\(\int F^{c (a+b x)} (d^2+2 d e x+e^2 x^2) \, dx\) [14]

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\frac {2 e^2 F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {2 e F^{c (a+b x)} (d+e x)}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)^2}{b c \log (F)} \]

[Out]

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
)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {27, 2207, 2225} \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\frac {2 e^2 F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {2 e (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {(d+e x)^2 F^{c (a+b x)}}{b c \log (F)} \]

[In]

Int[F^(c*(a + b*x))*(d^2 + 2*d*e*x + e^2*x^2),x]

[Out]

(2*e^2*F^(c*(a + b*x)))/(b^3*c^3*Log[F]^3) - (2*e*F^(c*(a + b*x))*(d + e*x))/(b^2*c^2*Log[F]^2) + (F^(c*(a + b
*x))*(d + e*x)^2)/(b*c*Log[F])

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2207

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] - Dist[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]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int F^{c (a+b x)} (d+e x)^2 \, dx \\ & = \frac {F^{c (a+b x)} (d+e x)^2}{b c \log (F)}-\frac {(2 e) \int F^{c (a+b x)} (d+e x) \, dx}{b c \log (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)}+\frac {\left (2 e^2\right ) \int F^{c (a+b x)} \, dx}{b^2 c^2 \log ^2(F)} \\ & = \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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\frac {F^{c (a+b x)} \left (2 e^2-2 b c e (d+e x) \log (F)+b^2 c^2 (d+e x)^2 \log ^2(F)\right )}{b^3 c^3 \log ^3(F)} \]

[In]

Integrate[F^(c*(a + b*x))*(d^2 + 2*d*e*x + e^2*x^2),x]

[Out]

(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*Log[F]^2))/(b^3*c^3*Log[F]^3)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15

method	result	size
gosper	\(\frac {\left (e^{2} x^{2} c^{2} b^{2} \ln \left (F \right )^{2}+2 \ln \left (F \right )^{2} b^{2} c^{2} d e x +\ln \left (F \right )^{2} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) b c \,e^{2} x -2 \ln \left (F \right ) b c e d +2 e^{2}\right ) F^{c \left (b x +a \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}\)	\(91\)
risch	\(\frac {\left (e^{2} x^{2} c^{2} b^{2} \ln \left (F \right )^{2}+2 \ln \left (F \right )^{2} b^{2} c^{2} d e x +\ln \left (F \right )^{2} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) b c \,e^{2} x -2 \ln \left (F \right ) b c e d +2 e^{2}\right ) F^{c \left (b x +a \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}\)	\(91\)
norman	\(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) b c e d +2 e^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {e^{2} x^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c b \ln \left (F \right )}+\frac {2 e \left (\ln \left (F \right ) b c d -e \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{2} b^{2} \ln \left (F \right )^{2}}\)	\(112\)
meijerg	\(-\frac {F^{c a} e^{2} \left (2-\frac {\left (3 b^{2} c^{2} x^{2} \ln \left (F \right )^{2}-6 b c x \ln \left (F \right )+6\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{3}\right )}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {2 F^{c a} e d \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{c^{2} b^{2} \ln \left (F \right )^{2}}-\frac {F^{c a} d^{2} \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{c b \ln \left (F \right )}\)	\(127\)
parallelrisch	\(\frac {x^{2} F^{c \left (b x +a \right )} e^{2} c^{2} b^{2} \ln \left (F \right )^{2}+2 \ln \left (F \right )^{2} x \,F^{c \left (b x +a \right )} b^{2} c^{2} d e +\ln \left (F \right )^{2} F^{c \left (b x +a \right )} b^{2} c^{2} d^{2}-2 \ln \left (F \right ) x \,F^{c \left (b x +a \right )} b c \,e^{2}-2 \ln \left (F \right ) F^{c \left (b x +a \right )} b c d e +2 F^{c \left (b x +a \right )} e^{2}}{c^{3} b^{3} \ln \left (F \right )^{3}}\)	\(136\)

[In]

int(F^(c*(b*x+a))*(e^2*x^2+2*d*e*x+d^2),x,method=_RETURNVERBOSE)

[Out]

(e^2*x^2*c^2*b^2*ln(F)^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))/c^3/b^3/ln(F)^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\frac {{\left ({\left (b^{2} c^{2} e^{2} x^{2} + 2 \, b^{2} c^{2} d e x + b^{2} c^{2} d^{2}\right )} \log \left (F\right )^{2} + 2 \, e^{2} - 2 \, {\left (b c e^{2} x + b c d e\right )} \log \left (F\right )\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3}} \]

[In]

integrate(F^(c*(b*x+a))*(e^2*x^2+2*d*e*x+d^2),x, algorithm="fricas")

[Out]

((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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.68 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\begin {cases} \frac {F^{c \left (a + b x\right )} \left (b^{2} c^{2} d^{2} \log {\left (F \right )}^{2} + 2 b^{2} c^{2} d e x \log {\left (F \right )}^{2} + b^{2} c^{2} e^{2} x^{2} \log {\left (F \right )}^{2} - 2 b c d e \log {\left (F \right )} - 2 b c e^{2} x \log {\left (F \right )} + 2 e^{2}\right )}{b^{3} c^{3} \log {\left (F \right )}^{3}} & \text {for}\: b^{3} c^{3} \log {\left (F \right )}^{3} \neq 0 \\d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(c*(b*x+a))*(e**2*x**2+2*d*e*x+d**2),x)

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.56 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\frac {F^{b c x + a c} d^{2}}{b c \log \left (F\right )} + \frac {2 \, {\left (F^{a c} b c x \log \left (F\right ) - F^{a c}\right )} F^{b c x} d e}{b^{2} c^{2} \log \left (F\right )^{2}} + \frac {{\left (F^{a c} b^{2} c^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c} b c x \log \left (F\right ) + 2 \, F^{a c}\right )} F^{b c x} e^{2}}{b^{3} c^{3} \log \left (F\right )^{3}} \]

[In]

integrate(F^(c*(b*x+a))*(e^2*x^2+2*d*e*x+d^2),x, algorithm="maxima")

[Out]

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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 2214, normalized size of antiderivative = 28.03 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\text {Too large to display} \]

[In]

integrate(F^(c*(b*x+a))*(e^2*x^2+2*d*e*x+d^2),x, algorithm="giac")

[Out]

-((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*s
gn(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(a
bs(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)/((p
i^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*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)*(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) + 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)*(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))*sin(-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))*e^(
b*c*x*log(abs(F)) + a*c*log(abs(F))) - 2*I*((-I*pi^2*b^2*c^2*e^2*x^2*sgn(F) + 2*pi*b^2*c^2*e^2*x^2*log(abs(F))
*sgn(F) + I*pi^2*b^2*c^2*e^2*x^2 - 2*pi*b^2*c^2*e^2*x^2*log(abs(F)) - 2*I*b^2*c^2*e^2*x^2*log(abs(F))^2 - 2*I*
pi^2*b^2*c^2*d*e*x*sgn(F) + 4*pi*b^2*c^2*d*e*x*log(abs(F))*sgn(F) + 2*I*pi^2*b^2*c^2*d*e*x - 4*pi*b^2*c^2*d*e*
x*log(abs(F)) - 4*I*b^2*c^2*d*e*x*log(abs(F))^2 - I*pi^2*b^2*c^2*d^2*sgn(F) + 2*pi*b^2*c^2*d^2*log(abs(F))*sgn
(F) + I*pi^2*b^2*c^2*d^2 - 2*pi*b^2*c^2*d^2*log(abs(F)) - 2*I*b^2*c^2*d^2*log(abs(F))^2 - 2*pi*b*c*e^2*x*sgn(F
) + 2*pi*b*c*e^2*x + 4*I*b*c*e^2*x*log(abs(F)) - 2*pi*b*c*d*e*sgn(F) + 2*pi*b*c*d*e + 4*I*b*c*d*e*log(abs(F))
- 4*I*e^2)*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c)/(-4*I*pi^3*b^3*c^3*
sgn(F) + 12*pi^2*b^3*c^3*log(abs(F))*sgn(F) + 12*I*pi*b^3*c^3*log(abs(F))^2*sgn(F) + 4*I*pi^3*b^3*c^3 - 12*pi^
2*b^3*c^3*log(abs(F)) - 12*I*pi*b^3*c^3*log(abs(F))^2 + 8*b^3*c^3*log(abs(F))^3) - (-I*pi^2*b^2*c^2*e^2*x^2*sg
n(F) - 2*pi*b^2*c^2*e^2*x^2*log(abs(F))*sgn(F) + I*pi^2*b^2*c^2*e^2*x^2 + 2*pi*b^2*c^2*e^2*x^2*log(abs(F)) - 2
*I*b^2*c^2*e^2*x^2*log(abs(F))^2 - 2*I*pi^2*b^2*c^2*d*e*x*sgn(F) - 4*pi*b^2*c^2*d*e*x*log(abs(F))*sgn(F) + 2*I
*pi^2*b^2*c^2*d*e*x + 4*pi*b^2*c^2*d*e*x*log(abs(F)) - 4*I*b^2*c^2*d*e*x*log(abs(F))^2 - I*pi^2*b^2*c^2*d^2*sg
n(F) - 2*pi*b^2*c^2*d^2*log(abs(F))*sgn(F) + I*pi^2*b^2*c^2*d^2 + 2*pi*b^2*c^2*d^2*log(abs(F)) - 2*I*b^2*c^2*d
^2*log(abs(F))^2 + 2*pi*b*c*e^2*x*sgn(F) - 2*pi*b*c*e^2*x + 4*I*b*c*e^2*x*log(abs(F)) + 2*pi*b*c*d*e*sgn(F) -
2*pi*b*c*d*e + 4*I*b*c*d*e*log(abs(F)) - 4*I*e^2)*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sg
n(F) + 1/2*I*pi*a*c)/(4*I*pi^3*b^3*c^3*sgn(F) + 12*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 12*I*pi*b^3*c^3*log(abs(F
))^2*sgn(F) - 4*I*pi^3*b^3*c^3 - 12*pi^2*b^3*c^3*log(abs(F)) + 12*I*pi*b^3*c^3*log(abs(F))^2 + 8*b^3*c^3*log(a
bs(F))^3))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right ) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (b^2\,c^2\,d^2\,{\ln \left (F\right )}^2+2\,b^2\,c^2\,d\,e\,x\,{\ln \left (F\right )}^2+b^2\,c^2\,e^2\,x^2\,{\ln \left (F\right )}^2-2\,b\,c\,d\,e\,\ln \left (F\right )-2\,b\,c\,e^2\,x\,\ln \left (F\right )+2\,e^2\right )}{b^3\,c^3\,{\ln \left (F\right )}^3} \]

[In]

int(F^(c*(a + b*x))*(d^2 + e^2*x^2 + 2*d*e*x),x)

[Out]

(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)