3.161 \(\int \frac{\tanh ^3(c+d x)}{(a+b \text{sech}^2(c+d x))^3} \, dx\)
Optimal. Leaf size=81 \[ -\frac{b (a+b)}{4 a^3 d \left (a \cosh ^2(c+d x)+b\right )^2}+\frac{a+2 b}{2 a^3 d \left (a \cosh ^2(c+d x)+b\right )}+\frac{\log \left (a \cosh ^2(c+d x)+b\right )}{2 a^3 d} \]
[Out]
-(b*(a + b))/(4*a^3*d*(b + a*Cosh[c + d*x]^2)^2) + (a + 2*b)/(2*a^3*d*(b + a*Cosh[c + d*x]^2)) + Log[b + a*Cos
h[c + d*x]^2]/(2*a^3*d)
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Rubi [A] time = 0.126136, antiderivative size = 81, 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, 446, 77} \[ -\frac{b (a+b)}{4 a^3 d \left (a \cosh ^2(c+d x)+b\right )^2}+\frac{a+2 b}{2 a^3 d \left (a \cosh ^2(c+d x)+b\right )}+\frac{\log \left (a \cosh ^2(c+d x)+b\right )}{2 a^3 d} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[c + d*x]^3/(a + b*Sech[c + d*x]^2)^3,x]
[Out]
-(b*(a + b))/(4*a^3*d*(b + a*Cosh[c + d*x]^2)^2) + (a + 2*b)/(2*a^3*d*(b + a*Cosh[c + d*x]^2)) + Log[b + a*Cos
h[c + d*x]^2]/(2*a^3*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 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 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]))))
Rubi steps
\begin{align*} \int \frac{\tanh ^3(c+d x)}{\left (a+b \text{sech}^2(c+d x)\right )^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (1-x^2\right )}{\left (b+a x^2\right )^3} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x) x}{(b+a x)^3} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{b (a+b)}{a^2 (b+a x)^3}+\frac{a+2 b}{a^2 (b+a x)^2}-\frac{1}{a^2 (b+a x)}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{b (a+b)}{4 a^3 d \left (b+a \cosh ^2(c+d x)\right )^2}+\frac{a+2 b}{2 a^3 d \left (b+a \cosh ^2(c+d x)\right )}+\frac{\log \left (b+a \cosh ^2(c+d x)\right )}{2 a^3 d}\\ \end{align*}
Mathematica [A] time = 1.39039, size = 131, normalized size = 1.62 \[ \frac{2 \left (a^2+3 a b+3 b^2\right )+a^2 \cosh ^2(2 (c+d x)) \log (a \cosh (2 (c+d x))+a+2 b)+(a+2 b)^2 \log (a \cosh (2 (c+d x))+a+2 b)+2 a (a+2 b) \cosh (2 (c+d x)) (\log (a \cosh (2 (c+d x))+a+2 b)+1)}{2 a^3 d (a \cosh (2 (c+d x))+a+2 b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[c + d*x]^3/(a + b*Sech[c + d*x]^2)^3,x]
[Out]
(2*(a^2 + 3*a*b + 3*b^2) + (a + 2*b)^2*Log[a + 2*b + a*Cosh[2*(c + d*x)]] + a^2*Cosh[2*(c + d*x)]^2*Log[a + 2*
b + a*Cosh[2*(c + d*x)]] + 2*a*(a + 2*b)*Cosh[2*(c + d*x)]*(1 + Log[a + 2*b + a*Cosh[2*(c + d*x)]]))/(2*a^3*d*
(a + 2*b + a*Cosh[2*(c + d*x)])^2)
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Maple [B] time = 0.093, size = 672, normalized size = 8.3 \begin{align*} \text{result too large to display} \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)^3,x)
[Out]
-1/d/a^3*ln(tanh(1/2*d*x+1/2*c)+1)-2/d/a/(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)^2*tanh(1/2*d*x+1/2*c)^6-2/d*b/a^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)^2*tanh(1/2*d*x+1/2*c)^6-4/d/(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)^2/(a+b)*tanh(1/2*d*
x+1/2*c)^4-4/d*b/a/(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)^2/(a+b)*tanh(1/2*d*x+1/2*c)^4+4/d*b^2/a^2/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*ta
nh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)*tanh(1/2*d*x+1/2*c)^4-2/d/a/(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)^2*tanh(1/2*d*x+1/2*c)^2-2
/d*b/a^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)^2*tanh(1/2*d*x+1/2*c)^2+1/2/d/a^3*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^3*ln(tanh(1/2*d*x+1/2*c)-1)
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Maxima [B] time = 1.29911, size = 282, normalized size = 3.48 \begin{align*} \frac{2 \,{\left ({\left (a^{2} + 2 \, a b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \,{\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} +{\left (a^{2} + 2 \, a b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} e^{\left (-8 \, d x - 8 \, c\right )} + a^{5} + 4 \,{\left (a^{5} + 2 \, a^{4} b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \,{\left (3 \, a^{5} + 8 \, a^{4} b + 8 \, a^{3} b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \,{\left (a^{5} + 2 \, a^{4} b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} + \frac{d x + c}{a^{3} 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^{3} 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)^3,x, algorithm="maxima")
[Out]
2*((a^2 + 2*a*b)*e^(-2*d*x - 2*c) + 2*(a^2 + 3*a*b + 3*b^2)*e^(-4*d*x - 4*c) + (a^2 + 2*a*b)*e^(-6*d*x - 6*c))
/((a^5*e^(-8*d*x - 8*c) + a^5 + 4*(a^5 + 2*a^4*b)*e^(-2*d*x - 2*c) + 2*(3*a^5 + 8*a^4*b + 8*a^3*b^2)*e^(-4*d*x
- 4*c) + 4*(a^5 + 2*a^4*b)*e^(-6*d*x - 6*c))*d) + (d*x + c)/(a^3*d) + 1/2*log(2*(a + 2*b)*e^(-2*d*x - 2*c) +
a*e^(-4*d*x - 4*c) + a)/(a^3*d)
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Fricas [B] time = 2.4993, size = 4297, normalized size = 53.05 \begin{align*} \text{result too large to display} \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)^3,x, algorithm="fricas")
[Out]
-1/2*(2*a^2*d*x*cosh(d*x + c)^8 + 16*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + 2*a^2*d*x*sinh(d*x + c)^8 + 4*(2*
(a^2 + 2*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^6 + 4*(14*a^2*d*x*cosh(d*x + c)^2 + 2*(a^2 + 2*a*b)*d*x - a^2 -
2*a*b)*sinh(d*x + c)^6 + 8*(14*a^2*d*x*cosh(d*x + c)^3 + 3*(2*(a^2 + 2*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c))
*sinh(d*x + c)^5 + 4*((3*a^2 + 8*a*b + 8*b^2)*d*x - 2*a^2 - 6*a*b - 6*b^2)*cosh(d*x + c)^4 + 4*(35*a^2*d*x*cos
h(d*x + c)^4 + (3*a^2 + 8*a*b + 8*b^2)*d*x + 15*(2*(a^2 + 2*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^2 - 2*a^2 -
6*a*b - 6*b^2)*sinh(d*x + c)^4 + 2*a^2*d*x + 16*(7*a^2*d*x*cosh(d*x + c)^5 + 5*(2*(a^2 + 2*a*b)*d*x - a^2 - 2*
a*b)*cosh(d*x + c)^3 + ((3*a^2 + 8*a*b + 8*b^2)*d*x - 2*a^2 - 6*a*b - 6*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 +
4*(2*(a^2 + 2*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^2 + 4*(14*a^2*d*x*cosh(d*x + c)^6 + 15*(2*(a^2 + 2*a*b)*d*
x - a^2 - 2*a*b)*cosh(d*x + c)^4 + 2*(a^2 + 2*a*b)*d*x + 6*((3*a^2 + 8*a*b + 8*b^2)*d*x - 2*a^2 - 6*a*b - 6*b^
2)*cosh(d*x + c)^2 - a^2 - 2*a*b)*sinh(d*x + c)^2 - (a^2*cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7
+ a^2*sinh(d*x + c)^8 + 4*(a^2 + 2*a*b)*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x +
c)^6 + 8*(7*a^2*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*a^2 + 8*a*b + 8*b^2)*c
osh(d*x + c)^4 + 2*(35*a^2*cosh(d*x + c)^4 + 30*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 3*a^2 + 8*a*b + 8*b^2)*sinh(d*
x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5 + 10*(a^2 + 2*a*b)*cosh(d*x + c)^3 + (3*a^2 + 8*a*b + 8*b^2)*cosh(d*x + c)
)*sinh(d*x + c)^3 + 4*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 4*(7*a^2*cosh(d*x + c)^6 + 15*(a^2 + 2*a*b)*cosh(d*x + c
)^4 + 3*(3*a^2 + 8*a*b + 8*b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*(a^2*cosh(d*x + c)^7
+ 3*(a^2 + 2*a*b)*cosh(d*x + c)^5 + (3*a^2 + 8*a*b + 8*b^2)*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sin
h(d*x + c))*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*(2*a^2*d*x*cosh(d*x + c)^7 + 3*(2*(a^2 + 2*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c
)^5 + 2*((3*a^2 + 8*a*b + 8*b^2)*d*x - 2*a^2 - 6*a*b - 6*b^2)*cosh(d*x + c)^3 + (2*(a^2 + 2*a*b)*d*x - a^2 - 2
*a*b)*cosh(d*x + c))*sinh(d*x + c))/(a^5*d*cosh(d*x + c)^8 + 8*a^5*d*cosh(d*x + c)*sinh(d*x + c)^7 + a^5*d*sin
h(d*x + c)^8 + 4*(a^5 + 2*a^4*b)*d*cosh(d*x + c)^6 + 4*(7*a^5*d*cosh(d*x + c)^2 + (a^5 + 2*a^4*b)*d)*sinh(d*x
+ c)^6 + a^5*d + 2*(3*a^5 + 8*a^4*b + 8*a^3*b^2)*d*cosh(d*x + c)^4 + 8*(7*a^5*d*cosh(d*x + c)^3 + 3*(a^5 + 2*a
^4*b)*d*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^5*d*cosh(d*x + c)^4 + 30*(a^5 + 2*a^4*b)*d*cosh(d*x + c)^2 +
(3*a^5 + 8*a^4*b + 8*a^3*b^2)*d)*sinh(d*x + c)^4 + 4*(a^5 + 2*a^4*b)*d*cosh(d*x + c)^2 + 8*(7*a^5*d*cosh(d*x +
c)^5 + 10*(a^5 + 2*a^4*b)*d*cosh(d*x + c)^3 + (3*a^5 + 8*a^4*b + 8*a^3*b^2)*d*cosh(d*x + c))*sinh(d*x + c)^3
+ 4*(7*a^5*d*cosh(d*x + c)^6 + 15*(a^5 + 2*a^4*b)*d*cosh(d*x + c)^4 + 3*(3*a^5 + 8*a^4*b + 8*a^3*b^2)*d*cosh(d
*x + c)^2 + (a^5 + 2*a^4*b)*d)*sinh(d*x + c)^2 + 8*(a^5*d*cosh(d*x + c)^7 + 3*(a^5 + 2*a^4*b)*d*cosh(d*x + c)^
5 + (3*a^5 + 8*a^4*b + 8*a^3*b^2)*d*cosh(d*x + c)^3 + (a^5 + 2*a^4*b)*d*cosh(d*x + c))*sinh(d*x + c))
________________________________________________________________________________________
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)**3,x)
[Out]
Timed out
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Giac [B] time = 2.75303, size = 236, normalized size = 2.91 \begin{align*} -\frac{\frac{4 \, d x}{a^{3}} - \frac{2 \, \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^{3}} + \frac{3 \, a e^{\left (8 \, d x + 8 \, c\right )} + 4 \, a e^{\left (6 \, d x + 6 \, c\right )} + 8 \, b e^{\left (6 \, d x + 6 \, c\right )} + 2 \, a e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a}{{\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 )}^{2} a^{2}}}{4 \, 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)^3,x, algorithm="giac")
[Out]
-1/4*(4*d*x/a^3 - 2*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^3 + (3*a*e^(8*d*x
+ 8*c) + 4*a*e^(6*d*x + 6*c) + 8*b*e^(6*d*x + 6*c) + 2*a*e^(4*d*x + 4*c) + 4*a*e^(2*d*x + 2*c) + 8*b*e^(2*d*x
+ 2*c) + 3*a)/((a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c) + a)^2*a^2))/d