Integrand size = 20, antiderivative size = 126 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\frac {e^{-\frac {1+4 a b f n \log (F)}{4 b^2 f n^2 \log (F)}} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\frac {1}{n}+2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]
1/2*(e*x+d)*erfi(1/2*(1/n+2*a*b*f*ln(F)+2*b^2*f*ln(F)*ln(c*(e*x+d)^n))/b/f ^(1/2)/ln(F)^(1/2))*Pi^(1/2)/b/e/exp(1/4*(1+4*a*b*f*n*ln(F))/b^2/f/n^2/ln( F))/n/((c*(e*x+d)^n)^(1/n))/f^(1/2)/ln(F)^(1/2)
Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\frac {e^{-\frac {1+4 a b f n \log (F)}{4 b^2 f n^2 \log (F)}} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {1+2 b f n \log (F) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 b \sqrt {f} n \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}} \]
(Sqrt[Pi]*(d + e*x)*Erfi[(1 + 2*b*f*n*Log[F]*(a + b*Log[c*(d + e*x)^n]))/( 2*b*Sqrt[f]*n*Sqrt[Log[F]])])/(2*b*e*E^((1 + 4*a*b*f*n*Log[F])/(4*b^2*f*n^ 2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*Sqrt[Log[F]])
Time = 0.46 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2710, 2706, 2725, 2664, 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.
| \(\displaystyle \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx\) |
| \(\Big \downarrow \) 2710 |
| \(\displaystyle (d+e x)^{-2 a b f n \log (F)} \left (c (d+e x)^n\right )^{2 a b f \log (F)} \int F^{f a^2+b^2 f \log ^2\left (c (d+e x)^n\right )} (d+e x)^{2 a b f n \log (F)}dx\) |
| \(\Big \downarrow \) 2706 |
| \(\displaystyle \frac {(d+e x) \left (c (d+e x)^n\right )^{-1/n} \int F^{f a^2+2 b f \log \left (c (d+e x)^n\right ) a+b^2 f \log ^2\left (c (d+e x)^n\right )} \left (c (d+e x)^n\right )^{\frac {1}{n}}d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2725 |
| \(\displaystyle \frac {(d+e x) \left (c (d+e x)^n\right )^{-1/n} \int \exp \left (f \log (F) a^2+b^2 f \log (F) \log ^2\left (c (d+e x)^n\right )+\left (2 a b f \log (F)+\frac {1}{n}\right ) \log \left (c (d+e x)^n\right )\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2664 |
| \(\displaystyle \frac {(d+e x) \left (c (d+e x)^n\right )^{-1/n} e^{-\frac {4 a b f n \log (F)+1}{4 b^2 f n^2 \log (F)}} \int \exp \left (\frac {\left (2 f \log (F) \log \left (c (d+e x)^n\right ) b^2+2 a f \log (F) b+\frac {1}{n}\right )^2}{4 b^2 f \log (F)}\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
| \(\Big \downarrow \) 2633 |
| \(\displaystyle \frac {\sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} e^{-\frac {4 a b f n \log (F)+1}{4 b^2 f n^2 \log (F)}} \text {erfi}\left (\frac {2 a b f \log (F)+2 b^2 f \log (F) \log \left (c (d+e x)^n\right )+\frac {1}{n}}{2 b \sqrt {f} \sqrt {\log (F)}}\right )}{2 b e \sqrt {f} n \sqrt {\log (F)}}\) |
(Sqrt[Pi]*(d + e*x)*Erfi[(n^(-1) + 2*a*b*f*Log[F] + 2*b^2*f*Log[F]*Log[c*( d + e*x)^n])/(2*b*Sqrt[f]*Sqrt[Log[F]])])/(2*b*e*E^((1 + 4*a*b*f*n*Log[F]) /(4*b^2*f*n^2*Log[F]))*Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*Sqrt[Log[F]])
3.7.5.3.1 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_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ (4*c)) Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*(( g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[(g + h*x)^(m + 1)/(h*n*(c*(d + e*x)^n)^((m + 1)/n)) Subst[Int[E^(a*f*Log[F] + ((m + 1)*x)/n + b*f*Log[F] *x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.)), x _Symbol] :> Simp[((c*(d + e*x)^n)^(2*a*b*f*Log[F])/(d + e*x)^(2*a*b*f*n*Log [F]))*Int[(d + e*x)^(2*a*b*f*n*Log[F])*F^(a^2*f + b^2*f*Log[c*(d + e*x)^n]^ 2), x], x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && !IntegerQ[2*a*b*f*Log[ F]]
Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x], x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]
\[\int F^{f {\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )}^{2}}d x\]
Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.04 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2} f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 2 \, a b f n \log \left (F\right ) + 1\right )} \sqrt {-b^{2} f n^{2} \log \left (F\right )}}{2 \, b^{2} f n^{2} \log \left (F\right )}\right ) e^{\left (-\frac {4 \, b^{2} f n \log \left (F\right ) \log \left (c\right ) + 4 \, a b f n \log \left (F\right ) + 1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, b e n} \]
-1/2*sqrt(pi)*sqrt(-b^2*f*n^2*log(F))*erf(1/2*(2*b^2*f*n^2*log(e*x + d)*lo g(F) + 2*b^2*f*n*log(F)*log(c) + 2*a*b*f*n*log(F) + 1)*sqrt(-b^2*f*n^2*log (F))/(b^2*f*n^2*log(F)))*e^(-1/4*(4*b^2*f*n*log(F)*log(c) + 4*a*b*f*n*log( F) + 1)/(b^2*f*n^2*log(F)))/(b*e*n)
Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (116) = 232\).
Time = 10.64 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.63 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\begin {cases} \frac {2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} a b d f n \log {\left (F \right )}}{e} - 2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} a b f n x \log {\left (F \right )} - \frac {2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} d f n^{2} \log {\left (F \right )}}{e} - \frac {2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} d f n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{e} + 2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} f n^{2} x \log {\left (F \right )} - 2 F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b^{2} f n x \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )} + \frac {F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d}{e} + F^{a^{2} f + 2 a b f \log {\left (c \left (d + e x\right )^{n} \right )} + b^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} x & \text {for}\: e \neq 0 \\F^{f \left (a + b \log {\left (c d^{n} \right )}\right )^{2}} x & \text {otherwise} \end {cases} \]
Piecewise((2*F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*a*b*d*f*n*log(F)/e - 2*F**(a**2*f + 2*a*b*f*log(c*(d + e*x)** n) + b**2*f*log(c*(d + e*x)**n)**2)*a*b*f*n*x*log(F) - 2*F**(a**2*f + 2*a* b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*b**2*d*f*n**2*log (F)/e - 2*F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x )**n)**2)*b**2*d*f*n*log(F)*log(c*(d + e*x)**n)/e + 2*F**(a**2*f + 2*a*b*f *log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*b**2*f*n**2*x*log(F) - 2*F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n) **2)*b**2*f*n*x*log(F)*log(c*(d + e*x)**n) + F**(a**2*f + 2*a*b*f*log(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*d/e + F**(a**2*f + 2*a*b*f*lo g(c*(d + e*x)**n) + b**2*f*log(c*(d + e*x)**n)**2)*x, Ne(e, 0)), (F**(f*(a + b*log(c*d**n))**2)*x, True))
\[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\int { F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f} \,d x } \]
Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.93 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {-f \log \left (F\right )} b n \log \left (e x + d\right ) - \sqrt {-f \log \left (F\right )} b \log \left (c\right ) - \sqrt {-f \log \left (F\right )} a - \frac {\sqrt {-f \log \left (F\right )}}{2 \, b f n \log \left (F\right )}\right ) e^{\left (-\frac {a}{b n} - \frac {1}{4 \, b^{2} f n^{2} \log \left (F\right )}\right )}}{2 \, \sqrt {-f \log \left (F\right )} b c^{\left (\frac {1}{n}\right )} e n} \]
-1/2*sqrt(pi)*erf(-sqrt(-f*log(F))*b*n*log(e*x + d) - sqrt(-f*log(F))*b*lo g(c) - sqrt(-f*log(F))*a - 1/2*sqrt(-f*log(F))/(b*f*n*log(F)))*e^(-a/(b*n) - 1/4/(b^2*f*n^2*log(F)))/(sqrt(-f*log(F))*b*c^(1/n)*e*n)
Timed out. \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx=\int {\mathrm {e}}^{f\,\ln \left (F\right )\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2} \,d x \]