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]

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

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Rubi [A]  time = 0.13, antiderivative size = 81, 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, 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 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 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)}{\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*}

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Mathematica [A]  time = 1.57, 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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fricas [B]  time = 0.45, size = 1753, normalized size = 21.64 \[ \text {result too large to display} \]

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))

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giac [B]  time = 1.22, size = 175, normalized size = 2.16 \[ -\frac {\frac {4 \, d x}{a^{3}} - \frac {2 \, \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{a^{3}} + \frac {3 \, a e^{\left (8 \, d x + 8 \, c\right )} + 4 \, a e^{\left (6 \, d x + 6 \, c\right )} + 8 \, b e^{\left (6 \, d x + 6 \, c\right )} + 2 \, a e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}^{2} a^{2}}}{4 \, d} \]

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

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maple [B]  time = 0.40, size = 672, normalized size = 8.30 \[ \text {result too large to display} \]

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)-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/a^2*b/(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/a*b/(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/a^2*b^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/(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/a^2*b/(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)

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maxima [B]  time = 0.34, size = 209, normalized size = 2.58 \[ \frac {2 \, {\left ({\left (a^{2} + 2 \, a b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{2} + 2 \, a b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} e^{\left (-8 \, d x - 8 \, c\right )} + a^{5} + 4 \, {\left (a^{5} + 2 \, a^{4} b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} + 8 \, a^{4} b + 8 \, a^{3} b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} + 2 \, a^{4} b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} + \frac {d x + c}{a^{3} d} + \frac {\log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, a^{3} d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6\,{\mathrm {tanh}\left (c+d\,x\right )}^3}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cosh(c + d*x)^6*tanh(c + d*x)^3)/(b + a*cosh(c + d*x)^2)^3, x)

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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*sech(d*x+c)**2)**3,x)

[Out]

Timed out

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