Integrand size = 21, antiderivative size = 136 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac {4 b^2 F^{a+\frac {b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}-\frac {4 b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {5}{2}}(F)}{15 d} \]
1/5*F^(a+b/(d*x+c)^2)*(d*x+c)^5/d+2/15*b*F^(a+b/(d*x+c)^2)*(d*x+c)^3*ln(F) /d+4/15*b^2*F^(a+b/(d*x+c)^2)*(d*x+c)*ln(F)^2/d-4/15*b^(5/2)*F^a*erfi(b^(1 /2)*ln(F)^(1/2)/(d*x+c))*ln(F)^(5/2)*Pi^(1/2)/d
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {F^a \left (-4 b^{5/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {5}{2}}(F)+F^{\frac {b}{(c+d x)^2}} (c+d x) \left (3 (c+d x)^4+2 b (c+d x)^2 \log (F)+4 b^2 \log ^2(F)\right )\right )}{15 d} \]
(F^a*(-4*b^(5/2)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)]*Log[F]^(5 /2) + F^(b/(c + d*x)^2)*(c + d*x)*(3*(c + d*x)^4 + 2*b*(c + d*x)^2*Log[F] + 4*b^2*Log[F]^2)))/(15*d)
Time = 0.46 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2643, 2643, 2635, 2640, 2633}
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.
| \(\displaystyle \int (c+d x)^4 F^{a+\frac {b}{(c+d x)^2}} \, dx\) |
| \(\Big \downarrow \) 2643 |
| \(\displaystyle \frac {2}{5} b \log (F) \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2dx+\frac {(c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{5 d}\) |
| \(\Big \downarrow \) 2643 |
| \(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \int F^{a+\frac {b}{(c+d x)^2}}dx+\frac {(c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{3 d}\right )+\frac {(c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{5 d}\) |
| \(\Big \downarrow \) 2635 |
| \(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (2 b \log (F) \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2}dx+\frac {(c+d x) F^{a+\frac {b}{(c+d x)^2}}}{d}\right )+\frac {(c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{3 d}\right )+\frac {(c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{5 d}\) |
| \(\Big \downarrow \) 2640 |
| \(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (\frac {(c+d x) F^{a+\frac {b}{(c+d x)^2}}}{d}-\frac {2 b \log (F) \int F^{a+\frac {b}{(c+d x)^2}}d\frac {1}{c+d x}}{d}\right )+\frac {(c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{3 d}\right )+\frac {(c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{5 d}\) |
| \(\Big \downarrow \) 2633 |
| \(\displaystyle \frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (\frac {(c+d x) F^{a+\frac {b}{(c+d x)^2}}}{d}-\frac {\sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{d}\right )+\frac {(c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{3 d}\right )+\frac {(c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{5 d}\) |
(F^(a + b/(c + d*x)^2)*(c + d*x)^5)/(5*d) + (2*b*Log[F]*((F^(a + b/(c + d* x)^2)*(c + d*x)^3)/(3*d) + (2*b*((F^(a + b/(c + d*x)^2)*(c + d*x))/d - (Sq rt[b]*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)]*Sqrt[Log[F]])/d) *Log[F])/3))/5
3.4.30.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x] - Simp[b*n*Log[F] Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n] && ILtQ[n, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[1/(d*(m + 1)) Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1)]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))) , x] - Simp[b*n*(Log[F]/(m + 1)) Int[(c + d*x)^(m + n)*F^(a + b*(c + d*x) ^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[ -4, (m + 1)/n, 5] && IntegerQ[n] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))
Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(118)=236\).
Time = 0.74 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.38
| method | result | size |
| risch | \(\frac {F^{a} d^{4} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{5}}{5}+F^{a} d^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{4}+2 F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{3}+2 F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x^{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{5}}{5 d}+\frac {2 F^{a} d^{2} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x^{3}}{15}+\frac {2 F^{a} d b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{2}}{5}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x}{5}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3}}{15 d}+\frac {4 F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} x}{15}+\frac {4 F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c}{15 d}-\frac {4 F^{a} b^{3} \ln \left (F \right )^{3} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{15 d \sqrt {-b \ln \left (F \right )}}\) | \(324\) |
1/5*F^a*d^4*F^(b/(d*x+c)^2)*x^5+F^a*d^3*F^(b/(d*x+c)^2)*c*x^4+2*F^a*d^2*F^ (b/(d*x+c)^2)*c^2*x^3+2*F^a*d*F^(b/(d*x+c)^2)*c^3*x^2+F^a*F^(b/(d*x+c)^2)* c^4*x+1/5*F^a/d*F^(b/(d*x+c)^2)*c^5+2/15*F^a*d^2*b*ln(F)*F^(b/(d*x+c)^2)*x ^3+2/5*F^a*d*b*ln(F)*F^(b/(d*x+c)^2)*c*x^2+2/5*F^a*b*ln(F)*F^(b/(d*x+c)^2) *c^2*x+2/15*F^a/d*b*ln(F)*F^(b/(d*x+c)^2)*c^3+4/15*F^a*b^2*ln(F)^2*F^(b/(d *x+c)^2)*x+4/15*F^a/d*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c-4/15*F^a/d*b^3*ln(F)^3 *Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d*x+c))
Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.48 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {4 \, \sqrt {\pi } F^{a} b^{2} d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) \log \left (F\right )^{2} + {\left (3 \, d^{5} x^{5} + 15 \, c d^{4} x^{4} + 30 \, c^{2} d^{3} x^{3} + 30 \, c^{3} d^{2} x^{2} + 15 \, c^{4} d x + 3 \, c^{5} + 4 \, {\left (b^{2} d x + b^{2} c\right )} \log \left (F\right )^{2} + 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{15 \, d} \]
1/15*(4*sqrt(pi)*F^a*b^2*d*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/( d*x + c))*log(F)^2 + (3*d^5*x^5 + 15*c*d^4*x^4 + 30*c^2*d^3*x^3 + 30*c^3*d ^2*x^2 + 15*c^4*d*x + 3*c^5 + 4*(b^2*d*x + b^2*c)*log(F)^2 + 2*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F))*F^((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/d
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{4}\, dx \]
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\int { {\left (d x + c\right )}^{4} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
1/15*(3*F^a*d^4*x^5 + 15*F^a*c*d^3*x^4 + 2*(15*F^a*c^2*d^2 + F^a*b*d^2*log (F))*x^3 + 6*(5*F^a*c^3*d + F^a*b*c*d*log(F))*x^2 + (15*F^a*c^4 + 6*F^a*b* c^2*log(F) + 4*F^a*b^2*log(F)^2)*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + inte grate(2/15*(4*F^a*b^3*d*x*log(F)^3 - 3*F^a*b*c^5*log(F) - 2*F^a*b^2*c^3*lo g(F)^2)*F^(b/(d^2*x^2 + 2*c*d*x + c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)
\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\int { {\left (d x + c\right )}^{4} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
Time = 0.68 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.22 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^5}{5\,d}+\frac {4\,F^a\,\sqrt {\pi }\,{\left (c+d\,x\right )}^5\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{5/2}}{15\,d}+\frac {2\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3}{15\,d}+\frac {4\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \left (F\right )}^2\,\left (c+d\,x\right )}{15\,d}-\frac {4\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}}\right )\,{\left (c+d\,x\right )}^5\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{5/2}}{15\,d} \]
(F^a*F^(b/(c + d*x)^2)*(c + d*x)^5)/(5*d) + (4*F^a*pi^(1/2)*(c + d*x)^5*(- (b*log(F))/(c + d*x)^2)^(5/2))/(15*d) + (2*F^a*F^(b/(c + d*x)^2)*b*log(F)* (c + d*x)^3)/(15*d) + (4*F^a*F^(b/(c + d*x)^2)*b^2*log(F)^2*(c + d*x))/(15 *d) - (4*F^a*pi^(1/2)*erfc((-(b*log(F))/(c + d*x)^2)^(1/2))*(c + d*x)^5*(- (b*log(F))/(c + d*x)^2)^(5/2))/(15*d)