Integrand size = 17, antiderivative size = 79 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\frac {2 e^2 F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {2 e F^{c (a+b x)} (d+e x)}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)^2}{b c \log (F)} \] Output:
2*e^2*F^(c*(b*x+a))/b^3/c^3/ln(F)^3-2*e*F^(c*(b*x+a))*(e*x+d)/b^2/c^2/ln(F )^2+F^(c*(b*x+a))*(e*x+d)^2/b/c/ln(F)
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\frac {F^{c (a+b x)} \left (2 e^2-2 b c e (d+e x) \log (F)+b^2 c^2 (d+e x)^2 \log ^2(F)\right )}{b^3 c^3 \log ^3(F)} \] Input:
Integrate[F^(c*(a + b*x))*(d + e*x)^2,x]
Output:
(F^(c*(a + b*x))*(2*e^2 - 2*b*c*e*(d + e*x)*Log[F] + b^2*c^2*(d + e*x)^2*L og[F]^2))/(b^3*c^3*Log[F]^3)
Time = 0.44 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2607, 2607, 2624}
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.
\(\displaystyle \int (d+e x)^2 F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {(d+e x)^2 F^{c (a+b x)}}{b c \log (F)}-\frac {2 e \int F^{c (a+b x)} (d+e x)dx}{b c \log (F)}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {(d+e x)^2 F^{c (a+b x)}}{b c \log (F)}-\frac {2 e \left (\frac {(d+e x) F^{c (a+b x)}}{b c \log (F)}-\frac {e \int F^{c (a+b x)}dx}{b c \log (F)}\right )}{b c \log (F)}\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle \frac {(d+e x)^2 F^{c (a+b x)}}{b c \log (F)}-\frac {2 e \left (\frac {(d+e x) F^{c (a+b x)}}{b c \log (F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}\right )}{b c \log (F)}\) |
Input:
Int[F^(c*(a + b*x))*(d + e*x)^2,x]
Output:
(F^(c*(a + b*x))*(d + e*x)^2)/(b*c*Log[F]) - (2*e*(-((e*F^(c*(a + b*x)))/( b^2*c^2*Log[F]^2)) + (F^(c*(a + b*x))*(d + e*x))/(b*c*Log[F])))/(b*c*Log[F ])
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m _.), x_Symbol] :> Simp[(c + d*x)^m*((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Simp[d*(m/(f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x)))^ n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2* m] && !TrueQ[$UseGamma]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15
method | result | size |
gosper | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} e^{2} x^{2}+2 \ln \left (F \right )^{2} b^{2} c^{2} d e x +\ln \left (F \right )^{2} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) b c \,e^{2} x -2 \ln \left (F \right ) b c e d +2 e^{2}\right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(91\) |
risch | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} e^{2} x^{2}+2 \ln \left (F \right )^{2} b^{2} c^{2} d e x +\ln \left (F \right )^{2} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) b c \,e^{2} x -2 \ln \left (F \right ) b c e d +2 e^{2}\right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(91\) |
orering | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} e^{2} x^{2}+2 \ln \left (F \right )^{2} b^{2} c^{2} d e x +\ln \left (F \right )^{2} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) b c \,e^{2} x -2 \ln \left (F \right ) b c e d +2 e^{2}\right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(91\) |
norman | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) b c e d +2 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {e^{2} x^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right ) b c}+\frac {2 e \left (b c d \ln \left (F \right )-e \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2} c^{2}}\) | \(112\) |
meijerg | \(-\frac {F^{a c} e^{2} \left (2-\frac {\left (3 b^{2} c^{2} x^{2} \ln \left (F \right )^{2}-6 b c x \ln \left (F \right )+6\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{3}\right )}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {2 F^{a c} e d \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{\ln \left (F \right )^{2} b^{2} c^{2}}-\frac {F^{a c} d^{2} \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{\ln \left (F \right ) b c}\) | \(127\) |
parallelrisch | \(\frac {\ln \left (F \right )^{2} x^{2} F^{c \left (b x +a \right )} b^{2} c^{2} e^{2}+2 \ln \left (F \right )^{2} x \,F^{c \left (b x +a \right )} b^{2} c^{2} d e +\ln \left (F \right )^{2} F^{c \left (b x +a \right )} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) x \,F^{c \left (b x +a \right )} b c \,e^{2}-2 F^{c \left (b x +a \right )} \ln \left (F \right ) b c d e +2 F^{c \left (b x +a \right )} e^{2}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(136\) |
Input:
int(F^(c*(b*x+a))*(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
(ln(F)^2*b^2*c^2*e^2*x^2+2*ln(F)^2*b^2*c^2*d*e*x+ln(F)^2*b^2*c^2*d^2-2*ln( F)*b*c*e^2*x-2*ln(F)*b*c*e*d+2*e^2)*F^(c*(b*x+a))/ln(F)^3/b^3/c^3
Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\frac {{\left ({\left (b^{2} c^{2} e^{2} x^{2} + 2 \, b^{2} c^{2} d e x + b^{2} c^{2} d^{2}\right )} \log \left (F\right )^{2} + 2 \, e^{2} - 2 \, {\left (b c e^{2} x + b c d e\right )} \log \left (F\right )\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3}} \] Input:
integrate(F^((b*x+a)*c)*(e*x+d)^2,x, algorithm="fricas")
Output:
((b^2*c^2*e^2*x^2 + 2*b^2*c^2*d*e*x + b^2*c^2*d^2)*log(F)^2 + 2*e^2 - 2*(b *c*e^2*x + b*c*d*e)*log(F))*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3)
Time = 0.07 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.68 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\begin {cases} \frac {F^{c \left (a + b x\right )} \left (b^{2} c^{2} d^{2} \log {\left (F \right )}^{2} + 2 b^{2} c^{2} d e x \log {\left (F \right )}^{2} + b^{2} c^{2} e^{2} x^{2} \log {\left (F \right )}^{2} - 2 b c d e \log {\left (F \right )} - 2 b c e^{2} x \log {\left (F \right )} + 2 e^{2}\right )}{b^{3} c^{3} \log {\left (F \right )}^{3}} & \text {for}\: b^{3} c^{3} \log {\left (F \right )}^{3} \neq 0 \\d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(F**((b*x+a)*c)*(e*x+d)**2,x)
Output:
Piecewise((F**(c*(a + b*x))*(b**2*c**2*d**2*log(F)**2 + 2*b**2*c**2*d*e*x* log(F)**2 + b**2*c**2*e**2*x**2*log(F)**2 - 2*b*c*d*e*log(F) - 2*b*c*e**2* x*log(F) + 2*e**2)/(b**3*c**3*log(F)**3), Ne(b**3*c**3*log(F)**3, 0)), (d* *2*x + d*e*x**2 + e**2*x**3/3, True))
Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.56 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\frac {F^{b c x + a c} d^{2}}{b c \log \left (F\right )} + \frac {2 \, {\left (F^{a c} b c x \log \left (F\right ) - F^{a c}\right )} F^{b c x} d e}{b^{2} c^{2} \log \left (F\right )^{2}} + \frac {{\left (F^{a c} b^{2} c^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c} b c x \log \left (F\right ) + 2 \, F^{a c}\right )} F^{b c x} e^{2}}{b^{3} c^{3} \log \left (F\right )^{3}} \] Input:
integrate(F^((b*x+a)*c)*(e*x+d)^2,x, algorithm="maxima")
Output:
F^(b*c*x + a*c)*d^2/(b*c*log(F)) + 2*(F^(a*c)*b*c*x*log(F) - F^(a*c))*F^(b *c*x)*d*e/(b^2*c^2*log(F)^2) + (F^(a*c)*b^2*c^2*x^2*log(F)^2 - 2*F^(a*c)*b *c*x*log(F) + 2*F^(a*c))*F^(b*c*x)*e^2/(b^3*c^3*log(F)^3)
Result contains complex when optimal does not.
Time = 0.17 (sec) , antiderivative size = 2214, normalized size of antiderivative = 28.03 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\text {Too large to display} \] Input:
integrate(F^((b*x+a)*c)*(e*x+d)^2,x, algorithm="giac")
Output:
-((2*(pi*b^2*c^2*e^2*x^2*log(abs(F))*sgn(F) - pi*b^2*c^2*e^2*x^2*log(abs(F )) + 2*pi*b^2*c^2*d*e*x*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*d*e*x*log(abs(F) ) + pi*b^2*c^2*d^2*log(abs(F))*sgn(F) - pi*b^2*c^2*d^2*log(abs(F)) - pi*b* c*e^2*x*sgn(F) + pi*b*c*e^2*x - pi*b*c*d*e*sgn(F) + pi*b*c*d*e)*(pi^3*b^3* c^3*sgn(F) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c ^3*log(abs(F))^2)/((pi^3*b^3*c^3*sgn(F) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F ) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c^3*log(abs (F))*sgn(F) - 3*pi^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3)^2) - ( pi^2*b^2*c^2*e^2*x^2*sgn(F) - pi^2*b^2*c^2*e^2*x^2 + 2*b^2*c^2*e^2*x^2*log (abs(F))^2 + 2*pi^2*b^2*c^2*d*e*x*sgn(F) - 2*pi^2*b^2*c^2*d*e*x + 4*b^2*c^ 2*d*e*x*log(abs(F))^2 + pi^2*b^2*c^2*d^2*sgn(F) - pi^2*b^2*c^2*d^2 + 2*b^2 *c^2*d^2*log(abs(F))^2 - 4*b*c*e^2*x*log(abs(F)) - 4*b*c*d*e*log(abs(F)) + 4*e^2)*(3*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3)/((pi^3*b^3*c^3*sgn(F) - 3*pi*b^3*c^3*log(abs(F))^ 2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c^3* log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3) ^2))*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi* a*c) - ((pi^2*b^2*c^2*e^2*x^2*sgn(F) - pi^2*b^2*c^2*e^2*x^2 + 2*b^2*c^2*e^ 2*x^2*log(abs(F))^2 + 2*pi^2*b^2*c^2*d*e*x*sgn(F) - 2*pi^2*b^2*c^2*d*e*x + 4*b^2*c^2*d*e*x*log(abs(F))^2 + pi^2*b^2*c^2*d^2*sgn(F) - pi^2*b^2*c^2...
Time = 22.81 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (b^2\,c^2\,d^2\,{\ln \left (F\right )}^2+2\,b^2\,c^2\,d\,e\,x\,{\ln \left (F\right )}^2+b^2\,c^2\,e^2\,x^2\,{\ln \left (F\right )}^2-2\,b\,c\,d\,e\,\ln \left (F\right )-2\,b\,c\,e^2\,x\,\ln \left (F\right )+2\,e^2\right )}{b^3\,c^3\,{\ln \left (F\right )}^3} \] Input:
int(F^(c*(a + b*x))*(d + e*x)^2,x)
Output:
(F^(a*c + b*c*x)*(2*e^2 + b^2*c^2*d^2*log(F)^2 - 2*b*c*e^2*x*log(F) + b^2* c^2*e^2*x^2*log(F)^2 - 2*b*c*d*e*log(F) + 2*b^2*c^2*d*e*x*log(F)^2))/(b^3* c^3*log(F)^3)
Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int F^{c (a+b x)} (d+e x)^2 \, dx=\frac {f^{b c x +a c} \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2} d^{2}+2 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d e x +\mathrm {log}\left (f \right )^{2} b^{2} c^{2} e^{2} x^{2}-2 \,\mathrm {log}\left (f \right ) b c d e -2 \,\mathrm {log}\left (f \right ) b c \,e^{2} x +2 e^{2}\right )}{\mathrm {log}\left (f \right )^{3} b^{3} c^{3}} \] Input:
int(F^((b*x+a)*c)*(e*x+d)^2,x)
Output:
(f**(a*c + b*c*x)*(log(f)**2*b**2*c**2*d**2 + 2*log(f)**2*b**2*c**2*d*e*x + log(f)**2*b**2*c**2*e**2*x**2 - 2*log(f)*b*c*d*e - 2*log(f)*b*c*e**2*x + 2*e**2))/(log(f)**3*b**3*c**3)