Integrand size = 23, antiderivative size = 236 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx=\frac {a^3 (c+d x)^2}{2 d}-\frac {3 a^2 b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}-\frac {3 a b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}-\frac {b^3 d \left (F^{e g+f g x}\right )^{3 n}}{9 f^2 g^2 n^2 \log ^2(F)}+\frac {3 a^2 b \left (F^{e g+f g x}\right )^n (c+d x)}{f g n \log (F)}+\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n} (c+d x)}{2 f g n \log (F)}+\frac {b^3 \left (F^{e g+f g x}\right )^{3 n} (c+d x)}{3 f g n \log (F)} \]
1/2*a^3*(d*x+c)^2/d-3*a^2*b*d*(F^(f*g*x+e*g))^n/f^2/g^2/n^2/ln(F)^2-3/4*a* b^2*d*(F^(f*g*x+e*g))^(2*n)/f^2/g^2/n^2/ln(F)^2-1/9*b^3*d*(F^(f*g*x+e*g))^ (3*n)/f^2/g^2/n^2/ln(F)^2+3*a^2*b*(F^(f*g*x+e*g))^n*(d*x+c)/f/g/n/ln(F)+3/ 2*a*b^2*(F^(f*g*x+e*g))^(2*n)*(d*x+c)/f/g/n/ln(F)+1/3*b^3*(F^(f*g*x+e*g))^ (3*n)*(d*x+c)/f/g/n/ln(F)
Time = 0.45 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.68 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx=\frac {-b d \left (F^{g (e+f x)}\right )^n \left (108 a^2+27 a b \left (F^{g (e+f x)}\right )^n+4 b^2 \left (F^{g (e+f x)}\right )^{2 n}\right )+6 b f \left (F^{g (e+f x)}\right )^n \left (18 a^2+9 a b \left (F^{g (e+f x)}\right )^n+2 b^2 \left (F^{g (e+f x)}\right )^{2 n}\right ) g n (c+d x) \log (F)+18 a^3 f^2 g^2 n^2 x (2 c+d x) \log ^2(F)}{36 f^2 g^2 n^2 \log ^2(F)} \]
(-(b*d*(F^(g*(e + f*x)))^n*(108*a^2 + 27*a*b*(F^(g*(e + f*x)))^n + 4*b^2*( F^(g*(e + f*x)))^(2*n))) + 6*b*f*(F^(g*(e + f*x)))^n*(18*a^2 + 9*a*b*(F^(g *(e + f*x)))^n + 2*b^2*(F^(g*(e + f*x)))^(2*n))*g*n*(c + d*x)*Log[F] + 18* a^3*f^2*g^2*n^2*x*(2*c + d*x)*Log[F]^2)/(36*f^2*g^2*n^2*Log[F]^2)
Time = 0.44 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.
| \(\displaystyle \int (c+d x) \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx\) |
| \(\Big \downarrow \) 2614 |
| \(\displaystyle \int \left (a^3 (c+d x)+3 a^2 b (c+d x) \left (F^{e g+f g x}\right )^n+3 a b^2 (c+d x) \left (F^{e g+f g x}\right )^{2 n}+b^3 (c+d x) \left (F^{e g+f g x}\right )^{3 n}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 (c+d x)^2}{2 d}+\frac {3 a^2 b (c+d x) \left (F^{e g+f g x}\right )^n}{f g n \log (F)}-\frac {3 a^2 b d \left (F^{e g+f g x}\right )^n}{f^2 g^2 n^2 \log ^2(F)}+\frac {3 a b^2 (c+d x) \left (F^{e g+f g x}\right )^{2 n}}{2 f g n \log (F)}-\frac {3 a b^2 d \left (F^{e g+f g x}\right )^{2 n}}{4 f^2 g^2 n^2 \log ^2(F)}+\frac {b^3 (c+d x) \left (F^{e g+f g x}\right )^{3 n}}{3 f g n \log (F)}-\frac {b^3 d \left (F^{e g+f g x}\right )^{3 n}}{9 f^2 g^2 n^2 \log ^2(F)}\) |
(a^3*(c + d*x)^2)/(2*d) - (3*a^2*b*d*(F^(e*g + f*g*x))^n)/(f^2*g^2*n^2*Log [F]^2) - (3*a*b^2*d*(F^(e*g + f*g*x))^(2*n))/(4*f^2*g^2*n^2*Log[F]^2) - (b ^3*d*(F^(e*g + f*g*x))^(3*n))/(9*f^2*g^2*n^2*Log[F]^2) + (3*a^2*b*(F^(e*g + f*g*x))^n*(c + d*x))/(f*g*n*Log[F]) + (3*a*b^2*(F^(e*g + f*g*x))^(2*n)*( c + d*x))/(2*f*g*n*Log[F]) + (b^3*(F^(e*g + f*g*x))^(3*n)*(c + d*x))/(3*f* g*n*Log[F])
3.1.41.3.1 Defintions of rubi rules used
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 = 0.35 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.11
| method | result | size |
| parallelrisch | \(\frac {18 a^{3} d \,x^{2} n^{2} g^{2} f^{2} \ln \left (F \right )^{2}+36 a^{3} c x \,n^{2} g^{2} f^{2} \ln \left (F \right )^{2}+12 x \left (F^{g \left (f x +e \right )}\right )^{3 n} b^{3} d n g f \ln \left (F \right )+54 x \left (F^{g \left (f x +e \right )}\right )^{2 n} a \,b^{2} d n g f \ln \left (F \right )+12 \ln \left (F \right ) \left (F^{g \left (f x +e \right )}\right )^{3 n} b^{3} c f g n +108 x \left (F^{g \left (f x +e \right )}\right )^{n} a^{2} b d n g f \ln \left (F \right )+54 \ln \left (F \right ) \left (F^{g \left (f x +e \right )}\right )^{2 n} a \,b^{2} c f g n +108 \ln \left (F \right ) \left (F^{g \left (f x +e \right )}\right )^{n} a^{2} b c f g n -4 \left (F^{g \left (f x +e \right )}\right )^{3 n} b^{3} d -27 \left (F^{g \left (f x +e \right )}\right )^{2 n} a \,b^{2} d -108 \left (F^{g \left (f x +e \right )}\right )^{n} a^{2} b d}{36 n^{2} g^{2} f^{2} \ln \left (F \right )^{2}}\) | \(263\) |
1/36*(18*a^3*d*x^2*n^2*g^2*f^2*ln(F)^2+36*a^3*c*x*n^2*g^2*f^2*ln(F)^2+12*x *((F^(g*(f*x+e)))^n)^3*b^3*d*n*g*f*ln(F)+54*x*((F^(g*(f*x+e)))^n)^2*a*b^2* d*n*g*f*ln(F)+12*ln(F)*((F^(g*(f*x+e)))^n)^3*b^3*c*f*g*n+108*x*(F^(g*(f*x+ e)))^n*a^2*b*d*n*g*f*ln(F)+54*ln(F)*((F^(g*(f*x+e)))^n)^2*a*b^2*c*f*g*n+10 8*ln(F)*(F^(g*(f*x+e)))^n*a^2*b*c*f*g*n-4*((F^(g*(f*x+e)))^n)^3*b^3*d-27*( (F^(g*(f*x+e)))^n)^2*a*b^2*d-108*(F^(g*(f*x+e)))^n*a^2*b*d)/n^2/g^2/f^2/ln (F)^2
Time = 0.25 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.81 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx=\frac {18 \, {\left (a^{3} d f^{2} g^{2} n^{2} x^{2} + 2 \, a^{3} c f^{2} g^{2} n^{2} x\right )} \log \left (F\right )^{2} - 4 \, {\left (b^{3} d - 3 \, {\left (b^{3} d f g n x + b^{3} c f g n\right )} \log \left (F\right )\right )} F^{3 \, f g n x + 3 \, e g n} - 27 \, {\left (a b^{2} d - 2 \, {\left (a b^{2} d f g n x + a b^{2} c f g n\right )} \log \left (F\right )\right )} F^{2 \, f g n x + 2 \, e g n} - 108 \, {\left (a^{2} b d - {\left (a^{2} b d f g n x + a^{2} b c f g n\right )} \log \left (F\right )\right )} F^{f g n x + e g n}}{36 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \]
1/36*(18*(a^3*d*f^2*g^2*n^2*x^2 + 2*a^3*c*f^2*g^2*n^2*x)*log(F)^2 - 4*(b^3 *d - 3*(b^3*d*f*g*n*x + b^3*c*f*g*n)*log(F))*F^(3*f*g*n*x + 3*e*g*n) - 27* (a*b^2*d - 2*(a*b^2*d*f*g*n*x + a*b^2*c*f*g*n)*log(F))*F^(2*f*g*n*x + 2*e* g*n) - 108*(a^2*b*d - (a^2*b*d*f*g*n*x + a^2*b*c*f*g*n)*log(F))*F^(f*g*n*x + e*g*n))/(f^2*g^2*n^2*log(F)^2)
Time = 1.79 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.56 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx=\begin {cases} \left (a + b\right )^{3} \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: F = 1 \wedge f = 0 \wedge g = 0 \wedge n = 0 \\\left (a + b \left (F^{e g}\right )^{n}\right )^{3} \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: f = 0 \\\left (a + b\right )^{3} \left (c x + \frac {d x^{2}}{2}\right ) & \text {for}\: F = 1 \vee g = 0 \vee n = 0 \\a^{3} c x + \frac {a^{3} d x^{2}}{2} + \frac {3 a^{2} b c \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} + \frac {3 a^{2} b d x \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} - \frac {3 a^{2} b d \left (F^{e g + f g x}\right )^{n}}{f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}} + \frac {3 a b^{2} c \left (F^{e g + f g x}\right )^{2 n}}{2 f g n \log {\left (F \right )}} + \frac {3 a b^{2} d x \left (F^{e g + f g x}\right )^{2 n}}{2 f g n \log {\left (F \right )}} - \frac {3 a b^{2} d \left (F^{e g + f g x}\right )^{2 n}}{4 f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}} + \frac {b^{3} c \left (F^{e g + f g x}\right )^{3 n}}{3 f g n \log {\left (F \right )}} + \frac {b^{3} d x \left (F^{e g + f g x}\right )^{3 n}}{3 f g n \log {\left (F \right )}} - \frac {b^{3} d \left (F^{e g + f g x}\right )^{3 n}}{9 f^{2} g^{2} n^{2} \log {\left (F \right )}^{2}} & \text {otherwise} \end {cases} \]
Piecewise(((a + b)**3*(c*x + d*x**2/2), Eq(F, 1) & Eq(f, 0) & Eq(g, 0) & E q(n, 0)), ((a + b*(F**(e*g))**n)**3*(c*x + d*x**2/2), Eq(f, 0)), ((a + b)* *3*(c*x + d*x**2/2), Eq(F, 1) | Eq(g, 0) | Eq(n, 0)), (a**3*c*x + a**3*d*x **2/2 + 3*a**2*b*c*(F**(e*g + f*g*x))**n/(f*g*n*log(F)) + 3*a**2*b*d*x*(F* *(e*g + f*g*x))**n/(f*g*n*log(F)) - 3*a**2*b*d*(F**(e*g + f*g*x))**n/(f**2 *g**2*n**2*log(F)**2) + 3*a*b**2*c*(F**(e*g + f*g*x))**(2*n)/(2*f*g*n*log( F)) + 3*a*b**2*d*x*(F**(e*g + f*g*x))**(2*n)/(2*f*g*n*log(F)) - 3*a*b**2*d *(F**(e*g + f*g*x))**(2*n)/(4*f**2*g**2*n**2*log(F)**2) + b**3*c*(F**(e*g + f*g*x))**(3*n)/(3*f*g*n*log(F)) + b**3*d*x*(F**(e*g + f*g*x))**(3*n)/(3* f*g*n*log(F)) - b**3*d*(F**(e*g + f*g*x))**(3*n)/(9*f**2*g**2*n**2*log(F)* *2), True))
Time = 0.23 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.14 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx=\frac {1}{2} \, a^{3} d x^{2} + a^{3} c x + \frac {3 \, F^{f g n x + e g n} a^{2} b c}{f g n \log \left (F\right )} + \frac {3 \, F^{2 \, f g n x + 2 \, e g n} a b^{2} c}{2 \, f g n \log \left (F\right )} + \frac {F^{3 \, f g n x + 3 \, e g n} b^{3} c}{3 \, f g n \log \left (F\right )} + \frac {3 \, {\left (F^{e g n} f g n x \log \left (F\right ) - F^{e g n}\right )} F^{f g n x} a^{2} b d}{f^{2} g^{2} n^{2} \log \left (F\right )^{2}} + \frac {3 \, {\left (2 \, F^{2 \, e g n} f g n x \log \left (F\right ) - F^{2 \, e g n}\right )} F^{2 \, f g n x} a b^{2} d}{4 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} + \frac {{\left (3 \, F^{3 \, e g n} f g n x \log \left (F\right ) - F^{3 \, e g n}\right )} F^{3 \, f g n x} b^{3} d}{9 \, f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \]
1/2*a^3*d*x^2 + a^3*c*x + 3*F^(f*g*n*x + e*g*n)*a^2*b*c/(f*g*n*log(F)) + 3 /2*F^(2*f*g*n*x + 2*e*g*n)*a*b^2*c/(f*g*n*log(F)) + 1/3*F^(3*f*g*n*x + 3*e *g*n)*b^3*c/(f*g*n*log(F)) + 3*(F^(e*g*n)*f*g*n*x*log(F) - F^(e*g*n))*F^(f *g*n*x)*a^2*b*d/(f^2*g^2*n^2*log(F)^2) + 3/4*(2*F^(2*e*g*n)*f*g*n*x*log(F) - F^(2*e*g*n))*F^(2*f*g*n*x)*a*b^2*d/(f^2*g^2*n^2*log(F)^2) + 1/9*(3*F^(3 *e*g*n)*f*g*n*x*log(F) - F^(3*e*g*n))*F^(3*f*g*n*x)*b^3*d/(f^2*g^2*n^2*log (F)^2)
Result contains complex when optimal does not.
Time = 0.51 (sec) , antiderivative size = 3528, normalized size of antiderivative = 14.95 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx=\text {Too large to display} \]
1/2*a^3*d*x^2 + a^3*c*x + 1/9*(2*((3*b^3*d*f*g*n*x*log(abs(F)) + 3*b^3*c*f *g*n*log(abs(F)) - b^3*d)*(pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2* f^2*g^2*n^2*log(abs(F))^2)/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi *f^2*g^2*n^2*log(abs(F)))^2) + 3*(pi*b^3*d*f*g*n*x*sgn(F) - pi*b^3*d*f*g*n *x + pi*b^3*c*f*g*n*sgn(F) - pi*b^3*c*f*g*n)*(pi*f^2*g^2*n^2*log(abs(F))*s gn(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*sgn(F) - pi^2*f^2*g ^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2*n^2*log(abs(F))*sg n(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*cos(-3/2*pi*f*g*n*x*sgn(F) + 3/2*pi *f*g*n*x - 3/2*pi*e*g*n*sgn(F) + 3/2*pi*e*g*n) + (3*(pi*b^3*d*f*g*n*x*sgn( F) - pi*b^3*d*f*g*n*x + pi*b^3*c*f*g*n*sgn(F) - pi*b^3*c*f*g*n)*(pi^2*f^2* g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)/((pi^2*f^ 2*g^2*n^2*sgn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*( pi*f^2*g^2*n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2) - 4*(3* b^3*d*f*g*n*x*log(abs(F)) + 3*b^3*c*f*g*n*log(abs(F)) - b^3*d)*(pi*f^2*g^2 *n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))/((pi^2*f^2*g^2*n^2*s gn(F) - pi^2*f^2*g^2*n^2 + 2*f^2*g^2*n^2*log(abs(F))^2)^2 + 4*(pi*f^2*g^2* n^2*log(abs(F))*sgn(F) - pi*f^2*g^2*n^2*log(abs(F)))^2))*sin(-3/2*pi*f*g*n *x*sgn(F) + 3/2*pi*f*g*n*x - 3/2*pi*e*g*n*sgn(F) + 3/2*pi*e*g*n))*e^(3*f*g *n*x*log(abs(F)) + 3*e*g*n*log(abs(F))) - 1/18*I*((3*pi*b^3*d*f*g*n*x*s...
Time = 0.37 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.93 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 (c+d x) \, dx=\frac {a^3\,d\,x^2}{2}-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{2\,n}\,\left (\frac {3\,a\,b^2\,\left (d-2\,c\,f\,g\,n\,\ln \left (F\right )\right )}{4\,f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {3\,a\,b^2\,d\,x}{2\,f\,g\,n\,\ln \left (F\right )}\right )-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{3\,n}\,\left (\frac {b^3\,\left (d-3\,c\,f\,g\,n\,\ln \left (F\right )\right )}{9\,f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {b^3\,d\,x}{3\,f\,g\,n\,\ln \left (F\right )}\right )-{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n\,\left (\frac {3\,a^2\,b\,\left (d-c\,f\,g\,n\,\ln \left (F\right )\right )}{f^2\,g^2\,n^2\,{\ln \left (F\right )}^2}-\frac {3\,a^2\,b\,d\,x}{f\,g\,n\,\ln \left (F\right )}\right )+a^3\,c\,x \]
(a^3*d*x^2)/2 - (F^(f*g*x)*F^(e*g))^(2*n)*((3*a*b^2*(d - 2*c*f*g*n*log(F)) )/(4*f^2*g^2*n^2*log(F)^2) - (3*a*b^2*d*x)/(2*f*g*n*log(F))) - (F^(f*g*x)* F^(e*g))^(3*n)*((b^3*(d - 3*c*f*g*n*log(F)))/(9*f^2*g^2*n^2*log(F)^2) - (b ^3*d*x)/(3*f*g*n*log(F))) - (F^(f*g*x)*F^(e*g))^n*((3*a^2*b*(d - c*f*g*n*l og(F)))/(f^2*g^2*n^2*log(F)^2) - (3*a^2*b*d*x)/(f*g*n*log(F))) + a^3*c*x