3.81 \(\int \cosh ^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=63 \[ \frac{(a+b) \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{(3 a-b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{1}{8} x (3 a-b) \]

[Out]

((3*a - b)*x)/8 + ((3*a - b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + ((a + b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(4*d
)

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Rubi [A]  time = 0.0496557, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3675, 385, 199, 206} \[ \frac{(a+b) \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{(3 a-b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{1}{8} x (3 a-b) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

((3*a - b)*x)/8 + ((3*a - b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + ((a + b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(4*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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 ^4(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 )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac{(3 a-b) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{(3 a-b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac{(3 a-b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{1}{8} (3 a-b) x+\frac{(3 a-b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.162677, size = 44, normalized size = 0.7 \[ \frac{(a+b) \sinh (4 (c+d x))+12 a (c+d x)+8 a \sinh (2 (c+d x))-4 b d x}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

(-4*b*d*x + 12*a*(c + d*x) + 8*a*Sinh[2*(c + d*x)] + (a + b)*Sinh[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.039, size = 82, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( b \left ({\frac{\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{8}}-{\frac{dx}{8}}-{\frac{c}{8}} \right ) +a \left ( \left ({\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{3\,\cosh \left ( dx+c \right ) }{8}} \right ) \sinh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2),x)

[Out]

1/d*(b*(1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8*d*x-1/8*c)+a*((1/4*cosh(d*x+c)^3+3/8*cos
h(d*x+c))*sinh(d*x+c)+3/8*d*x+3/8*c))

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Maxima [A]  time = 1.14363, size = 140, normalized size = 2.22 \begin{align*} \frac{1}{64} \, a{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac{1}{64} \, b{\left (\frac{8 \,{\left (d x + c\right )}}{d} - \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

1/64*a*(24*x + e^(4*d*x + 4*c)/d + 8*e^(2*d*x + 2*c)/d - 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) - 1/64*b*(
8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d)

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Fricas [A]  time = 1.82498, size = 169, normalized size = 2.68 \begin{align*} \frac{{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (3 \, a - b\right )} d x +{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + 4 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

1/8*((a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a - b)*d*x + ((a + b)*cosh(d*x + c)^3 + 4*a*cosh(d*x + c))*sin
h(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**4*(a+b*tanh(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.27921, size = 144, normalized size = 2.29 \begin{align*} \frac{8 \,{\left (3 \, a - b\right )} d x -{\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} - 6 \, b e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} +{\left (a e^{\left (4 \, d x + 12 \, c\right )} + b e^{\left (4 \, d x + 12 \, c\right )} + 8 \, a e^{\left (2 \, d x + 10 \, c\right )}\right )} e^{\left (-8 \, c\right )}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

1/64*(8*(3*a - b)*d*x - (18*a*e^(4*d*x + 4*c) - 6*b*e^(4*d*x + 4*c) + 8*a*e^(2*d*x + 2*c) + a + b)*e^(-4*d*x -
4*c) + (a*e^(4*d*x + 12*c) + b*e^(4*d*x + 12*c) + 8*a*e^(2*d*x + 10*c))*e^(-8*c))/d