3.52 \(\int F^{c (a+b x)} (d+e x+f x^2) \, dx\)
Optimal. Leaf size=135 \[ \frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f x F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e x F^{c (a+b x)}}{b c \log (F)}+\frac {f x^2 F^{c (a+b x)}}{b c \log (F)} \]
[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)
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Rubi [A] time = 0.10, antiderivative size = 135, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used =
{2196, 2194, 2176} \[ -\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}-\frac {2 f x F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {2 f F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}+\frac {e x F^{c (a+b x)}}{b c \log (F)}+\frac {f x^2 F^{c (a+b x)}}{b c \log (F)} \]
Antiderivative was successfully verified.
[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 2176
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] && !$UseGamma === True
Rule 2194
Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]
Rule 2196
Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] && !$UseGamma === True
Rubi steps
\begin {align*} \int F^{c (a+b x)} \left (d+e x+f x^2\right ) \, dx &=\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*}
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Mathematica [A] time = 0.08, size = 56, normalized size = 0.41 \[ \frac {F^{c (a+b x)} \left (b^2 c^2 \log ^2(F) (d+x (e+f x))-b c \log (F) (e+2 f x)+2 f\right )}{b^3 c^3 \log ^3(F)} \]
Antiderivative was successfully verified.
[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)
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fricas [A] time = 0.42, size = 74, normalized size = 0.55 \[ \frac {{\left ({\left (b^{2} c^{2} f x^{2} + b^{2} c^{2} e x + b^{2} c^{2} d\right )} \log \relax (F)^{2} - {\left (2 \, b c f x + b c e\right )} \log \relax (F) + 2 \, f\right )} F^{b c x + a c}}{b^{3} c^{3} \log \relax (F)^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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giac [C] time = 0.75, size = 2490, normalized size = 18.44 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*(f*x^2+e*x+d),x, algorithm="giac")
[Out]
(2*((pi*b^2*c^2*log(abs(F))*sgn(F) - pi*b^2*c^2*log(abs(F)))*(pi*b*c*x*sgn(F) - pi*b*c*x)/((pi^2*b^2*c^2*sgn(F
) - pi^2*b^2*c^2 + 2*b^2*c^2*log(abs(F))^2)^2 + 4*(pi*b^2*c^2*log(abs(F))*sgn(F) - pi*b^2*c^2*log(abs(F)))^2)
+ (pi^2*b^2*c^2*sgn(F) - pi^2*b^2*c^2 + 2*b^2*c^2*log(abs(F))^2)*(b*c*x*log(abs(F)) - 1)/((pi^2*b^2*c^2*sgn(F)
- pi^2*b^2*c^2 + 2*b^2*c^2*log(abs(F))^2)^2 + 4*(pi*b^2*c^2*log(abs(F))*sgn(F) - pi*b^2*c^2*log(abs(F)))^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*sgn(F) - pi^2*b^2*c
^2 + 2*b^2*c^2*log(abs(F))^2)*(pi*b*c*x*sgn(F) - pi*b*c*x)/((pi^2*b^2*c^2*sgn(F) - pi^2*b^2*c^2 + 2*b^2*c^2*lo
g(abs(F))^2)^2 + 4*(pi*b^2*c^2*log(abs(F))*sgn(F) - pi*b^2*c^2*log(abs(F)))^2) - 4*(pi*b^2*c^2*log(abs(F))*sgn
(F) - pi*b^2*c^2*log(abs(F)))*(b*c*x*log(abs(F)) - 1)/((pi^2*b^2*c^2*sgn(F) - pi^2*b^2*c^2 + 2*b^2*c^2*log(abs
(F))^2)^2 + 4*(pi*b^2*c^2*log(abs(F))*sgn(F) - pi*b^2*c^2*log(abs(F)))^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)) + 1) - 1/2*I*((2*pi*b*c*x*sgn(F
) - 2*pi*b*c*x - 4*I*b*c*x*log(abs(F)) + 4*I)*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)/(2*pi^2*b^2*c^2*sgn(F) + 4*I*pi*b^2*c^2*log(abs(F))*sgn(F) - 2*pi^2*b^2*c^2 - 4*I*pi*b^2*c^2*l
og(abs(F)) + 4*b^2*c^2*log(abs(F))^2) + (2*pi*b*c*x*sgn(F) - 2*pi*b*c*x + 4*I*b*c*x*log(abs(F)) - 4*I)*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)/(2*pi^2*b^2*c^2*sgn(F) - 4*I*pi*b^2*
c^2*log(abs(F))*sgn(F) - 2*pi^2*b^2*c^2 + 4*I*pi*b^2*c^2*log(abs(F)) + 4*b^2*c^2*log(abs(F))^2))*e^(b*c*x*log(
abs(F)) + a*c*log(abs(F)) + 1) + (((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*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)) + 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) - 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*b^2*c^2*f*x^2*log(abs(F))*sgn(F) - pi*b^2*c
^2*f*x^2*log(abs(F)) + pi*b^2*c^2*d*log(abs(F))*sgn(F) - pi*b^2*c^2*d*log(abs(F)) - pi*b*c*f*x*sgn(F) + pi*b*c
*f*x)/((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^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^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*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*lo
g(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(a
bs(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*(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*b^2*c^2*f*x^2
*log(abs(F))*sgn(F) - pi*b^2*c^2*f*x^2*log(abs(F)) + pi*b^2*c^2*d*log(abs(F))*sgn(F) - pi*b^2*c^2*d*log(abs(F)
) - pi*b*c*f*x*sgn(F) + pi*b*c*f*x)/((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*l
og(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))) + 1/2*I*((4*I*pi^2*b^2*c^2*f*x^2*sgn(F) - 8*pi*b^2*c^2*f*x^2*log(abs(F))*sgn(F) - 4*I*p
i^2*b^2*c^2*f*x^2 + 8*pi*b^2*c^2*f*x^2*log(abs(F)) + 8*I*b^2*c^2*f*x^2*log(abs(F))^2 + 4*I*pi^2*b^2*c^2*d*sgn(
F) - 8*pi*b^2*c^2*d*log(abs(F))*sgn(F) - 4*I*pi^2*b^2*c^2*d + 8*pi*b^2*c^2*d*log(abs(F)) + 8*I*b^2*c^2*d*log(a
bs(F))^2 + 8*pi*b*c*f*x*sgn(F) - 8*pi*b*c*f*x - 16*I*b*c*f*x*log(abs(F)) + 16*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))*s
gn(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) - (4*I*pi^2*b^2*c^2*f*x^2*sgn(F) + 8*pi*b^2*c^2*f*x^2*log(abs(F))*s
gn(F) - 4*I*pi^2*b^2*c^2*f*x^2 - 8*pi*b^2*c^2*f*x^2*log(abs(F)) + 8*I*b^2*c^2*f*x^2*log(abs(F))^2 + 4*I*pi^2*b
^2*c^2*d*sgn(F) + 8*pi*b^2*c^2*d*log(abs(F))*sgn(F) - 4*I*pi^2*b^2*c^2*d - 8*pi*b^2*c^2*d*log(abs(F)) + 8*I*b^
2*c^2*d*log(abs(F))^2 - 8*pi*b*c*f*x*sgn(F) + 8*pi*b*c*f*x - 16*I*b*c*f*x*log(abs(F)) + 16*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)) + 1
2*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)))
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maple [A] time = 0.01, size = 80, normalized size = 0.59 \[ \frac {\left (b^{2} c^{2} f \,x^{2} \ln \relax (F )^{2}+b^{2} c^{2} e x \ln \relax (F )^{2}+b^{2} c^{2} d \ln \relax (F )^{2}-2 b c f x \ln \relax (F )-b c e \ln \relax (F )+2 f \right ) F^{\left (b x +a \right ) c}}{b^{3} c^{3} \ln \relax (F )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^((b*x+a)*c)*(f*x^2+e*x+d),x)
[Out]
(f*x^2*b^2*c^2*ln(F)^2+ln(F)^2*b^2*c^2*e*x+b^2*c^2*ln(F)^2*d-2*ln(F)*b*c*f*x-ln(F)*b*c*e+2*f)*F^((b*x+a)*c)/b^
3/c^3/ln(F)^3
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maxima [A] time = 0.70, size = 117, normalized size = 0.87 \[ \frac {F^{b c x + a c} d}{b c \log \relax (F)} + \frac {{\left (F^{a c} b c x \log \relax (F) - F^{a c}\right )} F^{b c x} e}{b^{2} c^{2} \log \relax (F)^{2}} + \frac {{\left (F^{a c} b^{2} c^{2} x^{2} \log \relax (F)^{2} - 2 \, F^{a c} b c x \log \relax (F) + 2 \, F^{a c}\right )} F^{b c x} f}{b^{3} c^{3} \log \relax (F)^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [B] time = 3.40, size = 80, normalized size = 0.59 \[ \frac {F^{a\,c+b\,c\,x}\,\left (f\,b^2\,c^2\,x^2\,{\ln \relax (F)}^2+e\,b^2\,c^2\,x\,{\ln \relax (F)}^2+d\,b^2\,c^2\,{\ln \relax (F)}^2-2\,f\,b\,c\,x\,\ln \relax (F)-e\,b\,c\,\ln \relax (F)+2\,f\right )}{b^3\,c^3\,{\ln \relax (F)}^3} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [A] time = 0.17, size = 116, normalized size = 0.86 \[ \begin {cases} \frac {F^{c \left (a + b x\right )} \left (b^{2} c^{2} d \log {\relax (F )}^{2} + b^{2} c^{2} e x \log {\relax (F )}^{2} + b^{2} c^{2} f x^{2} \log {\relax (F )}^{2} - b c e \log {\relax (F )} - 2 b c f x \log {\relax (F )} + 2 f\right )}{b^{3} c^{3} \log {\relax (F )}^{3}} & \text {for}\: b^{3} c^{3} \log {\relax (F )}^{3} \neq 0 \\d x + \frac {e x^{2}}{2} + \frac {f x^{3}}{3} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[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))
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