3.83 \(\int \cosh ^2(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=33 \[ \frac{(a+b) \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac{1}{2} x (a-b) \]

[Out]

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

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Rubi [A]  time = 0.0390378, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3675, 385, 206} \[ \frac{(a+b) \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac{1}{2} x (a-b) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]^2*(a + b*Tanh[c + d*x]^2),x]

[Out]

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

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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])

Rubi steps

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

Mathematica [A]  time = 0.056917, size = 32, normalized size = 0.97 \[ \frac{2 (a-b) (c+d x)+(a+b) \sinh (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]^2*(a + b*Tanh[c + d*x]^2),x]

[Out]

(2*(a - b)*(c + d*x) + (a + b)*Sinh[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.033, size = 54, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ( b \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) +a \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2),x)

[Out]

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

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Maxima [B]  time = 1.11015, size = 93, normalized size = 2.82 \begin{align*} \frac{1}{8} \, a{\left (4 \, x + \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{8} \, b{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) - 1/8*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d)

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Fricas [A]  time = 1.86134, size = 80, normalized size = 2.42 \begin{align*} \frac{{\left (a - b\right )} d x +{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \cosh ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**2*(a+b*tanh(d*x+c)**2),x)

[Out]

Integral((a + b*tanh(c + d*x)**2)*cosh(c + d*x)**2, x)

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Giac [B]  time = 1.21928, size = 109, normalized size = 3.3 \begin{align*} \frac{4 \,{\left (a - b\right )} d x -{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a e^{\left (2 \, d x + 4 \, c\right )} + b e^{\left (2 \, d x + 4 \, c\right )}\right )} e^{\left (-2 \, c\right )}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(a - b)*d*x - (2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)*e^(-2*d*x - 2*c) + (a*e^(2*d*x + 4*c)
+ b*e^(2*d*x + 4*c))*e^(-2*c))/d