### 3.92 $$\int \cosh (c+d x) (a+b \tanh ^2(c+d x))^2 \, dx$$

Optimal. Leaf size=60 $\frac{(a+b)^2 \sinh (c+d x)}{d}-\frac{b (4 a+3 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^2 \tanh (c+d x) \text{sech}(c+d x)}{2 d}$

[Out]

-(b*(4*a + 3*b)*ArcTan[Sinh[c + d*x]])/(2*d) + ((a + b)^2*Sinh[c + d*x])/d + (b^2*Sech[c + d*x]*Tanh[c + d*x])
/(2*d)

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Rubi [A]  time = 0.0824405, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.19, Rules used = {3676, 390, 385, 203} $\frac{(a+b)^2 \sinh (c+d x)}{d}-\frac{b (4 a+3 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^2 \tanh (c+d x) \text{sech}(c+d x)}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(b*(4*a + 3*b)*ArcTan[Sinh[c + d*x]])/(2*d) + ((a + b)^2*Sinh[c + d*x])/d + (b^2*Sech[c + d*x]*Tanh[c + d*x])
/(2*d)

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.178763, size = 54, normalized size = 0.9 $\frac{2 (a+b)^2 \sinh (c+d x)+b \left (b \tanh (c+d x) \text{sech}(c+d x)-(4 a+3 b) \tan ^{-1}(\sinh (c+d x))\right )}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(a + b)^2*Sinh[c + d*x] + b*(-((4*a + 3*b)*ArcTan[Sinh[c + d*x]]) + b*Sech[c + d*x]*Tanh[c + d*x]))/(2*d)

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Maple [B]  time = 0.043, size = 122, normalized size = 2. \begin{align*}{\frac{{a}^{2}\sinh \left ( dx+c \right ) }{d}}+2\,{\frac{ab\sinh \left ( dx+c \right ) }{d}}-4\,{\frac{ab\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{b}^{2}\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{b}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}-3\,{\frac{{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^2*sinh(d*x+c)+2*a*b*sinh(d*x+c)/d-4/d*a*b*arctan(exp(d*x+c))+1/d*b^2*sinh(d*x+c)^3/cosh(d*x+c)^2+3/d*b^2
*sinh(d*x+c)/cosh(d*x+c)^2-3/2/d*b^2*sech(d*x+c)*tanh(d*x+c)-3/d*b^2*arctan(exp(d*x+c))

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Maxima [B]  time = 1.74926, size = 205, normalized size = 3.42 \begin{align*} \frac{1}{2} \, b^{2}{\left (\frac{6 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{e^{\left (-d x - c\right )}}{d} + \frac{4 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + 1}{d{\left (e^{\left (-d x - c\right )} + 2 \, e^{\left (-3 \, d x - 3 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )}\right )}}\right )} + a b{\left (\frac{4 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{e^{\left (d x + c\right )}}{d} - \frac{e^{\left (-d x - c\right )}}{d}\right )} + \frac{a^{2} \sinh \left (d x + c\right )}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*(6*arctan(e^(-d*x - c))/d - e^(-d*x - c)/d + (4*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) + 1)/(d*(e^(-d*x -
c) + 2*e^(-3*d*x - 3*c) + e^(-5*d*x - 5*c)))) + a*b*(4*arctan(e^(-d*x - c))/d + e^(d*x + c)/d - e^(-d*x - c)/
d) + a^2*sinh(d*x + c)/d

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Fricas [B]  time = 1.99283, size = 1967, normalized size = 32.78 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b
+ b^2)*sinh(d*x + c)^6 + (a^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^4 + (15*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2
+ 2*a*b + 3*b^2)*sinh(d*x + c)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 + 2*a*b + 3*b^2)*cosh(d*x
+ c))*sinh(d*x + c)^3 - (a^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^2 + (15*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 6*(a
^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^2 - a^2 - 2*a*b - 3*b^2)*sinh(d*x + c)^2 - a^2 - 2*a*b - b^2 - 2*((4*a*b + 3
*b^2)*cosh(d*x + c)^5 + 5*(4*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 + (4*a*b + 3*b^2)*sinh(d*x + c)^5 + 2*
(4*a*b + 3*b^2)*cosh(d*x + c)^3 + 2*(5*(4*a*b + 3*b^2)*cosh(d*x + c)^2 + 4*a*b + 3*b^2)*sinh(d*x + c)^3 + 2*(5
*(4*a*b + 3*b^2)*cosh(d*x + c)^3 + 3*(4*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + (4*a*b + 3*b^2)*cosh(d*x
+ c) + (5*(4*a*b + 3*b^2)*cosh(d*x + c)^4 + 6*(4*a*b + 3*b^2)*cosh(d*x + c)^2 + 4*a*b + 3*b^2)*sinh(d*x + c))
*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 2*(a^2 + 2*a*b + 3*b^2)*co
sh(d*x + c)^3 - (a^2 + 2*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sin
h(d*x + c)^4 + d*sinh(d*x + c)^5 + 2*d*cosh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + 2*(5*d*
cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + d*cosh(d*x + c) + (5*d*cosh(d*x + c)^4 + 6*d*cosh(d*x +
c)^2 + d)*sinh(d*x + c))

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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 )^{2} \cosh{\left (c + d x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.66563, size = 185, normalized size = 3.08 \begin{align*} -\frac{2 \,{\left (4 \, a b e^{c} + 3 \, b^{2} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )} +{\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-d x - c\right )} -{\left (a^{2} e^{\left (d x + 8 \, c\right )} + 2 \, a b e^{\left (d x + 8 \, c\right )} + b^{2} e^{\left (d x + 8 \, c\right )}\right )} e^{\left (-7 \, c\right )} - \frac{2 \,{\left (b^{2} e^{\left (3 \, d x + 3 \, c\right )} - b^{2} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{2 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(4*a*b*e^c + 3*b^2*e^c)*arctan(e^(d*x + c))*e^(-c) + (a^2 + 2*a*b + b^2)*e^(-d*x - c) - (a^2*e^(d*x +
8*c) + 2*a*b*e^(d*x + 8*c) + b^2*e^(d*x + 8*c))*e^(-7*c) - 2*(b^2*e^(3*d*x + 3*c) - b^2*e^(d*x + c))/(e^(2*d*x
+ 2*c) + 1)^2)/d