3.894 $$\int F^{c (a+b x)} (f+i f \sinh (d+e x)) \, dx$$

Optimal. Leaf size=106 $-\frac{i b c f \log (F) \sinh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac{i e f \cosh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac{f F^{a c+b c x}}{b c \log (F)}$

[Out]

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

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Rubi [A]  time = 0.179155, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {6741, 12, 6742, 2194, 5474} $-\frac{i b c f \log (F) \sinh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac{i e f \cosh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac{f F^{a c+b c x}}{b c \log (F)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6741

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

Rule 12

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

Rule 6742

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

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 5474

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

Rubi steps

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

Mathematica [A]  time = 0.612028, size = 93, normalized size = 0.88 $\frac{f F^{c (a+b x)} \left (i b^2 c^2 \log ^2(F) \sinh (d+e x)+b^2 c^2 \log ^2(F)-i b c e \log (F) \cosh (d+e x)-e^2\right )}{b c \log (F) (b c \log (F)-e) (b c \log (F)+e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 141, normalized size = 1.3 \begin{align*}{\frac{f \left ( -i \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{2}{{\rm e}^{2\,ex+2\,d}}+i \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{2}-2\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{c}^{2}{{\rm e}^{ex+d}}+i\ln \left ( F \right ) bce{{\rm e}^{2\,ex+2\,d}}+i\ln \left ( F \right ) bce+2\,{e}^{2}{{\rm e}^{ex+d}} \right ){{\rm e}^{-ex-d}}{F}^{c \left ( bx+a \right ) }}{2\,bc\ln \left ( F \right ) \left ( e-bc\ln \left ( F \right ) \right ) \left ( e+bc\ln \left ( F \right ) \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03752, size = 119, normalized size = 1.12 \begin{align*} \frac{1}{2} i \, f{\left (\frac{F^{a c} e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{b c \log \left (F\right ) + e} - \frac{F^{a c} e^{\left (b c x \log \left (F\right ) - e x\right )}}{b c e^{d} \log \left (F\right ) - e e^{d}}\right )} + \frac{F^{b c x + a c} f}{b c \log \left (F\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*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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Fricas [A]  time = 1.33784, size = 317, normalized size = 2.99 \begin{align*} -\frac{{\left (2 \, e^{2} f e^{\left (e x + d\right )} -{\left (i \, b^{2} c^{2} f e^{\left (2 \, e x + 2 \, d\right )} + 2 \, b^{2} c^{2} f e^{\left (e x + d\right )} - i \, b^{2} c^{2} f\right )} \log \left (F\right )^{2} -{\left (-i \, b c e f e^{\left (2 \, e x + 2 \, d\right )} - i \, b c e f\right )} \log \left (F\right )\right )} F^{b c x + a c}}{2 \,{\left (b^{3} c^{3} e^{\left (e x + d\right )} \log \left (F\right )^{3} - b c e^{2} e^{\left (e x + d\right )} \log \left (F\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*e^2*f*e^(e*x + d) - (I*b^2*c^2*f*e^(2*e*x + 2*d) + 2*b^2*c^2*f*e^(e*x + d) - I*b^2*c^2*f)*log(F)^2 - (
-I*b*c*e*f*e^(2*e*x + 2*d) - I*b*c*e*f)*log(F))*F^(b*c*x + a*c)/(b^3*c^3*e^(e*x + d)*log(F)^3 - b*c*e^2*e^(e*x
+ d)*log(F))

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Sympy [A]  time = 17.4982, size = 400, normalized size = 3.77 \begin{align*} \begin{cases} f x + \frac{i f \cosh{\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{i f \cosh{\left (d + e x \right )}}{e}\right ) & \text{for}\: b = 0 \\f x + \frac{i f \cosh{\left (d + e x \right )}}{e} & \text{for}\: c = 0 \\\frac{i F^{a c} F^{b c x} b^{2} c^{2} f \log{\left (F \right )}^{2} \sinh{\left (d + e x \right )}}{b^{3} c^{3} \log{\left (F \right )}^{3} - b c e^{2} \log{\left (F \right )}} + \frac{F^{a c} F^{b c x} b^{2} c^{2} f \log{\left (F \right )}^{2}}{b^{3} c^{3} \log{\left (F \right )}^{3} - b c e^{2} \log{\left (F \right )}} - \frac{i F^{a c} F^{b c x} b c e f \log{\left (F \right )} \cosh{\left (d + e x \right )}}{b^{3} c^{3} \log{\left (F \right )}^{3} - b c e^{2} \log{\left (F \right )}} - \frac{F^{a c} F^{b c x} e^{2} f}{b^{3} c^{3} \log{\left (F \right )}^{3} - b c e^{2} \log{\left (F \right )}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.25489, size = 1214, normalized size = 11.45 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(f+I*f*sinh(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))) - ((pi*b*c*sgn(F) - pi*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)/((pi*b*c*sgn(F) - pi*b*c)^2 + 4*(b*c*log(abs(F)) + e)^2) + 2*(b*c*log(abs(F)) + e)*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*(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) + ((pi*b*c*sgn(F) - pi*b*c)*f*c
os(-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) + 2*(b*c*log(abs(F)) - e)*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*(-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*p
i*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)