Integrand size = 15, antiderivative size = 415 \[ \int f^{\frac {c}{(a+b x)^2}} x^4 \, dx=\frac {a^4 f^{\frac {c}{(a+b x)^2}} (a+b x)}{b^5}-\frac {2 a^3 f^{\frac {c}{(a+b x)^2}} (a+b x)^2}{b^5}+\frac {2 a^2 f^{\frac {c}{(a+b x)^2}} (a+b x)^3}{b^5}-\frac {a f^{\frac {c}{(a+b x)^2}} (a+b x)^4}{b^5}+\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)^5}{5 b^5}-\frac {a^4 \sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b^5}+\frac {4 a^2 c f^{\frac {c}{(a+b x)^2}} (a+b x) \log (f)}{b^5}-\frac {a c f^{\frac {c}{(a+b x)^2}} (a+b x)^2 \log (f)}{b^5}+\frac {2 c f^{\frac {c}{(a+b x)^2}} (a+b x)^3 \log (f)}{15 b^5}+\frac {2 a^3 c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f)}{b^5}-\frac {4 a^2 c^{3/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \log ^{\frac {3}{2}}(f)}{b^5}+\frac {4 c^2 f^{\frac {c}{(a+b x)^2}} (a+b x) \log ^2(f)}{15 b^5}+\frac {a c^2 \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log ^2(f)}{b^5}-\frac {4 c^{5/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \log ^{\frac {5}{2}}(f)}{15 b^5} \]
a^4*f^(c/(b*x+a)^2)*(b*x+a)/b^5-2*a^3*f^(c/(b*x+a)^2)*(b*x+a)^2/b^5+2*a^2* f^(c/(b*x+a)^2)*(b*x+a)^3/b^5-a*f^(c/(b*x+a)^2)*(b*x+a)^4/b^5+1/5*f^(c/(b* x+a)^2)*(b*x+a)^5/b^5+4*a^2*c*f^(c/(b*x+a)^2)*(b*x+a)*ln(f)/b^5-a*c*f^(c/( b*x+a)^2)*(b*x+a)^2*ln(f)/b^5+2/15*c*f^(c/(b*x+a)^2)*(b*x+a)^3*ln(f)/b^5+2 *a^3*c*Ei(c*ln(f)/(b*x+a)^2)*ln(f)/b^5+4/15*c^2*f^(c/(b*x+a)^2)*(b*x+a)*ln (f)^2/b^5+a*c^2*Ei(c*ln(f)/(b*x+a)^2)*ln(f)^2/b^5-4*a^2*c^(3/2)*erfi(c^(1/ 2)*ln(f)^(1/2)/(b*x+a))*ln(f)^(3/2)*Pi^(1/2)/b^5-4/15*c^(5/2)*erfi(c^(1/2) *ln(f)^(1/2)/(b*x+a))*ln(f)^(5/2)*Pi^(1/2)/b^5-a^4*erfi(c^(1/2)*ln(f)^(1/2 )/(b*x+a))*c^(1/2)*Pi^(1/2)*ln(f)^(1/2)/b^5
Time = 0.15 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.47 \[ \int f^{\frac {c}{(a+b x)^2}} x^4 \, dx=\frac {a f^{\frac {c}{(a+b x)^2}} \left (3 a^4+47 a^2 c \log (f)+4 c^2 \log ^2(f)\right )}{15 b^5}+\frac {15 a c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right ) \log (f) \left (2 a^2+c \log (f)\right )-\sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)} \left (15 a^4+60 a^2 c \log (f)+4 c^2 \log ^2(f)\right )+b f^{\frac {c}{(a+b x)^2}} x \left (3 b^4 x^4+c \left (36 a^2-9 a b x+2 b^2 x^2\right ) \log (f)+4 c^2 \log ^2(f)\right )}{15 b^5} \]
(a*f^(c/(a + b*x)^2)*(3*a^4 + 47*a^2*c*Log[f] + 4*c^2*Log[f]^2))/(15*b^5) + (15*a*c*ExpIntegralEi[(c*Log[f])/(a + b*x)^2]*Log[f]*(2*a^2 + c*Log[f]) - Sqrt[c]*Sqrt[Pi]*Erfi[(Sqrt[c]*Sqrt[Log[f]])/(a + b*x)]*Sqrt[Log[f]]*(15 *a^4 + 60*a^2*c*Log[f] + 4*c^2*Log[f]^2) + b*f^(c/(a + b*x)^2)*x*(3*b^4*x^ 4 + c*(36*a^2 - 9*a*b*x + 2*b^2*x^2)*Log[f] + 4*c^2*Log[f]^2))/(15*b^5)
Time = 0.79 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2656, 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.
| \(\displaystyle \int x^4 f^{\frac {c}{(a+b x)^2}} \, dx\) |
| \(\Big \downarrow \) 2656 |
| \(\displaystyle \int \left (\frac {a^4 f^{\frac {c}{(a+b x)^2}}}{b^4}-\frac {4 a^3 (a+b x) f^{\frac {c}{(a+b x)^2}}}{b^4}+\frac {6 a^2 (a+b x)^2 f^{\frac {c}{(a+b x)^2}}}{b^4}+\frac {(a+b x)^4 f^{\frac {c}{(a+b x)^2}}}{b^4}-\frac {4 a (a+b x)^3 f^{\frac {c}{(a+b x)^2}}}{b^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {\pi } a^4 \sqrt {c} \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b^5}+\frac {a^4 (a+b x) f^{\frac {c}{(a+b x)^2}}}{b^5}+\frac {2 a^3 c \log (f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right )}{b^5}-\frac {2 a^3 (a+b x)^2 f^{\frac {c}{(a+b x)^2}}}{b^5}-\frac {4 \sqrt {\pi } a^2 c^{3/2} \log ^{\frac {3}{2}}(f) \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b^5}+\frac {2 a^2 (a+b x)^3 f^{\frac {c}{(a+b x)^2}}}{b^5}+\frac {4 a^2 c \log (f) (a+b x) f^{\frac {c}{(a+b x)^2}}}{b^5}-\frac {4 \sqrt {\pi } c^{5/2} \log ^{\frac {5}{2}}(f) \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{15 b^5}+\frac {a c^2 \log ^2(f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^2}\right )}{b^5}+\frac {4 c^2 \log ^2(f) (a+b x) f^{\frac {c}{(a+b x)^2}}}{15 b^5}+\frac {(a+b x)^5 f^{\frac {c}{(a+b x)^2}}}{5 b^5}-\frac {a (a+b x)^4 f^{\frac {c}{(a+b x)^2}}}{b^5}+\frac {2 c \log (f) (a+b x)^3 f^{\frac {c}{(a+b x)^2}}}{15 b^5}-\frac {a c \log (f) (a+b x)^2 f^{\frac {c}{(a+b x)^2}}}{b^5}\) |
(a^4*f^(c/(a + b*x)^2)*(a + b*x))/b^5 - (2*a^3*f^(c/(a + b*x)^2)*(a + b*x) ^2)/b^5 + (2*a^2*f^(c/(a + b*x)^2)*(a + b*x)^3)/b^5 - (a*f^(c/(a + b*x)^2) *(a + b*x)^4)/b^5 + (f^(c/(a + b*x)^2)*(a + b*x)^5)/(5*b^5) - (a^4*Sqrt[c] *Sqrt[Pi]*Erfi[(Sqrt[c]*Sqrt[Log[f]])/(a + b*x)]*Sqrt[Log[f]])/b^5 + (4*a^ 2*c*f^(c/(a + b*x)^2)*(a + b*x)*Log[f])/b^5 - (a*c*f^(c/(a + b*x)^2)*(a + b*x)^2*Log[f])/b^5 + (2*c*f^(c/(a + b*x)^2)*(a + b*x)^3*Log[f])/(15*b^5) + (2*a^3*c*ExpIntegralEi[(c*Log[f])/(a + b*x)^2]*Log[f])/b^5 - (4*a^2*c^(3/ 2)*Sqrt[Pi]*Erfi[(Sqrt[c]*Sqrt[Log[f]])/(a + b*x)]*Log[f]^(3/2))/b^5 + (4* c^2*f^(c/(a + b*x)^2)*(a + b*x)*Log[f]^2)/(15*b^5) + (a*c^2*ExpIntegralEi[ (c*Log[f])/(a + b*x)^2]*Log[f]^2)/b^5 - (4*c^(5/2)*Sqrt[Pi]*Erfi[(Sqrt[c]* Sqrt[Log[f]])/(a + b*x)]*Log[f]^(5/2))/(15*b^5)
3.3.24.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(Px_), x_Symbol] :> Int[ ExpandLinearProduct[F^(a + b*(c + d*x)^n), Px, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[Px, x]
Time = 0.28 (sec) , antiderivative size = 343, normalized size of antiderivative = 0.83
| method | result | size |
| risch | \(\frac {f^{\frac {c}{\left (b x +a \right )^{2}}} x^{5}}{5}+\frac {2 f^{\frac {c}{\left (b x +a \right )^{2}}} \ln \left (f \right ) c \,x^{3}}{15 b^{2}}-\frac {3 f^{\frac {c}{\left (b x +a \right )^{2}}} \ln \left (f \right ) a c \,x^{2}}{5 b^{3}}+\frac {4 f^{\frac {c}{\left (b x +a \right )^{2}}} \ln \left (f \right )^{2} c^{2} x}{15 b^{4}}+\frac {12 f^{\frac {c}{\left (b x +a \right )^{2}}} \ln \left (f \right ) a^{2} c x}{5 b^{4}}-\frac {4 \ln \left (f \right )^{3} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-c \ln \left (f \right )}}{b x +a}\right ) c^{3}}{15 b^{5} \sqrt {-c \ln \left (f \right )}}-\frac {4 \ln \left (f \right )^{2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-c \ln \left (f \right )}}{b x +a}\right ) a^{2} c^{2}}{b^{5} \sqrt {-c \ln \left (f \right )}}-\frac {\ln \left (f \right ) \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-c \ln \left (f \right )}}{b x +a}\right ) a^{4} c}{b^{5} \sqrt {-c \ln \left (f \right )}}+\frac {4 f^{\frac {c}{\left (b x +a \right )^{2}}} \ln \left (f \right )^{2} a \,c^{2}}{15 b^{5}}+\frac {47 f^{\frac {c}{\left (b x +a \right )^{2}}} \ln \left (f \right ) a^{3} c}{15 b^{5}}+\frac {f^{\frac {c}{\left (b x +a \right )^{2}}} a^{5}}{5 b^{5}}-\frac {\ln \left (f \right )^{2} \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{\left (b x +a \right )^{2}}\right ) a \,c^{2}}{b^{5}}-\frac {2 \ln \left (f \right ) \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{\left (b x +a \right )^{2}}\right ) a^{3} c}{b^{5}}\) | \(343\) |
1/5*f^(c/(b*x+a)^2)*x^5+2/15/b^2*f^(c/(b*x+a)^2)*ln(f)*c*x^3-3/5/b^3*f^(c/ (b*x+a)^2)*ln(f)*a*c*x^2+4/15/b^4*f^(c/(b*x+a)^2)*ln(f)^2*c^2*x+12/5/b^4*f ^(c/(b*x+a)^2)*ln(f)*a^2*c*x-4/15/b^5/(-c*ln(f))^(1/2)*ln(f)^3*Pi^(1/2)*er f((-c*ln(f))^(1/2)/(b*x+a))*c^3-4/b^5/(-c*ln(f))^(1/2)*ln(f)^2*Pi^(1/2)*er f((-c*ln(f))^(1/2)/(b*x+a))*a^2*c^2-1/b^5/(-c*ln(f))^(1/2)*ln(f)*Pi^(1/2)* erf((-c*ln(f))^(1/2)/(b*x+a))*a^4*c+4/15/b^5*f^(c/(b*x+a)^2)*ln(f)^2*a*c^2 +47/15/b^5*f^(c/(b*x+a)^2)*ln(f)*a^3*c+1/5/b^5*f^(c/(b*x+a)^2)*a^5-1/b^5*l n(f)^2*Ei(1,-c*ln(f)/(b*x+a)^2)*a*c^2-2/b^5*ln(f)*Ei(1,-c*ln(f)/(b*x+a)^2) *a^3*c
Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.48 \[ \int f^{\frac {c}{(a+b x)^2}} x^4 \, dx=\frac {\sqrt {\pi } {\left (15 \, a^{4} b + 60 \, a^{2} b c \log \left (f\right ) + 4 \, b c^{2} \log \left (f\right )^{2}\right )} \sqrt {-\frac {c \log \left (f\right )}{b^{2}}} \operatorname {erf}\left (\frac {b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}}}{b x + a}\right ) + {\left (3 \, b^{5} x^{5} + 3 \, a^{5} + 4 \, {\left (b c^{2} x + a c^{2}\right )} \log \left (f\right )^{2} + {\left (2 \, b^{3} c x^{3} - 9 \, a b^{2} c x^{2} + 36 \, a^{2} b c x + 47 \, a^{3} c\right )} \log \left (f\right )\right )} f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 15 \, {\left (2 \, a^{3} c \log \left (f\right ) + a c^{2} \log \left (f\right )^{2}\right )} {\rm Ei}\left (\frac {c \log \left (f\right )}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{15 \, b^{5}} \]
1/15*(sqrt(pi)*(15*a^4*b + 60*a^2*b*c*log(f) + 4*b*c^2*log(f)^2)*sqrt(-c*l og(f)/b^2)*erf(b*sqrt(-c*log(f)/b^2)/(b*x + a)) + (3*b^5*x^5 + 3*a^5 + 4*( b*c^2*x + a*c^2)*log(f)^2 + (2*b^3*c*x^3 - 9*a*b^2*c*x^2 + 36*a^2*b*c*x + 47*a^3*c)*log(f))*f^(c/(b^2*x^2 + 2*a*b*x + a^2)) + 15*(2*a^3*c*log(f) + a *c^2*log(f)^2)*Ei(c*log(f)/(b^2*x^2 + 2*a*b*x + a^2)))/b^5
\[ \int f^{\frac {c}{(a+b x)^2}} x^4 \, dx=\int f^{\frac {c}{\left (a + b x\right )^{2}}} x^{4}\, dx \]
\[ \int f^{\frac {c}{(a+b x)^2}} x^4 \, dx=\int { f^{\frac {c}{{\left (b x + a\right )}^{2}}} x^{4} \,d x } \]
1/15*(3*b^4*x^5 + 2*b^2*c*x^3*log(f) - 9*a*b*c*x^2*log(f) + 4*(9*a^2*c*log (f) + c^2*log(f)^2)*x)*f^(c/(b^2*x^2 + 2*a*b*x + a^2))/b^4 - integrate(2/1 5*(18*a^5*c*log(f) + 2*a^3*c^2*log(f)^2 + 15*(2*a^3*b^2*c*log(f) + a*b^2*c ^2*log(f)^2)*x^2 + (45*a^4*b*c*log(f) - 30*a^2*b*c^2*log(f)^2 - 4*b*c^3*lo g(f)^3)*x)*f^(c/(b^2*x^2 + 2*a*b*x + a^2))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2* b^5*x + a^3*b^4), x)
\[ \int f^{\frac {c}{(a+b x)^2}} x^4 \, dx=\int { f^{\frac {c}{{\left (b x + a\right )}^{2}}} x^{4} \,d x } \]
Timed out. \[ \int f^{\frac {c}{(a+b x)^2}} x^4 \, dx=\int f^{\frac {c}{{\left (a+b\,x\right )}^2}}\,x^4 \,d x \]