∫ Fc(a+bx)(d + ex) dx [51]

3.1.51 $\int F^{c (a+b x)} (d+e x) \, dx$ [51]

Optimal. Leaf size=48 \[ -\frac {e F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} (d+e x)}{b c \log (F)} \]

[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)

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Rubi [A]
time = 0.01, 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 = {2207, 2225} \begin {gather*} \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)} \end {gather*}

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 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*} \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 = 0.01, size = 34, normalized size = 0.71 \begin {gather*} \frac {F^{c (a+b x)} (-e+b c (d+e x) \log (F))}{b^2 c^2 \log ^2(F)} \end {gather*}

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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Maple [A]
time = 0.00, size = 38, normalized size = 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	$\frac {\left (\ln \left (F \right ) b c d -e \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{2} b^{2} \ln \left (F \right )^{2}}+\frac {e x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c b \ln \left (F \right )}$	$56$
meijerg	$\frac {F^{c a} e \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{c^{2} b^{2} \ln \left (F \right )^{2}}-\frac {F^{c a} d \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{c b \ln \left (F \right )}$	$68$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Maxima [A]
time = 0.27, size = 62, normalized size = 1.29 \begin {gather*} \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 )} e^{\left (b c x \log \left (F\right ) + 1\right )}}{b^{2} c^{2} \log \left (F\right )^{2}} \end {gather*}

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))*e^(b*c*x*log(F) + 1)/(b^2*c^2*log(F)^2)

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Fricas [A]
time = 0.37, size = 40, normalized size = 0.83 \begin {gather*} \frac {{\left ({\left (b c x e + 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}} \end {gather*}

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

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Sympy [A]
time = 0.05, size = 60, normalized size = 1.25 \begin {gather*} \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} \end {gather*}

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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Giac [C] Result contains complex when optimal does not.
time = 2.32, size = 1079, normalized size = 22.48 \begin {gather*} \text {Too large to display} \end {gather*}

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*((pi*b*c*x*sgn(F)

- pi*b*c*x - 2*I*b*c*x*log(abs(F)) + 2*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)/(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*x*sgn(F) - pi*b*c*x + 2*I*b*c*x*log(abs(F)) - 2*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)/(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))*e^(b*c*x*log(abs(F)) + a*c*lo

g(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))) + I*(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))) - 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))))*e^(b*c*x*log(abs

(F)) + a*c*log(abs(F)))

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Mupad [B]
time = 0.00, size = 38, normalized size = 0.79 \begin {gather*} \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} \end {gather*}

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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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	\(\frac {\left (\ln \left (F \right ) b c d -e \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{2} b^{2} \ln \left (F \right )^{2}}+\frac {e x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c b \ln \left (F \right )}\)	\(56\)
meijerg	\(\frac {F^{c a} e \left (1-\frac {\left (-2 b c x \ln \left (F \right )+2\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{2}\right )}{c^{2} b^{2} \ln \left (F \right )^{2}}-\frac {F^{c a} d \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{c b \ln \left (F \right )}\)	\(68\)

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