Integrand size = 25, antiderivative size = 252 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {2 e f^2 p}{35 d x^5}+\frac {2 e^2 f^2 p}{21 d^2 x^3}-\frac {4 e f g p}{15 d x^3}-\frac {2 e^3 f^2 p}{7 d^3 x}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e g^2 p}{3 d x}-\frac {2 e^{7/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 d^{7/2}}+\frac {4 e^{5/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \]
-2/35*e*f^2*p/d/x^5+2/21*e^2*f^2*p/d^2/x^3-4/15*e*f*g*p/d/x^3-2/7*e^3*f^2* p/d^3/x+4/5*e^2*f*g*p/d^2/x-2/3*e*g^2*p/d/x-2/7*e^(7/2)*f^2*p*arctan(x*e^( 1/2)/d^(1/2))/d^(7/2)+4/5*e^(5/2)*f*g*p*arctan(x*e^(1/2)/d^(1/2))/d^(5/2)- 2/3*e^(3/2)*g^2*p*arctan(x*e^(1/2)/d^(1/2))/d^(3/2)-1/7*f^2*ln(c*(e*x^2+d) ^p)/x^7-2/5*f*g*ln(c*(e*x^2+d)^p)/x^5-1/3*g^2*ln(c*(e*x^2+d)^p)/x^3
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.64 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {2 e f^2 p \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\frac {e x^2}{d}\right )}{35 d x^5}-\frac {4 e f g p \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {e x^2}{d}\right )}{15 d x^3}-\frac {2 e g^2 p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {e x^2}{d}\right )}{3 d x}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3} \]
(-2*e*f^2*p*Hypergeometric2F1[-5/2, 1, -3/2, -((e*x^2)/d)])/(35*d*x^5) - ( 4*e*f*g*p*Hypergeometric2F1[-3/2, 1, -1/2, -((e*x^2)/d)])/(15*d*x^3) - (2* e*g^2*p*Hypergeometric2F1[-1/2, 1, 1/2, -((e*x^2)/d)])/(3*d*x) - (f^2*Log[ c*(d + e*x^2)^p])/(7*x^7) - (2*f*g*Log[c*(d + e*x^2)^p])/(5*x^5) - (g^2*Lo g[c*(d + e*x^2)^p])/(3*x^3)
Time = 0.40 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2926, 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.
| \(\displaystyle \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx\) |
| \(\Big \downarrow \) 2926 |
| \(\displaystyle \int \left (\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8}+\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{x^6}+\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 e^{7/2} f^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{7 d^{7/2}}+\frac {4 e^{5/2} f g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 d^{5/2}}-\frac {2 e^{3/2} g^2 p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^{3/2}}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{7 x^7}-\frac {2 f g \log \left (c \left (d+e x^2\right )^p\right )}{5 x^5}-\frac {g^2 \log \left (c \left (d+e x^2\right )^p\right )}{3 x^3}-\frac {2 e^3 f^2 p}{7 d^3 x}+\frac {2 e^2 f^2 p}{21 d^2 x^3}+\frac {4 e^2 f g p}{5 d^2 x}-\frac {2 e f^2 p}{35 d x^5}-\frac {4 e f g p}{15 d x^3}-\frac {2 e g^2 p}{3 d x}\) |
(-2*e*f^2*p)/(35*d*x^5) + (2*e^2*f^2*p)/(21*d^2*x^3) - (4*e*f*g*p)/(15*d*x ^3) - (2*e^3*f^2*p)/(7*d^3*x) + (4*e^2*f*g*p)/(5*d^2*x) - (2*e*g^2*p)/(3*d *x) - (2*e^(7/2)*f^2*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(7*d^(7/2)) + (4*e^(5/ 2)*f*g*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(5*d^(5/2)) - (2*e^(3/2)*g^2*p*ArcTa n[(Sqrt[e]*x)/Sqrt[d]])/(3*d^(3/2)) - (f^2*Log[c*(d + e*x^2)^p])/(7*x^7) - (2*f*g*Log[c*(d + e*x^2)^p])/(5*x^5) - (g^2*Log[c*(d + e*x^2)^p])/(3*x^3)
3.4.37.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b *Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c, d, e , f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] & & IntegerQ[s]
Time = 2.39 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.66
| method | result | size |
| parts | \(-\frac {g^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3 x^{3}}-\frac {2 f g \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5 x^{5}}-\frac {f^{2} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{7 x^{7}}-\frac {2 p e \left (-\frac {-35 g^{2} d^{2}+42 d e f g -15 e^{2} f^{2}}{d^{3} x}+\frac {3 f^{2}}{d \,x^{5}}+\frac {f \left (14 d g -5 e f \right )}{d^{2} x^{3}}+\frac {e \left (35 g^{2} d^{2}-42 d e f g +15 e^{2} f^{2}\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{d^{3} \sqrt {d e}}\right )}{105}\) | \(167\) |
| risch | \(-\frac {\left (35 g^{2} x^{4}+42 f g \,x^{2}+15 f^{2}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{105 x^{7}}-\frac {30 \ln \left (c \right ) d^{4} f^{2}-70 \sqrt {-d e}\, p e \ln \left (-e x +\sqrt {-d e}\right ) g^{2} d^{2} x^{7}+70 \sqrt {-d e}\, p e \ln \left (-e x -\sqrt {-d e}\right ) g^{2} d^{2} x^{7}+84 \sqrt {-d e}\, p \,e^{2} \ln \left (-e x +\sqrt {-d e}\right ) f g d \,x^{7}-84 \sqrt {-d e}\, p \,e^{2} \ln \left (-e x -\sqrt {-d e}\right ) f g d \,x^{7}+30 \sqrt {-d e}\, p \,e^{3} \ln \left (-e x -\sqrt {-d e}\right ) f^{2} x^{7}-35 i \pi \,d^{4} g^{2} x^{4} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+140 d^{3} e \,g^{2} p \,x^{6}+60 d \,e^{3} f^{2} p \,x^{6}-20 d^{2} e^{2} f^{2} p \,x^{4}+12 d^{3} e \,f^{2} p \,x^{2}-168 d^{2} e^{2} f g p \,x^{6}+56 d^{3} e f g p \,x^{4}+84 \ln \left (c \right ) d^{4} f g \,x^{2}-42 i \pi \,d^{4} f g \,x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+42 i \pi \,d^{4} f g \,x^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+42 i \pi \,d^{4} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+70 \ln \left (c \right ) d^{4} g^{2} x^{4}-30 \sqrt {-d e}\, p \,e^{3} \ln \left (-e x +\sqrt {-d e}\right ) f^{2} x^{7}-15 i \pi \,d^{4} f^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+35 i \pi \,d^{4} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+35 i \pi \,d^{4} g^{2} x^{4} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}-42 i \pi \,d^{4} f g \,x^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-15 i \pi \,d^{4} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-35 i \pi \,d^{4} g^{2} x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}+15 i \pi \,d^{4} f^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+15 i \pi \,d^{4} f^{2} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{210 d^{4} x^{7}}\) | \(784\) |
-1/3*g^2*ln(c*(e*x^2+d)^p)/x^3-2/5*f*g*ln(c*(e*x^2+d)^p)/x^5-1/7*f^2*ln(c* (e*x^2+d)^p)/x^7-2/105*p*e*(-1/d^3*(-35*d^2*g^2+42*d*e*f*g-15*e^2*f^2)/x+3 *f^2/d/x^5+f*(14*d*g-5*e*f)/d^2/x^3+e*(35*d^2*g^2-42*d*e*f*g+15*e^2*f^2)/d ^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2)))
Time = 0.34 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.70 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=\left [\frac {{\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{7} \sqrt {-\frac {e}{d}} \log \left (\frac {e x^{2} - 2 \, d x \sqrt {-\frac {e}{d}} - d}{e x^{2} + d}\right ) - 6 \, d^{2} e f^{2} p x^{2} - 2 \, {\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{6} + 2 \, {\left (5 \, d e^{2} f^{2} - 14 \, d^{2} e f g\right )} p x^{4} - {\left (35 \, d^{3} g^{2} p x^{4} + 42 \, d^{3} f g p x^{2} + 15 \, d^{3} f^{2} p\right )} \log \left (e x^{2} + d\right ) - {\left (35 \, d^{3} g^{2} x^{4} + 42 \, d^{3} f g x^{2} + 15 \, d^{3} f^{2}\right )} \log \left (c\right )}{105 \, d^{3} x^{7}}, -\frac {2 \, {\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{7} \sqrt {\frac {e}{d}} \arctan \left (x \sqrt {\frac {e}{d}}\right ) + 6 \, d^{2} e f^{2} p x^{2} + 2 \, {\left (15 \, e^{3} f^{2} - 42 \, d e^{2} f g + 35 \, d^{2} e g^{2}\right )} p x^{6} - 2 \, {\left (5 \, d e^{2} f^{2} - 14 \, d^{2} e f g\right )} p x^{4} + {\left (35 \, d^{3} g^{2} p x^{4} + 42 \, d^{3} f g p x^{2} + 15 \, d^{3} f^{2} p\right )} \log \left (e x^{2} + d\right ) + {\left (35 \, d^{3} g^{2} x^{4} + 42 \, d^{3} f g x^{2} + 15 \, d^{3} f^{2}\right )} \log \left (c\right )}{105 \, d^{3} x^{7}}\right ] \]
[1/105*((15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^7*sqrt(-e/d)*log((e *x^2 - 2*d*x*sqrt(-e/d) - d)/(e*x^2 + d)) - 6*d^2*e*f^2*p*x^2 - 2*(15*e^3* f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^6 + 2*(5*d*e^2*f^2 - 14*d^2*e*f*g)* p*x^4 - (35*d^3*g^2*p*x^4 + 42*d^3*f*g*p*x^2 + 15*d^3*f^2*p)*log(e*x^2 + d ) - (35*d^3*g^2*x^4 + 42*d^3*f*g*x^2 + 15*d^3*f^2)*log(c))/(d^3*x^7), -1/1 05*(2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g^2)*p*x^7*sqrt(e/d)*arctan(x* sqrt(e/d)) + 6*d^2*e*f^2*p*x^2 + 2*(15*e^3*f^2 - 42*d*e^2*f*g + 35*d^2*e*g ^2)*p*x^6 - 2*(5*d*e^2*f^2 - 14*d^2*e*f*g)*p*x^4 + (35*d^3*g^2*p*x^4 + 42* d^3*f*g*p*x^2 + 15*d^3*f^2*p)*log(e*x^2 + d) + (35*d^3*g^2*x^4 + 42*d^3*f* g*x^2 + 15*d^3*f^2)*log(c))/(d^3*x^7)]
Timed out. \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.34 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.82 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {2 \, {\left (15 \, e^{4} f^{2} p - 42 \, d e^{3} f g p + 35 \, d^{2} e^{2} g^{2} p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{105 \, \sqrt {d e} d^{3}} - \frac {{\left (35 \, g^{2} p x^{4} + 42 \, f g p x^{2} + 15 \, f^{2} p\right )} \log \left (e x^{2} + d\right )}{105 \, x^{7}} - \frac {30 \, e^{3} f^{2} p x^{6} - 84 \, d e^{2} f g p x^{6} + 70 \, d^{2} e g^{2} p x^{6} - 10 \, d e^{2} f^{2} p x^{4} + 28 \, d^{2} e f g p x^{4} + 35 \, d^{3} g^{2} x^{4} \log \left (c\right ) + 6 \, d^{2} e f^{2} p x^{2} + 42 \, d^{3} f g x^{2} \log \left (c\right ) + 15 \, d^{3} f^{2} \log \left (c\right )}{105 \, d^{3} x^{7}} \]
-2/105*(15*e^4*f^2*p - 42*d*e^3*f*g*p + 35*d^2*e^2*g^2*p)*arctan(e*x/sqrt( d*e))/(sqrt(d*e)*d^3) - 1/105*(35*g^2*p*x^4 + 42*f*g*p*x^2 + 15*f^2*p)*log (e*x^2 + d)/x^7 - 1/105*(30*e^3*f^2*p*x^6 - 84*d*e^2*f*g*p*x^6 + 70*d^2*e* g^2*p*x^6 - 10*d*e^2*f^2*p*x^4 + 28*d^2*e*f*g*p*x^4 + 35*d^3*g^2*x^4*log(c ) + 6*d^2*e*f^2*p*x^2 + 42*d^3*f*g*x^2*log(c) + 15*d^3*f^2*log(c))/(d^3*x^ 7)
Time = 1.66 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.59 \[ \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^8} \, dx=-\frac {\frac {6\,e\,f^2\,p}{d}+\frac {2\,e\,p\,x^4\,\left (35\,d^2\,g^2-42\,d\,e\,f\,g+15\,e^2\,f^2\right )}{d^3}+\frac {2\,e\,f\,p\,x^2\,\left (14\,d\,g-5\,e\,f\right )}{d^2}}{105\,x^5}-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {f^2}{7}+\frac {2\,f\,g\,x^2}{5}+\frac {g^2\,x^4}{3}\right )}{x^7}-\frac {2\,e^{3/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (35\,d^2\,g^2-42\,d\,e\,f\,g+15\,e^2\,f^2\right )}{105\,d^{7/2}} \]