\(\int F^{c (a+b x)} \sinh (d+e x) \, dx\) [324]

Optimal result

Integrand size = 16, antiderivative size = 75 \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=\frac {e F^{c (a+b x)} \cosh (d+e x)}{e^2-b^2 c^2 \log ^2(F)}-\frac {b c F^{c (a+b x)} \log (F) \sinh (d+e x)}{e^2-b^2 c^2 \log ^2(F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {5582} \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=\frac {e \cosh (d+e x) F^{c (a+b x)}}{e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \sinh (d+e x) F^{c (a+b x)}}{e^2-b^2 c^2 \log ^2(F)} \]

[In]

Int[F^(c*(a + b*x))*Sinh[d + e*x],x]

[Out]

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

Rule 5582

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*} \text {integral}& = \frac {e F^{c (a+b x)} \cosh (d+e x)}{e^2-b^2 c^2 \log ^2(F)}-\frac {b c F^{c (a+b x)} \log (F) \sinh (d+e x)}{e^2-b^2 c^2 \log ^2(F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=\frac {F^{c (a+b x)} (e \cosh (d+e x)-b c \log (F) \sinh (d+e x))}{(e-b c \log (F)) (e+b c \log (F))} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(\frac {F^{c \left (b x +a \right )} \left (\sinh \left (e x +d \right ) \ln \left (F \right ) b c -e \cosh \left (e x +d \right )\right )}{b^{2} c^{2} \ln \left (F \right )^{2}-e^{2}}\)	\(51\)
risch	\(\frac {\left (\ln \left (F \right ) b c \,{\mathrm e}^{2 e x +2 d}-b c \ln \left (F \right )-e \,{\mathrm e}^{2 e x +2 d}-e \right ) {\mathrm e}^{-e x -d} F^{c \left (b x +a \right )}}{2 \left (b c \ln \left (F \right )-e \right ) \left (e +b c \ln \left (F \right )\right )}\)	\(77\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (77) = 154\).

Time = 0.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.25 \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=-\frac {{\left (e \cosh \left (e x + d\right )^{2} - {\left (b c \log \left (F\right ) - e\right )} \sinh \left (e x + d\right )^{2} - {\left (b c \cosh \left (e x + d\right )^{2} - b c\right )} \log \left (F\right ) - 2 \, {\left (b c \cosh \left (e x + d\right ) \log \left (F\right ) - e \cosh \left (e x + d\right )\right )} \sinh \left (e x + d\right ) + e\right )} \cosh \left ({\left (b c x + a c\right )} \log \left (F\right )\right ) + {\left (e \cosh \left (e x + d\right )^{2} - {\left (b c \log \left (F\right ) - e\right )} \sinh \left (e x + d\right )^{2} - {\left (b c \cosh \left (e x + d\right )^{2} - b c\right )} \log \left (F\right ) - 2 \, {\left (b c \cosh \left (e x + d\right ) \log \left (F\right ) - e \cosh \left (e x + d\right )\right )} \sinh \left (e x + d\right ) + e\right )} \sinh \left ({\left (b c x + a c\right )} \log \left (F\right )\right )}{2 \, {\left (b^{2} c^{2} \cosh \left (e x + d\right ) \log \left (F\right )^{2} - e^{2} \cosh \left (e x + d\right ) + {\left (b^{2} c^{2} \log \left (F\right )^{2} - e^{2}\right )} \sinh \left (e x + d\right )\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (68) = 136\).

Time = 0.58 (sec) , antiderivative size = 323, normalized size of antiderivative = 4.31 \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=\begin {cases} x \sinh {\left (d \right )} & \text {for}\: F = 1 \wedge e = 0 \\F^{a c} x \sinh {\left (d \right )} & \text {for}\: b = 0 \wedge e = 0 \\x \sinh {\left (d \right )} & \text {for}\: c = 0 \wedge e = 0 \\- \frac {F^{a c + b c x} x \sinh {\left (b c x \log {\left (F \right )} - d \right )}}{2} + \frac {F^{a c + b c x} x \cosh {\left (b c x \log {\left (F \right )} - d \right )}}{2} + \frac {F^{a c + b c x} \sinh {\left (b c x \log {\left (F \right )} - d \right )}}{2 b c \log {\left (F \right )}} - \frac {F^{a c + b c x} \cosh {\left (b c x \log {\left (F \right )} - d \right )}}{b c \log {\left (F \right )}} & \text {for}\: e = - b c \log {\left (F \right )} \\\frac {F^{a c + b c x} x \sinh {\left (b c x \log {\left (F \right )} + d \right )}}{2} - \frac {F^{a c + b c x} x \cosh {\left (b c x \log {\left (F \right )} + d \right )}}{2} - \frac {F^{a c + b c x} \sinh {\left (b c x \log {\left (F \right )} + d \right )}}{2 b c \log {\left (F \right )}} + \frac {F^{a c + b c x} \cosh {\left (b c x \log {\left (F \right )} + d \right )}}{b c \log {\left (F \right )}} & \text {for}\: e = b c \log {\left (F \right )} \\\frac {F^{a c + b c x} b c \log {\left (F \right )} \sinh {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} - e^{2}} - \frac {F^{a c + b c x} e \cosh {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} - e^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*sinh(d), Eq(F, 1) & Eq(e, 0)), (F**(a*c)*x*sinh(d), Eq(b, 0) & Eq(e, 0)), (x*sinh(d), Eq(c, 0) &
Eq(e, 0)), (-F**(a*c + b*c*x)*x*sinh(b*c*x*log(F) - d)/2 + F**(a*c + b*c*x)*x*cosh(b*c*x*log(F) - d)/2 + F**(a
*c + b*c*x)*sinh(b*c*x*log(F) - d)/(2*b*c*log(F)) - F**(a*c + b*c*x)*cosh(b*c*x*log(F) - d)/(b*c*log(F)), Eq(e
, -b*c*log(F))), (F**(a*c + b*c*x)*x*sinh(b*c*x*log(F) + d)/2 - F**(a*c + b*c*x)*x*cosh(b*c*x*log(F) + d)/2 -
F**(a*c + b*c*x)*sinh(b*c*x*log(F) + d)/(2*b*c*log(F)) + F**(a*c + b*c*x)*cosh(b*c*x*log(F) + d)/(b*c*log(F)),
Eq(e, b*c*log(F))), (F**(a*c + b*c*x)*b*c*log(F)*sinh(d + e*x)/(b**2*c**2*log(F)**2 - e**2) - F**(a*c + b*c*x
)*e*cosh(d + e*x)/(b**2*c**2*log(F)**2 - e**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{2 \, {\left (b c \log \left (F\right ) + e\right )}} - \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - e x\right )}}{2 \, {\left (b c e^{d} \log \left (F\right ) - e e^{d}\right )}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 598, normalized size of antiderivative = 7.97 \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int F^{c (a+b x)} \sinh (d+e x) \, dx=\frac {F^{a\,c+b\,c\,x}\,{\mathrm {e}}^{-d-e\,x}\,\left (e+e\,{\mathrm {e}}^{2\,d+2\,e\,x}+b\,c\,\ln \left (F\right )-b\,c\,{\mathrm {e}}^{2\,d+2\,e\,x}\,\ln \left (F\right )\right )}{2\,\left (e^2-b^2\,c^2\,{\ln \left (F\right )}^2\right )} \]

[In]

int(F^(c*(a + b*x))*sinh(d + e*x),x)

[Out]

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