3.151 \(\int \frac {\tanh ^2(c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 \sqrt {b} d \sqrt {a+b}}+\frac {x}{a^2}-\frac {\tanh (c+d x)}{2 a d \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

x/a^2-1/2*(a+2*b)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/a^2/d/b^(1/2)/(a+b)^(1/2)-1/2*tanh(d*x+c)/a/d/(a+b-
b*tanh(d*x+c)^2)

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Rubi [A]  time = 0.17, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 = {4141, 1975, 471, 522, 206, 208} \[ -\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 \sqrt {b} d \sqrt {a+b}}+\frac {x}{a^2}-\frac {\tanh (c+d x)}{2 a d \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a^2 - ((a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*a^2*Sqrt[b]*Sqrt[a + b]*d) - Tanh[c + d*x]
/(2*a*d*(a + b - b*Tanh[c + d*x]^2))

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 208

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

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, 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 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4141

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

Rubi steps

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

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Mathematica [B]  time = 4.63, size = 326, normalized size = 3.84 \[ \frac {\text {sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b)^2 \left (\frac {\left (a^2+8 a b+8 b^2\right ) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{a^2 b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}+\frac {16 x}{a^2}+\frac {\left (a^3-6 a^2 b-24 a b^2-16 b^3\right ) (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )}{a^2 b d (a+b)^{3/2} \sqrt {b (\cosh (c)-\sinh (c))^4}}-\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{3/2} d (a+b)^{3/2}}+\frac {a \sinh (2 (c+d x))}{b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}\right )}{64 \left (a+b \text {sech}^2(c+d x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x]^4*((16*x)/a^2 - ((a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sq
rt[a + b]])/(b^(3/2)*(a + b)^(3/2)*d) + ((a^3 - 6*a^2*b - 24*a*b^2 - 16*b^3)*ArcTanh[(Sech[d*x]*(Cosh[2*c] - S
inh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(Cosh[2*c]
- Sinh[2*c]))/(a^2*b*(a + b)^(3/2)*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + ((a^2 + 8*a*b + 8*b^2)*Sech[2*c]*((a +
2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/(a^2*b*(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)])) + (a*Sinh[2*(c + d*x)])/(b*
(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)]))))/(64*(a + b*Sech[c + d*x]^2)^2)

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fricas [B]  time = 0.47, size = 1846, normalized size = 21.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 + 16*(a^2*b + a*b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(a^2*b
+ a*b^2)*d*x*sinh(d*x + c)^4 + 4*a^2*b + 4*a*b^2 + 4*(a^2*b + a*b^2)*d*x + 4*(a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2
*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c)^2 + 4*(6*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^2 + a^2*b + 3*a*b^2 + 2*b^
3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*sinh(d*x + c)^2 + ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d
*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 4*a*b + 4*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2
+ 2*a*b)*cosh(d*x + c)^2 + a^2 + 4*a*b + 4*b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)
^3 + (a^2 + 4*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a*b + b^2)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh
(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 +
a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*si
nh(d*x + c) + 4*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(a*b +
b^2))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^
2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d
*x + c) + a)) + 8*(2*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^3 + (a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b
^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^4*b + a^3*b^2)*d*cosh(d*x +
c)*sinh(d*x + c)^3 + (a^4*b + a^3*b^2)*d*sinh(d*x + c)^4 + 2*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c)^
2 + 2*(3*(a^4*b + a^3*b^2)*d*cosh(d*x + c)^2 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d)*sinh(d*x + c)^2 + (a^4*b + a
^3*b^2)*d + 4*((a^4*b + a^3*b^2)*d*cosh(d*x + c)^3 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x
+ c)), 1/2*(2*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 + 8*(a^2*b + a*b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(
a^2*b + a*b^2)*d*x*sinh(d*x + c)^4 + 2*a^2*b + 2*a*b^2 + 2*(a^2*b + a*b^2)*d*x + 2*(a^2*b + 3*a*b^2 + 2*b^3 +
2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c)^2 + 2*(6*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^2 + a^2*b + 3*a*b^2
+ 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*sinh(d*x + c)^2 - ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*
cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 4*a*b + 4*b^2)*cosh(d*x + c)^2 + 2*(3
*(a^2 + 2*a*b)*cosh(d*x + c)^2 + a^2 + 4*a*b + 4*b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*
x + c)^3 + (a^2 + 4*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-a*b - b^2)*arctan(1/2*(a*cosh(d*x + c)^2
+ 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-a*b - b^2)/(a*b + b^2)) + 4*(2*(a^2*b +
a*b^2)*d*x*cosh(d*x + c)^3 + (a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c))*sinh(
d*x + c))/((a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^4*b + a^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4*b
+ a^3*b^2)*d*sinh(d*x + c)^4 + 2*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^4*b + a^3*b^2)*d*
cosh(d*x + c)^2 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d)*sinh(d*x + c)^2 + (a^4*b + a^3*b^2)*d + 4*((a^4*b + a^3*b
^2)*d*cosh(d*x + c)^3 + (a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [A]  time = 0.72, size = 147, normalized size = 1.73 \[ -\frac {\frac {{\left (a e^{\left (2 \, c\right )} + 2 \, b e^{\left (2 \, c\right )}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right ) e^{\left (-2 \, c\right )}}{\sqrt {-a b - b^{2}} a^{2}} - \frac {2 \, d x}{a^{2}} - \frac {2 \, {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.35, size = 411, normalized size = 4.84 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d a \sqrt {b}\, \sqrt {a +b}}-\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d a \sqrt {b}\, \sqrt {a +b}}+\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{2} \sqrt {a +b}}-\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{2} \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)+1/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/a/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2
*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*tanh(1/2*d*x+1/2*c)^3-1/d/a/(tanh(1/2*d
*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*tanh(1/2*d*x+1/
2*c)+1/4/d/a/b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2
))-1/4/d/a/b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))
+1/2/d/a^2*b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))
-1/2/d/a^2*b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))

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maxima [B]  time = 0.55, size = 597, normalized size = 7.02 \[ -\frac {{\left (a^{2} + 6 \, a b + 4 \, b^{2}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} + \frac {{\left (a^{2} + 6 \, a b + 4 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} + \frac {a^{2} + 2 \, a b + {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{4 \, {\left (a^{4} + a^{3} b + {\left (a^{4} + a^{3} b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} d} - \frac {a^{2} + 2 \, a b + {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, {\left (a^{4} + a^{3} b + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + a^{3} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {{\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a}{2 \, {\left (a^{3} + a^{2} b + 2 \, {\left (a^{3} + 3 \, a^{2} b + 2 \, a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + a^{2} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {\log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} {\left (a + b\right )} d} + \frac {\log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{4 \, a^{2} d} - \frac {\log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{4 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(a^2 + 6*a*b + 4*b^2)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b
+ 2*sqrt((a + b)*b)))/((a^3 + a^2*b)*sqrt((a + b)*b)*d) + 1/16*(a^2 + 6*a*b + 4*b^2)*log((a*e^(-2*d*x - 2*c)
+ a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^3 + a^2*b)*sqrt((a + b)
*b)*d) + 1/4*(a^2 + 2*a*b + (a^2 + 8*a*b + 8*b^2)*e^(2*d*x + 2*c))/((a^4 + a^3*b + (a^4 + a^3*b)*e^(4*d*x + 4*
c) + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*e^(2*d*x + 2*c))*d) - 1/4*(a^2 + 2*a*b + (a^2 + 8*a*b + 8*b^2)*e^(-2*d*x -
2*c))/((a^4 + a^3*b + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*e^(-2*d*x - 2*c) + (a^4 + a^3*b)*e^(-4*d*x - 4*c))*d) - 1/
2*((a + 2*b)*e^(-2*d*x - 2*c) + a)/((a^3 + a^2*b + 2*(a^3 + 3*a^2*b + 2*a*b^2)*e^(-2*d*x - 2*c) + (a^3 + a^2*b
)*e^(-4*d*x - 4*c))*d) + 1/8*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a +
2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*(a + b)*d) + 1/4*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c
) + a)/(a^2*d) - 1/4*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/(a^2*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^2\,\left ({\mathrm {cosh}\left (c+d\,x\right )}^2-1\right )}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cosh(c + d*x)^2*(cosh(c + d*x)^2 - 1))/(b + a*cosh(c + d*x)^2)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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