Integrand size = 23, antiderivative size = 115 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx=\frac {d (f+g x)^3}{3 g}+\frac {2 e \left (F^{a c+b c x}\right )^n g^2}{b^3 c^3 n^3 \log ^3(F)}-\frac {2 e \left (F^{a c+b c x}\right )^n g (f+g x)}{b^2 c^2 n^2 \log ^2(F)}+\frac {e \left (F^{a c+b c x}\right )^n (f+g x)^2}{b c n \log (F)} \] Output:
1/3*d*(g*x+f)^3/g+2*e*(F^(b*c*x+a*c))^n*g^2/b^3/c^3/n^3/ln(F)^3-2*e*(F^(b* c*x+a*c))^n*g*(g*x+f)/b^2/c^2/n^2/ln(F)^2+e*(F^(b*c*x+a*c))^n*(g*x+f)^2/b/ c/n/ln(F)
Time = 0.51 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.79 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx=d f^2 x+d f g x^2+\frac {1}{3} d g^2 x^3+\frac {e \left (F^{c (a+b x)}\right )^n \left (2 g^2-2 b c g n (f+g x) \log (F)+b^2 c^2 n^2 (f+g x)^2 \log ^2(F)\right )}{b^3 c^3 n^3 \log ^3(F)} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)*(f + g*x)^2,x]
Output:
d*f^2*x + d*f*g*x^2 + (d*g^2*x^3)/3 + (e*(F^(c*(a + b*x)))^n*(2*g^2 - 2*b* c*g*n*(f + g*x)*Log[F] + b^2*c^2*n^2*(f + g*x)^2*Log[F]^2))/(b^3*c^3*n^3*L og[F]^3)
Time = 0.56 (sec) , antiderivative size = 115, 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.
| \(\displaystyle \int (f+g x)^2 \left (e \left (F^{c (a+b x)}\right )^n+d\right ) \, dx\) |
| \(\Big \downarrow \) 2614 |
| \(\displaystyle \int \left (e (f+g x)^2 \left (F^{a c+b c x}\right )^n+d (f+g x)^2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 e g^2 \left (F^{a c+b c x}\right )^n}{b^3 c^3 n^3 \log ^3(F)}-\frac {2 e g (f+g x) \left (F^{a c+b c x}\right )^n}{b^2 c^2 n^2 \log ^2(F)}+\frac {e (f+g x)^2 \left (F^{a c+b c x}\right )^n}{b c n \log (F)}+\frac {d (f+g x)^3}{3 g}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)*(f + g*x)^2,x]
Output:
(d*(f + g*x)^3)/(3*g) + (2*e*(F^(a*c + b*c*x))^n*g^2)/(b^3*c^3*n^3*Log[F]^ 3) - (2*e*(F^(a*c + b*c*x))^n*g*(f + g*x))/(b^2*c^2*n^2*Log[F]^2) + (e*(F^ (a*c + b*c*x))^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 = 0.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.42
| method | result | size |
| norman | \(d \,f^{2} x +d f g \,x^{2}+\frac {e \left (\ln \left (F \right )^{2} b^{2} c^{2} f^{2} n^{2}-2 \ln \left (F \right ) b c f g n +2 g^{2}\right ) {\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right )^{3} b^{3} c^{3} n^{3}}+\frac {e \,g^{2} x^{2} {\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) b c n}+\frac {d \,g^{2} x^{3}}{3}+\frac {2 e g \left (\ln \left (F \right ) b c f n -g \right ) x \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right )^{2} b^{2} c^{2} n^{2}}\) | \(163\) |
| parallelrisch | \(\frac {d \,g^{2} x^{3} \ln \left (F \right )^{3} b^{3} c^{3} n^{3}+3 d f g \,x^{2} \ln \left (F \right )^{3} b^{3} c^{3} n^{3}+3 d \,f^{2} x \ln \left (F \right )^{3} b^{3} c^{3} n^{3}+3 x^{2} \left (F^{c \left (b x +a \right )}\right )^{n} e \,g^{2} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+6 \ln \left (F \right )^{2} x \left (F^{c \left (b x +a \right )}\right )^{n} b^{2} c^{2} e f g \,n^{2}+3 \ln \left (F \right )^{2} \left (F^{c \left (b x +a \right )}\right )^{n} b^{2} c^{2} e \,f^{2} n^{2}-6 \ln \left (F \right ) x \left (F^{c \left (b x +a \right )}\right )^{n} b c e \,g^{2} n -6 \ln \left (F \right ) \left (F^{c \left (b x +a \right )}\right )^{n} b c e f g n +6 \left (F^{c \left (b x +a \right )}\right )^{n} e \,g^{2}}{3 \ln \left (F \right )^{3} b^{3} c^{3} n^{3}}\) | \(233\) |
| orering | \(\frac {\left (\ln \left (F \right )^{3} b^{3} c^{3} g^{3} n^{3} x^{4}+4 \ln \left (F \right )^{3} b^{3} c^{3} f \,g^{2} n^{3} x^{3}+6 \ln \left (F \right )^{3} b^{3} c^{3} f^{2} g \,n^{3} x^{2}+3 \ln \left (F \right )^{3} b^{3} c^{3} f^{3} n^{3} x +2 \ln \left (F \right )^{2} b^{2} c^{2} g^{3} n^{2} x^{3}+6 \ln \left (F \right )^{2} b^{2} c^{2} f \,g^{2} n^{2} x^{2}+9 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} g \,n^{2} x +3 \ln \left (F \right )^{2} b^{2} c^{2} f^{3} n^{2}-6 g^{3} \ln \left (F \right ) b c n \,x^{2}-18 g^{2} \ln \left (F \right ) b c n x f -6 g \ln \left (F \right ) b c \,f^{2} n +18 g^{3} x +6 g^{2} f \right ) \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}{3 \ln \left (F \right )^{3} b^{3} c^{3} n^{3} \left (g x +f \right )}-\frac {x \left (g^{2} x^{2} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+3 \ln \left (F \right )^{2} b^{2} c^{2} f g \,n^{2} x +3 \ln \left (F \right )^{2} b^{2} c^{2} f^{2} n^{2}-3 \ln \left (F \right ) b c \,g^{2} n x -6 \ln \left (F \right ) b c f g n +6 g^{2}\right ) \left (e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right ) b c n \left (g x +f \right )^{2}+2 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \left (g x +f \right ) g \right )}{3 \ln \left (F \right )^{3} b^{3} c^{3} n^{3} \left (g x +f \right )^{2}}\) | \(406\) |
Input:
int((d+e*(F^(c*(b*x+a)))^n)*(g*x+f)^2,x,method=_RETURNVERBOSE)
Output:
d*f^2*x+d*f*g*x^2+e*(ln(F)^2*b^2*c^2*f^2*n^2-2*ln(F)*b*c*f*g*n+2*g^2)/ln(F )^3/b^3/c^3/n^3*exp(n*ln(exp(c*(b*x+a)*ln(F))))+1/ln(F)/b/c/n*e*g^2*x^2*ex p(n*ln(exp(c*(b*x+a)*ln(F))))+1/3*d*g^2*x^3+2*e*g*(ln(F)*b*c*f*n-g)/ln(F)^ 2/b^2/c^2/n^2*x*exp(n*ln(exp(c*(b*x+a)*ln(F))))
Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.44 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx=\frac {{\left (b^{3} c^{3} d g^{2} n^{3} x^{3} + 3 \, b^{3} c^{3} d f g n^{3} x^{2} + 3 \, b^{3} c^{3} d f^{2} n^{3} x\right )} \log \left (F\right )^{3} + 3 \, {\left (2 \, e g^{2} + {\left (b^{2} c^{2} e g^{2} n^{2} x^{2} + 2 \, b^{2} c^{2} e f g n^{2} x + b^{2} c^{2} e f^{2} n^{2}\right )} \log \left (F\right )^{2} - 2 \, {\left (b c e g^{2} n x + b c e f g n\right )} \log \left (F\right )\right )} F^{b c n x + a c n}}{3 \, b^{3} c^{3} n^{3} \log \left (F\right )^{3}} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^2,x, algorithm="fricas")
Output:
1/3*((b^3*c^3*d*g^2*n^3*x^3 + 3*b^3*c^3*d*f*g*n^3*x^2 + 3*b^3*c^3*d*f^2*n^ 3*x)*log(F)^3 + 3*(2*e*g^2 + (b^2*c^2*e*g^2*n^2*x^2 + 2*b^2*c^2*e*f*g*n^2* x + b^2*c^2*e*f^2*n^2)*log(F)^2 - 2*(b*c*e*g^2*n*x + b*c*e*f*g*n)*log(F))* F^(b*c*n*x + a*c*n))/(b^3*c^3*n^3*log(F)^3)
Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (110) = 220\).
Time = 0.71 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.56 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx=\begin {cases} \left (d + e\right ) \left (f^{2} x + f g x^{2} + \frac {g^{2} x^{3}}{3}\right ) & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge n = 0 \\\left (d + e \left (F^{a c}\right )^{n}\right ) \left (f^{2} x + f g x^{2} + \frac {g^{2} x^{3}}{3}\right ) & \text {for}\: b = 0 \\\left (d + e\right ) \left (f^{2} x + f g x^{2} + \frac {g^{2} x^{3}}{3}\right ) & \text {for}\: F = 1 \vee c = 0 \vee n = 0 \\d f^{2} x + d f g x^{2} + \frac {d g^{2} x^{3}}{3} + \frac {e f^{2} \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {2 e f g x \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {e g^{2} x^{2} \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} - \frac {2 e f g \left (F^{a c + b c x}\right )^{n}}{b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} - \frac {2 e g^{2} x \left (F^{a c + b c x}\right )^{n}}{b^{2} c^{2} n^{2} \log {\left (F \right )}^{2}} + \frac {2 e g^{2} \left (F^{a c + b c x}\right )^{n}}{b^{3} c^{3} n^{3} \log {\left (F \right )}^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((d+e*(F**((b*x+a)*c))**n)*(g*x+f)**2,x)
Output:
Piecewise(((d + e)*(f**2*x + f*g*x**2 + g**2*x**3/3), Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(n, 0)), ((d + e*(F**(a*c))**n)*(f**2*x + f*g*x**2 + g**2*x **3/3), Eq(b, 0)), ((d + e)*(f**2*x + f*g*x**2 + g**2*x**3/3), Eq(F, 1) | Eq(c, 0) | Eq(n, 0)), (d*f**2*x + d*f*g*x**2 + d*g**2*x**3/3 + e*f**2*(F** (a*c + b*c*x))**n/(b*c*n*log(F)) + 2*e*f*g*x*(F**(a*c + b*c*x))**n/(b*c*n* log(F)) + e*g**2*x**2*(F**(a*c + b*c*x))**n/(b*c*n*log(F)) - 2*e*f*g*(F**( a*c + b*c*x))**n/(b**2*c**2*n**2*log(F)**2) - 2*e*g**2*x*(F**(a*c + b*c*x) )**n/(b**2*c**2*n**2*log(F)**2) + 2*e*g**2*(F**(a*c + b*c*x))**n/(b**3*c** 3*n**3*log(F)**3), True))
Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.49 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx=\frac {1}{3} \, d g^{2} x^{3} + d f g x^{2} + d f^{2} x + \frac {F^{b c n x + a c n} e f^{2}}{b c n \log \left (F\right )} + \frac {2 \, {\left (F^{a c n} b c n x \log \left (F\right ) - F^{a c n}\right )} F^{b c n x} e f g}{b^{2} c^{2} n^{2} \log \left (F\right )^{2}} + \frac {{\left (F^{a c n} b^{2} c^{2} n^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c n} b c n x \log \left (F\right ) + 2 \, F^{a c n}\right )} F^{b c n x} e g^{2}}{b^{3} c^{3} n^{3} \log \left (F\right )^{3}} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^2,x, algorithm="maxima")
Output:
1/3*d*g^2*x^3 + d*f*g*x^2 + d*f^2*x + F^(b*c*n*x + a*c*n)*e*f^2/(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)*e*f*g/(b^2*c^2 *n^2*log(F)^2) + (F^(a*c*n)*b^2*c^2*n^2*x^2*log(F)^2 - 2*F^(a*c*n)*b*c*n*x *log(F) + 2*F^(a*c*n))*F^(b*c*n*x)*e*g^2/(b^3*c^3*n^3*log(F)^3)
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 2716, normalized size of antiderivative = 23.62 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx=\text {Too large to display} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^2,x, algorithm="giac")
Output:
1/3*d*g^2*x^3 + d*f*g*x^2 + d*f^2*x - ((2*(pi*b^2*c^2*e*g^2*n^2*x^2*log(ab s(F))*sgn(F) - pi*b^2*c^2*e*g^2*n^2*x^2*log(abs(F)) + 2*pi*b^2*c^2*e*f*g*n ^2*x*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*e*f*g*n^2*x*log(abs(F)) + pi*b^2*c^ 2*e*f^2*n^2*log(abs(F))*sgn(F) - pi*b^2*c^2*e*f^2*n^2*log(abs(F)) - pi*b*c *e*g^2*n*x*sgn(F) + pi*b*c*e*g^2*n*x - pi*b*c*e*f*g*n*sgn(F) + pi*b*c*e*f* g*n)*(pi^3*b^3*c^3*n^3*sgn(F) - 3*pi*b^3*c^3*n^3*log(abs(F))^2*sgn(F) - pi ^3*b^3*c^3*n^3 + 3*pi*b^3*c^3*n^3*log(abs(F))^2)/((pi^3*b^3*c^3*n^3*sgn(F) - 3*pi*b^3*c^3*n^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3*n^3 + 3*pi*b^3*c^3 *n^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c^3*n^3*log(abs(F))*sgn(F) - 3*pi^2*b^ 3*c^3*n^3*log(abs(F)) + 2*b^3*c^3*n^3*log(abs(F))^3)^2) - (pi^2*b^2*c^2*e* g^2*n^2*x^2*sgn(F) - pi^2*b^2*c^2*e*g^2*n^2*x^2 + 2*b^2*c^2*e*g^2*n^2*x^2* log(abs(F))^2 + 2*pi^2*b^2*c^2*e*f*g*n^2*x*sgn(F) - 2*pi^2*b^2*c^2*e*f*g*n ^2*x + 4*b^2*c^2*e*f*g*n^2*x*log(abs(F))^2 + pi^2*b^2*c^2*e*f^2*n^2*sgn(F) - pi^2*b^2*c^2*e*f^2*n^2 + 2*b^2*c^2*e*f^2*n^2*log(abs(F))^2 - 4*b*c*e*g^ 2*n*x*log(abs(F)) - 4*b*c*e*f*g*n*log(abs(F)) + 4*e*g^2)*(3*pi^2*b^3*c^3*n ^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*n^3*log(abs(F)) + 2*b^3*c^3*n^3*log (abs(F))^3)/((pi^3*b^3*c^3*n^3*sgn(F) - 3*pi*b^3*c^3*n^3*log(abs(F))^2*sgn (F) - pi^3*b^3*c^3*n^3 + 3*pi*b^3*c^3*n^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c ^3*n^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*n^3*log(abs(F)) + 2*b^3*c^3*n^3 *log(abs(F))^3)^2))*cos(-1/2*pi*b*c*n*x*sgn(F) + 1/2*pi*b*c*n*x - 1/2*p...
Time = 22.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.17 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx={\left (F^{b\,c\,x}\,F^{a\,c}\right )}^n\,\left (\frac {e\,\left (b^2\,c^2\,f^2\,n^2\,{\ln \left (F\right )}^2-2\,b\,c\,f\,g\,n\,\ln \left (F\right )+2\,g^2\right )}{b^3\,c^3\,n^3\,{\ln \left (F\right )}^3}+\frac {e\,g^2\,x^2}{b\,c\,n\,\ln \left (F\right )}-\frac {2\,e\,g\,x\,\left (g-b\,c\,f\,n\,\ln \left (F\right )\right )}{b^2\,c^2\,n^2\,{\ln \left (F\right )}^2}\right )+d\,f^2\,x+\frac {d\,g^2\,x^3}{3}+d\,f\,g\,x^2 \] Input:
int((f + g*x)^2*(d + e*(F^(c*(a + b*x)))^n),x)
Output:
(F^(b*c*x)*F^(a*c))^n*((e*(2*g^2 + b^2*c^2*f^2*n^2*log(F)^2 - 2*b*c*f*g*n* log(F)))/(b^3*c^3*n^3*log(F)^3) + (e*g^2*x^2)/(b*c*n*log(F)) - (2*e*g*x*(g - b*c*f*n*log(F)))/(b^2*c^2*n^2*log(F)^2)) + d*f^2*x + (d*g^2*x^3)/3 + d* f*g*x^2
Time = 0.20 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.07 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^2 \, dx=\frac {3 f^{b c n x +a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} e \,f^{2} n^{2}+6 f^{b c n x +a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} e f g \,n^{2} x +3 f^{b c n x +a c n} \mathrm {log}\left (f \right )^{2} b^{2} c^{2} e \,g^{2} n^{2} x^{2}-6 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c e f g n -6 f^{b c n x +a c n} \mathrm {log}\left (f \right ) b c e \,g^{2} n x +6 f^{b c n x +a c n} e \,g^{2}+3 \mathrm {log}\left (f \right )^{3} b^{3} c^{3} d \,f^{2} n^{3} x +3 \mathrm {log}\left (f \right )^{3} b^{3} c^{3} d f g \,n^{3} x^{2}+\mathrm {log}\left (f \right )^{3} b^{3} c^{3} d \,g^{2} n^{3} x^{3}}{3 \mathrm {log}\left (f \right )^{3} b^{3} c^{3} n^{3}} \] Input:
int((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^2,x)
Output:
(3*f**(a*c*n + b*c*n*x)*log(f)**2*b**2*c**2*e*f**2*n**2 + 6*f**(a*c*n + b* c*n*x)*log(f)**2*b**2*c**2*e*f*g*n**2*x + 3*f**(a*c*n + b*c*n*x)*log(f)**2 *b**2*c**2*e*g**2*n**2*x**2 - 6*f**(a*c*n + b*c*n*x)*log(f)*b*c*e*f*g*n - 6*f**(a*c*n + b*c*n*x)*log(f)*b*c*e*g**2*n*x + 6*f**(a*c*n + b*c*n*x)*e*g* *2 + 3*log(f)**3*b**3*c**3*d*f**2*n**3*x + 3*log(f)**3*b**3*c**3*d*f*g*n** 3*x**2 + log(f)**3*b**3*c**3*d*g**2*n**3*x**3)/(3*log(f)**3*b**3*c**3*n**3 )