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]

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

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Rubi [A]  time = 0.09, antiderivative size = 51, 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, 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 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])

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 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 )^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*}

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Mathematica [A]  time = 0.76, 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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fricas [B]  time = 0.42, size = 485, normalized size = 9.51 \[ -\frac {2 \, a d x \cosh \left (d x + c\right )^{4} + 8 \, a d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, a d x \sinh \left (d x + c\right )^{4} + 2 \, a d x + 4 \, {\left ({\left (a + 2 \, b\right )} d x - a - b\right )} \cosh \left (d x + c\right )^{2} + 4 \, {\left (3 \, a d x \cosh \left (d x + c\right )^{2} + {\left (a + 2 \, b\right )} d x - a - b\right )} \sinh \left (d x + c\right )^{2} - {\left (a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} + 2 \, {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a \cosh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a \cosh \left (d x + c\right )^{3} + {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a\right )} \log \left (\frac {2 \, {\left (a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )}}{\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right ) + 8 \, {\left (a d x \cosh \left (d x + c\right )^{3} + {\left ({\left (a + 2 \, b\right )} d x - a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left (a^{3} d \cosh \left (d x + c\right )^{4} + 4 \, a^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{3} d \sinh \left (d x + c\right )^{4} + a^{3} d + 2 \, {\left (a^{3} + 2 \, a^{2} b\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} d \cosh \left (d x + c\right )^{2} + {\left (a^{3} + 2 \, a^{2} b\right )} d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a^{3} d \cosh \left (d x + c\right )^{3} + {\left (a^{3} + 2 \, a^{2} b\right )} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]

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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giac [B]  time = 0.89, size = 121, normalized size = 2.37 \[ -\frac {\frac {2 \, d x}{a^{2}} + \frac {e^{\left (4 \, d x + 4 \, c\right )} - 2 \, e^{\left (2 \, d x + 2 \, c\right )} + 1}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} a} - \frac {\log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{a^{2}}}{2 \, d} \]

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

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maple [B]  time = 0.35, size = 186, normalized size = 3.65 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {\ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}{2 d \,a^{2}} \]

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)-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)

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maxima [B]  time = 0.38, size = 108, normalized size = 2.12 \[ \frac {2 \, {\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} e^{\left (-4 \, d x - 4 \, c\right )} + a^{3} + 2 \, {\left (a^{3} + 2 \, a^{2} b\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} + \frac {d x + c}{a^{2} d} + \frac {\log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, a^{2} d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\mathrm {tanh}\left (c+d\,x\right )}^3}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cosh(c + d*x)^4*tanh(c + d*x)^3)/(b + a*cosh(c + d*x)^2)^2, 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)**2,x)

[Out]

Timed out

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