Integrand size = 23, antiderivative size = 156 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=\frac {d^2 (f+g x)^2}{2 g}-\frac {2 d e \left (F^{a c+b c x}\right )^n g}{b^2 c^2 n^2 \log ^2(F)}-\frac {e^2 \left (F^{a c+b c x}\right )^{2 n} g}{4 b^2 c^2 n^2 \log ^2(F)}+\frac {2 d e \left (F^{a c+b c x}\right )^n (f+g x)}{b c n \log (F)}+\frac {e^2 \left (F^{a c+b c x}\right )^{2 n} (f+g x)}{2 b c n \log (F)} \] Output:
1/2*d^2*(g*x+f)^2/g-2*d*e*(F^(b*c*x+a*c))^n*g/b^2/c^2/n^2/ln(F)^2-1/4*e^2* (F^(b*c*x+a*c))^(2*n)*g/b^2/c^2/n^2/ln(F)^2+2*d*e*(F^(b*c*x+a*c))^n*(g*x+f )/b/c/n/ln(F)+1/2*e^2*(F^(b*c*x+a*c))^(2*n)*(g*x+f)/b/c/n/ln(F)
Time = 0.60 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=\frac {-e \left (F^{c (a+b x)}\right )^n \left (8 d+e \left (F^{c (a+b x)}\right )^n\right ) g+2 b c e \left (F^{c (a+b x)}\right )^n \left (4 d+e \left (F^{c (a+b x)}\right )^n\right ) n (f+g x) \log (F)+2 b^2 c^2 d^2 n^2 x (2 f+g x) \log ^2(F)}{4 b^2 c^2 n^2 \log ^2(F)} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)^2*(f + g*x),x]
Output:
(-(e*(F^(c*(a + b*x)))^n*(8*d + e*(F^(c*(a + b*x)))^n)*g) + 2*b*c*e*(F^(c* (a + b*x)))^n*(4*d + e*(F^(c*(a + b*x)))^n)*n*(f + g*x)*Log[F] + 2*b^2*c^2 *d^2*n^2*x*(2*f + g*x)*Log[F]^2)/(4*b^2*c^2*n^2*Log[F]^2)
Time = 0.60 (sec) , antiderivative size = 156, 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.
| \(\displaystyle \int (f+g x) \left (e \left (F^{c (a+b x)}\right )^n+d\right )^2 \, dx\) |
| \(\Big \downarrow \) 2614 |
| \(\displaystyle \int \left (2 d e (f+g x) \left (F^{a c+b c x}\right )^n+e^2 (f+g x) \left (F^{a c+b c x}\right )^{2 n}+d^2 (f+g x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 d e g \left (F^{a c+b c x}\right )^n}{b^2 c^2 n^2 \log ^2(F)}-\frac {e^2 g \left (F^{a c+b c x}\right )^{2 n}}{4 b^2 c^2 n^2 \log ^2(F)}+\frac {2 d e (f+g x) \left (F^{a c+b c x}\right )^n}{b c n \log (F)}+\frac {e^2 (f+g x) \left (F^{a c+b c x}\right )^{2 n}}{2 b c n \log (F)}+\frac {d^2 (f+g x)^2}{2 g}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)^2*(f + g*x),x]
Output:
(d^2*(f + g*x)^2)/(2*g) - (2*d*e*(F^(a*c + b*c*x))^n*g)/(b^2*c^2*n^2*Log[F ]^2) - (e^2*(F^(a*c + b*c*x))^(2*n)*g)/(4*b^2*c^2*n^2*Log[F]^2) + (2*d*e*( F^(a*c + b*c*x))^n*(f + g*x))/(b*c*n*Log[F]) + (e^2*(F^(a*c + b*c*x))^(2*n )*(f + g*x))/(2*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 = 0.22 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.12
| method | result | size |
| norman | \(d^{2} f x +\frac {d^{2} g \,x^{2}}{2}+\frac {e^{2} \left (2 \ln \left (F \right ) b c f n -g \right ) {\mathrm e}^{2 n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{4 \ln \left (F \right )^{2} b^{2} c^{2} n^{2}}+\frac {2 d e \left (\ln \left (F \right ) b c f n -g \right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right )^{2} b^{2} c^{2} n^{2}}+\frac {e^{2} g x \,{\mathrm e}^{2 n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{2 \ln \left (F \right ) b c n}+\frac {2 d e g x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) b c n}\) | \(174\) |
| parallelrisch | \(\frac {2 d^{2} g \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+4 d^{2} f x \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+2 \ln \left (F \right ) \left (F^{c \left (b x +a \right )}\right )^{2 n} b c \,e^{2} g n x +8 \ln \left (F \right ) \left (F^{c \left (b x +a \right )}\right )^{n} b c d e g n x +2 \ln \left (F \right ) \left (F^{c \left (b x +a \right )}\right )^{2 n} b c \,e^{2} f n +8 \ln \left (F \right ) \left (F^{c \left (b x +a \right )}\right )^{n} b c d e f n -\left (F^{c \left (b x +a \right )}\right )^{2 n} e^{2} g -8 \left (F^{c \left (b x +a \right )}\right )^{n} d e g}{4 \ln \left (F \right )^{2} b^{2} c^{2} n^{2}}\) | \(186\) |
| orering | \(\frac {\left (2 \ln \left (F \right )^{3} b^{3} c^{3} g^{3} n^{3} x^{4}+8 \ln \left (F \right )^{3} b^{3} c^{3} f \,g^{2} n^{3} x^{3}+10 \ln \left (F \right )^{3} b^{3} c^{3} f^{2} g \,n^{3} x^{2}+4 \ln \left (F \right )^{3} b^{3} c^{3} f^{3} n^{3} x +3 \ln \left (F \right )^{2} b^{2} c^{2} g^{3} n^{2} x^{3}+15 \ln \left (F \right )^{2} b^{2} c^{2} f \,g^{2} n^{2} x^{2}+18 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} g \,n^{2} x +6 \ln \left (F \right )^{2} b^{2} c^{2} f^{3} n^{2}-12 g^{3} \ln \left (F \right ) b c n \,x^{2}-15 g^{2} \ln \left (F \right ) b c n x f -5 g \ln \left (F \right ) b c \,f^{2} n -9 g^{3} x -3 g^{2} f \right ) {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{2}}{4 \ln \left (F \right )^{3} b^{3} c^{3} n^{3} \left (g x +f \right )^{2}}-\frac {\left (3 \ln \left (F \right )^{2} b^{2} c^{2} g^{2} n^{2} x^{3}+9 \ln \left (F \right )^{2} b^{2} c^{2} f g \,n^{2} x^{2}+6 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} n^{2} x -5 \ln \left (F \right ) b c \,g^{2} n \,x^{2}-\ln \left (F \right ) b c f g n x +2 \ln \left (F \right ) b c \,f^{2} n -9 g^{2} x -3 g f \right ) \left (2 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \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 )}^{2} g \right )}{4 \ln \left (F \right )^{3} b^{3} c^{3} n^{3} \left (g x +f \right )^{2}}+\frac {x \left (\ln \left (F \right ) b c g n x +2 \ln \left (F \right ) b c f n -3 g \right ) \left (2 e^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n} \ln \left (F \right )^{2} b^{2} c^{2} n^{2} \left (g x +f \right )+4 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) g e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right ) b c n +2 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \left (g x +f \right ) e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}\right )}{4 \ln \left (F \right )^{3} b^{3} c^{3} n^{3} \left (g x +f \right )}\) | \(602\) |
Input:
int((d+e*(F^(c*(b*x+a)))^n)^2*(g*x+f),x,method=_RETURNVERBOSE)
Output:
d^2*f*x+1/2*d^2*g*x^2+1/4*e^2*(2*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))))^2+2*d*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))))+1/2/ln(F)/b/c/n*e^2*g*x*exp(n*ln(exp(c*(b* x+a)*ln(F))))^2+2/ln(F)/b/c/n*d*e*g*x*exp(n*ln(exp(c*(b*x+a)*ln(F))))
Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.89 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=\frac {2 \, {\left (b^{2} c^{2} d^{2} g n^{2} x^{2} + 2 \, b^{2} c^{2} d^{2} f n^{2} x\right )} \log \left (F\right )^{2} - {\left (e^{2} g - 2 \, {\left (b c e^{2} g n x + b c e^{2} f n\right )} \log \left (F\right )\right )} F^{2 \, b c n x + 2 \, a c n} - 8 \, {\left (d e g - {\left (b c d e g n x + b c d e f n\right )} \log \left (F\right )\right )} F^{b c n x + a c n}}{4 \, b^{2} c^{2} n^{2} \log \left (F\right )^{2}} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f),x, algorithm="fricas")
Output:
1/4*(2*(b^2*c^2*d^2*g*n^2*x^2 + 2*b^2*c^2*d^2*f*n^2*x)*log(F)^2 - (e^2*g - 2*(b*c*e^2*g*n*x + b*c*e^2*f*n)*log(F))*F^(2*b*c*n*x + 2*a*c*n) - 8*(d*e* g - (b*c*d*e*g*n*x + b*c*d*e*f*n)*log(F))*F^(b*c*n*x + a*c*n))/(b^2*c^2*n^ 2*log(F)^2)
Time = 0.81 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.65 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=\begin {cases} \left (d + e\right )^{2} \left (f x + \frac {g x^{2}}{2}\right ) & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge n = 0 \\\left (d + e \left (F^{a c}\right )^{n}\right )^{2} \left (f x + \frac {g x^{2}}{2}\right ) & \text {for}\: b = 0 \\\left (d + e\right )^{2} \left (f x + \frac {g x^{2}}{2}\right ) & \text {for}\: F = 1 \vee c = 0 \vee n = 0 \\d^{2} f x + \frac {d^{2} g x^{2}}{2} + \frac {2 d e f \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {2 d e g x \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {e^{2} f \left (F^{a c + b c x}\right )^{2 n}}{2 b c n \log {\left (F \right )}} + \frac {e^{2} g x \left (F^{a c + b c x}\right )^{2 n}}{2 b c n \log {\left (F \right )}} - \frac {2 d e g \left (F^{a c + b c x}\right )^{n}}{b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} - \frac {e^{2} g \left (F^{a c + b c x}\right )^{2 n}}{4 b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((d+e*(F**((b*x+a)*c))**n)**2*(g*x+f),x)
Output:
Piecewise(((d + e)**2*(f*x + g*x**2/2), Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & E q(n, 0)), ((d + e*(F**(a*c))**n)**2*(f*x + g*x**2/2), Eq(b, 0)), ((d + e)* *2*(f*x + g*x**2/2), Eq(F, 1) | Eq(c, 0) | Eq(n, 0)), (d**2*f*x + d**2*g*x **2/2 + 2*d*e*f*(F**(a*c + b*c*x))**n/(b*c*n*log(F)) + 2*d*e*g*x*(F**(a*c + b*c*x))**n/(b*c*n*log(F)) + e**2*f*(F**(a*c + b*c*x))**(2*n)/(2*b*c*n*lo g(F)) + e**2*g*x*(F**(a*c + b*c*x))**(2*n)/(2*b*c*n*log(F)) - 2*d*e*g*(F** (a*c + b*c*x))**n/(b**2*c**2*n**2*log(F)**2) - e**2*g*(F**(a*c + b*c*x))** (2*n)/(4*b**2*c**2*n**2*log(F)**2), True))
Time = 0.06 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.14 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=\frac {1}{2} \, d^{2} g x^{2} + d^{2} f x + \frac {2 \, F^{b c n x + a c n} d e f}{b c n \log \left (F\right )} + \frac {F^{2 \, b c n x + 2 \, a c n} e^{2} f}{2 \, b c n \log \left (F\right )} + \frac {2 \, {\left (F^{a c n} b c n x \log \left (F\right ) - F^{a c n}\right )} F^{b c n x} d e g}{b^{2} c^{2} n^{2} \log \left (F\right )^{2}} + \frac {{\left (2 \, F^{2 \, a c n} b c n x \log \left (F\right ) - F^{2 \, a c n}\right )} F^{2 \, b c n x} e^{2} g}{4 \, b^{2} c^{2} n^{2} \log \left (F\right )^{2}} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f),x, algorithm="maxima")
Output:
1/2*d^2*g*x^2 + d^2*f*x + 2*F^(b*c*n*x + a*c*n)*d*e*f/(b*c*n*log(F)) + 1/2 *F^(2*b*c*n*x + 2*a*c*n)*e^2*f/(b*c*n*log(F)) + 2*(F^(a*c*n)*b*c*n*x*log(F ) - F^(a*c*n))*F^(b*c*n*x)*d*e*g/(b^2*c^2*n^2*log(F)^2) + 1/4*(2*F^(2*a*c* n)*b*c*n*x*log(F) - F^(2*a*c*n))*F^(2*b*c*n*x)*e^2*g/(b^2*c^2*n^2*log(F)^2 )
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 2284, normalized size of antiderivative = 14.64 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=\text {Too large to display} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f),x, algorithm="giac")
Output:
1/2*d^2*g*x^2 + d^2*f*x + 1/2*((2*(pi*b*c*e^2*g*n*x*sgn(F) - pi*b*c*e^2*g* n*x + pi*b*c*e^2*f*n*sgn(F) - pi*b*c*e^2*f*n)*(pi*b^2*c^2*n^2*log(abs(F))* sgn(F) - pi*b^2*c^2*n^2*log(abs(F)))/((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))*s gn(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)*(2*b*c*e^2*g*n*x*log(abs(F)) + 2 *b*c*e^2*f*n*log(abs(F)) - e^2*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))*cos(-pi*b*c*n*x*sgn(F) + pi*b*c*n*x - pi*a*c*n*sgn(F) + 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^2*g*n*x*sgn(F) - pi*b*c*e^2*g*n *x + pi*b*c*e^2*f*n*sgn(F) - pi*b*c*e^2*f*n)/((pi^2*b^2*c^2*n^2*sgn(F) - p i^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(a bs(F))*sgn(F) - pi*b^2*c^2*n^2*log(abs(F)))^2) - 2*(pi*b^2*c^2*n^2*log(abs (F))*sgn(F) - pi*b^2*c^2*n^2*log(abs(F)))*(2*b*c*e^2*g*n*x*log(abs(F)) + 2 *b*c*e^2*f*n*log(abs(F)) - e^2*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(-pi*b*c*n*x*sgn(F) + pi*b*c*n*x - pi*a*c*n*sgn(F) + pi*a*c*n))*e^(2*b*c*n*x*log(abs(F)) + 2*a*c*n*log(abs(F ))) - 1/4*I*((pi*b*c*e^2*g*n*x*sgn(F) - pi*b*c*e^2*g*n*x - 2*I*b*c*e^2*...
Time = 23.07 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.94 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=d^2\,f\,x-{\left (F^{b\,c\,x}\,F^{a\,c}\right )}^{2\,n}\,\left (\frac {e^2\,\left (g-2\,b\,c\,f\,n\,\ln \left (F\right )\right )}{4\,b^2\,c^2\,n^2\,{\ln \left (F\right )}^2}-\frac {e^2\,g\,x}{2\,b\,c\,n\,\ln \left (F\right )}\right )-{\left (F^{b\,c\,x}\,F^{a\,c}\right )}^n\,\left (\frac {2\,d\,e\,\left (g-b\,c\,f\,n\,\ln \left (F\right )\right )}{b^2\,c^2\,n^2\,{\ln \left (F\right )}^2}-\frac {2\,d\,e\,g\,x}{b\,c\,n\,\ln \left (F\right )}\right )+\frac {d^2\,g\,x^2}{2} \] Input:
int((f + g*x)*(d + e*(F^(c*(a + b*x)))^n)^2,x)
Output:
d^2*f*x - (F^(b*c*x)*F^(a*c))^(2*n)*((e^2*(g - 2*b*c*f*n*log(F)))/(4*b^2*c ^2*n^2*log(F)^2) - (e^2*g*x)/(2*b*c*n*log(F))) - (F^(b*c*x)*F^(a*c))^n*((2 *d*e*(g - b*c*f*n*log(F)))/(b^2*c^2*n^2*log(F)^2) - (2*d*e*g*x)/(b*c*n*log (F))) + (d^2*g*x^2)/2
Time = 0.17 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.22 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x) \, dx=\frac {2 f^{2 b c n x +2 a c n} \mathrm {log}\left (f \right ) b c \,e^{2} f n +2 f^{2 b c n x +2 a c n} \mathrm {log}\left (f \right ) b c \,e^{2} g n x -f^{2 b c n x +2 a c n} e^{2} g +8 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c d e f n +8 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c d e g n x -8 f^{b c n x +a c n} d e g +4 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d^{2} f \,n^{2} x +2 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d^{2} g \,n^{2} x^{2}}{4 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} n^{2}} \] Input:
int((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f),x)
Output:
(2*f**(2*a*c*n + 2*b*c*n*x)*log(f)*b*c*e**2*f*n + 2*f**(2*a*c*n + 2*b*c*n* x)*log(f)*b*c*e**2*g*n*x - f**(2*a*c*n + 2*b*c*n*x)*e**2*g + 8*f**(a*c*n + b*c*n*x)*log(f)*b*c*d*e*f*n + 8*f**(a*c*n + b*c*n*x)*log(f)*b*c*d*e*g*n*x - 8*f**(a*c*n + b*c*n*x)*d*e*g + 4*log(f)**2*b**2*c**2*d**2*f*n**2*x + 2* log(f)**2*b**2*c**2*d**2*g*n**2*x**2)/(4*log(f)**2*b**2*c**2*n**2)