3.5 \(\int F^{c (a+b x)} (d+e x) \, dx\)
Optimal. Leaf size=48 \[ \frac {(d+e x) F^{c (a+b x)}}{b c \log (F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)} \]
[Out]
-e*F^(c*(b*x+a))/b^2/c^2/ln(F)^2+F^(c*(b*x+a))*(e*x+d)/b/c/ln(F)
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Rubi [A] time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00,
number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used =
{2176, 2194} \[ \frac {(d+e x) F^{c (a+b x)}}{b c \log (F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)} \]
Antiderivative was successfully verified.
[In]
Int[F^(c*(a + b*x))*(d + e*x),x]
[Out]
-((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])
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]
Rubi steps
\begin {align*} \int F^{c (a+b x)} (d+e x) \, dx &=\frac {F^{c (a+b x)} (d+e x)}{b c \log (F)}-\frac {e \int F^{c (a+b x)} \, dx}{b c \log (F)}\\ &=-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)}{b c \log (F)}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 34, normalized size = 0.71 \[ \frac {F^{c (a+b x)} (b c \log (F) (d+e x)-e)}{b^2 c^2 \log ^2(F)} \]
Antiderivative was successfully verified.
[In]
Integrate[F^(c*(a + b*x))*(d + e*x),x]
[Out]
(F^(c*(a + b*x))*(-e + b*c*(d + e*x)*Log[F]))/(b^2*c^2*Log[F]^2)
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fricas [A] time = 0.82, size = 38, normalized size = 0.79 \[ \frac {{\left ({\left (b c e x + b c d\right )} \log \relax (F) - e\right )} F^{b c x + a c}}{b^{2} c^{2} \log \relax (F)^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*(e*x+d),x, algorithm="fricas")
[Out]
((b*c*e*x + b*c*d)*log(F) - e)*F^(b*c*x + a*c)/(b^2*c^2*log(F)^2)
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giac [C] time = 0.67, size = 1083, normalized size = 22.56 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*(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) + 2*(2*b*c*d*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)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2) - (pi*b*c*sgn(F) - pi*b*c)*d*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)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F)
- pi*b*c)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - 1/2*I*(-2*I*d*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)/(I*pi*b*c*sgn(F) - I*pi*b*c + 2*b*c*log(abs(F))) + 2*I*d*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)/(-I*pi*b*c*sgn(F) + I*pi*b*c + 2*b*c*log(a
bs(F))))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F)))
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maple [A] time = 0.01, size = 38, normalized size = 0.79 \[ \frac {\left (b c e x \ln \relax (F )+b c d \ln \relax (F )-e \right ) F^{\left (b x +a \right ) c}}{b^{2} c^{2} \ln \relax (F )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^((b*x+a)*c)*(e*x+d),x)
[Out]
(ln(F)*b*c*e*x+ln(F)*b*c*d-e)*F^((b*x+a)*c)/b^2/c^2/ln(F)^2
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maxima [A] time = 0.49, size = 60, normalized size = 1.25 \[ \frac {F^{b c x + a c} d}{b c \log \relax (F)} + \frac {{\left (F^{a c} b c x \log \relax (F) - F^{a c}\right )} F^{b c x} e}{b^{2} c^{2} \log \relax (F)^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))*(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)
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mupad [B] time = 3.37, size = 38, normalized size = 0.79 \[ \frac {F^{a\,c+b\,c\,x}\,\left (b\,c\,d\,\ln \relax (F)-e+b\,c\,e\,x\,\ln \relax (F)\right )}{b^2\,c^2\,{\ln \relax (F)}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(a + b*x))*(d + e*x),x)
[Out]
(F^(a*c + b*c*x)*(b*c*d*log(F) - e + b*c*e*x*log(F)))/(b^2*c^2*log(F)^2)
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sympy [A] time = 0.14, size = 60, normalized size = 1.25 \[ \begin {cases} \frac {F^{c \left (a + b x\right )} \left (b c d \log {\relax (F )} + b c e x \log {\relax (F )} - e\right )}{b^{2} c^{2} \log {\relax (F )}^{2}} & \text {for}\: b^{2} c^{2} \log {\relax (F )}^{2} \neq 0 \\d x + \frac {e x^{2}}{2} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F**(c*(b*x+a))*(e*x+d),x)
[Out]
Piecewise((F**(c*(a + b*x))*(b*c*d*log(F) + b*c*e*x*log(F) - e)/(b**2*c**2*log(F)**2), Ne(b**2*c**2*log(F)**2,
0)), (d*x + e*x**2/2, True))
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