### 3.107 $$\int \frac{\cosh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=77 $\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a+b)^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 d (a+b)}+\frac{x (a+3 b)}{2 (a+b)^2}$

[Out]

((a + 3*b)*x)/(2*(a + b)^2) + (b^(3/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a + b)^2*d) + (Cosh[
c + d*x]*Sinh[c + d*x])/(2*(a + b)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.102965, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {3675, 414, 522, 206, 205} $\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a+b)^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 d (a+b)}+\frac{x (a+3 b)}{2 (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((a + 3*b)*x)/(2*(a + b)^2) + (b^(3/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a + b)^2*d) + (Cosh[
c + d*x]*Sinh[c + d*x])/(2*(a + b)*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 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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 205

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

Rubi steps

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

Mathematica [A]  time = 0.147657, size = 77, normalized size = 1. $\frac{4 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )+2 \sqrt{a} (a+3 b) (c+d x)+\sqrt{a} (a+b) \sinh (2 (c+d x))}{4 \sqrt{a} d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.083, size = 608, normalized size = 7.9 \begin{align*} \text{result too large to display} \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/(2*b+2*a)/(tanh(1/2*d*x+1/2*c)+1)^2+2/d/(4*a+4*b)/(tanh(1/2*d*x+1/2*c)+1)+1/2/d/(a+b)^2*ln(tanh(1/2*d*x+1
/2*c)+1)*a+3/2/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)+1)*b-1/d*a*b^2/(a+b)^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2
*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d*b^2/(a+b)^2/((2*(b*(a+b))^
(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-1/d*b^3/(a+b)^2/(b*(a
+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/
2))-1/d*a*b^2/(a+b)^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*
(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/d*b^2/(a+b)^2/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)
/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/d*b^3/(a+b)^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arct
an(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+1/d/(2*b+2*a)/(tanh(1/2*d*x+1/2*c)-1)^2+2/d/(4*a
+4*b)/(tanh(1/2*d*x+1/2*c)-1)-1/2/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)-1)*a-3/2/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)-1
)*b

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.28545, size = 2520, normalized size = 32.73 \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)^2/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/8*(4*(a + 3*b)*d*x*cosh(d*x + c)^2 + (a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a
+ b)*sinh(d*x + c)^4 + 2*(2*(a + 3*b)*d*x + 3*(a + b)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^2
+ 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2)*sqrt(-b/a)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 +
4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh
(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((
a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^2
+ 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 - a*b)*sqrt(-b/a))/((a + b)*c
osh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2
+ 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))
*sinh(d*x + c) + a + b)) + 4*(2*(a + 3*b)*d*x*cosh(d*x + c) + (a + b)*cosh(d*x + c)^3)*sinh(d*x + c) - a - b)/
((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2 + 2*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c) + (a^2 + 2*a*b +
b^2)*d*sinh(d*x + c)^2), 1/8*(4*(a + 3*b)*d*x*cosh(d*x + c)^2 + (a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x +
c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(2*(a + 3*b)*d*x + 3*(a + b)*cosh(d*x + c)^2)*sinh(d*x + c)^
2 + 8*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2)*sqrt(b/a)*arctan(1/2*((a + b)*
cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(b/a)/b) + 4*(2
*(a + 3*b)*d*x*cosh(d*x + c) + (a + b)*cosh(d*x + c)^3)*sinh(d*x + c) - a - b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x
+ c)^2 + 2*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c) + (a^2 + 2*a*b + b^2)*d*sinh(d*x + c)^2)]

________________________________________________________________________________________

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

________________________________________________________________________________________

Giac [B]  time = 1.85299, size = 232, normalized size = 3.01 \begin{align*} \frac{\frac{4 \,{\left (a + 3 \, b\right )} d x}{a^{2} + 2 \, a b + b^{2}} + \frac{8 \, b^{2} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b}} - \frac{{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-2 \, d x\right )}}{a^{2} e^{\left (2 \, c\right )} + 2 \, a b e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}} + \frac{e^{\left (2 \, d x + 8 \, c\right )}}{a e^{\left (6 \, c\right )} + b e^{\left (6 \, 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 + 3*b)*d*x/(a^2 + 2*a*b + b^2) + 8*b^2*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sq
rt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)) - (2*a*e^(2*d*x + 2*c) + 6*b*e^(2*d*x + 2*c) + a + b)*e^(-2*d*x)/(a^2
*e^(2*c) + 2*a*b*e^(2*c) + b^2*e^(2*c)) + e^(2*d*x + 8*c)/(a*e^(6*c) + b*e^(6*c)))/d