a^{5} b^{3} d^{4} e^{2} - 10 \, a^{6} b^{2} d^{3} e^{3} + 5 \, a^{7} b d^{2} e^{4} - a^{8} d e^{5} + {\left (b^{8} d^{5} e - 5 \, a b^{7} d^{4} e^{2} + 10 \, a^{2} b^{6} d^{3} e^{3} - 10 \, a^{3} b^{5} d^{2} e^{4} + 5 \, a^{4} b^{4} d e^{5} - a^{5} b^{3} e^{6}\right )} x^{4} + {\left (b^{8} d^{6} - 2 \, a b^{7} d^{5} e - 5 \, a^{2} b^{6} d^{4} e^{2} + 20 \, a^{3} b^{5} d^{3} e^{3} - 25 \, a^{4} b^{4} d^{2} e^{4} + 14 \, a^{5} b^{3} d e^{5} - 3 \, a^{6} b^{2} e^{6}\right )} x^{3} + 3 \, {\left (a b^{7} d^{6} - 4 \, a^{2} b^{6} d^{5} e + 5 \, a^{3} b^{5} d^{4} e^{2} - 5 \, a^{5} b^{3} d^{2} e^{4} + 4 \, a^{6} b^{2} d e^{5} - a^{7} b e^{6}\right )} x^{2} + {\left (3 \, a^{2} b^{6} d^{6} - 14 \, a^{3} b^{5} d^{5} e + 25 \, a^{4} b^{4} d^{4} e^{2} - 20 \, a^{5} b^{3} d^{3} e^{3} + 5 \, a^{6} b^{2} d^{2} e^{4} + 2 \, a^{7} b d e^{5} - a^{8} e^{6}\right )} x\right )}} \] Input:
integrate((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fric as")
Output:
-1/3*(b^4*d^4 - 6*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 10*a^3*b*d*e^3 - 3*a^ 4*e^4 + 12*(b^4*d*e^3 - a*b^3*e^4)*x^3 + 6*(b^4*d^2*e^2 + 4*a*b^3*d*e^3 - 5*a^2*b^2*e^4)*x^2 - 2*(b^4*d^3*e - 9*a*b^3*d^2*e^2 - 3*a^2*b^2*d*e^3 + 11 *a^3*b*e^4)*x + 12*(b^4*e^4*x^4 + a^3*b*d*e^3 + (b^4*d*e^3 + 3*a*b^3*e^4)* x^3 + 3*(a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + (3*a^2*b^2*d*e^3 + a^3*b*e^4)*x) *log(b*x + a) - 12*(b^4*e^4*x^4 + a^3*b*d*e^3 + (b^4*d*e^3 + 3*a*b^3*e^4)* x^3 + 3*(a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + (3*a^2*b^2*d*e^3 + a^3*b*e^4)*x) *log(e*x + d))/(a^3*b^5*d^6 - 5*a^4*b^4*d^5*e + 10*a^5*b^3*d^4*e^2 - 10*a^ 6*b^2*d^3*e^3 + 5*a^7*b*d^2*e^4 - a^8*d*e^5 + (b^8*d^5*e - 5*a*b^7*d^4*e^2 + 10*a^2*b^6*d^3*e^3 - 10*a^3*b^5*d^2*e^4 + 5*a^4*b^4*d*e^5 - a^5*b^3*e^6 )*x^4 + (b^8*d^6 - 2*a*b^7*d^5*e - 5*a^2*b^6*d^4*e^2 + 20*a^3*b^5*d^3*e^3 - 25*a^4*b^4*d^2*e^4 + 14*a^5*b^3*d*e^5 - 3*a^6*b^2*e^6)*x^3 + 3*(a*b^7*d^ 6 - 4*a^2*b^6*d^5*e + 5*a^3*b^5*d^4*e^2 - 5*a^5*b^3*d^2*e^4 + 4*a^6*b^2*d* e^5 - a^7*b*e^6)*x^2 + (3*a^2*b^6*d^6 - 14*a^3*b^5*d^5*e + 25*a^4*b^4*d^4* e^2 - 20*a^5*b^3*d^3*e^3 + 5*a^6*b^2*d^2*e^4 + 2*a^7*b*d*e^5 - a^8*e^6)*x)
Timed out. \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)/(e*x+d)**2/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
Output:
Timed out
Exception generated. \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxi ma")
Output:
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*e-b*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (192) = 384\).
Time = 0.18 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.65 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 \, b^{2} e^{3} \log \left ({\left | b x + a \right |}\right )}{b^{6} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} b e^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {4 \, b e^{4} \log \left ({\left | e x + d \right |}\right )}{b^{5} d^{5} e \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {b^{4} d^{4} - 6 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 10 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} + 4 \, a b^{3} d e^{3} - 5 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (b^{4} d^{3} e - 9 \, a b^{3} d^{2} e^{2} - 3 \, a^{2} b^{2} d e^{3} + 11 \, a^{3} b e^{4}\right )} x}{3 \, {\left (b d - a e\right )}^{5} {\left (b x + a\right )}^{3} {\left (e x + d\right )} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac ")
Output:
-4*b^2*e^3*log(abs(b*x + a))/(b^6*d^5*sgn(b*x + a) - 5*a*b^5*d^4*e*sgn(b*x + a) + 10*a^2*b^4*d^3*e^2*sgn(b*x + a) - 10*a^3*b^3*d^2*e^3*sgn(b*x + a) + 5*a^4*b^2*d*e^4*sgn(b*x + a) - a^5*b*e^5*sgn(b*x + a)) + 4*b*e^4*log(abs (e*x + d))/(b^5*d^5*e*sgn(b*x + a) - 5*a*b^4*d^4*e^2*sgn(b*x + a) + 10*a^2 *b^3*d^3*e^3*sgn(b*x + a) - 10*a^3*b^2*d^2*e^4*sgn(b*x + a) + 5*a^4*b*d*e^ 5*sgn(b*x + a) - a^5*e^6*sgn(b*x + a)) - 1/3*(b^4*d^4 - 6*a*b^3*d^3*e + 18 *a^2*b^2*d^2*e^2 - 10*a^3*b*d*e^3 - 3*a^4*e^4 + 12*(b^4*d*e^3 - a*b^3*e^4) *x^3 + 6*(b^4*d^2*e^2 + 4*a*b^3*d*e^3 - 5*a^2*b^2*e^4)*x^2 - 2*(b^4*d^3*e - 9*a*b^3*d^2*e^2 - 3*a^2*b^2*d*e^3 + 11*a^3*b*e^4)*x)/((b*d - a*e)^5*(b*x + a)^3*(e*x + d)*sgn(b*x + a))
Timed out. \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {a+b\,x}{{\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:
int((a + b*x)/((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)
Output:
int((a + b*x)/((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)
Time = 0.28 (sec) , antiderivative size = 1236, normalized size of antiderivative = 4.75 \[ \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
int((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
Output:
(36*log(a + b*x)*a**4*b*d*e**4 + 36*log(a + b*x)*a**4*b*e**5*x + 12*log(a + b*x)*a**3*b**2*d**2*e**3 + 120*log(a + b*x)*a**3*b**2*d*e**4*x + 108*log (a + b*x)*a**3*b**2*e**5*x**2 + 36*log(a + b*x)*a**2*b**3*d**2*e**3*x + 14 4*log(a + b*x)*a**2*b**3*d*e**4*x**2 + 108*log(a + b*x)*a**2*b**3*e**5*x** 3 + 36*log(a + b*x)*a*b**4*d**2*e**3*x**2 + 72*log(a + b*x)*a*b**4*d*e**4* x**3 + 36*log(a + b*x)*a*b**4*e**5*x**4 + 12*log(a + b*x)*b**5*d**2*e**3*x **3 + 12*log(a + b*x)*b**5*d*e**4*x**4 - 36*log(d + e*x)*a**4*b*d*e**4 - 3 6*log(d + e*x)*a**4*b*e**5*x - 12*log(d + e*x)*a**3*b**2*d**2*e**3 - 120*l og(d + e*x)*a**3*b**2*d*e**4*x - 108*log(d + e*x)*a**3*b**2*e**5*x**2 - 36 *log(d + e*x)*a**2*b**3*d**2*e**3*x - 144*log(d + e*x)*a**2*b**3*d*e**4*x* *2 - 108*log(d + e*x)*a**2*b**3*e**5*x**3 - 36*log(d + e*x)*a*b**4*d**2*e* *3*x**2 - 72*log(d + e*x)*a*b**4*d*e**4*x**3 - 36*log(d + e*x)*a*b**4*e**5 *x**4 - 12*log(d + e*x)*b**5*d**2*e**3*x**3 - 12*log(d + e*x)*b**5*d*e**4* x**4 - 9*a**5*e**5 - 21*a**4*b*d*e**4 - 54*a**4*b*e**5*x + 32*a**3*b**2*d* *2*e**3 + 20*a**3*b**2*d*e**4*x - 54*a**3*b**2*e**5*x**2 + 24*a**2*b**3*d* *2*e**3*x + 42*a**2*b**3*d*e**4*x**2 - 3*a*b**4*d**4*e + 12*a*b**4*d**3*e* *2*x + 6*a*b**4*d**2*e**3*x**2 + 12*a*b**4*e**5*x**4 + b**5*d**5 - 2*b**5* d**4*e*x + 6*b**5*d**3*e**2*x**2 - 12*b**5*d*e**4*x**4)/(3*(3*a**9*d*e**6 + 3*a**9*e**7*x - 14*a**8*b*d**2*e**5 - 5*a**8*b*d*e**6*x + 9*a**8*b*e**7* x**2 + 25*a**7*b**2*d**3*e**4 - 17*a**7*b**2*d**2*e**5*x - 33*a**7*b**2...