3.4.31 \(\int e^{c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \, dx\) [331]

Optimal. Leaf size=74 \[ \frac {e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{4 b c}-\frac {1}{2} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)} \]

[Out]

1/4*exp(2*c*(b*x+a))*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)/b/c-1/2*x*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(
1/2)

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Rubi [A]
time = 0.08, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6852, 2320, 12, 14} \begin {gather*} \frac {e^{2 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{4 b c}-\frac {1}{2} x \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*Sqrt[Sinh[a*c + b*c*x]^2],x]

[Out]

(E^(2*c*(a + b*x))*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/(4*b*c) - (x*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c +
b*c*x]^2])/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6852

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m)^FracPart[p]/v^(m*FracPart[p])), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int e^{c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \, dx &=\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \sinh (a c+b c x) \, dx\\ &=\frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {-1+x^2}{2 x} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {-1+x^2}{x} \, dx,x,e^{c (a+b x)}\right )}{2 b c}\\ &=\frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \left (-\frac {1}{x}+x\right ) \, dx,x,e^{c (a+b x)}\right )}{2 b c}\\ &=\frac {e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{4 b c}-\frac {1}{2} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 48, normalized size = 0.65 \begin {gather*} \frac {\left (e^{2 c (a+b x)}-2 b c x\right ) \text {csch}(c (a+b x)) \sqrt {\sinh ^2(c (a+b x))}}{4 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))*Sqrt[Sinh[a*c + b*c*x]^2],x]

[Out]

((E^(2*c*(a + b*x)) - 2*b*c*x)*Csch[c*(a + b*x)]*Sqrt[Sinh[c*(a + b*x)]^2])/(4*b*c)

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Maple [A]
time = 1.88, size = 89, normalized size = 1.20

method	result	size
default	\(\frac {\frac {\sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\, \cosh \left (c \left (b x +a \right )\right )}{2}-\frac {\ln \left (\cosh \left (c \left (b x +a \right )\right )+\sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\right )}{2}+\frac {\sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\, \left (\cosh ^{2}\left (c \left (b x +a \right )\right )\right )}{2 \sinh \left (c \left (b x +a \right )\right )}}{b c}\)	\(89\)
risch	\(-\frac {x \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{c \left (b x +a \right )}}{2 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{3 c \left (b x +a \right )}}{4 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(1/2*(sinh(c*(b*x+a))^2)^(1/2)*cosh(c*(b*x+a))-1/2*ln(cosh(c*(b*x+a))+(sinh(c*(b*x+a))^2)^(1/2))+1/2*(sinh(c*(
b*x+a))^2)^(1/2)*cosh(c*(b*x+a))^2/sinh(c*(b*x+a)))/b/c

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Maxima [A]
time = 0.50, size = 36, normalized size = 0.49 \begin {gather*} -\frac {b c x + a c}{2 \, b c} + \frac {e^{\left (2 \, b c x + 2 \, a c\right )}}{4 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*(b*c*x + a*c)/(b*c) + 1/4*e^(2*b*c*x + 2*a*c)/(b*c)

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Fricas [A]
time = 0.35, size = 66, normalized size = 0.89 \begin {gather*} -\frac {{\left (2 \, b c x - 1\right )} \cosh \left (b c x + a c\right ) - {\left (2 \, b c x + 1\right )} \sinh \left (b c x + a c\right )}{4 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/4*((2*b*c*x - 1)*cosh(b*c*x + a*c) - (2*b*c*x + 1)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sinh(b*c
*x + a*c))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (68) = 136\).
time = 2.83, size = 275, normalized size = 3.72 \begin {gather*} \begin {cases} \frac {\sqrt {\sinh ^{2}{\left (b c x + \log {\left (- e^{- b c x} \right )} \right )}} \log {\left (- e^{- b c x} \right )}}{b c} & \text {for}\: a = \frac {\log {\left (- e^{- b c x} \right )}}{c} \\- \frac {\sqrt {\sinh ^{2}{\left (b c x + \log {\left (e^{- b c x} \right )} \right )}} \log {\left (e^{- b c x} \right )}}{b c} & \text {for}\: a = \frac {\log {\left (e^{- b c x} \right )}}{c} \\0 & \text {for}\: c = 0 \\x \sqrt {\sinh ^{2}{\left (a c \right )}} e^{a c} & \text {for}\: b = 0 \\\frac {x \sqrt {\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x}}{2} - \frac {x \sqrt {\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x} \cosh {\left (a c + b c x \right )}}{2 \sinh {\left (a c + b c x \right )}} - \frac {\sqrt {\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x}}{2 b c} + \frac {\sqrt {\sinh ^{2}{\left (a c + b c x \right )}} e^{a c} e^{b c x} \cosh {\left (a c + b c x \right )}}{b c \sinh {\left (a c + b c x \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)**2)**(1/2),x)

[Out]

Piecewise((sqrt(sinh(b*c*x + log(-exp(-b*c*x)))**2)*log(-exp(-b*c*x))/(b*c), Eq(a, log(-exp(-b*c*x))/c)), (-sq
rt(sinh(b*c*x + log(exp(-b*c*x)))**2)*log(exp(-b*c*x))/(b*c), Eq(a, log(exp(-b*c*x))/c)), (0, Eq(c, 0)), (x*sq
rt(sinh(a*c)**2)*exp(a*c), Eq(b, 0)), (x*sqrt(sinh(a*c + b*c*x)**2)*exp(a*c)*exp(b*c*x)/2 - x*sqrt(sinh(a*c +
b*c*x)**2)*exp(a*c)*exp(b*c*x)*cosh(a*c + b*c*x)/(2*sinh(a*c + b*c*x)) - sqrt(sinh(a*c + b*c*x)**2)*exp(a*c)*e
xp(b*c*x)/(2*b*c) + sqrt(sinh(a*c + b*c*x)**2)*exp(a*c)*exp(b*c*x)*cosh(a*c + b*c*x)/(b*c*sinh(a*c + b*c*x)),
True))

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Giac [A]
time = 0.43, size = 71, normalized size = 0.96 \begin {gather*} -\frac {1}{2} \, x \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \frac {e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )}{4 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(1/2),x, algorithm="giac")

[Out]

-1/2*x*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) + 1/4*e^(2*b*c*x + 2*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c
))/(b*c)

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Mupad [B]
time = 0.63, size = 77, normalized size = 1.04 \begin {gather*} -\frac {\left (x\,{\mathrm {e}}^{a\,c+b\,c\,x}-\frac {{\mathrm {e}}^{3\,a\,c+3\,b\,c\,x}}{2\,b\,c}\right )\,\sqrt {{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}-\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}}{{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(a + b*x))*(sinh(a*c + b*c*x)^2)^(1/2),x)

[Out]

-((x*exp(a*c + b*c*x) - exp(3*a*c + 3*b*c*x)/(2*b*c))*((exp(a*c + b*c*x)/2 - exp(- a*c - b*c*x)/2)^2)^(1/2))/(
exp(2*a*c + 2*b*c*x) - 1)

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