3.149 \(\int \frac{\tanh ^4(c+d x)}{(a+b \text{sech}^2(c+d x))^2} \, dx\)

Optimal. Leaf size=91 \[ \frac{(a-2 b) \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 b^{3/2} d}+\frac{x}{a^2}-\frac{(a+b) \tanh (c+d x)}{2 a b d \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

x/a^2 + ((a - 2*b)*Sqrt[a + b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*a^2*b^(3/2)*d) - ((a + b)*Tanh
[c + d*x])/(2*a*b*d*(a + b - b*Tanh[c + d*x]^2))

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Rubi [A]  time = 0.194069, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4141, 1975, 470, 522, 206, 208} \[ \frac{(a-2 b) \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 b^{3/2} d}+\frac{x}{a^2}-\frac{(a+b) \tanh (c+d x)}{2 a b d \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[c + d*x]^4/(a + b*Sech[c + d*x]^2)^2,x]

[Out]

x/a^2 + ((a - 2*b)*Sqrt[a + b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*a^2*b^(3/2)*d) - ((a + b)*Tanh
[c + d*x])/(2*a*b*d*(a + b - b*Tanh[c + d*x]^2))

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tanh ^4(c+d x)}{\left (a+b \text{sech}^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right ) \left (a+b \left (1-x^2\right )\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right ) \left (a+b-b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{(a+b) \tanh (c+d x)}{2 a b d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{a+b+(-a+b) x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 a b d}\\ &=-\frac{(a+b) \tanh (c+d x)}{2 a b d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{a^2 d}+\frac{((a-2 b) (a+b)) \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{2 a^2 b d}\\ &=\frac{x}{a^2}+\frac{(a-2 b) \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 b^{3/2} d}-\frac{(a+b) \tanh (c+d x)}{2 a b d \left (a+b-b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 2.10837, size = 228, normalized size = 2.51 \[ \frac{\text{sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\frac{\left (a^2-a b-2 b^2\right ) (\cosh (2 c)-\sinh (2 c)) (a \cosh (2 (c+d x))+a+2 b) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )}{b d \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}+2 x (a \cosh (2 (c+d x))+a+2 b)+\frac{(a+b) \text{sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{b d}\right )}{8 a^2 \left (a+b \text{sech}^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[c + d*x]^4/(a + b*Sech[c + d*x]^2)^2,x]

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^4*(2*x*(a + 2*b + a*Cosh[2*(c + d*x)]) + ((a^2 - a*b - 2*b^2)*A
rcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Co
sh[c] - Sinh[c])^4])]*(a + 2*b + a*Cosh[2*(c + d*x)])*(Cosh[2*c] - Sinh[2*c]))/(b*Sqrt[a + b]*d*Sqrt[b*(Cosh[c
] - Sinh[c])^4]) + ((a + b)*Sech[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/(b*d)))/(8*a^2*(a + b*Sech[c + d*
x]^2)^2)

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Maple [B]  time = 0.086, size = 667, normalized size = 7.3 \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)^4/(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/(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)/b*tanh(1/2*d*x+1/2*c)^3-1/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)*tanh(1/2*d*x+1/2*c)^3-1/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)/b*tanh(1/2*d*x+1/2*c)-1/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)*tanh
(1/2*d*x+1/2*c)+1/4/d/b^(3/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(
a+b)^(1/2))-1/4/d/a/b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+
b)^(1/2))-1/2/d*b^(1/2)/a^2/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+
b)^(1/2))-1/4/d/b^(3/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(
1/2))+1/4/d/a/b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/
2))+1/2/d*b^(1/2)/a^2/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/
2))-1/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.38887, size = 3748, normalized size = 41.19 \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)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(4*a*b*d*x*cosh(d*x + c)^4 + 16*a*b*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 4*a*b*d*x*sinh(d*x + c)^4 + 4*a*b
*d*x + 4*(2*(a*b + 2*b^2)*d*x + a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^2 + 4*(6*a*b*d*x*cosh(d*x + c)^2 + 2*(a*b +
2*b^2)*d*x + a^2 + 3*a*b + 2*b^2)*sinh(d*x + c)^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 - 4*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 - 2*a*b)*c
osh(d*x + c)^2 + a^2 - 4*b^2)*sinh(d*x + c)^2 + a^2 - 2*a*b + 4*((a^2 - 2*a*b)*cosh(d*x + c)^3 + (a^2 - 4*b^2)
*cosh(d*x + c))*sinh(d*x + c))*sqrt((a + b)/b)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3
+ a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c
)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*(a*b*cosh(
d*x + c)^2 + 2*a*b*cosh(d*x + c)*sinh(d*x + c) + a*b*sinh(d*x + c)^2 + a*b + 2*b^2)*sqrt((a + b)/b))/(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)) +
4*a^2 + 4*a*b + 8*(2*a*b*d*x*cosh(d*x + c)^3 + (2*(a*b + 2*b^2)*d*x + a^2 + 3*a*b + 2*b^2)*cosh(d*x + c))*sinh
(d*x + c))/(a^3*b*d*cosh(d*x + c)^4 + 4*a^3*b*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*b*d*sinh(d*x + c)^4 + a^3*
b*d + 2*(a^3*b + 2*a^2*b^2)*d*cosh(d*x + c)^2 + 2*(3*a^3*b*d*cosh(d*x + c)^2 + (a^3*b + 2*a^2*b^2)*d)*sinh(d*x
+ c)^2 + 4*(a^3*b*d*cosh(d*x + c)^3 + (a^3*b + 2*a^2*b^2)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*a*b*d*x*cos
h(d*x + c)^4 + 8*a*b*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*a*b*d*x*sinh(d*x + c)^4 + 2*a*b*d*x + 2*(2*(a*b + 2
*b^2)*d*x + a^2 + 3*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(6*a*b*d*x*cosh(d*x + c)^2 + 2*(a*b + 2*b^2)*d*x + a^2 +
3*a*b + 2*b^2)*sinh(d*x + c)^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 - 4*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 - 2*a*b)*cosh(d*x + c)^2 + a^2
- 4*b^2)*sinh(d*x + c)^2 + a^2 - 2*a*b + 4*((a^2 - 2*a*b)*cosh(d*x + c)^3 + (a^2 - 4*b^2)*cosh(d*x + c))*sinh
(d*x + c))*sqrt(-(a + b)/b)*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^
2 + a + 2*b)*sqrt(-(a + b)/b)/(a + b)) + 2*a^2 + 2*a*b + 4*(2*a*b*d*x*cosh(d*x + c)^3 + (2*(a*b + 2*b^2)*d*x +
a^2 + 3*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))/(a^3*b*d*cosh(d*x + c)^4 + 4*a^3*b*d*cosh(d*x + c)*sinh(d*
x + c)^3 + a^3*b*d*sinh(d*x + c)^4 + a^3*b*d + 2*(a^3*b + 2*a^2*b^2)*d*cosh(d*x + c)^2 + 2*(3*a^3*b*d*cosh(d*x
+ c)^2 + (a^3*b + 2*a^2*b^2)*d)*sinh(d*x + c)^2 + 4*(a^3*b*d*cosh(d*x + c)^3 + (a^3*b + 2*a^2*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 ^{4}{\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)**4/(a+b*sech(d*x+c)**2)**2,x)

[Out]

Integral(tanh(c + d*x)**4/(a + b*sech(c + d*x)**2)**2, x)

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Giac [B]  time = 2.5214, size = 252, normalized size = 2.77 \begin{align*} \frac{\frac{2 \, d x}{a^{2}} + \frac{{\left (a^{2} e^{\left (2 \, c\right )} - a b e^{\left (2 \, c\right )} - 2 \, b^{2} e^{\left (2 \, c\right )}\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right ) e^{\left (-2 \, c\right )}}{\sqrt{-a b - b^{2}} a^{2} b} + \frac{2 \,{\left (a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + a b\right )}}{{\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 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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