Integrand size = 28, antiderivative size = 231 \[ \int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a+b x}{3 (b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{2 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]
1/3*(b*x+a)/(-a*e+b*d)/(e*x+d)^3/((b*x+a)^2)^(1/2)+1/2*b*(b*x+a)/(-a*e+b*d )^2/(e*x+d)^2/((b*x+a)^2)^(1/2)+b^2*(b*x+a)/(-a*e+b*d)^3/(e*x+d)/((b*x+a)^ 2)^(1/2)+b^3*(b*x+a)*ln(b*x+a)/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)-b^3*(b*x+a)* ln(e*x+d)/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)
Time = 1.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \left (2 (b d-a e)^3+3 b (b d-a e)^2 (d+e x)+6 b^2 (b d-a e) (d+e x)^2+6 b^3 (d+e x)^3 \log (a+b x)-6 b^3 (d+e x)^3 \log (d+e x)\right )}{6 (b d-a e)^4 \sqrt {(a+b x)^2} (d+e x)^3} \]
((a + b*x)*(2*(b*d - a*e)^3 + 3*b*(b*d - a*e)^2*(d + e*x) + 6*b^2*(b*d - a *e)*(d + e*x)^2 + 6*b^3*(d + e*x)^3*Log[a + b*x] - 6*b^3*(d + e*x)^3*Log[d + e*x]))/(6*(b*d - a*e)^4*Sqrt[(a + b*x)^2]*(d + e*x)^3)
Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.57, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 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.
\(\displaystyle \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {b (a+b x) \int \frac {1}{b (a+b x) (d+e x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x) (d+e x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {b^4}{(b d-a e)^4 (a+b x)}-\frac {e b^3}{(b d-a e)^4 (d+e x)}-\frac {e b^2}{(b d-a e)^3 (d+e x)^2}-\frac {e b}{(b d-a e)^2 (d+e x)^3}-\frac {e}{(b d-a e) (d+e x)^4}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (\frac {b^3 \log (a+b x)}{(b d-a e)^4}-\frac {b^3 \log (d+e x)}{(b d-a e)^4}+\frac {b^2}{(d+e x) (b d-a e)^3}+\frac {b}{2 (d+e x)^2 (b d-a e)^2}+\frac {1}{3 (d+e x)^3 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(1/(3*(b*d - a*e)*(d + e*x)^3) + b/(2*(b*d - a*e)^2*(d + e*x)^2 ) + b^2/((b*d - a*e)^3*(d + e*x)) + (b^3*Log[a + b*x])/(b*d - a*e)^4 - (b^ 3*Log[d + e*x])/(b*d - a*e)^4))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.16.95.3.1 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_.), 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_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.40 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.11
method | result | size |
default | \(\frac {\left (b x +a \right ) \left (6 \ln \left (b x +a \right ) x^{3} b^{3} e^{3}-6 \ln \left (e x +d \right ) b^{3} e^{3} x^{3}+18 \ln \left (b x +a \right ) b^{3} d \,e^{2} x^{2}-18 \ln \left (e x +d \right ) b^{3} d \,e^{2} x^{2}+18 \ln \left (b x +a \right ) x \,b^{3} d^{2} e -18 \ln \left (e x +d \right ) b^{3} d^{2} e x -6 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+6 \ln \left (b x +a \right ) b^{3} d^{3}-6 \ln \left (e x +d \right ) b^{3} d^{3}+3 a^{2} b \,e^{3} x -18 x a \,b^{2} d \,e^{2}+15 b^{3} d^{2} e x -2 a^{3} e^{3}+9 a^{2} b d \,e^{2}-18 a \,b^{2} d^{2} e +11 b^{3} d^{3}\right )}{6 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{4} \left (e x +d \right )^{3}}\) | \(256\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{2} e^{2} x^{2}}{a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}}+\frac {\left (a e -5 b d \right ) b e x}{2 a^{3} e^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}}-\frac {2 a^{2} e^{2}-7 a b d e +11 b^{2} d^{2}}{6 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}\right )}{\left (b x +a \right ) \left (e x +d \right )^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} \ln \left (e x +d \right )}{\left (b x +a \right ) \left (e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} \ln \left (-b x -a \right )}{\left (b x +a \right ) \left (e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\) | \(348\) |
1/6*(b*x+a)*(6*ln(b*x+a)*x^3*b^3*e^3-6*ln(e*x+d)*b^3*e^3*x^3+18*ln(b*x+a)* b^3*d*e^2*x^2-18*ln(e*x+d)*b^3*d*e^2*x^2+18*ln(b*x+a)*x*b^3*d^2*e-18*ln(e* x+d)*b^3*d^2*e*x-6*x^2*a*b^2*e^3+6*x^2*b^3*d*e^2+6*ln(b*x+a)*b^3*d^3-6*ln( e*x+d)*b^3*d^3+3*a^2*b*e^3*x-18*x*a*b^2*d*e^2+15*b^3*d^2*e*x-2*a^3*e^3+9*a ^2*b*d*e^2-18*a*b^2*d^2*e+11*b^3*d^3)/((b*x+a)^2)^(1/2)/(a*e-b*d)^4/(e*x+d )^3
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (172) = 344\).
Time = 0.36 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.84 \[ \int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {11 \, b^{3} d^{3} - 18 \, a b^{2} d^{2} e + 9 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 3 \, {\left (5 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, b^{3} d e^{2} x^{2} + 3 \, b^{3} d^{2} e x + b^{3} d^{3}\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, b^{3} d e^{2} x^{2} + 3 \, b^{3} d^{2} e x + b^{3} d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (b^{4} d^{7} - 4 \, a b^{3} d^{6} e + 6 \, a^{2} b^{2} d^{5} e^{2} - 4 \, a^{3} b d^{4} e^{3} + a^{4} d^{3} e^{4} + {\left (b^{4} d^{4} e^{3} - 4 \, a b^{3} d^{3} e^{4} + 6 \, a^{2} b^{2} d^{2} e^{5} - 4 \, a^{3} b d e^{6} + a^{4} e^{7}\right )} x^{3} + 3 \, {\left (b^{4} d^{5} e^{2} - 4 \, a b^{3} d^{4} e^{3} + 6 \, a^{2} b^{2} d^{3} e^{4} - 4 \, a^{3} b d^{2} e^{5} + a^{4} d e^{6}\right )} x^{2} + 3 \, {\left (b^{4} d^{6} e - 4 \, a b^{3} d^{5} e^{2} + 6 \, a^{2} b^{2} d^{4} e^{3} - 4 \, a^{3} b d^{3} e^{4} + a^{4} d^{2} e^{5}\right )} x\right )}} \]
1/6*(11*b^3*d^3 - 18*a*b^2*d^2*e + 9*a^2*b*d*e^2 - 2*a^3*e^3 + 6*(b^3*d*e^ 2 - a*b^2*e^3)*x^2 + 3*(5*b^3*d^2*e - 6*a*b^2*d*e^2 + a^2*b*e^3)*x + 6*(b^ 3*e^3*x^3 + 3*b^3*d*e^2*x^2 + 3*b^3*d^2*e*x + b^3*d^3)*log(b*x + a) - 6*(b ^3*e^3*x^3 + 3*b^3*d*e^2*x^2 + 3*b^3*d^2*e*x + b^3*d^3)*log(e*x + d))/(b^4 *d^7 - 4*a*b^3*d^6*e + 6*a^2*b^2*d^5*e^2 - 4*a^3*b*d^4*e^3 + a^4*d^3*e^4 + (b^4*d^4*e^3 - 4*a*b^3*d^3*e^4 + 6*a^2*b^2*d^2*e^5 - 4*a^3*b*d*e^6 + a^4* e^7)*x^3 + 3*(b^4*d^5*e^2 - 4*a*b^3*d^4*e^3 + 6*a^2*b^2*d^3*e^4 - 4*a^3*b* d^2*e^5 + a^4*d*e^6)*x^2 + 3*(b^4*d^6*e - 4*a*b^3*d^5*e^2 + 6*a^2*b^2*d^4* e^3 - 4*a^3*b*d^3*e^4 + a^4*d^2*e^5)*x)
\[ \int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\left (d + e x\right )^{4} \sqrt {\left (a + b x\right )^{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Exception raised: ValueError} \]
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*e-b*d>0)', see `assume?` for m ore detail
Time = 0.27 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.09 \[ \int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {1}{6} \, {\left (\frac {6 \, b^{4} \log \left ({\left | b x + a \right |}\right )}{b^{5} d^{4} - 4 \, a b^{4} d^{3} e + 6 \, a^{2} b^{3} d^{2} e^{2} - 4 \, a^{3} b^{2} d e^{3} + a^{4} b e^{4}} - \frac {6 \, b^{3} e \log \left ({\left | e x + d \right |}\right )}{b^{4} d^{4} e - 4 \, a b^{3} d^{3} e^{2} + 6 \, a^{2} b^{2} d^{2} e^{3} - 4 \, a^{3} b d e^{4} + a^{4} e^{5}} + \frac {11 \, b^{3} d^{3} - 18 \, a b^{2} d^{2} e + 9 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 3 \, {\left (5 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{{\left (b d - a e\right )}^{4} {\left (e x + d\right )}^{3}}\right )} \mathrm {sgn}\left (b x + a\right ) \]
1/6*(6*b^4*log(abs(b*x + a))/(b^5*d^4 - 4*a*b^4*d^3*e + 6*a^2*b^3*d^2*e^2 - 4*a^3*b^2*d*e^3 + a^4*b*e^4) - 6*b^3*e*log(abs(e*x + d))/(b^4*d^4*e - 4* a*b^3*d^3*e^2 + 6*a^2*b^2*d^2*e^3 - 4*a^3*b*d*e^4 + a^4*e^5) + (11*b^3*d^3 - 18*a*b^2*d^2*e + 9*a^2*b*d*e^2 - 2*a^3*e^3 + 6*(b^3*d*e^2 - a*b^2*e^3)* x^2 + 3*(5*b^3*d^2*e - 6*a*b^2*d*e^2 + a^2*b*e^3)*x)/((b*d - a*e)^4*(e*x + d)^3))*sgn(b*x + a)
Timed out. \[ \int \frac {1}{(d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^4} \,d x \]