Integrand size = 24, antiderivative size = 244 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {20 a^3}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^6}{4 b^7 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^5}{b^7 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 a^4}{2 b^7 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a x (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 a^2 (a+b x) \log (a+b x)}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}} \]
20*a^3/b^7/((b*x+a)^2)^(1/2)-1/4*a^6/b^7/(b*x+a)^3/((b*x+a)^2)^(1/2)+2*a^5 /b^7/(b*x+a)^2/((b*x+a)^2)^(1/2)-15/2*a^4/b^7/(b*x+a)/((b*x+a)^2)^(1/2)-5* a*x*(b*x+a)/b^6/((b*x+a)^2)^(1/2)+1/2*x^2*(b*x+a)/b^5/((b*x+a)^2)^(1/2)+15 *a^2*(b*x+a)*ln(b*x+a)/b^7/((b*x+a)^2)^(1/2)
Time = 1.13 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.26 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\frac {b x \left (-a^2 b^5 x^5 \sqrt {(a+b x)^2}+a b^6 x^6 \sqrt {(a+b x)^2}+\sqrt {a^2} b^5 x^5 \left (-2 a^2+b^2 x^2\right )+10 a^5 b^2 x^2 \left (26 \sqrt {a^2}-11 \sqrt {(a+b x)^2}\right )+30 a^6 b x \left (7 \sqrt {a^2}-5 \sqrt {(a+b x)^2}\right )+5 a^4 b^3 x^3 \left (25 \sqrt {a^2}-3 \sqrt {(a+b x)^2}\right )+60 a^7 \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )+3 a^3 b^4 x^4 \left (4 \sqrt {a^2}+\sqrt {(a+b x)^2}\right )\right )}{(a+b x)^3 \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )}-120 a^4 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )}{4 a^2 b^7} \]
((b*x*(-(a^2*b^5*x^5*Sqrt[(a + b*x)^2]) + a*b^6*x^6*Sqrt[(a + b*x)^2] + Sq rt[a^2]*b^5*x^5*(-2*a^2 + b^2*x^2) + 10*a^5*b^2*x^2*(26*Sqrt[a^2] - 11*Sqr t[(a + b*x)^2]) + 30*a^6*b*x*(7*Sqrt[a^2] - 5*Sqrt[(a + b*x)^2]) + 5*a^4*b ^3*x^3*(25*Sqrt[a^2] - 3*Sqrt[(a + b*x)^2]) + 60*a^7*(Sqrt[a^2] - Sqrt[(a + b*x)^2]) + 3*a^3*b^4*x^4*(4*Sqrt[a^2] + Sqrt[(a + b*x)^2])))/((a + b*x)^ 3*(a^2 + a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2])) - 120*a^4*ArcTanh[(b*x)/(Sq rt[a^2] - Sqrt[(a + b*x)^2])])/(4*a^2*b^7)
Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.50, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1102, 27, 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.
\(\displaystyle \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {x^6}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {x^6}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {a^6}{b^6 (a+b x)^5}-\frac {6 a^5}{b^6 (a+b x)^4}+\frac {15 a^4}{b^6 (a+b x)^3}-\frac {20 a^3}{b^6 (a+b x)^2}+\frac {15 a^2}{b^6 (a+b x)}-\frac {5 a}{b^6}+\frac {x}{b^5}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {a^6}{4 b^7 (a+b x)^4}+\frac {2 a^5}{b^7 (a+b x)^3}-\frac {15 a^4}{2 b^7 (a+b x)^2}+\frac {20 a^3}{b^7 (a+b x)}+\frac {15 a^2 \log (a+b x)}{b^7}-\frac {5 a x}{b^6}+\frac {x^2}{2 b^5}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((-5*a*x)/b^6 + x^2/(2*b^5) - a^6/(4*b^7*(a + b*x)^4) + (2*a^5) /(b^7*(a + b*x)^3) - (15*a^4)/(2*b^7*(a + b*x)^2) + (20*a^3)/(b^7*(a + b*x )) + (15*a^2*Log[a + b*x])/b^7))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.2.98.3.1 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 [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[((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 = 2.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.48
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{2} b \,x^{2}-5 a x \right )}{\left (b x +a \right ) b^{6}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (20 a^{3} b^{2} x^{3}+\frac {105 a^{4} b \,x^{2}}{2}+47 a^{5} x +\frac {57 a^{6}}{4 b}\right )}{\left (b x +a \right )^{5} b^{6}}+\frac {15 \sqrt {\left (b x +a \right )^{2}}\, a^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{7}}\) | \(118\) |
default | \(\frac {\left (2 b^{6} x^{6}+60 \ln \left (b x +a \right ) a^{2} b^{4} x^{4}-12 a \,x^{5} b^{5}+240 \ln \left (b x +a \right ) a^{3} b^{3} x^{3}-68 a^{2} x^{4} b^{4}+360 \ln \left (b x +a \right ) a^{4} b^{2} x^{2}-32 a^{3} x^{3} b^{3}+240 \ln \left (b x +a \right ) a^{5} b x +132 a^{4} x^{2} b^{2}+60 \ln \left (b x +a \right ) a^{6}+168 a^{5} x b +57 a^{6}\right ) \left (b x +a \right )}{4 b^{7} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(158\) |
((b*x+a)^2)^(1/2)/(b*x+a)*(1/2*b*x^2-5*a*x)/b^6+((b*x+a)^2)^(1/2)/(b*x+a)^ 5*(20*a^3*b^2*x^3+105/2*a^4*b*x^2+47*a^5*x+57/4/b*a^6)/b^6+15*((b*x+a)^2)^ (1/2)/(b*x+a)/b^7*a^2*ln(b*x+a)
Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.66 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {2 \, b^{6} x^{6} - 12 \, a b^{5} x^{5} - 68 \, a^{2} b^{4} x^{4} - 32 \, a^{3} b^{3} x^{3} + 132 \, a^{4} b^{2} x^{2} + 168 \, a^{5} b x + 57 \, a^{6} + 60 \, {\left (a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{3} + 6 \, a^{4} b^{2} x^{2} + 4 \, a^{5} b x + a^{6}\right )} \log \left (b x + a\right )}{4 \, {\left (b^{11} x^{4} + 4 \, a b^{10} x^{3} + 6 \, a^{2} b^{9} x^{2} + 4 \, a^{3} b^{8} x + a^{4} b^{7}\right )}} \]
1/4*(2*b^6*x^6 - 12*a*b^5*x^5 - 68*a^2*b^4*x^4 - 32*a^3*b^3*x^3 + 132*a^4* b^2*x^2 + 168*a^5*b*x + 57*a^6 + 60*(a^2*b^4*x^4 + 4*a^3*b^3*x^3 + 6*a^4*b ^2*x^2 + 4*a^5*b*x + a^6)*log(b*x + a))/(b^11*x^4 + 4*a*b^10*x^3 + 6*a^2*b ^9*x^2 + 4*a^3*b^8*x + a^4*b^7)
\[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{6}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.52 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {2 \, b^{6} x^{6} - 12 \, a b^{5} x^{5} - 68 \, a^{2} b^{4} x^{4} - 32 \, a^{3} b^{3} x^{3} + 132 \, a^{4} b^{2} x^{2} + 168 \, a^{5} b x + 57 \, a^{6}}{4 \, {\left (b^{11} x^{4} + 4 \, a b^{10} x^{3} + 6 \, a^{2} b^{9} x^{2} + 4 \, a^{3} b^{8} x + a^{4} b^{7}\right )}} + \frac {15 \, a^{2} \log \left (b x + a\right )}{b^{7}} \]
1/4*(2*b^6*x^6 - 12*a*b^5*x^5 - 68*a^2*b^4*x^4 - 32*a^3*b^3*x^3 + 132*a^4* b^2*x^2 + 168*a^5*b*x + 57*a^6)/(b^11*x^4 + 4*a*b^10*x^3 + 6*a^2*b^9*x^2 + 4*a^3*b^8*x + a^4*b^7) + 15*a^2*log(b*x + a)/b^7
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.45 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {15 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{7} \mathrm {sgn}\left (b x + a\right )} + \frac {b^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a b^{4} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{10}} + \frac {80 \, a^{3} b^{3} x^{3} + 210 \, a^{4} b^{2} x^{2} + 188 \, a^{5} b x + 57 \, a^{6}}{4 \, {\left (b x + a\right )}^{4} b^{7} \mathrm {sgn}\left (b x + a\right )} \]
15*a^2*log(abs(b*x + a))/(b^7*sgn(b*x + a)) + 1/2*(b^5*x^2*sgn(b*x + a) - 10*a*b^4*x*sgn(b*x + a))/b^10 + 1/4*(80*a^3*b^3*x^3 + 210*a^4*b^2*x^2 + 18 8*a^5*b*x + 57*a^6)/((b*x + a)^4*b^7*sgn(b*x + a))
Timed out. \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^6}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]