3.898 \(\int F^{c (a+b x)} (f+f \cosh (d+e x)) \, dx\)

Optimal. Leaf size=101 \[ \frac {e f \sinh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}-\frac {b c f \log (F) \cosh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac {f F^{a c+b c x}}{b c \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6741, 12, 6742, 2194, 5475} \[ \frac {e f \sinh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}-\frac {b c f \log (F) \cosh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac {f F^{a c+b c x}}{b c \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5475

Int[Cosh[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b*x))
*Cosh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sinh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2)
, x] /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 - b^2*c^2*Log[F]^2, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int F^{c (a+b x)} (f+f \cosh (d+e x)) \, dx &=\int f F^{a c+b c x} (1+\cosh (d+e x)) \, dx\\ &=f \int F^{a c+b c x} (1+\cosh (d+e x)) \, dx\\ &=f \int \left (F^{a c+b c x}+F^{a c+b c x} \cosh (d+e x)\right ) \, dx\\ &=f \int F^{a c+b c x} \, dx+f \int F^{a c+b c x} \cosh (d+e x) \, dx\\ &=\frac {f F^{a c+b c x}}{b c \log (F)}-\frac {b c f F^{a c+b c x} \cosh (d+e x) \log (F)}{e^2-b^2 c^2 \log ^2(F)}+\frac {e f F^{a c+b c x} \sinh (d+e x)}{e^2-b^2 c^2 \log ^2(F)}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 88, normalized size = 0.87 \[ \frac {f F^{c (a+b x)} \left (b^2 c^2 \log ^2(F) \cosh (d+e x)+b^2 c^2 \log ^2(F)-b c e \log (F) \sinh (d+e x)-e^2\right )}{b c \log (F) (b c \log (F)-e) (b c \log (F)+e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 430, normalized size = 4.26 \[ -\frac {{\left (2 \, e^{2} f \cosh \left (e x + d\right ) - {\left (b^{2} c^{2} f \cosh \left (e x + d\right )^{2} + 2 \, b^{2} c^{2} f \cosh \left (e x + d\right ) + b^{2} c^{2} f\right )} \log \relax (F)^{2} - {\left (b^{2} c^{2} f \log \relax (F)^{2} - b c e f \log \relax (F)\right )} \sinh \left (e x + d\right )^{2} + {\left (b c e f \cosh \left (e x + d\right )^{2} - b c e f\right )} \log \relax (F) + 2 \, {\left (b c e f \cosh \left (e x + d\right ) \log \relax (F) + e^{2} f - {\left (b^{2} c^{2} f \cosh \left (e x + d\right ) + b^{2} c^{2} f\right )} \log \relax (F)^{2}\right )} \sinh \left (e x + d\right )\right )} \cosh \left ({\left (b c x + a c\right )} \log \relax (F)\right ) + {\left (2 \, e^{2} f \cosh \left (e x + d\right ) - {\left (b^{2} c^{2} f \cosh \left (e x + d\right )^{2} + 2 \, b^{2} c^{2} f \cosh \left (e x + d\right ) + b^{2} c^{2} f\right )} \log \relax (F)^{2} - {\left (b^{2} c^{2} f \log \relax (F)^{2} - b c e f \log \relax (F)\right )} \sinh \left (e x + d\right )^{2} + {\left (b c e f \cosh \left (e x + d\right )^{2} - b c e f\right )} \log \relax (F) + 2 \, {\left (b c e f \cosh \left (e x + d\right ) \log \relax (F) + e^{2} f - {\left (b^{2} c^{2} f \cosh \left (e x + d\right ) + b^{2} c^{2} f\right )} \log \relax (F)^{2}\right )} \sinh \left (e x + d\right )\right )} \sinh \left ({\left (b c x + a c\right )} \log \relax (F)\right )}{2 \, {\left (b^{3} c^{3} \cosh \left (e x + d\right ) \log \relax (F)^{3} - b c e^{2} \cosh \left (e x + d\right ) \log \relax (F) + {\left (b^{3} c^{3} \log \relax (F)^{3} - b c e^{2} \log \relax (F)\right )} \sinh \left (e x + d\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((2*e^2*f*cosh(e*x + d) - (b^2*c^2*f*cosh(e*x + d)^2 + 2*b^2*c^2*f*cosh(e*x + d) + b^2*c^2*f)*log(F)^2 -
(b^2*c^2*f*log(F)^2 - b*c*e*f*log(F))*sinh(e*x + d)^2 + (b*c*e*f*cosh(e*x + d)^2 - b*c*e*f)*log(F) + 2*(b*c*e*
f*cosh(e*x + d)*log(F) + e^2*f - (b^2*c^2*f*cosh(e*x + d) + b^2*c^2*f)*log(F)^2)*sinh(e*x + d))*cosh((b*c*x +
a*c)*log(F)) + (2*e^2*f*cosh(e*x + d) - (b^2*c^2*f*cosh(e*x + d)^2 + 2*b^2*c^2*f*cosh(e*x + d) + b^2*c^2*f)*lo
g(F)^2 - (b^2*c^2*f*log(F)^2 - b*c*e*f*log(F))*sinh(e*x + d)^2 + (b*c*e*f*cosh(e*x + d)^2 - b*c*e*f)*log(F) +
2*(b*c*e*f*cosh(e*x + d)*log(F) + e^2*f - (b^2*c^2*f*cosh(e*x + d) + b^2*c^2*f)*log(F)^2)*sinh(e*x + d))*sinh(
(b*c*x + a*c)*log(F)))/(b^3*c^3*cosh(e*x + d)*log(F)^3 - b*c*e^2*cosh(e*x + d)*log(F) + (b^3*c^3*log(F)^3 - b*
c*e^2*log(F))*sinh(e*x + d))

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giac [C]  time = 0.23, size = 900, normalized size = 8.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 135, normalized size = 1.34 \[ \frac {f \left (\ln \relax (F )^{2} b^{2} c^{2} {\mathrm e}^{2 e x +2 d}+2 \ln \relax (F )^{2} b^{2} c^{2} {\mathrm e}^{e x +d}+b^{2} c^{2} \ln \relax (F )^{2}-\ln \relax (F ) b c e \,{\mathrm e}^{2 e x +2 d}+\ln \relax (F ) b c e -2 e^{2} {\mathrm e}^{e x +d}\right ) {\mathrm e}^{-e x -d} F^{c \left (b x +a \right )}}{2 b c \ln \relax (F ) \left (b c \ln \relax (F )-e \right ) \left (e +b c \ln \relax (F )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f*(ln(F)^2*b^2*c^2*exp(2*e*x+2*d)+2*ln(F)^2*b^2*c^2*exp(e*x+d)+b^2*c^2*ln(F)^2-ln(F)*b*c*e*exp(2*e*x+2*d)+
ln(F)*b*c*e-2*e^2*exp(e*x+d))/b/c/ln(F)/(b*c*ln(F)-e)*exp(-e*x-d)/(e+b*c*ln(F))*F^(c*(b*x+a))

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maxima [A]  time = 0.33, size = 87, normalized size = 0.86 \[ \frac {1}{2} \, f {\left (\frac {F^{a c} e^{\left (b c x \log \relax (F) + e x + d\right )}}{b c \log \relax (F) + e} + \frac {F^{a c} e^{\left (b c x \log \relax (F) - e x\right )}}{b c e^{d} \log \relax (F) - e e^{d}}\right )} + \frac {F^{b c x + a c} f}{b c \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f*(F^(a*c)*e^(b*c*x*log(F) + e*x + d)/(b*c*log(F) + e) + F^(a*c)*e^(b*c*x*log(F) - e*x)/(b*c*e^d*log(F) -
e*e^d)) + F^(b*c*x + a*c)*f/(b*c*log(F))

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mupad [B]  time = 1.74, size = 134, normalized size = 1.33 \[ -\frac {F^{b\,c\,x}\,F^{a\,c}\,f\,{\mathrm {e}}^{-d-e\,x}\,\left (b^2\,c^2\,{\ln \relax (F)}^2-2\,e^2\,{\mathrm {e}}^{d+e\,x}+b\,c\,e\,\ln \relax (F)+2\,b^2\,c^2\,{\mathrm {e}}^{d+e\,x}\,{\ln \relax (F)}^2+b^2\,c^2\,{\mathrm {e}}^{2\,d+2\,e\,x}\,{\ln \relax (F)}^2-b\,c\,e\,{\mathrm {e}}^{2\,d+2\,e\,x}\,\ln \relax (F)\right )}{2\,b\,c\,\ln \relax (F)\,\left (e^2-b^2\,c^2\,{\ln \relax (F)}^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.46, size = 391, normalized size = 3.87 \[ \begin {cases} f x + \frac {f \sinh {\left (d + e x \right )}}{e} & \text {for}\: F = 1 \\\tilde {\infty } e^{2} f \left (e^{- \frac {e}{b c}}\right )^{a c} \left (e^{- \frac {e}{b c}}\right )^{b c x} \sinh {\left (d + e x \right )} + \tilde {\infty } e^{2} f \left (e^{- \frac {e}{b c}}\right )^{a c} \left (e^{- \frac {e}{b c}}\right )^{b c x} \cosh {\left (d + e x \right )} & \text {for}\: F = e^{- \frac {e}{b c}} \\\tilde {\infty } e^{2} f \left (e^{\frac {e}{b c}}\right )^{a c} \left (e^{\frac {e}{b c}}\right )^{b c x} \sinh {\left (d + e x \right )} + \tilde {\infty } e^{2} f \left (e^{\frac {e}{b c}}\right )^{a c} \left (e^{\frac {e}{b c}}\right )^{b c x} \cosh {\left (d + e x \right )} & \text {for}\: F = e^{\frac {e}{b c}} \\F^{a c} \left (f x + \frac {f \sinh {\left (d + e x \right )}}{e}\right ) & \text {for}\: b = 0 \\f x + \frac {f \sinh {\left (d + e x \right )}}{e} & \text {for}\: c = 0 \\\frac {F^{a c} F^{b c x} b^{2} c^{2} f \log {\relax (F )}^{2} \cosh {\left (d + e x \right )}}{b^{3} c^{3} \log {\relax (F )}^{3} - b c e^{2} \log {\relax (F )}} + \frac {F^{a c} F^{b c x} b^{2} c^{2} f \log {\relax (F )}^{2}}{b^{3} c^{3} \log {\relax (F )}^{3} - b c e^{2} \log {\relax (F )}} - \frac {F^{a c} F^{b c x} b c e f \log {\relax (F )} \sinh {\left (d + e x \right )}}{b^{3} c^{3} \log {\relax (F )}^{3} - b c e^{2} \log {\relax (F )}} - \frac {F^{a c} F^{b c x} e^{2} f}{b^{3} c^{3} \log {\relax (F )}^{3} - b c e^{2} \log {\relax (F )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((f*x + f*sinh(d + e*x)/e, Eq(F, 1)), (zoo*e**2*f*exp(-e/(b*c))**(a*c)*exp(-e/(b*c))**(b*c*x)*sinh(d
+ e*x) + zoo*e**2*f*exp(-e/(b*c))**(a*c)*exp(-e/(b*c))**(b*c*x)*cosh(d + e*x), Eq(F, exp(-e/(b*c)))), (zoo*e**
2*f*exp(e/(b*c))**(a*c)*exp(e/(b*c))**(b*c*x)*sinh(d + e*x) + zoo*e**2*f*exp(e/(b*c))**(a*c)*exp(e/(b*c))**(b*
c*x)*cosh(d + e*x), Eq(F, exp(e/(b*c)))), (F**(a*c)*(f*x + f*sinh(d + e*x)/e), Eq(b, 0)), (f*x + f*sinh(d + e*
x)/e, Eq(c, 0)), (F**(a*c)*F**(b*c*x)*b**2*c**2*f*log(F)**2*cosh(d + e*x)/(b**3*c**3*log(F)**3 - b*c*e**2*log(
F)) + F**(a*c)*F**(b*c*x)*b**2*c**2*f*log(F)**2/(b**3*c**3*log(F)**3 - b*c*e**2*log(F)) - F**(a*c)*F**(b*c*x)*
b*c*e*f*log(F)*sinh(d + e*x)/(b**3*c**3*log(F)**3 - b*c*e**2*log(F)) - F**(a*c)*F**(b*c*x)*e**2*f/(b**3*c**3*l
og(F)**3 - b*c*e**2*log(F)), True))

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