3.140 \(\int \frac {\tanh ^3(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=45 \[ \frac {(a+b) \log \left (a \cosh ^2(c+d x)+b\right )}{2 a b d}-\frac {\log (\cosh (c+d x))}{b d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4138, 446, 72} \[ \frac {(a+b) \log \left (a \cosh ^2(c+d x)+b\right )}{2 a b d}-\frac {\log (\cosh (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[c + d*x]^3/(a + b*Sech[c + d*x]^2),x]

[Out]

-(Log[Cosh[c + d*x]]/(b*d)) + ((a + b)*Log[b + a*Cosh[c + d*x]^2])/(2*a*b*d)

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

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

Rule 4138

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 41, normalized size = 0.91 \[ \frac {(a+b) \log \left (a \cosh ^2(c+d x)+b\right )-2 a \log (\cosh (c+d x))}{2 a b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[c + d*x]^3/(a + b*Sech[c + d*x]^2),x]

[Out]

(-2*a*Log[Cosh[c + d*x]] + (a + b)*Log[b + a*Cosh[c + d*x]^2])/(2*a*b*d)

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fricas [B]  time = 0.46, size = 112, normalized size = 2.49 \[ -\frac {2 \, b d x - {\left (a + b\right )} \log \left (\frac {2 \, {\left (a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )}}{\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right ) + 2 \, a \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{2 \, a b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(2*b*d*x - (a + b)*log(2*(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + a + 2*b)/(cosh(d*x + c)^2 - 2*cosh(d*x
+ c)*sinh(d*x + c) + sinh(d*x + c)^2)) + 2*a*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/(a*b*d)

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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)^3/(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 ValueDone

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maple [B]  time = 0.33, size = 196, normalized size = 4.36 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}+\frac {\ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}{2 d b}+\frac {\ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}{2 d a}-\frac {\ln \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.47, size = 77, normalized size = 1.71 \[ \frac {d x + c}{a d} + \frac {{\left (a + b\right )} \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, a b d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.65, size = 238, normalized size = 5.29 \[ \frac {\ln \left (a\,b+3\,a^2+6\,a^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+3\,a^2\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}+4\,b^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+14\,a\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a\,b\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}\right )\,\left (a+b\right )}{2\,a\,b\,d}-\frac {\ln \left (21\,a\,b^4+108\,a^4\,b+27\,a^5+2\,b^5+82\,a^2\,b^3+144\,a^3\,b^2+27\,a^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+21\,a\,b^4\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+108\,a^4\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+82\,a^2\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+144\,a^3\,b^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )}{b\,d}-\frac {x}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(c + d*x)^3/(a + b/cosh(c + d*x)^2),x)

[Out]

(log(a*b + 3*a^2 + 6*a^2*exp(2*c)*exp(2*d*x) + 3*a^2*exp(4*c)*exp(4*d*x) + 4*b^2*exp(2*c)*exp(2*d*x) + 14*a*b*
exp(2*c)*exp(2*d*x) + a*b*exp(4*c)*exp(4*d*x))*(a + b))/(2*a*b*d) - log(21*a*b^4 + 108*a^4*b + 27*a^5 + 2*b^5
+ 82*a^2*b^3 + 144*a^3*b^2 + 27*a^5*exp(2*c)*exp(2*d*x) + 2*b^5*exp(2*c)*exp(2*d*x) + 21*a*b^4*exp(2*c)*exp(2*
d*x) + 108*a^4*b*exp(2*c)*exp(2*d*x) + 82*a^2*b^3*exp(2*c)*exp(2*d*x) + 144*a^3*b^2*exp(2*c)*exp(2*d*x))/(b*d)
- x/a

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{3}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)**3/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(tanh(c + d*x)**3/(a + b*sech(c + d*x)**2), x)

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