\(\int F^{c (a+b x)} (d+e x) \, dx\) [51]

Optimal result

Integrand size = 15, antiderivative size = 48 \[ \int F^{c (a+b x)} (d+e x) \, dx=-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)}{b c \log (F)} \]

[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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2207, 2225} \[ \int F^{c (a+b x)} (d+e x) \, dx=\frac {(d+e x) F^{c (a+b x)}}{b c \log (F)}-\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)} \]

[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 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}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.71 \[ \int F^{c (a+b x)} (d+e x) \, dx=\frac {F^{c (a+b x)} (-e+b c (d+e x) \log (F))}{b^2 c^2 \log ^2(F)} \]

[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)

Maple [A] (verified)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79

method	result	size
gosper	\(\frac {\left (\ln \left (F \right ) b c e x +\ln \left (F \right ) b c d -e \right ) F^{c \left (b x +a \right )}}{c^{2} b^{2} \ln \left (F \right )^{2}}\)	\(38\)
risch	\(\frac {\left (\ln \left (F \right ) b c e x +\ln \left (F \right ) b c d -e \right ) F^{c \left (b x +a \right )}}{c^{2} b^{2} \ln \left (F \right )^{2}}\)	\(38\)
norman	\(\frac {\left (\ln \left (F \right ) b c d -e \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{2} b^{2} \ln \left (F \right )^{2}}+\frac {e x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c b \ln \left (F \right )}\)	\(56\)
parallelrisch	\(\frac {x \,F^{c \left (b x +a \right )} e c b \ln \left (F \right )+\ln \left (F \right ) F^{c \left (b x +a \right )} b c d -F^{c \left (b x +a \right )} e}{c^{2} b^{2} \ln \left (F \right )^{2}}\)	\(56\)
meijerg	\(\frac {F^{c a} e \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{c^{2} b^{2} \ln \left (F \right )^{2}}-\frac {F^{c a} d \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{c b \ln \left (F \right )}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int F^{c (a+b x)} (d+e x) \, dx=\frac {{\left ({\left (b c e x + b c d\right )} \log \left (F\right ) - e\right )} F^{b c x + a c}}{b^{2} c^{2} \log \left (F\right )^{2}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.25 \[ \int F^{c (a+b x)} (d+e x) \, dx=\begin {cases} \frac {F^{c \left (a + b x\right )} \left (b c d \log {\left (F \right )} + b c e x \log {\left (F \right )} - e\right )}{b^{2} c^{2} \log {\left (F \right )}^{2}} & \text {for}\: b^{2} c^{2} \log {\left (F \right )}^{2} \neq 0 \\d x + \frac {e x^{2}}{2} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.25 \[ \int F^{c (a+b x)} (d+e x) \, dx=\frac {F^{b c x + a c} d}{b c \log \left (F\right )} + \frac {{\left (F^{a c} b c x \log \left (F\right ) - F^{a c}\right )} F^{b c x} e}{b^{2} c^{2} \log \left (F\right )^{2}} \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 898, normalized size of antiderivative = 18.71 \[ \int F^{c (a+b x)} (d+e x) \, dx=\text {Too large to display} \]

[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*e*x*sgn(F) - pi*b*c*e*x + pi*b*c*d*sgn(F)
- pi*b*c*d)/((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*e*x*log
(abs(F)) + b*c*d*log(abs(F)) - e)/((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*sg
n(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*e*x*sgn(F) - pi*b
*c*e*x + pi*b*c*d*sgn(F) - pi*b*c*d)/((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) - 4*(pi*b^2*c^2*log(abs(F))*sgn(F) - pi*b^2*c^2*log(a
bs(F)))*(b*c*e*x*log(abs(F)) + b*c*d*log(abs(F)) - e)/((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))) - 1/2*I*((pi*b*c*e*x*sgn(F) -
pi*b*c*e*x - 2*I*b*c*e*x*log(abs(F)) + pi*b*c*d*sgn(F) - pi*b*c*d - 2*I*b*c*d*log(abs(F)) + 2*I*e)*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)/(pi^2*b^2*c^2*sgn(F) + 2*I*pi*b^2*c^2*log
(abs(F))*sgn(F) - pi^2*b^2*c^2 - 2*I*pi*b^2*c^2*log(abs(F)) + 2*b^2*c^2*log(abs(F))^2) + (pi*b*c*e*x*sgn(F) -
pi*b*c*e*x + 2*I*b*c*e*x*log(abs(F)) + pi*b*c*d*sgn(F) - pi*b*c*d + 2*I*b*c*d*log(abs(F)) - 2*I*e)*e^(-1/2*I*p
i*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)/(pi^2*b^2*c^2*sgn(F) - 2*I*pi*b^2*c^2*lo
g(abs(F))*sgn(F) - pi^2*b^2*c^2 + 2*I*pi*b^2*c^2*log(abs(F)) + 2*b^2*c^2*log(abs(F))^2))*e^(b*c*x*log(abs(F))
+ a*c*log(abs(F)))

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int F^{c (a+b x)} (d+e x) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (b\,c\,d\,\ln \left (F\right )-e+b\,c\,e\,x\,\ln \left (F\right )\right )}{b^2\,c^2\,{\ln \left (F\right )}^2} \]

[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)