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

Optimal. Leaf size=72 \[ \frac {a}{2 b d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^2}+\frac {\log (\cosh (c+d x))}{d (a+b)^2} \]

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

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Rubi [A]  time = 0.12, antiderivative size = 72, 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 = {3670, 446, 77} \[ \frac {a}{2 b d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^2}+\frac {\log (\cosh (c+d x))}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.53, size = 57, normalized size = 0.79 \[ \frac {\frac {a (a+b)}{b \left (a+b \tanh ^2(c+d x)\right )}+\log \left (a+b \tanh ^2(c+d x)\right )+2 \log (\cosh (c+d x))}{2 d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 629, normalized size = 8.74 \[ -\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 - a\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 - a\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 - a\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)^3/(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 - a)*cosh(d*x + c)^2 + 4*(3*(a + b)*d*x*cosh(d*x + c)^2 + (a - b)*d*x - a
)*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 - a)*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 = 0.32, size = 149, normalized size = 2.07 \[ \frac {\frac {\log \left ({\left | a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2}{{\left (a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )} {\left (a + b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^3/(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 = 0.10, size = 118, normalized size = 1.64 \[ -\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 d \left (a +b \right )^{2}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 d \left (a +b \right )^{2}}+\frac {\ln \left (a +b \left (\tanh ^{2}\left (d x +c \right )\right )\right )}{2 \left (a +b \right )^{2} d}+\frac {a^{2}}{2 d \left (a +b \right )^{2} b \left (a +b \left (\tanh ^{2}\left (d x +c \right )\right )\right )}+\frac {a}{2 d \left (a +b \right )^{2} \left (a +b \left (\tanh ^{2}\left (d x +c \right )\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [B]  time = 0.34, size = 170, normalized size = 2.36 \[ \frac {2 \, a e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*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 = 0.43, size = 210, normalized size = 2.92 \[ -\frac {-a^2+a\,b\,\left (-1+\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,2{}\mathrm {i}\right )+b^2\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,2{}\mathrm {i}}{2\,d\,a^3\,b+2\,d\,a^2\,b^2\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\,d\,a^2\,b^2+4\,d\,a\,b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^2+2\,d\,a\,b^3+2\,d\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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