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

Optimal. Leaf size=51 \[ \frac{a+b}{2 a^2 d \left (a \cosh ^2(c+d x)+b\right )}+\frac{\log \left (a \cosh ^2(c+d x)+b\right )}{2 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0898654, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4138, 444, 43} \[ \frac{a+b}{2 a^2 d \left (a \cosh ^2(c+d x)+b\right )}+\frac{\log \left (a \cosh ^2(c+d x)+b\right )}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.710901, size = 81, normalized size = 1.59 \[ \frac{(a+2 b) \log (a \cosh (2 (c+d x))+a+2 b)+a \cosh (2 (c+d x)) \log (a \cosh (2 (c+d x))+a+2 b)+2 (a+b)}{2 a^2 d (a \cosh (2 (c+d x))+a+2 b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.076, size = 186, normalized size = 3.7 \begin{align*} -{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{da \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+b \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+{\frac{1}{2\,d{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }-{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.23022, size = 146, normalized size = 2.86 \begin{align*} \frac{2 \,{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} e^{\left (-4 \, d x - 4 \, c\right )} + a^{3} + 2 \,{\left (a^{3} + 2 \, a^{2} b\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} + \frac{d x + c}{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 )}{2 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a + b)*e^(-2*d*x - 2*c)/((a^3*e^(-4*d*x - 4*c) + a^3 + 2*(a^3 + 2*a^2*b)*e^(-2*d*x - 2*c))*d) + (d*x + c)/(
a^2*d) + 1/2*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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Fricas [B]  time = 2.2051, size = 1260, normalized size = 24.71 \begin{align*} -\frac{2 \, a d x \cosh \left (d x + c\right )^{4} + 8 \, a d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, a d x \sinh \left (d x + c\right )^{4} + 2 \, a d x + 4 \,{\left ({\left (a + 2 \, b\right )} d x - a - b\right )} \cosh \left (d x + c\right )^{2} + 4 \,{\left (3 \, a d x \cosh \left (d x + c\right )^{2} +{\left (a + 2 \, b\right )} d x - a - b\right )} \sinh \left (d x + c\right )^{2} -{\left (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 )} \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 ) + 8 \,{\left (a d x \cosh \left (d x + c\right )^{3} +{\left ({\left (a + 2 \, b\right )} d x - a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \,{\left (a^{3} d \cosh \left (d x + c\right )^{4} + 4 \, a^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{3} d \sinh \left (d x + c\right )^{4} + a^{3} d + 2 \,{\left (a^{3} + 2 \, a^{2} b\right )} d \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{3} d \cosh \left (d x + c\right )^{2} +{\left (a^{3} + 2 \, a^{2} b\right )} d\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (a^{3} d \cosh \left (d x + c\right )^{3} +{\left (a^{3} + 2 \, a^{2} b\right )} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 2.13644, size = 163, normalized size = 3.2 \begin{align*} -\frac{\frac{2 \, d x}{a^{2}} + \frac{e^{\left (4 \, d x + 4 \, c\right )} - 2 \, e^{\left (2 \, d x + 2 \, c\right )} + 1}{{\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} - \frac{\log \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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