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

Optimal. Leaf size=76 \[ \frac{\left (\frac{1}{a^2}-\frac{1}{b^2}\right ) \log \left (a \cosh ^2(c+d x)+b\right )}{2 d}+\frac{(a+b)^2}{2 a^2 b d \left (a \cosh ^2(c+d x)+b\right )}+\frac{\log (\cosh (c+d x))}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.116779, antiderivative size = 76, 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} \[ \frac{\left (\frac{1}{a^2}-\frac{1}{b^2}\right ) \log \left (a \cosh ^2(c+d x)+b\right )}{2 d}+\frac{(a+b)^2}{2 a^2 b d \left (a \cosh ^2(c+d x)+b\right )}+\frac{\log (\cosh (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.444285, size = 109, normalized size = 1.43 \[ \frac{\text{sech}^4(c+d x) (a \cosh (2 c+2 d x)+a+2 b)^2 \left (\left (\frac{1}{a^2}-\frac{1}{b^2}\right ) \log \left (a \cosh ^2(c+d x)+b\right )+\frac{(a+b)^2}{a^2 b \left (a \cosh ^2(c+d x)+b\right )}+\frac{2 \log (\cosh (c+d x))}{b^2}\right )}{8 d \left (a+b \text{sech}^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*c + 2*d*x])^2*((a + b)^2/(a^2*b*(b + a*Cosh[c + d*x]^2)) + (2*Log[Cosh[c + d*x]])/b^2 + (
a^(-2) - b^(-2))*Log[b + a*Cosh[c + d*x]^2])*Sech[c + d*x]^4)/(8*d*(a + b*Sech[c + d*x]^2)^2)

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Maple [B]  time = 0.083, size = 351, normalized size = 4.6 \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}}{bd \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{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 ) }-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 ) }+{\frac{1}{d{b}^{2}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{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)^2,x)

[Out]

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

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Maxima [B]  time = 1.63365, size = 208, normalized size = 2.74 \begin{align*} \frac{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} b e^{\left (-4 \, d x - 4 \, c\right )} + a^{3} b + 2 \,{\left (a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} + \frac{d x + c}{a^{2} d} + \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{2} d} - \frac{{\left (a^{2} - 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^{2} 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.70322, size = 2044, normalized size = 26.89 \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)^2,x, algorithm="fricas")

[Out]

-1/2*(2*a*b^2*d*x*cosh(d*x + c)^4 + 8*a*b^2*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*a*b^2*d*x*sinh(d*x + c)^4 +
2*a*b^2*d*x - 4*(a^2*b + 2*a*b^2 + b^3 - (a*b^2 + 2*b^3)*d*x)*cosh(d*x + c)^2 + 4*(3*a*b^2*d*x*cosh(d*x + c)^2
- a^2*b - 2*a*b^2 - b^3 + (a*b^2 + 2*b^3)*d*x)*sinh(d*x + c)^2 + ((a^3 - a*b^2)*cosh(d*x + c)^4 + 4*(a^3 - a*
b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 - a*b^2)*sinh(d*x + c)^4 + a^3 - a*b^2 + 2*(a^3 + 2*a^2*b - a*b^2 -
2*b^3)*cosh(d*x + c)^2 + 2*(a^3 + 2*a^2*b - a*b^2 - 2*b^3 + 3*(a^3 - a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 +
4*((a^3 - a*b^2)*cosh(d*x + c)^3 + (a^3 + 2*a^2*b - a*b^2 - 2*b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*(a*cos
h(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^3*cosh(d*x + c)^4 + 4*a^3*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*sinh(d*x + c)^4 + a^3 + 2*(a^3 + 2*a^2
*b)*cosh(d*x + c)^2 + 2*(3*a^3*cosh(d*x + c)^2 + a^3 + 2*a^2*b)*sinh(d*x + c)^2 + 4*(a^3*cosh(d*x + c)^3 + (a^
3 + 2*a^2*b)*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*(a*b^2*d*x
*cosh(d*x + c)^3 - (a^2*b + 2*a*b^2 + b^3 - (a*b^2 + 2*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/(a^3*b^2*d*cosh
(d*x + c)^4 + 4*a^3*b^2*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*b^2*d*sinh(d*x + c)^4 + a^3*b^2*d + 2*(a^3*b^2 +
2*a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*a^3*b^2*d*cosh(d*x + c)^2 + (a^3*b^2 + 2*a^2*b^3)*d)*sinh(d*x + c)^2 + 4*
(a^3*b^2*d*cosh(d*x + c)^3 + (a^3*b^2 + 2*a^2*b^3)*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 )}}{\left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right )^{2}}\, 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)**2,x)

[Out]

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

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Giac [B]  time = 2.99672, size = 281, normalized size = 3.7 \begin{align*} -\frac{\frac{2 \, d x}{a^{2}} - \frac{2 \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{b^{2}} + \frac{{\left (a^{2} - 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^{2} b^{2}} - \frac{a^{2} e^{\left (4 \, d x + 4 \, c\right )} - b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} - b^{2}}{{\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}}}{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)^2,x, algorithm="giac")

[Out]

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