Integrand size = 21, antiderivative size = 183 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=-\frac {105 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)}+\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)} \]
105/16*F^(a+b/(d*x+c)^2)/b^4/d/(d*x+c)/ln(F)^4-35/8*F^(a+b/(d*x+c)^2)/b^3/ d/(d*x+c)^3/ln(F)^3+7/4*F^(a+b/(d*x+c)^2)/b^2/d/(d*x+c)^5/ln(F)^2-1/2*F^(a +b/(d*x+c)^2)/b/d/(d*x+c)^7/ln(F)-105/32*F^a*erfi(b^(1/2)*ln(F)^(1/2)/(d*x +c))*Pi^(1/2)/b^(9/2)/d/ln(F)^(9/2)
Time = 0.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.69 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\frac {F^a \left (-105 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )+\frac {2 \sqrt {b} F^{\frac {b}{(c+d x)^2}} \sqrt {\log (F)} \left (105 (c+d x)^6-70 b (c+d x)^4 \log (F)+28 b^2 (c+d x)^2 \log ^2(F)-8 b^3 \log ^3(F)\right )}{(c+d x)^7}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)} \]
(F^a*(-105*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)] + (2*Sqrt[b]*F^ (b/(c + d*x)^2)*Sqrt[Log[F]]*(105*(c + d*x)^6 - 70*b*(c + d*x)^4*Log[F] + 28*b^2*(c + d*x)^2*Log[F]^2 - 8*b^3*Log[F]^3))/(c + d*x)^7))/(32*b^(9/2)*d *Log[F]^(9/2))
Time = 0.63 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2641, 2641, 2641, 2641, 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.
| \(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx\) |
| \(\Big \downarrow \) 2641 |
| \(\displaystyle -\frac {7 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8}dx}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7}\) |
| \(\Big \downarrow \) 2641 |
| \(\displaystyle -\frac {7 \left (-\frac {5 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6}dx}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7}\) |
| \(\Big \downarrow \) 2641 |
| \(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4}dx}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7}\) |
| \(\Big \downarrow \) 2641 |
| \(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (-\frac {\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2}dx}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7}\) |
| \(\Big \downarrow \) 2640 |
| \(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\int F^{a+\frac {b}{(c+d x)^2}}d\frac {1}{c+d x}}{2 b d \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7}\) |
| \(\Big \downarrow \) 2633 |
| \(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac {3}{2}}(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7}\) |
-1/2*F^(a + b/(c + d*x)^2)/(b*d*(c + d*x)^7*Log[F]) - (7*(-1/2*F^(a + b/(c + d*x)^2)/(b*d*(c + d*x)^5*Log[F]) - (5*(-1/2*F^(a + b/(c + d*x)^2)/(b*d* (c + d*x)^3*Log[F]) - (3*((F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d *x)])/(4*b^(3/2)*d*Log[F]^(3/2)) - F^(a + b/(c + d*x)^2)/(2*b*d*(c + d*x)* Log[F])))/(2*b*Log[F])))/(2*b*Log[F])))/(2*b*Log[F])
3.4.37.3.1 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_))*((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 - n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*L og[F])), x] - Simp[(m - n + 1)/(b*n*Log[F]) 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[0, (m + 1)/n, 5] && IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n , 0])
Time = 4.23 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(-\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 d \left (d x +c \right )^{7} b \ln \left (F \right )}+\frac {7 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 d \,b^{2} \ln \left (F \right )^{2} \left (d x +c \right )^{5}}-\frac {35 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{8 d \,b^{3} \ln \left (F \right )^{3} \left (d x +c \right )^{3}}+\frac {105 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{16 d \,b^{4} \ln \left (F \right )^{4} \left (d x +c \right )}-\frac {105 F^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{32 d \,b^{4} \ln \left (F \right )^{4} \sqrt {-b \ln \left (F \right )}}\) | \(175\) |
-1/2*F^a/d*F^(b/(d*x+c)^2)/(d*x+c)^7/b/ln(F)+7/4*F^a/d/b^2/ln(F)^2*F^(b/(d *x+c)^2)/(d*x+c)^5-35/8*F^a/d/b^3/ln(F)^3*F^(b/(d*x+c)^2)/(d*x+c)^3+105/16 *F^a/d/b^4/ln(F)^4*F^(b/(d*x+c)^2)/(d*x+c)-105/32*F^a/d/b^4/ln(F)^4*Pi^(1/ 2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (163) = 326\).
Time = 0.28 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.40 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\frac {105 \, \sqrt {\pi } {\left (d^{8} x^{7} + 7 \, c d^{7} x^{6} + 21 \, c^{2} d^{6} x^{5} + 35 \, c^{3} d^{5} x^{4} + 35 \, c^{4} d^{4} x^{3} + 21 \, c^{5} d^{3} x^{2} + 7 \, c^{6} d^{2} x + c^{7} d\right )} F^{a} \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) - 2 \, {\left (8 \, b^{4} \log \left (F\right )^{4} - 28 \, {\left (b^{3} d^{2} x^{2} + 2 \, b^{3} c d x + b^{3} c^{2}\right )} \log \left (F\right )^{3} + 70 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} - 105 \, {\left (b d^{6} x^{6} + 6 \, b c d^{5} x^{5} + 15 \, b c^{2} d^{4} x^{4} + 20 \, b c^{3} d^{3} x^{3} + 15 \, b c^{4} d^{2} x^{2} + 6 \, b c^{5} d x + b c^{6}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{32 \, {\left (b^{5} d^{8} x^{7} + 7 \, b^{5} c d^{7} x^{6} + 21 \, b^{5} c^{2} d^{6} x^{5} + 35 \, b^{5} c^{3} d^{5} x^{4} + 35 \, b^{5} c^{4} d^{4} x^{3} + 21 \, b^{5} c^{5} d^{3} x^{2} + 7 \, b^{5} c^{6} d^{2} x + b^{5} c^{7} d\right )} \log \left (F\right )^{5}} \]
1/32*(105*sqrt(pi)*(d^8*x^7 + 7*c*d^7*x^6 + 21*c^2*d^6*x^5 + 35*c^3*d^5*x^ 4 + 35*c^4*d^4*x^3 + 21*c^5*d^3*x^2 + 7*c^6*d^2*x + c^7*d)*F^a*sqrt(-b*log (F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c)) - 2*(8*b^4*log(F)^4 - 28*(b^ 3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2)*log(F)^3 + 70*(b^2*d^4*x^4 + 4*b^2*c*d^ 3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*log(F)^2 - 105*(b*d^6 *x^6 + 6*b*c*d^5*x^5 + 15*b*c^2*d^4*x^4 + 20*b*c^3*d^3*x^3 + 15*b*c^4*d^2* x^2 + 6*b*c^5*d*x + b*c^6)*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)))/((b^5*d^8*x^7 + 7*b^5*c*d^7*x^6 + 21*b^5*c^2*d ^6*x^5 + 35*b^5*c^3*d^5*x^4 + 35*b^5*c^4*d^4*x^3 + 21*b^5*c^5*d^3*x^2 + 7* b^5*c^6*d^2*x + b^5*c^7*d)*log(F)^5)
Timed out. \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\text {Timed out} \]
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{10}} \,d x } \]
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{10}} \,d x } \]
Time = 1.25 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.87 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=-\frac {\frac {F^a\,\left (105\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )-\frac {210\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\sqrt {b\,\ln \left (F\right )}}{c+d\,x}\right )}{32\,\sqrt {b\,\ln \left (F\right )}}-\frac {7\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \left (F\right )}^2}{4\,{\left (c+d\,x\right )}^5}+\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^3\,{\ln \left (F\right )}^3}{2\,{\left (c+d\,x\right )}^7}+\frac {35\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )}{8\,{\left (c+d\,x\right )}^3}}{b^4\,d\,{\ln \left (F\right )}^4} \]
-((F^a*(105*pi^(1/2)*erfi((b*log(F))/((b*log(F))^(1/2)*(c + d*x))) - (210* F^(b/(c + d*x)^2)*(b*log(F))^(1/2))/(c + d*x)))/(32*(b*log(F))^(1/2)) - (7 *F^a*F^(b/(c + d*x)^2)*b^2*log(F)^2)/(4*(c + d*x)^5) + (F^a*F^(b/(c + d*x) ^2)*b^3*log(F)^3)/(2*(c + d*x)^7) + (35*F^a*F^(b/(c + d*x)^2)*b*log(F))/(8 *(c + d*x)^3))/(b^4*d*log(F)^4)