3.138 \(\int \frac{\tanh ^5(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=70 \[ \frac{(a+b)^2 \log \left (a \cosh ^2(c+d x)+b\right )}{2 a b^2 d}-\frac{(a+2 b) \log (\cosh (c+d x))}{b^2 d}-\frac{\text{sech}^2(c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.111549, antiderivative size = 70, 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, 88} \[ \frac{(a+b)^2 \log \left (a \cosh ^2(c+d x)+b\right )}{2 a b^2 d}-\frac{(a+2 b) \log (\cosh (c+d x))}{b^2 d}-\frac{\text{sech}^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[c + d*x]^5/(a + b*Sech[c + d*x]^2),x]

[Out]

-(((a + 2*b)*Log[Cosh[c + d*x]])/(b^2*d)) + ((a + b)^2*Log[b + a*Cosh[c + d*x]^2])/(2*a*b^2*d) - Sech[c + d*x]
^2/(2*b*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\tanh ^5(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^3 \left (b+a x^2\right )} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^2 (b+a x)} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{b x^2}+\frac{-a-2 b}{b^2 x}+\frac{(a+b)^2}{b^2 (b+a x)}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{(a+2 b) \log (\cosh (c+d x))}{b^2 d}+\frac{(a+b)^2 \log \left (b+a \cosh ^2(c+d x)\right )}{2 a b^2 d}-\frac{\text{sech}^2(c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.289927, size = 98, normalized size = 1.4 \[ -\frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (a b \text{sech}^2(c+d x)+(a+b)^2 \left (-\log \left (a \sinh ^2(c+d x)+a+b\right )\right )+2 a (a+2 b) \log (\cosh (c+d x))\right )}{4 a b^2 d \left (a+b \text{sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[c + d*x]^5/(a + b*Sech[c + d*x]^2),x]

[Out]

-((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(2*a*(a + 2*b)*Log[Cosh[c + d*x]] - (a + b)^2*Log[a + b + a*
Sinh[c + d*x]^2] + a*b*Sech[c + d*x]^2))/(4*a*b^2*d*(a + b*Sech[c + d*x]^2))

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Maple [B]  time = 0.062, size = 331, normalized size = 4.7 \begin{align*} -{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{a}{2\,d{b}^{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}{bd}\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}{2\,da}\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}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-2\,{\frac{1}{bd \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{a}{d{b}^{2}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) }-2\,{\frac{\ln \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }{bd}}+2\,{\frac{1}{bd \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.76718, size = 177, normalized size = 2.53 \begin{align*} \frac{d x + c}{a d} - \frac{2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (2 \, b e^{\left (-2 \, d x - 2 \, c\right )} + b e^{\left (-4 \, d x - 4 \, c\right )} + b\right )} d} - \frac{{\left (a + 2 \, b\right )} \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{2} d} + \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \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 b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.66465, size = 1895, normalized size = 27.07 \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)^5/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(2*b^2*d*x*cosh(d*x + c)^4 + 8*b^2*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*b^2*d*x*sinh(d*x + c)^4 + 2*b^2*
d*x + 4*(b^2*d*x + a*b)*cosh(d*x + c)^2 + 4*(3*b^2*d*x*cosh(d*x + c)^2 + b^2*d*x + a*b)*sinh(d*x + c)^2 - ((a^
2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*s
inh(d*x + c)^4 + 2*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*
b + b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*co
sh(d*x + c))*sinh(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)) + 2*((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*
sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b)*cosh(d*
x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(
d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*(b^2*d*x*cosh(d*x + c)^3 + (
b^2*d*x + a*b)*cosh(d*x + c))*sinh(d*x + c))/(a*b^2*d*cosh(d*x + c)^4 + 4*a*b^2*d*cosh(d*x + c)*sinh(d*x + c)^
3 + a*b^2*d*sinh(d*x + c)^4 + 2*a*b^2*d*cosh(d*x + c)^2 + a*b^2*d + 2*(3*a*b^2*d*cosh(d*x + c)^2 + a*b^2*d)*si
nh(d*x + c)^2 + 4*(a*b^2*d*cosh(d*x + c)^3 + a*b^2*d*cosh(d*x + c))*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{5}{\left (c + d x \right )}}{a + b \operatorname{sech}^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)**5/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(tanh(c + d*x)**5/(a + b*sech(c + d*x)**2), x)

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Giac [B]  time = 2.24284, size = 240, normalized size = 3.43 \begin{align*} -\frac{\frac{2 \, d x}{a} + \frac{2 \,{\left (a e^{\left (2 \, c\right )} + 2 \, b e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, c\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{b^{2}} - \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \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 b^{2}} - \frac{3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a + 6 \, b}{b^{2}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

-1/2*(2*d*x/a + 2*(a*e^(2*c) + 2*b*e^(2*c))*e^(-2*c)*log(e^(2*d*x + 2*c) + 1)/b^2 - (a^2 + 2*a*b + b^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*b^2) - (3*a*e^(4*d*x + 4*c) + 6*b*e^(4*d*x
+ 4*c) + 6*a*e^(2*d*x + 2*c) + 8*b*e^(2*d*x + 2*c) + 3*a + 6*b)/(b^2*(e^(2*d*x + 2*c) + 1)^2))/d