Integrand size = 21, antiderivative size = 77 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\frac {d (f+g x)^2}{2 g}-\frac {e \left (F^{a c+b c x}\right )^n g}{b^2 c^2 n^2 \log ^2(F)}+\frac {e \left (F^{a c+b c x}\right )^n (f+g x)}{b c n \log (F)} \] Output:
1/2*d*(g*x+f)^2/g-e*(F^(b*c*x+a*c))^n*g/b^2/c^2/n^2/ln(F)^2+e*(F^(b*c*x+a* c))^n*(g*x+f)/b/c/n/ln(F)
Time = 0.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\frac {1}{2} d x (2 f+g x)-\frac {e \left (F^{c (a+b x)}\right )^n g}{b^2 c^2 n^2 \log ^2(F)}+\frac {e \left (F^{c (a+b x)}\right )^n (f+g x)}{b c n \log (F)} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)*(f + g*x),x]
Output:
(d*x*(2*f + g*x))/2 - (e*(F^(c*(a + b*x)))^n*g)/(b^2*c^2*n^2*Log[F]^2) + ( e*(F^(c*(a + b*x)))^n*(f + g*x))/(b*c*n*Log[F])
Time = 0.42 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2614, 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.
\(\displaystyle \int (f+g x) \left (e \left (F^{c (a+b x)}\right )^n+d\right ) \, dx\) |
\(\Big \downarrow \) 2614 |
\(\displaystyle \int \left (e (f+g x) \left (F^{a c+b c x}\right )^n+d (f+g x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e g \left (F^{a c+b c x}\right )^n}{b^2 c^2 n^2 \log ^2(F)}+\frac {e (f+g x) \left (F^{a c+b c x}\right )^n}{b c n \log (F)}+\frac {d (f+g x)^2}{2 g}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)*(f + g*x),x]
Output:
(d*(f + g*x)^2)/(2*g) - (e*(F^(a*c + b*c*x))^n*g)/(b^2*c^2*n^2*Log[F]^2) + (e*(F^(a*c + b*c*x))^n*(f + g*x))/(b*c*n*Log[F])
Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*(F ^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09
method | result | size |
norman | \(d f x +\frac {e \left (\ln \left (F \right ) b c f n -g \right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right )^{2} b^{2} c^{2} n^{2}}+\frac {e g x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) b c n}+\frac {d g \,x^{2}}{2}\) | \(84\) |
parallelrisch | \(\frac {d g \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+2 d f x \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+2 \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right ) b c e g n x +2 \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right ) b c e f n -2 \left (F^{c \left (b x +a \right )}\right )^{n} e g}{2 \ln \left (F \right )^{2} b^{2} c^{2} n^{2}}\) | \(110\) |
orering | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} g^{2} n^{2} x^{3}+3 \ln \left (F \right )^{2} b^{2} c^{2} f g \,n^{2} x^{2}+2 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} n^{2} x +\ln \left (F \right ) b c \,g^{2} n \,x^{2}+4 \ln \left (F \right ) b c f g n x +2 \ln \left (F \right ) b c \,f^{2} n -4 g^{2} x -2 g f \right ) \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}{2 \ln \left (F \right )^{2} b^{2} c^{2} n^{2} \left (g x +f \right )}-\frac {x \left (\ln \left (F \right ) b c g n x +2 \ln \left (F \right ) b c f n -2 g \right ) \left (\left (g x +f \right ) e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right ) b c n +\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) g \right )}{2 \ln \left (F \right )^{2} b^{2} c^{2} n^{2} \left (g x +f \right )}\) | \(225\) |
Input:
int((d+e*(F^(c*(b*x+a)))^n)*(g*x+f),x,method=_RETURNVERBOSE)
Output:
d*f*x+e*(ln(F)*b*c*f*n-g)/ln(F)^2/b^2/c^2/n^2*exp(n*ln(exp(c*(b*x+a)*ln(F) )))+1/ln(F)/b/c/n*e*g*x*exp(n*ln(exp(c*(b*x+a)*ln(F))))+1/2*d*g*x^2
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\frac {{\left (b^{2} c^{2} d g n^{2} x^{2} + 2 \, b^{2} c^{2} d f n^{2} x\right )} \log \left (F\right )^{2} - 2 \, {\left (e g - {\left (b c e g n x + b c e f n\right )} \log \left (F\right )\right )} F^{b c n x + a c n}}{2 \, b^{2} c^{2} n^{2} \log \left (F\right )^{2}} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f),x, algorithm="fricas")
Output:
1/2*((b^2*c^2*d*g*n^2*x^2 + 2*b^2*c^2*d*f*n^2*x)*log(F)^2 - 2*(e*g - (b*c* e*g*n*x + b*c*e*f*n)*log(F))*F^(b*c*n*x + a*c*n))/(b^2*c^2*n^2*log(F)^2)
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (68) = 136\).
Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.87 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\begin {cases} \left (d + e\right ) \left (f x + \frac {g x^{2}}{2}\right ) & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge n = 0 \\\left (d + e \left (F^{a c}\right )^{n}\right ) \left (f x + \frac {g x^{2}}{2}\right ) & \text {for}\: b = 0 \\\left (d + e\right ) \left (f x + \frac {g x^{2}}{2}\right ) & \text {for}\: F = 1 \vee c = 0 \vee n = 0 \\d f x + \frac {d g x^{2}}{2} + \frac {e f \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {e g x \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} - \frac {e g \left (F^{a c + b c x}\right )^{n}}{b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((d+e*(F**((b*x+a)*c))**n)*(g*x+f),x)
Output:
Piecewise(((d + e)*(f*x + g*x**2/2), Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(n , 0)), ((d + e*(F**(a*c))**n)*(f*x + g*x**2/2), Eq(b, 0)), ((d + e)*(f*x + g*x**2/2), Eq(F, 1) | Eq(c, 0) | Eq(n, 0)), (d*f*x + d*g*x**2/2 + e*f*(F* *(a*c + b*c*x))**n/(b*c*n*log(F)) + e*g*x*(F**(a*c + b*c*x))**n/(b*c*n*log (F)) - e*g*(F**(a*c + b*c*x))**n/(b**2*c**2*n**2*log(F)**2), True))
Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\frac {1}{2} \, d g x^{2} + d f x + \frac {F^{b c n x + a c n} e f}{b c n \log \left (F\right )} + \frac {{\left (F^{a c n} b c n x \log \left (F\right ) - F^{a c n}\right )} F^{b c n x} e g}{b^{2} c^{2} n^{2} \log \left (F\right )^{2}} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f),x, algorithm="maxima")
Output:
1/2*d*g*x^2 + d*f*x + F^(b*c*n*x + a*c*n)*e*f/(b*c*n*log(F)) + (F^(a*c*n)* b*c*n*x*log(F) - F^(a*c*n))*F^(b*c*n*x)*e*g/(b^2*c^2*n^2*log(F)^2)
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 1101, normalized size of antiderivative = 14.30 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\text {Too large to display} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f),x, algorithm="giac")
Output:
1/2*d*g*x^2 + d*f*x + (2*((pi*b^2*c^2*n^2*log(abs(F))*sgn(F) - pi*b^2*c^2* n^2*log(abs(F)))*(pi*b*c*e*g*n*x*sgn(F) - pi*b*c*e*g*n*x + pi*b*c*e*f*n*sg n(F) - pi*b*c*e*f*n)/((pi^2*b^2*c^2*n^2*sgn(F) - pi^2*b^2*c^2*n^2 + 2*b^2* c^2*n^2*log(abs(F))^2)^2 + 4*(pi*b^2*c^2*n^2*log(abs(F))*sgn(F) - pi*b^2*c ^2*n^2*log(abs(F)))^2) + (pi^2*b^2*c^2*n^2*sgn(F) - pi^2*b^2*c^2*n^2 + 2*b ^2*c^2*n^2*log(abs(F))^2)*(b*c*e*g*n*x*log(abs(F)) + b*c*e*f*n*log(abs(F)) - e*g)/((pi^2*b^2*c^2*n^2*sgn(F) - pi^2*b^2*c^2*n^2 + 2*b^2*c^2*n^2*log(a bs(F))^2)^2 + 4*(pi*b^2*c^2*n^2*log(abs(F))*sgn(F) - pi*b^2*c^2*n^2*log(ab s(F)))^2))*cos(-1/2*pi*b*c*n*x*sgn(F) + 1/2*pi*b*c*n*x - 1/2*pi*a*c*n*sgn( F) + 1/2*pi*a*c*n) + ((pi^2*b^2*c^2*n^2*sgn(F) - pi^2*b^2*c^2*n^2 + 2*b^2* c^2*n^2*log(abs(F))^2)*(pi*b*c*e*g*n*x*sgn(F) - pi*b*c*e*g*n*x + pi*b*c*e* f*n*sgn(F) - pi*b*c*e*f*n)/((pi^2*b^2*c^2*n^2*sgn(F) - pi^2*b^2*c^2*n^2 + 2*b^2*c^2*n^2*log(abs(F))^2)^2 + 4*(pi*b^2*c^2*n^2*log(abs(F))*sgn(F) - pi *b^2*c^2*n^2*log(abs(F)))^2) - 4*(pi*b^2*c^2*n^2*log(abs(F))*sgn(F) - pi*b ^2*c^2*n^2*log(abs(F)))*(b*c*e*g*n*x*log(abs(F)) + b*c*e*f*n*log(abs(F)) - e*g)/((pi^2*b^2*c^2*n^2*sgn(F) - pi^2*b^2*c^2*n^2 + 2*b^2*c^2*n^2*log(abs (F))^2)^2 + 4*(pi*b^2*c^2*n^2*log(abs(F))*sgn(F) - pi*b^2*c^2*n^2*log(abs( F)))^2))*sin(-1/2*pi*b*c*n*x*sgn(F) + 1/2*pi*b*c*n*x - 1/2*pi*a*c*n*sgn(F) + 1/2*pi*a*c*n))*e^(b*c*n*x*log(abs(F)) + a*c*n*log(abs(F))) - 1/2*I*((pi *b*c*e*g*n*x*sgn(F) - pi*b*c*e*g*n*x - 2*I*b*c*e*g*n*x*log(abs(F)) + pi...
Time = 22.85 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\frac {d\,g\,x^2}{2}-{\left (F^{b\,c\,x}\,F^{a\,c}\right )}^n\,\left (\frac {e\,\left (g-b\,c\,f\,n\,\ln \left (F\right )\right )}{b^2\,c^2\,n^2\,{\ln \left (F\right )}^2}-\frac {e\,g\,x}{b\,c\,n\,\ln \left (F\right )}\right )+d\,f\,x \] Input:
int((f + g*x)*(d + e*(F^(c*(a + b*x)))^n),x)
Output:
(d*g*x^2)/2 - (F^(b*c*x)*F^(a*c))^n*((e*(g - b*c*f*n*log(F)))/(b^2*c^2*n^2 *log(F)^2) - (e*g*x)/(b*c*n*log(F))) + d*f*x
Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.45 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x) \, dx=\frac {2 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c e f n +2 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c e g n x -2 f^{b c n x +a c n} e g +2 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d f \,n^{2} x +\mathrm {log}\left (f \right )^{2} b^{2} c^{2} d g \,n^{2} x^{2}}{2 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} n^{2}} \] Input:
int((d+e*(F^((b*x+a)*c))^n)*(g*x+f),x)
Output:
(2*f**(a*c*n + b*c*n*x)*log(f)*b*c*e*f*n + 2*f**(a*c*n + b*c*n*x)*log(f)*b *c*e*g*n*x - 2*f**(a*c*n + b*c*n*x)*e*g + 2*log(f)**2*b**2*c**2*d*f*n**2*x + log(f)**2*b**2*c**2*d*g*n**2*x**2)/(2*log(f)**2*b**2*c**2*n**2)