Integrand size = 19, antiderivative size = 85 \[ \int F^{a+b (c+d x)} (e+f x)^2 \, dx=\frac {2 f^2 F^{a+b c+b d x}}{b^3 d^3 \log ^3(F)}-\frac {2 f F^{a+b c+b d x} (e+f x)}{b^2 d^2 \log ^2(F)}+\frac {F^{a+b c+b d x} (e+f x)^2}{b d \log (F)} \]
2*f^2*F^(b*d*x+b*c+a)/b^3/d^3/ln(F)^3-2*f*F^(b*d*x+b*c+a)*(f*x+e)/b^2/d^2/ ln(F)^2+F^(b*d*x+b*c+a)*(f*x+e)^2/b/d/ln(F)
Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int F^{a+b (c+d x)} (e+f x)^2 \, dx=\frac {F^{a+b (c+d x)} \left (2 f^2-2 b d f (e+f x) \log (F)+b^2 d^2 (e+f x)^2 \log ^2(F)\right )}{b^3 d^3 \log ^3(F)} \]
(F^(a + b*(c + d*x))*(2*f^2 - 2*b*d*f*(e + f*x)*Log[F] + b^2*d^2*(e + f*x) ^2*Log[F]^2))/(b^3*d^3*Log[F]^3)
Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2618, 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.
\(\displaystyle \int (e+f x)^2 F^{a+b (c+d x)} \, dx\) |
\(\Big \downarrow \) 2618 |
\(\displaystyle \int (e+f x)^2 F^{a+b c+b d x}dx\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {(e+f x)^2 F^{a+b c+b d x}}{b d \log (F)}-\frac {2 f \int F^{a+b c+b d x} (e+f x)dx}{b d \log (F)}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle \frac {(e+f x)^2 F^{a+b c+b d x}}{b d \log (F)}-\frac {2 f \left (\frac {(e+f x) F^{a+b c+b d x}}{b d \log (F)}-\frac {f \int F^{a+b c+b d x}dx}{b d \log (F)}\right )}{b d \log (F)}\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle \frac {(e+f x)^2 F^{a+b c+b d x}}{b d \log (F)}-\frac {2 f \left (\frac {(e+f x) F^{a+b c+b d x}}{b d \log (F)}-\frac {f F^{a+b c+b d x}}{b^2 d^2 \log ^2(F)}\right )}{b d \log (F)}\) |
(F^(a + b*c + b*d*x)*(e + f*x)^2)/(b*d*Log[F]) - (2*f*(-((f*F^(a + b*c + b *d*x))/(b^2*d^2*Log[F]^2)) + (F^(a + b*c + b*d*x)*(e + f*x))/(b*d*Log[F])) )/(b*d*Log[F])
3.1.68.3.1 Defintions of rubi rules used
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[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m _.), x_Symbol] :> Int[(c + d*x)^m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, c, d, g, n, p}, x] && LinearQ[v, x] && !LinearMatchQ[v , x] && IntegerQ[m]
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 = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.09
method | result | size |
gosper | \(\frac {\left (\ln \left (F \right )^{2} b^{2} d^{2} f^{2} x^{2}+2 \ln \left (F \right )^{2} b^{2} d^{2} e f x +\ln \left (F \right )^{2} b^{2} d^{2} e^{2}-2 \ln \left (F \right ) b d \,f^{2} x -2 e f \ln \left (F \right ) b d +2 f^{2}\right ) F^{b d x +c b +a}}{b^{3} d^{3} \ln \left (F \right )^{3}}\) | \(93\) |
risch | \(\frac {\left (\ln \left (F \right )^{2} b^{2} d^{2} f^{2} x^{2}+2 \ln \left (F \right )^{2} b^{2} d^{2} e f x +\ln \left (F \right )^{2} b^{2} d^{2} e^{2}-2 \ln \left (F \right ) b d \,f^{2} x -2 e f \ln \left (F \right ) b d +2 f^{2}\right ) F^{b d x +c b +a}}{b^{3} d^{3} \ln \left (F \right )^{3}}\) | \(93\) |
norman | \(\frac {\left (\ln \left (F \right )^{2} b^{2} d^{2} e^{2}-2 e f \ln \left (F \right ) b d +2 f^{2}\right ) {\mathrm e}^{\left (a +b \left (d x +c \right )\right ) \ln \left (F \right )}}{b^{3} d^{3} \ln \left (F \right )^{3}}+\frac {f^{2} x^{2} {\mathrm e}^{\left (a +b \left (d x +c \right )\right ) \ln \left (F \right )}}{b d \ln \left (F \right )}+\frac {2 f \left (\ln \left (F \right ) b d e -f \right ) x \,{\mathrm e}^{\left (a +b \left (d x +c \right )\right ) \ln \left (F \right )}}{b^{2} d^{2} \ln \left (F \right )^{2}}\) | \(121\) |
meijerg | \(-\frac {F^{c b +a} f^{2} \left (2-\frac {\left (3 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}-6 b d x \ln \left (F \right )+6\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{3}\right )}{b^{3} d^{3} \ln \left (F \right )^{3}}+\frac {2 F^{c b +a} f e \left (1-\frac {\left (-2 b d x \ln \left (F \right )+2\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{2}\right )}{b^{2} d^{2} \ln \left (F \right )^{2}}-\frac {F^{c b +a} e^{2} \left (1-{\mathrm e}^{b d x \ln \left (F \right )}\right )}{b d \ln \left (F \right )}\) | \(133\) |
parallelrisch | \(\frac {\ln \left (F \right )^{2} x^{2} F^{b d x +c b +a} b^{2} d^{2} f^{2}+2 \ln \left (F \right )^{2} x \,F^{b d x +c b +a} b^{2} d^{2} e f +\ln \left (F \right )^{2} F^{b d x +c b +a} b^{2} d^{2} e^{2}-2 \ln \left (F \right ) x \,F^{b d x +c b +a} b d \,f^{2}-2 \ln \left (F \right ) F^{b d x +c b +a} b d e f +2 F^{b d x +c b +a} f^{2}}{b^{3} d^{3} \ln \left (F \right )^{3}}\) | \(148\) |
(ln(F)^2*b^2*d^2*f^2*x^2+2*ln(F)^2*b^2*d^2*e*f*x+ln(F)^2*b^2*d^2*e^2-2*ln( F)*b*d*f^2*x-2*e*f*ln(F)*b*d+2*f^2)*F^(b*d*x+b*c+a)/b^3/d^3/ln(F)^3
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)} (e+f x)^2 \, dx=\frac {{\left ({\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} d^{2} e f x + b^{2} d^{2} e^{2}\right )} \log \left (F\right )^{2} + 2 \, f^{2} - 2 \, {\left (b d f^{2} x + b d e f\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{b^{3} d^{3} \log \left (F\right )^{3}} \]
((b^2*d^2*f^2*x^2 + 2*b^2*d^2*e*f*x + b^2*d^2*e^2)*log(F)^2 + 2*f^2 - 2*(b *d*f^2*x + b*d*e*f)*log(F))*F^(b*d*x + b*c + a)/(b^3*d^3*log(F)^3)
Time = 0.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.58 \[ \int F^{a+b (c+d x)} (e+f x)^2 \, dx=\begin {cases} \frac {F^{a + b \left (c + d x\right )} \left (b^{2} d^{2} e^{2} \log {\left (F \right )}^{2} + 2 b^{2} d^{2} e f x \log {\left (F \right )}^{2} + b^{2} d^{2} f^{2} x^{2} \log {\left (F \right )}^{2} - 2 b d e f \log {\left (F \right )} - 2 b d f^{2} x \log {\left (F \right )} + 2 f^{2}\right )}{b^{3} d^{3} \log {\left (F \right )}^{3}} & \text {for}\: b^{3} d^{3} \log {\left (F \right )}^{3} \neq 0 \\e^{2} x + e f x^{2} + \frac {f^{2} x^{3}}{3} & \text {otherwise} \end {cases} \]
Piecewise((F**(a + b*(c + d*x))*(b**2*d**2*e**2*log(F)**2 + 2*b**2*d**2*e* f*x*log(F)**2 + b**2*d**2*f**2*x**2*log(F)**2 - 2*b*d*e*f*log(F) - 2*b*d*f **2*x*log(F) + 2*f**2)/(b**3*d**3*log(F)**3), Ne(b**3*d**3*log(F)**3, 0)), (e**2*x + e*f*x**2 + f**2*x**3/3, True))
Time = 0.19 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.58 \[ \int F^{a+b (c+d x)} (e+f x)^2 \, dx=\frac {F^{b d x + b c + a} e^{2}}{b d \log \left (F\right )} + \frac {2 \, {\left (F^{b c + a} b d x \log \left (F\right ) - F^{b c + a}\right )} F^{b d x} e f}{b^{2} d^{2} \log \left (F\right )^{2}} + \frac {{\left (F^{b c + a} b^{2} d^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{b c + a} b d x \log \left (F\right ) + 2 \, F^{b c + a}\right )} F^{b d x} f^{2}}{b^{3} d^{3} \log \left (F\right )^{3}} \]
F^(b*d*x + b*c + a)*e^2/(b*d*log(F)) + 2*(F^(b*c + a)*b*d*x*log(F) - F^(b* c + a))*F^(b*d*x)*e*f/(b^2*d^2*log(F)^2) + (F^(b*c + a)*b^2*d^2*x^2*log(F) ^2 - 2*F^(b*c + a)*b*d*x*log(F) + 2*F^(b*c + a))*F^(b*d*x)*f^2/(b^3*d^3*lo g(F)^3)
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 2264, normalized size of antiderivative = 26.64 \[ \int F^{a+b (c+d x)} (e+f x)^2 \, dx=\text {Too large to display} \]
-((2*(pi*b^2*d^2*f^2*x^2*log(abs(F))*sgn(F) - pi*b^2*d^2*f^2*x^2*log(abs(F )) + 2*pi*b^2*d^2*e*f*x*log(abs(F))*sgn(F) - 2*pi*b^2*d^2*e*f*x*log(abs(F) ) + pi*b^2*d^2*e^2*log(abs(F))*sgn(F) - pi*b^2*d^2*e^2*log(abs(F)) - pi*b* d*f^2*x*sgn(F) + pi*b*d*f^2*x - pi*b*d*e*f*sgn(F) + pi*b*d*e*f)*(pi^3*b^3* d^3*sgn(F) - 3*pi*b^3*d^3*log(abs(F))^2*sgn(F) - pi^3*b^3*d^3 + 3*pi*b^3*d ^3*log(abs(F))^2)/((pi^3*b^3*d^3*sgn(F) - 3*pi*b^3*d^3*log(abs(F))^2*sgn(F ) - pi^3*b^3*d^3 + 3*pi*b^3*d^3*log(abs(F))^2)^2 + (3*pi^2*b^3*d^3*log(abs (F))*sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^3*d^3*log(abs(F))^3)^2) - ( pi^2*b^2*d^2*f^2*x^2*sgn(F) - pi^2*b^2*d^2*f^2*x^2 + 2*b^2*d^2*f^2*x^2*log (abs(F))^2 + 2*pi^2*b^2*d^2*e*f*x*sgn(F) - 2*pi^2*b^2*d^2*e*f*x + 4*b^2*d^ 2*e*f*x*log(abs(F))^2 + pi^2*b^2*d^2*e^2*sgn(F) - pi^2*b^2*d^2*e^2 + 2*b^2 *d^2*e^2*log(abs(F))^2 - 4*b*d*f^2*x*log(abs(F)) - 4*b*d*e*f*log(abs(F)) + 4*f^2)*(3*pi^2*b^3*d^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^3*d^3*log(abs(F))^3)/((pi^3*b^3*d^3*sgn(F) - 3*pi*b^3*d^3*log(abs(F))^ 2*sgn(F) - pi^3*b^3*d^3 + 3*pi*b^3*d^3*log(abs(F))^2)^2 + (3*pi^2*b^3*d^3* log(abs(F))*sgn(F) - 3*pi^2*b^3*d^3*log(abs(F)) + 2*b^3*d^3*log(abs(F))^3) ^2))*cos(-1/2*pi*b*d*x*sgn(F) + 1/2*pi*b*d*x - 1/2*pi*b*c*sgn(F) + 1/2*pi* b*c - 1/2*pi*a*sgn(F) + 1/2*pi*a) - ((pi^2*b^2*d^2*f^2*x^2*sgn(F) - pi^2*b ^2*d^2*f^2*x^2 + 2*b^2*d^2*f^2*x^2*log(abs(F))^2 + 2*pi^2*b^2*d^2*e*f*x*sg n(F) - 2*pi^2*b^2*d^2*e*f*x + 4*b^2*d^2*e*f*x*log(abs(F))^2 + pi^2*b^2*...
Time = 0.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08 \[ \int F^{a+b (c+d x)} (e+f x)^2 \, dx=\frac {F^{a+b\,c+b\,d\,x}\,\left (b^2\,d^2\,e^2\,{\ln \left (F\right )}^2+2\,b^2\,d^2\,e\,f\,x\,{\ln \left (F\right )}^2+b^2\,d^2\,f^2\,x^2\,{\ln \left (F\right )}^2-2\,b\,d\,e\,f\,\ln \left (F\right )-2\,b\,d\,f^2\,x\,\ln \left (F\right )+2\,f^2\right )}{b^3\,d^3\,{\ln \left (F\right )}^3} \]