Integrand size = 20, antiderivative size = 118 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=\frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {1+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} n \sqrt {\log (F)}} \] Output:
1/2*F^(a*f)*Pi^(1/2)*(e*x+d)*erfi(1/2*(1+2*b*f*n*ln(F)*ln(c*(e*x+d)^n))/b^ (1/2)/f^(1/2)/n/ln(F)^(1/2))/b^(1/2)/e/exp(1/4/b/f/n^2/ln(F))/f^(1/2)/n/(( c*(e*x+d)^n)^(1/n))/ln(F)^(1/2)
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=\frac {e^{-\frac {1}{4 b f n^2 \log (F)}} F^{a f} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {1+2 b f n \log (F) \log \left (c (d+e x)^n\right )}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} n \sqrt {\log (F)}} \] Input:
Integrate[F^(f*(a + b*Log[c*(d + e*x)^n]^2)),x]
Output:
(F^(a*f)*Sqrt[Pi]*(d + e*x)*Erfi[(1 + 2*b*f*n*Log[F]*Log[c*(d + e*x)^n])/( 2*Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])])/(2*Sqrt[b]*e*E^(1/(4*b*f*n^2*Log[F]))* Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*Sqrt[Log[F]])
Time = 0.53 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2705, 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.
\(\displaystyle \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx\) |
\(\Big \downarrow \) 2705 |
\(\displaystyle \frac {(d+e x) \left (c (d+e x)^n\right )^{-1/n} \int F^{b f \log ^2\left (c (d+e x)^n\right )+a f} \left (c (d+e x)^n\right )^{\frac {1}{n}}d\log \left (c (d+e x)^n\right )}{e n}\) |
\(\Big \downarrow \) 2725 |
\(\displaystyle \frac {(d+e x) \left (c (d+e x)^n\right )^{-1/n} \int \exp \left (b f \log (F) \log ^2\left (c (d+e x)^n\right )+\frac {\log \left (c (d+e x)^n\right )}{n}+a f \log (F)\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
\(\Big \downarrow \) 2664 |
\(\displaystyle \frac {F^{a f} (d+e x) e^{-\frac {1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-1/n} \int \exp \left (\frac {\left (2 b f n \log (F) \log \left (c (d+e x)^n\right )+1\right )^2}{4 b f n^2 \log (F)}\right )d\log \left (c (d+e x)^n\right )}{e n}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {\sqrt {\pi } F^{a f} (d+e x) e^{-\frac {1}{4 b f n^2 \log (F)}} \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {2 b f n \log (F) \log \left (c (d+e x)^n\right )+1}{2 \sqrt {b} \sqrt {f} n \sqrt {\log (F)}}\right )}{2 \sqrt {b} e \sqrt {f} n \sqrt {\log (F)}}\) |
Input:
Int[F^(f*(a + b*Log[c*(d + e*x)^n]^2)),x]
Output:
(F^(a*f)*Sqrt[Pi]*(d + e*x)*Erfi[(1 + 2*b*f*n*Log[F]*Log[c*(d + e*x)^n])/( 2*Sqrt[b]*Sqrt[f]*n*Sqrt[Log[F]])])/(2*Sqrt[b]*e*E^(1/(4*b*f*n^2*Log[F]))* Sqrt[f]*n*(c*(d + e*x)^n)^n^(-1)*Sqrt[Log[F]])
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_.)), x _Symbol] :> Simp[(d + e*x)/(e*n*(c*(d + e*x)^n)^(1/n)) Subst[Int[E^(a*f*L og[F] + 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, n}, x]
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]
Timed out.
hanged
Input:
int(F^(f*(a+b*ln(c*(e*x+d)^n)^2)),x)
Output:
int(F^(f*(a+b*ln(c*(e*x+d)^n)^2)),x)
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.98 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=-\frac {\sqrt {\pi } \sqrt {-b f n^{2} \log \left (F\right )} \operatorname {erf}\left (\frac {{\left (2 \, b f n^{2} \log \left (e x + d\right ) \log \left (F\right ) + 2 \, b f n \log \left (F\right ) \log \left (c\right ) + 1\right )} \sqrt {-b f n^{2} \log \left (F\right )}}{2 \, b f n^{2} \log \left (F\right )}\right ) e^{\left (\frac {4 \, a b f^{2} n^{2} \log \left (F\right )^{2} - 4 \, b f n \log \left (F\right ) \log \left (c\right ) - 1}{4 \, b f n^{2} \log \left (F\right )}\right )}}{2 \, e n} \] Input:
integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2)),x, algorithm="fricas")
Output:
-1/2*sqrt(pi)*sqrt(-b*f*n^2*log(F))*erf(1/2*(2*b*f*n^2*log(e*x + d)*log(F) + 2*b*f*n*log(F)*log(c) + 1)*sqrt(-b*f*n^2*log(F))/(b*f*n^2*log(F)))*e^(1 /4*(4*a*b*f^2*n^2*log(F)^2 - 4*b*f*n*log(F)*log(c) - 1)/(b*f*n^2*log(F)))/ (e*n)
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (105) = 210\).
Time = 4.90 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.85 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=\begin {cases} - \frac {2 F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d f n^{2} \log {\left (F \right )}}{e} - \frac {2 F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b d f n \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )}}{e} + 2 F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b f n^{2} x \log {\left (F \right )} - 2 F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} b f n x \log {\left (F \right )} \log {\left (c \left (d + e x\right )^{n} \right )} + \frac {F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} d}{e} + F^{a f + b f \log {\left (c \left (d + e x\right )^{n} \right )}^{2}} x & \text {for}\: e \neq 0 \\F^{f \left (a + b \log {\left (c d^{n} \right )}^{2}\right )} x & \text {otherwise} \end {cases} \] Input:
integrate(F**(f*(a+b*ln(c*(e*x+d)**n)**2)),x)
Output:
Piecewise((-2*F**(a*f + b*f*log(c*(d + e*x)**n)**2)*b*d*f*n**2*log(F)/e - 2*F**(a*f + b*f*log(c*(d + e*x)**n)**2)*b*d*f*n*log(F)*log(c*(d + e*x)**n) /e + 2*F**(a*f + b*f*log(c*(d + e*x)**n)**2)*b*f*n**2*x*log(F) - 2*F**(a*f + b*f*log(c*(d + e*x)**n)**2)*b*f*n*x*log(F)*log(c*(d + e*x)**n) + F**(a* f + b*f*log(c*(d + e*x)**n)**2)*d/e + F**(a*f + b*f*log(c*(d + e*x)**n)**2 )*x, Ne(e, 0)), (F**(f*(a + b*log(c*d**n)**2))*x, True))
\[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=\int { F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a\right )} f} \,d x } \] Input:
integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2)),x, algorithm="maxima")
Output:
integrate(F^((b*log((e*x + d)^n*c)^2 + a)*f), x)
Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86 \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=-\frac {\sqrt {\pi } F^{a f} \operatorname {erf}\left (-\sqrt {-b f \log \left (F\right )} n \log \left (e x + d\right ) - \sqrt {-b f \log \left (F\right )} \log \left (c\right ) - \frac {\sqrt {-b f \log \left (F\right )}}{2 \, b f n \log \left (F\right )}\right ) e^{\left (-\frac {1}{4 \, b f n^{2} \log \left (F\right )}\right )}}{2 \, \sqrt {-b f \log \left (F\right )} c^{\left (\frac {1}{n}\right )} e n} \] Input:
integrate(F^(f*(a+b*log(c*(e*x+d)^n)^2)),x, algorithm="giac")
Output:
-1/2*sqrt(pi)*F^(a*f)*erf(-sqrt(-b*f*log(F))*n*log(e*x + d) - sqrt(-b*f*lo g(F))*log(c) - 1/2*sqrt(-b*f*log(F))/(b*f*n*log(F)))*e^(-1/4/(b*f*n^2*log( F)))/(sqrt(-b*f*log(F))*c^(1/n)*e*n)
Timed out. \[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=\int F^{b\,f\,{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2}\,F^{a\,f} \,d x \] Input:
int(F^(f*(a + b*log(c*(d + e*x)^n)^2)),x)
Output:
int(F^(b*f*log(c*(d + e*x)^n)^2)*F^(a*f), x)
\[ \int F^{f \left (a+b \log ^2\left (c (d+e x)^n\right )\right )} \, dx=f^{a f} \left (\int f^{\mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b f}d x \right ) \] Input:
int(F^(f*(a+b*log(c*(e*x+d)^n)^2)),x)
Output:
f**(a*f)*int(f**(log((d + e*x)**n*c)**2*b*f),x)