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

Optimal. Leaf size=68 \[ -\frac {1}{2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^2}+\frac {\log (\cosh (c+d x))}{d (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3670, 444, 44} \[ -\frac {1}{2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^2}+\frac {\log (\cosh (c+d x))}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Cosh[c + d*x]]/((a + b)^2*d) + Log[a + b*Tanh[c + d*x]^2]/(2*(a + b)^2*d) - 1/(2*(a + b)*d*(a + b*Tanh[c +
d*x]^2))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.47, size = 55, normalized size = 0.81 \[ -\frac {\frac {a+b}{a+b \tanh ^2(c+d x)}-\log \left (a+b \tanh ^2(c+d x)\right )-2 \log (\cosh (c+d x))}{2 d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(-2*Log[Cosh[c + d*x]] - Log[a + b*Tanh[c + d*x]^2] + (a + b)/(a + b*Tanh[c + d*x]^2))/((a + b)^2*d)

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fricas [B]  time = 0.44, size = 623, normalized size = 9.16 \[ -\frac {2 \, {\left (a + b\right )} d x \cosh \left (d x + c\right )^{4} + 8 \, {\left (a + b\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, {\left (a + b\right )} d x \sinh \left (d x + c\right )^{4} + 2 \, {\left (a + b\right )} d x + 4 \, {\left ({\left (a - b\right )} d x + b\right )} \cosh \left (d x + c\right )^{2} + 4 \, {\left (3 \, {\left (a + b\right )} d x \cosh \left (d x + c\right )^{2} + {\left (a - b\right )} d x + b\right )} \sinh \left (d x + c\right )^{2} - {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a + b\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a - b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + {\left (a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a + b\right )} \log \left (\frac {2 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - 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 ) + 8 \, {\left ({\left (a + b\right )} d x \cosh \left (d x + c\right )^{3} + {\left ({\left (a - b\right )} d x + b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d\right )} \sinh \left (d x + c\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d + 4 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{3} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a + b)*d*x*cosh(d*x + c)^4 + 8*(a + b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a + b)*d*x*sinh(d*x + c
)^4 + 2*(a + b)*d*x + 4*((a - b)*d*x + b)*cosh(d*x + c)^2 + 4*(3*(a + b)*d*x*cosh(d*x + c)^2 + (a - b)*d*x + b
)*sinh(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*(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)*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2
+ a - b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) + 8*((a + b)*d*x*cosh(d*x + c)^
3 + ((a - b)*d*x + b)*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)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d*x + c)
^4 + 2*(a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^2
+ (a^3 + a^2*b - a*b^2 - b^3)*d)*sinh(d*x + c)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d + 4*((a^3 + 3*a^2*b + 3*a
*b^2 + b^3)*d*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c))*sinh(d*x + c))

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giac [B]  time = 0.27, size = 149, normalized size = 2.19 \[ \frac {\frac {\log \left ({\left | a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2}{{\left (a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )} {\left (a + b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 113, normalized size = 1.66 \[ -\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 d \left (a +b \right )^{2}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 d \left (a +b \right )^{2}}+\frac {\ln \left (a +b \left (\tanh ^{2}\left (d x +c \right )\right )\right )}{2 \left (a +b \right )^{2} d}-\frac {a}{2 d \left (a +b \right )^{2} \left (a +b \left (\tanh ^{2}\left (d x +c \right )\right )\right )}-\frac {b}{2 d \left (a +b \right )^{2} \left (a +b \left (\tanh ^{2}\left (d x +c \right )\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 170, normalized size = 2.50 \[ -\frac {2 \, b e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.47, size = 129, normalized size = 1.90 \[ \frac {\frac {a\,x}{a^2+2\,a\,b+b^2}+\frac {b\,x\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{a^2+2\,a\,b+b^2}+\frac {b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{2\,a\,d\,\left (a+b\right )}}{b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a}+\frac {\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{2\,d\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )}{d\,{\left (a+b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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