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

Optimal. Leaf size=87 \[ -\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a d (a+b)^{5/2}}-\frac{\coth ^3(c+d x)}{3 d (a+b)}-\frac{(a+2 b) \coth (c+d x)}{d (a+b)^2}+\frac{x}{a} \]

[Out]

x/a - (b^(5/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*(a + b)^(5/2)*d) - ((a + 2*b)*Coth[c + d*x])/(
(a + b)^2*d) - Coth[c + d*x]^3/(3*(a + b)*d)

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Rubi [A]  time = 0.285895, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4141, 1975, 480, 583, 522, 206, 208} \[ -\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a d (a+b)^{5/2}}-\frac{\coth ^3(c+d x)}{3 d (a+b)}-\frac{(a+2 b) \coth (c+d x)}{d (a+b)^2}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a - (b^(5/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*(a + b)^(5/2)*d) - ((a + 2*b)*Coth[c + d*x])/(
(a + b)^2*d) - Coth[c + d*x]^3/(3*(a + b)*d)

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 480

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

Rule 583

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

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{\coth ^4(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^4 \left (1-x^2\right ) \left (a+b \left (1-x^2\right )\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^4 \left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{\coth ^3(c+d x)}{3 (a+b) d}+\frac{\operatorname{Subst}\left (\int \frac{3 (a+2 b)-3 b x^2}{x^2 \left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{3 (a+b) d}\\ &=-\frac{(a+2 b) \coth (c+d x)}{(a+b)^2 d}-\frac{\coth ^3(c+d x)}{3 (a+b) d}-\frac{\operatorname{Subst}\left (\int \frac{-3 \left (a^2+3 a b+3 b^2\right )+3 b (a+2 b) x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{3 (a+b)^2 d}\\ &=-\frac{(a+2 b) \coth (c+d x)}{(a+b)^2 d}-\frac{\coth ^3(c+d x)}{3 (a+b) d}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{a d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{a (a+b)^2 d}\\ &=\frac{x}{a}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a (a+b)^{5/2} d}-\frac{(a+2 b) \coth (c+d x)}{(a+b)^2 d}-\frac{\coth ^3(c+d x)}{3 (a+b) d}\\ \end{align*}

Mathematica [B]  time = 3.26882, size = 380, normalized size = 4.37 \[ \frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\frac{1}{8} \sqrt{a+b} \text{csch}(c) \sqrt{b (\cosh (c)-\sinh (c))^4} \text{csch}^3(c+d x) \left (-12 a^2 \sinh (2 c+d x)+8 a^2 \sinh (2 c+3 d x)-3 a^2 d x \cosh (2 c+3 d x)+3 a^2 d x \cosh (4 c+3 d x)-12 a^2 \sinh (d x)-18 a b \sinh (2 c+d x)+14 a b \sinh (2 c+3 d x)-6 a b d x \cosh (2 c+3 d x)+6 a b d x \cosh (4 c+3 d x)-9 d x (a+b)^2 \cosh (2 c+d x)-24 a b \sinh (d x)+9 d x (a+b)^2 \cosh (d x)-3 b^2 d x \cosh (2 c+3 d x)+3 b^2 d x \cosh (4 c+3 d x)\right )+3 b^3 (\sinh (2 c)-\cosh (2 c)) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )\right )}{6 a d (a+b)^{5/2} \sqrt{b (\cosh (c)-\sinh (c))^4} \left (a+b \text{sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(3*b^3*ArcTanh[(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*(Cosh[c] - Sinh[c])^4])]*(-Cosh[2*c] + Sinh[2*c]) + (Sqr
t[a + b]*Csch[c]*Csch[c + d*x]^3*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(9*(a + b)^2*d*x*Cosh[d*x] - 9*(a + b)^2*d*x*Co
sh[2*c + d*x] - 3*a^2*d*x*Cosh[2*c + 3*d*x] - 6*a*b*d*x*Cosh[2*c + 3*d*x] - 3*b^2*d*x*Cosh[2*c + 3*d*x] + 3*a^
2*d*x*Cosh[4*c + 3*d*x] + 6*a*b*d*x*Cosh[4*c + 3*d*x] + 3*b^2*d*x*Cosh[4*c + 3*d*x] - 12*a^2*Sinh[d*x] - 24*a*
b*Sinh[d*x] - 12*a^2*Sinh[2*c + d*x] - 18*a*b*Sinh[2*c + d*x] + 8*a^2*Sinh[2*c + 3*d*x] + 14*a*b*Sinh[2*c + 3*
d*x]))/8))/(6*a*(a + b)^(5/2)*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])

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Maple [B]  time = 0.085, size = 301, normalized size = 3.5 \begin{align*} -{\frac{a}{24\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{b}{24\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{5\,a}{8\,d \left ( a+b \right ) ^{2}}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{9\,b}{8\,d \left ( a+b \right ) ^{2}}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{2\,da}{b}^{{\frac{5}{2}}}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}}+{\frac{1}{2\,da}{b}^{{\frac{5}{2}}}\ln \left ( -\sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}-\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}}-{\frac{1}{24\,d \left ( a+b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{5\,a}{8\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{9\,b}{8\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^4/(a+b*sech(d*x+c)^2),x)

[Out]

-1/24/d/(a+b)^2*a*tanh(1/2*d*x+1/2*c)^3-1/24/d/(a+b)^2*b*tanh(1/2*d*x+1/2*c)^3-5/8/d/(a+b)^2*a*tanh(1/2*d*x+1/
2*c)-9/8/d/(a+b)^2*tanh(1/2*d*x+1/2*c)*b+1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)-1/2/d*b^(5/2)/a/(a+b)^(5/2)*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^(5/2)/a/(a+b)^(5/2)*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))-1/24/d/(a+b)/tanh(1/2*d*x+1/2*c)^3-5/8/d
/(a+b)^2/tanh(1/2*d*x+1/2*c)*a-9/8/d/(a+b)^2/tanh(1/2*d*x+1/2*c)*b-1/d/a*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(coth(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.62148, size = 6730, normalized size = 77.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(6*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 + 36*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^5 + 6
*(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^6 - 6*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c)^4 + 6*(
15*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 - 3*(a^2 + 2*a*b + b^2)*d*x - 4*a^2 - 6*a*b)*sinh(d*x + c)^4 + 24*(
5*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^3 - (3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c))*sinh(d*
x + c)^3 - 6*(a^2 + 2*a*b + b^2)*d*x + 6*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 8*a*b)*cosh(d*x + c)^2 + 6*(15*(
a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 3*(a^2 + 2*a*b + b^2)*d*x - 6*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*
a*b)*cosh(d*x + c)^2 + 4*a^2 + 8*a*b)*sinh(d*x + c)^2 + 3*(b^2*cosh(d*x + c)^6 + 6*b^2*cosh(d*x + c)*sinh(d*x
+ c)^5 + b^2*sinh(d*x + c)^6 - 3*b^2*cosh(d*x + c)^4 + 3*(5*b^2*cosh(d*x + c)^2 - b^2)*sinh(d*x + c)^4 + 3*b^2
*cosh(d*x + c)^2 + 4*(5*b^2*cosh(d*x + c)^3 - 3*b^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^2*cosh(d*x + c)^4
- 6*b^2*cosh(d*x + c)^2 + b^2)*sinh(d*x + c)^2 - b^2 + 6*(b^2*cosh(d*x + c)^5 - 2*b^2*cosh(d*x + c)^3 + b^2*co
sh(d*x + c))*sinh(d*x + c))*sqrt(b/(a + 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^2 + a*b)*
cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^
2)*sqrt(b/(a + 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)*c
osh(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)) - 16*a^2 - 28*a*b + 12*(3*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^5 - 2*(3*(a^2 + 2*a
*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c)^3 + (3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 8*a*b)*cosh(d*x + c))*si
nh(d*x + c))/((a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^6 + 6*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)*sinh(d*x +
c)^5 + (a^3 + 2*a^2*b + a*b^2)*d*sinh(d*x + c)^6 - 3*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^4 + 3*(5*(a^3 +
2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 - (a^3 + 2*a^2*b + a*b^2)*d)*sinh(d*x + c)^4 + 3*(a^3 + 2*a^2*b + a*b^2)*d*
cosh(d*x + c)^2 + 4*(5*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^3 - 3*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c))*
sinh(d*x + c)^3 + 3*(5*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^4 - 6*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2
+ (a^3 + 2*a^2*b + a*b^2)*d)*sinh(d*x + c)^2 - (a^3 + 2*a^2*b + a*b^2)*d + 6*((a^3 + 2*a^2*b + a*b^2)*d*cosh(
d*x + c)^5 - 2*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^3 + (a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c))*sinh(d*x +
c)), 1/3*(3*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 + 18*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^
5 + 3*(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^6 - 3*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c)^4
+ 3*(15*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 - 3*(a^2 + 2*a*b + b^2)*d*x - 4*a^2 - 6*a*b)*sinh(d*x + c)^4 +
12*(5*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^3 - (3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c))*si
nh(d*x + c)^3 - 3*(a^2 + 2*a*b + b^2)*d*x + 3*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 8*a*b)*cosh(d*x + c)^2 + 3*
(15*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 3*(a^2 + 2*a*b + b^2)*d*x - 6*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2
+ 6*a*b)*cosh(d*x + c)^2 + 4*a^2 + 8*a*b)*sinh(d*x + c)^2 - 3*(b^2*cosh(d*x + c)^6 + 6*b^2*cosh(d*x + c)*sinh
(d*x + c)^5 + b^2*sinh(d*x + c)^6 - 3*b^2*cosh(d*x + c)^4 + 3*(5*b^2*cosh(d*x + c)^2 - b^2)*sinh(d*x + c)^4 +
3*b^2*cosh(d*x + c)^2 + 4*(5*b^2*cosh(d*x + c)^3 - 3*b^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^2*cosh(d*x +
c)^4 - 6*b^2*cosh(d*x + c)^2 + b^2)*sinh(d*x + c)^2 - b^2 + 6*(b^2*cosh(d*x + c)^5 - 2*b^2*cosh(d*x + c)^3 + b
^2*cosh(d*x + c))*sinh(d*x + c))*sqrt(-b/(a + 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(-b/(a + b))/b) - 8*a^2 - 14*a*b + 6*(3*(a^2 + 2*a*b + b^2)*d*x*cosh(d*
x + c)^5 - 2*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c)^3 + (3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2
+ 8*a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^6 + 6*(a^3 + 2*a^2*b + a*b^2)*
d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^3 + 2*a^2*b + a*b^2)*d*sinh(d*x + c)^6 - 3*(a^3 + 2*a^2*b + a*b^2)*d*cosh
(d*x + c)^4 + 3*(5*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 - (a^3 + 2*a^2*b + a*b^2)*d)*sinh(d*x + c)^4 + 3*
(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 + 4*(5*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^3 - 3*(a^3 + 2*a^2*b
+ a*b^2)*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^4 - 6*(a^3 + 2*a^2*b
+ a*b^2)*d*cosh(d*x + c)^2 + (a^3 + 2*a^2*b + a*b^2)*d)*sinh(d*x + c)^2 - (a^3 + 2*a^2*b + a*b^2)*d + 6*((a^3
+ 2*a^2*b + a*b^2)*d*cosh(d*x + c)^5 - 2*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^3 + (a^3 + 2*a^2*b + 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{\coth ^{4}{\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(coth(d*x+c)**4/(a+b*sech(d*x+c)**2),x)

[Out]

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

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Giac [B]  time = 2.26468, size = 217, normalized size = 2.49 \begin{align*} -\frac{\frac{3 \, b^{3} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt{-a b - b^{2}}} - \frac{3 \, d x}{a} + \frac{2 \,{\left (6 \, a e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 12 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 7 \, b\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*b^3*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a^3 + 2*a^2*b + a*b^2)*sqrt(-a*b - b^
2)) - 3*d*x/a + 2*(6*a*e^(4*d*x + 4*c) + 9*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) - 12*b*e^(2*d*x + 2*c) + 4*
a + 7*b)/((a^2 + 2*a*b + b^2)*(e^(2*d*x + 2*c) - 1)^3))/d