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]

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

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Rubi [A]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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]

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 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 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])

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*}

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Mathematica [A]  time = 0.16, size = 77, normalized size = 1.00 \[ \frac {4 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )+2 \sqrt {a} (a+3 b) (c+d x)+\sqrt {a} (a+b) \sinh (2 (c+d x))}{4 \sqrt {a} d (a+b)^2} \]

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)

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fricas [B]  time = 0.45, size = 948, normalized size = 12.31 \[ \text {result too large to display} \]

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)]

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giac [B]  time = 0.56, size = 172, normalized size = 2.23 \[ \frac {\frac {4 \, {\left (a + 3 \, b\right )} d x}{a^{2} + 2 \, a b + b^{2}} + \frac {8 \, b^{2} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b}} - \frac {{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-2 \, d x\right )}}{a^{2} e^{\left (2 \, c\right )} + 2 \, a b e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}} + \frac {e^{\left (2 \, d x + 8 \, c\right )}}{a e^{\left (6 \, c\right )} + b e^{\left (6 \, c\right )}}}{8 \, d} \]

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

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maple [B]  time = 0.43, size = 608, normalized size = 7.90 \[ \frac {1}{d \left (2 b +2 a \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2}{d \left (4 a +4 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a}{2 d \left (a +b \right )^{2}}-\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b}{2 d \left (a +b \right )^{2}}-\frac {1}{d \left (2 b +2 a \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2}{d \left (4 a +4 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a}{2 d \left (a +b \right )^{2}}+\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b}{2 d \left (a +b \right )^{2}}-\frac {a \,b^{2} \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{d \left (a +b \right )^{2} \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}+\frac {b^{2} \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{d \left (a +b \right )^{2} \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}-\frac {b^{3} \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{d \left (a +b \right )^{2} \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}-\frac {a \,b^{2} \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{d \left (a +b \right )^{2} \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}-\frac {b^{2} \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{d \left (a +b \right )^{2} \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}-\frac {b^{3} \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{d \left (a +b \right )^{2} \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}} \]

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/(2*b+2*a)/(tanh(1/2*d*x+1/2*c)+1)^2+2/d/(4*a+4*b)/(tan
h(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*ta
nh(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)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2
))

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maxima [B]  time = 0.62, size = 316, normalized size = 4.10 \[ \frac {b \log \left ({\left (a + b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a - b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} - \frac {b \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} - \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} + \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} - \frac {b \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{2 \, \sqrt {a b} {\left (a + b\right )} d} + \frac {d x + c}{2 \, {\left (a + b\right )} d} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, {\left (a + b\right )} d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, {\left (a + b\right )} d} \]

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]

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

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mupad [B]  time = 1.93, size = 880, normalized size = 11.43 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d\,\left (a+b\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d\,\left (a+b\right )}+\frac {x\,\left (a+3\,b\right )}{2\,{\left (a+b\right )}^2}+\frac {\mathrm {atan}\left (\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {2\,\left (2\,b^3\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}+2\,a\,b^2\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}\right )}{d\,{\left (a+b\right )}^5\,\sqrt {b^3}\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}}-\frac {\left (a-b\right )\,\left (2\,a\,d\,{\left (b^3\right )}^{3/2}+b\,d\,{\left (b^3\right )}^{3/2}-a^4\,d\,\sqrt {b^3}-2\,a^3\,b\,d\,\sqrt {b^3}\right )}{b^2\,{\left (a+b\right )}^3\,\sqrt {a\,d^2\,{\left (a+b\right )}^4}\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}}\right )+\frac {\left (a-b\right )\,\left (4\,a\,d\,{\left (b^3\right )}^{3/2}+b\,d\,{\left (b^3\right )}^{3/2}+a^4\,d\,\sqrt {b^3}+4\,a^3\,b\,d\,\sqrt {b^3}+6\,a^2\,b^2\,d\,\sqrt {b^3}\right )}{b^2\,{\left (a+b\right )}^3\,\sqrt {a\,d^2\,{\left (a+b\right )}^4}\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}}\right )\,\left (\frac {a^4\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}}{2}+\frac {b^4\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}}{2}+3\,a^2\,b^2\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}+2\,a\,b^3\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}+2\,a^3\,b\,\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}\right )\right )\,\sqrt {b^3}}{\sqrt {a^5\,d^2+4\,a^4\,b\,d^2+6\,a^3\,b^2\,d^2+4\,a^2\,b^3\,d^2+a\,b^4\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*c + 2*d*x)/(8*d*(a + b)) - exp(- 2*c - 2*d*x)/(8*d*(a + b)) + (x*(a + 3*b))/(2*(a + b)^2) + (atan((exp(2
*c)*exp(2*d*x)*((2*(2*b^3*(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2) + 2*a*b^2*
(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2)))/(d*(a + b)^5*(b^3)^(1/2)*(3*a*b^2
+ 3*a^2*b + a^3 + b^3)*(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2)) - ((a - b)*(
2*a*d*(b^3)^(3/2) + b*d*(b^3)^(3/2) - a^4*d*(b^3)^(1/2) - 2*a^3*b*d*(b^3)^(1/2)))/(b^2*(a + b)^3*(a*d^2*(a + b
)^4)^(1/2)*(3*a*b^2 + 3*a^2*b + a^3 + b^3)*(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)
^(1/2))) + ((a - b)*(4*a*d*(b^3)^(3/2) + b*d*(b^3)^(3/2) + a^4*d*(b^3)^(1/2) + 4*a^3*b*d*(b^3)^(1/2) + 6*a^2*b
^2*d*(b^3)^(1/2)))/(b^2*(a + b)^3*(a*d^2*(a + b)^4)^(1/2)*(3*a*b^2 + 3*a^2*b + a^3 + b^3)*(a^5*d^2 + a*b^4*d^2
+ 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2)))*((a^4*(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d
^2 + 6*a^3*b^2*d^2)^(1/2))/2 + (b^4*(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2))
/2 + 3*a^2*b^2*(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2) + 2*a*b^3*(a^5*d^2 +
a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2) + 2*a^3*b*(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 +
4*a^2*b^3*d^2 + 6*a^3*b^2*d^2)^(1/2)))*(b^3)^(1/2))/(a^5*d^2 + a*b^4*d^2 + 4*a^4*b*d^2 + 4*a^2*b^3*d^2 + 6*a^3
*b^2*d^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh ^{2}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \]

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)

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