\(\int F^{c (a+b x)} (d+e x+f x^2) \, dx\) [52]

Optimal result

Integrand size = 20, antiderivative size = 135 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f F^{c (a+b x)} x}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e F^{c (a+b x)} x}{b c \log (F)}+\frac {f F^{c (a+b x)} x^2}{b c \log (F)} \]

[Out]

2*f*F^(c*(b*x+a))/b^3/c^3/ln(F)^3-e*F^(c*(b*x+a))/b^2/c^2/ln(F)^2-2*f*F^(c*(b*x+a))*x/b^2/c^2/ln(F)^2+d*F^(c*(
b*x+a))/b/c/ln(F)+e*F^(c*(b*x+a))*x/b/c/ln(F)+f*F^(c*(b*x+a))*x^2/b/c/ln(F)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2227, 2225, 2207} \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f x F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e x F^{c (a+b x)}}{b c \log (F)}+\frac {f x^2 F^{c (a+b x)}}{b c \log (F)} \]

[In]

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

[Out]

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

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]

Rule 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \left (d F^{c (a+b x)}+e F^{c (a+b x)} x+f F^{c (a+b x)} x^2\right ) \, dx \\ & = d \int F^{c (a+b x)} \, dx+e \int F^{c (a+b x)} x \, dx+f \int F^{c (a+b x)} x^2 \, dx \\ & = \frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e F^{c (a+b x)} x}{b c \log (F)}+\frac {f F^{c (a+b x)} x^2}{b c \log (F)}-\frac {e \int F^{c (a+b x)} \, dx}{b c \log (F)}-\frac {(2 f) \int F^{c (a+b x)} x \, dx}{b c \log (F)} \\ & = -\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f F^{c (a+b x)} x}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e F^{c (a+b x)} x}{b c \log (F)}+\frac {f F^{c (a+b x)} x^2}{b c \log (F)}+\frac {(2 f) \int F^{c (a+b x)} \, dx}{b^2 c^2 \log ^2(F)} \\ & = \frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f F^{c (a+b x)} x}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e F^{c (a+b x)} x}{b c \log (F)}+\frac {f F^{c (a+b x)} x^2}{b c \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.41 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {F^{c (a+b x)} \left (2 f-b c (e+2 f x) \log (F)+b^2 c^2 (d+x (e+f x)) \log ^2(F)\right )}{b^3 c^3 \log ^3(F)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.59

method	result	size
gosper	\(\frac {\left (f \,x^{2} c^{2} b^{2} \ln \left (F \right )^{2}+\ln \left (F \right )^{2} b^{2} c^{2} e x +c^{2} b^{2} \ln \left (F \right )^{2} d -2 \ln \left (F \right ) b c f x -\ln \left (F \right ) b c e +2 f \right ) F^{c \left (b x +a \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}\)	\(80\)
risch	\(\frac {\left (f \,x^{2} c^{2} b^{2} \ln \left (F \right )^{2}+\ln \left (F \right )^{2} b^{2} c^{2} e x +c^{2} b^{2} \ln \left (F \right )^{2} d -2 \ln \left (F \right ) b c f x -\ln \left (F \right ) b c e +2 f \right ) F^{c \left (b x +a \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}\)	\(80\)
norman	\(\frac {\left (c^{2} b^{2} \ln \left (F \right )^{2} d -\ln \left (F \right ) b c e +2 f \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {f \,x^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c b \ln \left (F \right )}+\frac {\left (\ln \left (F \right ) b c e -2 f \right ) x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{2} b^{2} \ln \left (F \right )^{2}}\)	\(103\)
meijerg	\(-\frac {F^{c a} f \left (2-\frac {\left (3 b^{2} c^{2} x^{2} \ln \left (F \right )^{2}-6 b c x \ln \left (F \right )+6\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{3}\right )}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {F^{c a} e \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{c^{2} b^{2} \ln \left (F \right )^{2}}-\frac {F^{c a} d \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{c b \ln \left (F \right )}\)	\(121\)
parallelrisch	\(\frac {x^{2} F^{c \left (b x +a \right )} f \,c^{2} b^{2} \ln \left (F \right )^{2}+\ln \left (F \right )^{2} x \,F^{c \left (b x +a \right )} b^{2} c^{2} e +\ln \left (F \right )^{2} F^{c \left (b x +a \right )} b^{2} c^{2} d -2 \ln \left (F \right ) x \,F^{c \left (b x +a \right )} b c f -\ln \left (F \right ) F^{c \left (b x +a \right )} b c e +2 F^{c \left (b x +a \right )} f}{c^{3} b^{3} \ln \left (F \right )^{3}}\)	\(125\)

[In]

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

[Out]

(f*x^2*c^2*b^2*ln(F)^2+ln(F)^2*b^2*c^2*e*x+c^2*b^2*ln(F)^2*d-2*ln(F)*b*c*f*x-ln(F)*b*c*e+2*f)*F^(c*(b*x+a))/c^
3/b^3/ln(F)^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.55 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {{\left ({\left (b^{2} c^{2} f x^{2} + b^{2} c^{2} e x + b^{2} c^{2} d\right )} \log \left (F\right )^{2} - {\left (2 \, b c f x + b c e\right )} \log \left (F\right ) + 2 \, f\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3}} \]

[In]

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

[Out]

((b^2*c^2*f*x^2 + b^2*c^2*e*x + b^2*c^2*d)*log(F)^2 - (2*b*c*f*x + b*c*e)*log(F) + 2*f)*F^(b*c*x + a*c)/(b^3*c
^3*log(F)^3)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.86 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\begin {cases} \frac {F^{c \left (a + b x\right )} \left (b^{2} c^{2} d \log {\left (F \right )}^{2} + b^{2} c^{2} e x \log {\left (F \right )}^{2} + b^{2} c^{2} f x^{2} \log {\left (F \right )}^{2} - b c e \log {\left (F \right )} - 2 b c f x \log {\left (F \right )} + 2 f\right )}{b^{3} c^{3} \log {\left (F \right )}^{3}} & \text {for}\: b^{3} c^{3} \log {\left (F \right )}^{3} \neq 0 \\d x + \frac {e x^{2}}{2} + \frac {f x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((F**(c*(a + b*x))*(b**2*c**2*d*log(F)**2 + b**2*c**2*e*x*log(F)**2 + b**2*c**2*f*x**2*log(F)**2 - b*
c*e*log(F) - 2*b*c*f*x*log(F) + 2*f)/(b**3*c**3*log(F)**3), Ne(b**3*c**3*log(F)**3, 0)), (d*x + e*x**2/2 + f*x
**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.87 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {F^{b c x + a c} d}{b c \log \left (F\right )} + \frac {{\left (F^{a c} b c x \log \left (F\right ) - F^{a c}\right )} F^{b c x} e}{b^{2} c^{2} \log \left (F\right )^{2}} + \frac {{\left (F^{a c} b^{2} c^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c} b c x \log \left (F\right ) + 2 \, F^{a c}\right )} F^{b c x} f}{b^{3} c^{3} \log \left (F\right )^{3}} \]

[In]

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

[Out]

F^(b*c*x + a*c)*d/(b*c*log(F)) + (F^(a*c)*b*c*x*log(F) - F^(a*c))*F^(b*c*x)*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)*f/(b^3*c^3*log(F)^3)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 2068, normalized size of antiderivative = 15.32 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

(((pi^2*b^2*c^2*f*x^2*sgn(F) - pi^2*b^2*c^2*f*x^2 + 2*b^2*c^2*f*x^2*log(abs(F))^2 + pi^2*b^2*c^2*e*x*sgn(F) -
pi^2*b^2*c^2*e*x + 2*b^2*c^2*e*x*log(abs(F))^2 + pi^2*b^2*c^2*d*sgn(F) - pi^2*b^2*c^2*d + 2*b^2*c^2*d*log(abs(
F))^2 - 4*b*c*f*x*log(abs(F)) - 2*b*c*e*log(abs(F)) + 4*f)*(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) - (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*pi*b^2*c^2*f*x^2*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*f*x^2*log(abs(F)) + 2*pi*b^2*c^2*e*x*log(
abs(F))*sgn(F) - 2*pi*b^2*c^2*e*x*log(abs(F)) + 2*pi*b^2*c^2*d*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*d*log(abs(F))
- 2*pi*b*c*f*x*sgn(F) + 2*pi*b*c*f*x - pi*b*c*e*sgn(F) + pi*b*c*e)/((pi^3*b^3*c^3*sgn(F) - 3*pi*b^3*c^3*log(a
bs(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^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(ab
s(F))^2)*(pi^2*b^2*c^2*f*x^2*sgn(F) - pi^2*b^2*c^2*f*x^2 + 2*b^2*c^2*f*x^2*log(abs(F))^2 + pi^2*b^2*c^2*e*x*sg
n(F) - pi^2*b^2*c^2*e*x + 2*b^2*c^2*e*x*log(abs(F))^2 + pi^2*b^2*c^2*d*sgn(F) - pi^2*b^2*c^2*d + 2*b^2*c^2*d*l
og(abs(F))^2 - 4*b*c*f*x*log(abs(F)) - 2*b*c*e*log(abs(F)) + 4*f)/((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) + (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*b^2*c^2*f*x^2*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*f*x^2*log(abs(F)) + 2*pi*b
^2*c^2*e*x*log(abs(F))*sgn(F) - 2*pi*b^2*c^2*e*x*log(abs(F)) + 2*pi*b^2*c^2*d*log(abs(F))*sgn(F) - 2*pi*b^2*c^
2*d*log(abs(F)) - 2*pi*b*c*f*x*sgn(F) + 2*pi*b*c*f*x - pi*b*c*e*sgn(F) + pi*b*c*e)/((pi^3*b^3*c^3*sgn(F) - 3*p
i*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))*sg
n(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*f*x^2*sgn(F) + 2
*pi*b^2*c^2*f*x^2*log(abs(F))*sgn(F) + I*pi^2*b^2*c^2*f*x^2 - 2*pi*b^2*c^2*f*x^2*log(abs(F)) - 2*I*b^2*c^2*f*x
^2*log(abs(F))^2 - I*pi^2*b^2*c^2*e*x*sgn(F) + 2*pi*b^2*c^2*e*x*log(abs(F))*sgn(F) + I*pi^2*b^2*c^2*e*x - 2*pi
*b^2*c^2*e*x*log(abs(F)) - 2*I*b^2*c^2*e*x*log(abs(F))^2 - I*pi^2*b^2*c^2*d*sgn(F) + 2*pi*b^2*c^2*d*log(abs(F)
)*sgn(F) + I*pi^2*b^2*c^2*d - 2*pi*b^2*c^2*d*log(abs(F)) - 2*I*b^2*c^2*d*log(abs(F))^2 - 2*pi*b*c*f*x*sgn(F) +
2*pi*b*c*f*x + 4*I*b*c*f*x*log(abs(F)) - pi*b*c*e*sgn(F) + pi*b*c*e + 2*I*b*c*e*log(abs(F)) - 4*I*f)*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*f*x^2*sgn(F) - 2*pi*b^2*c^2*f
*x^2*log(abs(F))*sgn(F) + I*pi^2*b^2*c^2*f*x^2 + 2*pi*b^2*c^2*f*x^2*log(abs(F)) - 2*I*b^2*c^2*f*x^2*log(abs(F)
)^2 - I*pi^2*b^2*c^2*e*x*sgn(F) - 2*pi*b^2*c^2*e*x*log(abs(F))*sgn(F) + I*pi^2*b^2*c^2*e*x + 2*pi*b^2*c^2*e*x*
log(abs(F)) - 2*I*b^2*c^2*e*x*log(abs(F))^2 - I*pi^2*b^2*c^2*d*sgn(F) - 2*pi*b^2*c^2*d*log(abs(F))*sgn(F) + I*
pi^2*b^2*c^2*d + 2*pi*b^2*c^2*d*log(abs(F)) - 2*I*b^2*c^2*d*log(abs(F))^2 + 2*pi*b*c*f*x*sgn(F) - 2*pi*b*c*f*x
+ 4*I*b*c*f*x*log(abs(F)) + pi*b*c*e*sgn(F) - pi*b*c*e + 2*I*b*c*e*log(abs(F)) - 4*I*f)*e^(-1/2*I*pi*b*c*x*sg
n(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))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)))

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.59 \[ \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (f\,b^2\,c^2\,x^2\,{\ln \left (F\right )}^2+e\,b^2\,c^2\,x\,{\ln \left (F\right )}^2+d\,b^2\,c^2\,{\ln \left (F\right )}^2-2\,f\,b\,c\,x\,\ln \left (F\right )-e\,b\,c\,\ln \left (F\right )+2\,f\right )}{b^3\,c^3\,{\ln \left (F\right )}^3} \]

[In]

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

[Out]

(F^(a*c + b*c*x)*(2*f - b*c*e*log(F) + b^2*c^2*d*log(F)^2 + b^2*c^2*f*x^2*log(F)^2 - 2*b*c*f*x*log(F) + b^2*c^
2*e*x*log(F)^2))/(b^3*c^3*log(F)^3)