3.139 \(\int \frac {\tanh ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=59 \[ -\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a b^{3/2} d}+\frac {x}{a}+\frac {\tanh (c+d x)}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 59, 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, 479, 522, 206, 208} \[ -\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a b^{3/2} d}+\frac {x}{a}+\frac {\tanh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a - ((a + b)^(3/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*b^(3/2)*d) + Tanh[c + d*x]/(b*d)

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 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && 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 ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right ) \left (a+b \left (1-x^2\right )\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\tanh (c+d x)}{b d}-\frac {\operatorname {Subst}\left (\int \frac {a+b+(-a-2 b) x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{b d}\\ &=\frac {\tanh (c+d x)}{b d}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{a d}-\frac {(a+b)^2 \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{a b d}\\ &=\frac {x}{a}-\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a b^{3/2} d}+\frac {\tanh (c+d x)}{b d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.09, size = 196, normalized size = 3.32 \[ \frac {\text {sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4} (a \text {sech}(c) \sinh (d x) \text {sech}(c+d x)+b d x)+(a+b)^2 (\sinh (2 c)-\cosh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )\right )}{2 a b d \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4} \left (a+b \text {sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*((a + b)^2*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[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]) +
Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(b*d*x + a*Sech[c]*Sech[c + d*x]*Sinh[d*x])))/(2*a*b*Sqrt[a + b]*d*(
a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 683, normalized size = 11.58 \[ \left [\frac {2 \, b d x \cosh \left (d x + c\right )^{2} + 4 \, b d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + 2 \, b d x \sinh \left (d x + c\right )^{2} + 2 \, b d x + {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a + b\right )} \sqrt {\frac {a + b}{b}} \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + a^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 4 \, {\left (a b \cosh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a b \sinh \left (d x + c\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {\frac {a + b}{b}}}{a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} + 2 \, {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a \cosh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a \cosh \left (d x + c\right )^{3} + {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a}\right ) - 4 \, a}{2 \, {\left (a b d \cosh \left (d x + c\right )^{2} + 2 \, a b d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a b d \sinh \left (d x + c\right )^{2} + a b d\right )}}, \frac {b d x \cosh \left (d x + c\right )^{2} + 2 \, b d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b d x \sinh \left (d x + c\right )^{2} + b d x - {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a + b\right )} \sqrt {-\frac {a + b}{b}} \arctan \left (\frac {{\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sqrt {-\frac {a + b}{b}}}{2 \, {\left (a + b\right )}}\right ) - 2 \, a}{a b d \cosh \left (d x + c\right )^{2} + 2 \, a b d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a b d \sinh \left (d x + c\right )^{2} + a b d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*b*d*x*cosh(d*x + c)^2 + 4*b*d*x*cosh(d*x + c)*sinh(d*x + c) + 2*b*d*x*sinh(d*x + c)^2 + 2*b*d*x + ((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((a + b)/b
)*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))*sinh(d*x + c) + 4*(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 + 2*b^2)*sqrt((a + b)/b))/(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)) - 4*a)/(a*b*d*cosh(d*x + c)^2 + 2*a*b*d*cosh(d*
x + c)*sinh(d*x + c) + a*b*d*sinh(d*x + c)^2 + a*b*d), (b*d*x*cosh(d*x + c)^2 + 2*b*d*x*cosh(d*x + c)*sinh(d*x
+ c) + b*d*x*sinh(d*x + c)^2 + b*d*x - ((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(-(a + b)/b)*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)/(a + b)) - 2*a)/(a*b*d*cosh(d*x + c)^2 + 2*a*b*d*cosh(d*x + c
)*sinh(d*x + c) + a*b*d*sinh(d*x + c)^2 + a*b*d)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument ValueError index.cc
index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd
Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument ValueDone

________________________________________________________________________________________

maple [B]  time = 0.35, size = 386, normalized size = 6.54 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}+\frac {a \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,b^{\frac {3}{2}} \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{d \sqrt {b}\, \sqrt {a +b}}+\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \sqrt {a +b}}-\frac {a \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,b^{\frac {3}{2}} \sqrt {a +b}}-\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{d \sqrt {b}\, \sqrt {a +b}}-\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \sqrt {a +b}}+\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.64, size = 637, normalized size = 10.80 \[ -\frac {{\left (a + 2 \, b\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 )}{8 \, \sqrt {{\left (a + b\right )} b} b d} + \frac {{\left (a + 2 \, b\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 )}{8 \, \sqrt {{\left (a + b\right )} b} b d} + \frac {3 \, a \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 \, \sqrt {{\left (a + b\right )} b} b d} + \frac {{\left (a + 2 \, b\right )} \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 )}{8 \, a b 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 \, b d} - \frac {{\left (a + 2 \, b\right )} \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 )}{8 \, a b 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 \, b d} - \frac {3 \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{4 \, b d} + \frac {3 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{4 \, b d} - \frac {{\left (a^{2} + 8 \, a b + 8 \, 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 )}{32 \, \sqrt {{\left (a + b\right )} b} a b d} + \frac {{\left (a^{2} + 8 \, a b + 8 \, 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 )}{32 \, \sqrt {{\left (a + b\right )} b} a b d} - \frac {5}{8 \, {\left (b e^{\left (2 \, d x + 2 \, c\right )} + b\right )} d} + \frac {11}{8 \, {\left (b e^{\left (-2 \, d x - 2 \, c\right )} + b\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(a + 2*b)*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)*b*d) + 1/8*(a + 2*b)*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)*b*d) + 3/16*a*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)*b*d) + 1/8*(a + 2*b)
*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/(a*b*d) + 1/4*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^
(2*d*x + 2*c) + a)/(b*d) - 1/8*(a + 2*b)*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/(a*b*d) -
1/4*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/(b*d) - 3/4*log(e^(2*d*x + 2*c) + 1)/(b*d) + 3/
4*log(e^(-2*d*x - 2*c) + 1)/(b*d) - 1/32*(a^2 + 8*a*b + 8*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)))/(sqrt((a + b)*b)*a*b*d) + 1/32*(a^2 + 8*a*b + 8*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)))/(s
qrt((a + b)*b)*a*b*d) - 5/8/((b*e^(2*d*x + 2*c) + b)*d) + 11/8/((b*e^(-2*d*x - 2*c) + b)*d)

________________________________________________________________________________________

mupad [B]  time = 1.87, size = 183, normalized size = 3.10 \[ \frac {x}{a}-\frac {2}{b\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {\ln \left (\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{a^2\,b}-\frac {2\,{\left (a+b\right )}^{3/2}\,\left (a+a\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^2\,b^{3/2}}\right )\,{\left (a+b\right )}^{3/2}}{2\,a\,b^{3/2}\,d}-\frac {\ln \left (\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{a^2\,b}+\frac {2\,{\left (a+b\right )}^{3/2}\,\left (a+a\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^2\,b^{3/2}}\right )\,{\left (a+b\right )}^{3/2}}{2\,a\,b^{3/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(c + d*x)^4/(a + b/cosh(c + d*x)^2),x)

[Out]

x/a - 2/(b*d*(exp(2*c + 2*d*x) + 1)) + (log((4*exp(2*c + 2*d*x)*(a + b)^2)/(a^2*b) - (2*(a + b)^(3/2)*(a + a*e
xp(2*c + 2*d*x) + 2*b*exp(2*c + 2*d*x)))/(a^2*b^(3/2)))*(a + b)^(3/2))/(2*a*b^(3/2)*d) - (log((4*exp(2*c + 2*d
*x)*(a + b)^2)/(a^2*b) + (2*(a + b)^(3/2)*(a + a*exp(2*c + 2*d*x) + 2*b*exp(2*c + 2*d*x)))/(a^2*b^(3/2)))*(a +
b)^(3/2))/(2*a*b^(3/2)*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{4}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)**4/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(tanh(c + d*x)**4/(a + b*sech(c + d*x)**2), x)

________________________________________________________________________________________