3.105 $$\int \frac{\cosh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=120 $\frac{x \left (3 a^2+10 a b+15 b^2\right )}{8 (a+b)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a+b)^3}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 d (a+b)}+\frac{(3 a+7 b) \sinh (c+d x) \cosh (c+d x)}{8 d (a+b)^2}$

[Out]

((3*a^2 + 10*a*b + 15*b^2)*x)/(8*(a + b)^3) + (b^(5/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a +
b)^3*d) + ((3*a + 7*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*(a + b)^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*(a +
b)*d)

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Rubi [A]  time = 0.168274, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.261, Rules used = {3675, 414, 527, 522, 206, 205} $\frac{x \left (3 a^2+10 a b+15 b^2\right )}{8 (a+b)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a+b)^3}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 d (a+b)}+\frac{(3 a+7 b) \sinh (c+d x) \cosh (c+d x)}{8 d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((3*a^2 + 10*a*b + 15*b^2)*x)/(8*(a + b)^3) + (b^(5/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a +
b)^3*d) + ((3*a + 7*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*(a + b)^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*(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 527

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

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 ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}+\frac{\operatorname{Subst}\left (\int \frac{3 a+4 b+3 b x^2}{\left (1-x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{4 (a+b) d}\\ &=\frac{(3 a+7 b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2+7 a b+8 b^2+b (3 a+7 b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 (a+b)^2 d}\\ &=\frac{(3 a+7 b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b)^3 d}+\frac{\left (3 a^2+10 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 (a+b)^3 d}\\ &=\frac{\left (3 a^2+10 a b+15 b^2\right ) x}{8 (a+b)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^3 d}+\frac{(3 a+7 b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}\\ \end{align*}

Mathematica [A]  time = 0.275048, size = 115, normalized size = 0.96 $\frac{\left (3 a^2+10 a b+15 b^2\right ) (c+d x)}{8 d (a+b)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a+b)^3}+\frac{(a+2 b) \sinh (2 (c+d x))}{4 d (a+b)^2}+\frac{\sinh (4 (c+d x))}{32 d (a+b)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((3*a^2 + 10*a*b + 15*b^2)*(c + d*x))/(8*(a + b)^3*d) + (b^(5/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqr
t[a]*(a + b)^3*d) + ((a + 2*b)*Sinh[2*(c + d*x)])/(4*(a + b)^2*d) + Sinh[4*(c + d*x)]/(32*(a + b)*d)

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Maple [B]  time = 0.088, size = 857, normalized size = 7.1 \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)^4/(a+b*tanh(d*x+c)^2),x)

[Out]

-1/2/d/(2*b+2*a)/(tanh(1/2*d*x+1/2*c)+1)^4+2/d/(4*a+4*b)/(tanh(1/2*d*x+1/2*c)+1)^3+5/8/d/(a+b)^2/(tanh(1/2*d*x
+1/2*c)+1)*a+9/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)+1)*b-7/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)+1)^2*a-11/8/d/(a+b)^2/
(tanh(1/2*d*x+1/2*c)+1)^2*b+3/8/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)+1)*a^2+5/4/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)+1
)*a*b+15/8/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)+1)*b^2-1/d*b^3/(a+b)^3*a/(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^3/(a+b)^3/((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^4/(a+b)^3/(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^3/(a+b)^3*a/(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^3/(a+b)^3/((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^4/(a+b)^3/(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))+1/2/d/(2*b+2*a)/(tanh(1/2*d*x+1/2*c)-1)^4+2/d/(4*a
+4*b)/(tanh(1/2*d*x+1/2*c)-1)^3+7/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)^2*a+11/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-
1)^2*b+5/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)*a+9/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)*b-3/8/d/(a+b)^3*ln(tanh(1
/2*d*x+1/2*c)-1)*a^2-5/4/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)-1)*a*b-15/8/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)-1)*b^2

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.57445, size = 5515, normalized size = 45.96 \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)^4/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/64*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*
b + b^2)*sinh(d*x + c)^8 + 8*(3*a^2 + 10*a*b + 15*b^2)*d*x*cosh(d*x + c)^4 + 8*(a^2 + 3*a*b + 2*b^2)*cosh(d*x
+ c)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 2*a^2 + 6*a*b + 4*b^2)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b
+ b^2)*cosh(d*x + c)^3 + 6*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*(a^2 + 2*a*b + b^2)*c
osh(d*x + c)^4 + 4*(3*a^2 + 10*a*b + 15*b^2)*d*x + 60*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^4 +
8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 4*(3*a^2 + 10*a*b + 15*b^2)*d*x*cosh(d*x + c) + 20*(a^2 + 3*a*b +
2*b^2)*cosh(d*x + c)^3)*sinh(d*x + c)^3 - 8*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*c
osh(d*x + c)^6 + 12*(3*a^2 + 10*a*b + 15*b^2)*d*x*cosh(d*x + c)^2 + 30*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^4 -
2*a^2 - 6*a*b - 4*b^2)*sinh(d*x + c)^2 + 32*(b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)^3*sinh(d*x + c) + 6*b^
2*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4)*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)*sin
h(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)*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*(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)*c
osh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - a^2 - 2*a*b - b^2 + 8*((a^2 + 2*a*b + b^2)*c
osh(d*x + c)^7 + 4*(3*a^2 + 10*a*b + 15*b^2)*d*x*cosh(d*x + c)^3 + 6*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^5 - 2
*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^4 + 4*(a
^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x +
c)^2*sinh(d*x + c)^2 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3
*a*b^2 + b^3)*d*sinh(d*x + c)^4), 1/64*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d*x +
c)*sinh(d*x + c)^7 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^8 + 8*(3*a^2 + 10*a*b + 15*b^2)*d*x*cosh(d*x + c)^4 +
8*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 2*a^2 + 6*a*b + 4*b^2)*si
nh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + 6*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c
)^5 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(3*a^2 + 10*a*b + 15*b^2)*d*x + 60*(a^2 + 3*a*b + 2*b^2)*c
osh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 4*(3*a^2 + 10*a*b + 15*b^2)*d*x*c
osh(d*x + c) + 20*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^3)*sinh(d*x + c)^3 - 8*(a^2 + 3*a*b + 2*b^2)*cosh(d*x +
c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 12*(3*a^2 + 10*a*b + 15*b^2)*d*x*cosh(d*x + c)^2 + 30*(a^2 +
3*a*b + 2*b^2)*cosh(d*x + c)^4 - 2*a^2 - 6*a*b - 4*b^2)*sinh(d*x + c)^2 + 64*(b^2*cosh(d*x + c)^4 + 4*b^2*cos
h(d*x + c)^3*sinh(d*x + c) + 6*b^2*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2
*sinh(d*x + c)^4)*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) - a^2 - 2*a*b - b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^7 + 4*(3*
a^2 + 10*a*b + 15*b^2)*d*x*cosh(d*x + c)^3 + 6*(a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^5 - 2*(a^2 + 3*a*b + 2*b^2)
*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2
+ b^3)*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^2*sinh(d*x + c)^2
+ 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d
*x + c)^4)]

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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 [B]  time = 2.79531, size = 440, normalized size = 3.67 \begin{align*} \frac{\frac{64 \, b^{3} \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^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b}} + \frac{8 \,{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} d x}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 16 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x\right )}}{a^{3} e^{\left (4 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, c\right )} + b^{3} e^{\left (4 \, c\right )}} + \frac{a e^{\left (4 \, d x + 20 \, c\right )} + b e^{\left (4 \, d x + 20 \, c\right )} + 8 \, a e^{\left (2 \, d x + 18 \, c\right )} + 16 \, b e^{\left (2 \, d x + 18 \, c\right )}}{a^{2} e^{\left (16 \, c\right )} + 2 \, a b e^{\left (16 \, c\right )} + b^{2} e^{\left (16 \, 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*(64*b^3*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*sqrt(a*b)) + 8*(3*a^2 + 10*a*b + 15*b^2)*d*x/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (18*a^2*e^(4*d*x + 4*c) +
60*a*b*e^(4*d*x + 4*c) + 90*b^2*e^(4*d*x + 4*c) + 8*a^2*e^(2*d*x + 2*c) + 24*a*b*e^(2*d*x + 2*c) + 16*b^2*e^(
2*d*x + 2*c) + a^2 + 2*a*b + b^2)*e^(-4*d*x)/(a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e^(4*c)) +
(a*e^(4*d*x + 20*c) + b*e^(4*d*x + 20*c) + 8*a*e^(2*d*x + 18*c) + 16*b*e^(2*d*x + 18*c))/(a^2*e^(16*c) + 2*a*
b*e^(16*c) + b^2*e^(16*c)))/d