3.72 \(\int \sinh (a+b x) \tanh ^3(a+b x) \, dx\)

Optimal. Leaf size=49 \[ \frac {3 \sinh (a+b x)}{2 b}-\frac {3 \tan ^{-1}(\sinh (a+b x))}{2 b}-\frac {\sinh (a+b x) \tanh ^2(a+b x)}{2 b} \]

[Out]

-3/2*arctan(sinh(b*x+a))/b+3/2*sinh(b*x+a)/b-1/2*sinh(b*x+a)*tanh(b*x+a)^2/b

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2592, 288, 321, 203} \[ \frac {3 \sinh (a+b x)}{2 b}-\frac {3 \tan ^{-1}(\sinh (a+b x))}{2 b}-\frac {\sinh (a+b x) \tanh ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]*Tanh[a + b*x]^3,x]

[Out]

(-3*ArcTan[Sinh[a + b*x]])/(2*b) + (3*Sinh[a + b*x])/(2*b) - (Sinh[a + b*x]*Tanh[a + b*x]^2)/(2*b)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rubi steps

\begin {align*} \int \sinh (a+b x) \tanh ^3(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\sinh (a+b x)\right )}{b}\\ &=-\frac {\sinh (a+b x) \tanh ^2(a+b x)}{2 b}+\frac {3 \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\sinh (a+b x)\right )}{2 b}\\ &=\frac {3 \sinh (a+b x)}{2 b}-\frac {\sinh (a+b x) \tanh ^2(a+b x)}{2 b}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (a+b x)\right )}{2 b}\\ &=-\frac {3 \tan ^{-1}(\sinh (a+b x))}{2 b}+\frac {3 \sinh (a+b x)}{2 b}-\frac {\sinh (a+b x) \tanh ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 48, normalized size = 0.98 \[ \frac {\sinh (a+b x) \tanh ^2(a+b x)}{b}-\frac {3 \left (\tan ^{-1}(\sinh (a+b x))-\tanh (a+b x) \text {sech}(a+b x)\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]*Tanh[a + b*x]^3,x]

[Out]

(Sinh[a + b*x]*Tanh[a + b*x]^2)/b - (3*(ArcTan[Sinh[a + b*x]] - Sech[a + b*x]*Tanh[a + b*x]))/(2*b)

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fricas [B]  time = 0.42, size = 463, normalized size = 9.45 \[ \frac {\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{4} + 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 6 \, {\left (\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{5} + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 2 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + {\left (5 \, \cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right )\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 1}{2 \, {\left (b \cosh \left (b x + a\right )^{5} + 5 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + b \sinh \left (b x + a\right )^{5} + 2 \, b \cosh \left (b x + a\right )^{3} + 2 \, {\left (5 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{3} + 2 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + b \cosh \left (b x + a\right ) + {\left (5 \, b \cosh \left (b x + a\right )^{4} + 6 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/2*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 + 1)*sinh(b*x
+ a)^4 + 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 +
6*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 6*(cosh(b*x + a)^5 + 5*cosh(b*x + a)*sinh(b*x + a)^4 + sinh(b*x + a)^
5 + 2*(5*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + 2*cosh(b*x + a)^3 + 2*(5*cosh(b*x + a)^3 + 3*cosh(b*x + a))*si
nh(b*x + a)^2 + (5*cosh(b*x + a)^4 + 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + cosh(b*x + a))*arctan(cosh(b*x + a
) + sinh(b*x + a)) - 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^5 + 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)
- 1)/(b*cosh(b*x + a)^5 + 5*b*cosh(b*x + a)*sinh(b*x + a)^4 + b*sinh(b*x + a)^5 + 2*b*cosh(b*x + a)^3 + 2*(5*
b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^3 + 2*(5*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^2 + b*cosh(
b*x + a) + (5*b*cosh(b*x + a)^4 + 6*b*cosh(b*x + a)^2 + b)*sinh(b*x + a))

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giac [A]  time = 0.14, size = 65, normalized size = 1.33 \[ \frac {\frac {2 \, {\left (e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} - 6 \, \arctan \left (e^{\left (b x + a\right )}\right ) + e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)^3,x, algorithm="giac")

[Out]

1/2*(2*(e^(3*b*x + 3*a) - e^(b*x + a))/(e^(2*b*x + 2*a) + 1)^2 - 6*arctan(e^(b*x + a)) + e^(b*x + a) - e^(-b*x
- a))/b

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maple [A]  time = 0.31, size = 70, normalized size = 1.43 \[ \frac {\sinh ^{3}\left (b x +a \right )}{b \cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{b \cosh \left (b x +a \right )^{2}}-\frac {3 \,\mathrm {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2 b}-\frac {3 \arctan \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)*tanh(b*x+a)^3,x)

[Out]

1/b*sinh(b*x+a)^3/cosh(b*x+a)^2+3/b*sinh(b*x+a)/cosh(b*x+a)^2-3/2*sech(b*x+a)*tanh(b*x+a)/b-3*arctan(exp(b*x+a
))/b

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maxima [B]  time = 0.41, size = 91, normalized size = 1.86 \[ \frac {3 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac {e^{\left (-b x - a\right )}}{2 \, b} + \frac {4 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} + 1}{2 \, b {\left (e^{\left (-b x - a\right )} + 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)^3,x, algorithm="maxima")

[Out]

3*arctan(e^(-b*x - a))/b - 1/2*e^(-b*x - a)/b + 1/2*(4*e^(-2*b*x - 2*a) - e^(-4*b*x - 4*a) + 1)/(b*(e^(-b*x -
a) + 2*e^(-3*b*x - 3*a) + e^(-5*b*x - 5*a)))

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mupad [B]  time = 1.46, size = 107, normalized size = 2.18 \[ \frac {{\mathrm {e}}^{a+b\,x}}{2\,b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {{\mathrm {e}}^{-a-b\,x}}{2\,b}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)*tanh(a + b*x)^3,x)

[Out]

exp(a + b*x)/(2*b) - (3*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(b^2)^(1/2) - exp(- a - b*x)/(2*b) - (2*exp(a +
b*x))/(b*(2*exp(2*a + 2*b*x) + exp(4*a + 4*b*x) + 1)) + exp(a + b*x)/(b*(exp(2*a + 2*b*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\left (a + b x \right )} \tanh ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)**3,x)

[Out]

Integral(sinh(a + b*x)*tanh(a + b*x)**3, x)

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