3.75 \(\int \sinh ^2(a+b x) \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=40 \[ \frac {3 \tanh (a+b x)}{2 b}+\frac {\sinh ^2(a+b x) \tanh (a+b x)}{2 b}-\frac {3 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2591, 288, 321, 206} \[ \frac {3 \tanh (a+b x)}{2 b}+\frac {\sinh ^2(a+b x) \tanh (a+b x)}{2 b}-\frac {3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^2*Tanh[a + b*x]^2,x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 2591

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 31, normalized size = 0.78 \[ \frac {-6 (a+b x)+\sinh (2 (a+b x))+4 \tanh (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^2*Tanh[a + b*x]^2,x]

[Out]

(-6*(a + b*x) + Sinh[2*(a + b*x)] + 4*Tanh[a + b*x])/(4*b)

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fricas [A]  time = 0.46, size = 54, normalized size = 1.35 \[ \frac {\sinh \left (b x + a\right )^{3} - 4 \, {\left (3 \, b x + 2\right )} \cosh \left (b x + a\right ) + 3 \, {\left (\cosh \left (b x + a\right )^{2} + 3\right )} \sinh \left (b x + a\right )}{8 \, b \cosh \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2*tanh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*(sinh(b*x + a)^3 - 4*(3*b*x + 2)*cosh(b*x + a) + 3*(cosh(b*x + a)^2 + 3)*sinh(b*x + a))/(b*cosh(b*x + a))

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giac [A]  time = 0.15, size = 68, normalized size = 1.70 \[ -\frac {12 \, b x - \frac {{\left (3 \, e^{\left (4 \, b x + 4 \, a\right )} - 14 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, a\right )}}{e^{\left (2 \, b x\right )} + e^{\left (4 \, b x + 2 \, a\right )}} - e^{\left (2 \, b x + 2 \, a\right )}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2*tanh(b*x+a)^2,x, algorithm="giac")

[Out]

-1/8*(12*b*x - (3*e^(4*b*x + 4*a) - 14*e^(2*b*x + 2*a) - 1)*e^(-2*a)/(e^(2*b*x) + e^(4*b*x + 2*a)) - e^(2*b*x
+ 2*a))/b

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maple [A]  time = 0.12, size = 39, normalized size = 0.98 \[ \frac {\frac {\sinh ^{3}\left (b x +a \right )}{2 \cosh \left (b x +a \right )}-\frac {3 b x}{2}-\frac {3 a}{2}+\frac {3 \tanh \left (b x +a \right )}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^2*tanh(b*x+a)^2,x)

[Out]

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

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maxima [A]  time = 0.42, size = 64, normalized size = 1.60 \[ -\frac {3 \, {\left (b x + a\right )}}{2 \, b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, b} + \frac {17 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1}{8 \, b {\left (e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2*tanh(b*x+a)^2,x, algorithm="maxima")

[Out]

-3/2*(b*x + a)/b - 1/8*e^(-2*b*x - 2*a)/b + 1/8*(17*e^(-2*b*x - 2*a) + 1)/(b*(e^(-2*b*x - 2*a) + e^(-4*b*x - 4
*a)))

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mupad [B]  time = 1.47, size = 50, normalized size = 1.25 \[ \frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{8\,b}-\frac {2}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}}{8\,b}-\frac {3\,x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^2*tanh(a + b*x)^2,x)

[Out]

exp(2*a + 2*b*x)/(8*b) - 2/(b*(exp(2*a + 2*b*x) + 1)) - exp(- 2*a - 2*b*x)/(8*b) - (3*x)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh ^{2}{\left (a + b x \right )} \tanh ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)**2*tanh(b*x+a)**2,x)

[Out]

Integral(sinh(a + b*x)**2*tanh(a + b*x)**2, x)

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