3.8 \(\int \frac{1}{(a+b \coth ^2(c+d x))^4} \, dx\)

Optimal. Leaf size=201 \[ \frac{b \left (19 a^2+16 a b+5 b^2\right ) \coth (c+d x)}{16 a^3 d (a+b)^3 \left (a+b \coth ^2(c+d x)\right )}-\frac{\sqrt{b} \left (35 a^2 b+35 a^3+21 a b^2+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{b}}\right )}{16 a^{7/2} d (a+b)^4}+\frac{b (11 a+5 b) \coth (c+d x)}{24 a^2 d (a+b)^2 \left (a+b \coth ^2(c+d x)\right )^2}+\frac{b \coth (c+d x)}{6 a d (a+b) \left (a+b \coth ^2(c+d x)\right )^3}+\frac{x}{(a+b)^4} \]

[Out]

x/(a + b)^4 - (Sqrt[b]*(35*a^3 + 35*a^2*b + 21*a*b^2 + 5*b^3)*ArcTan[(Sqrt[a]*Tanh[c + d*x])/Sqrt[b]])/(16*a^(
7/2)*(a + b)^4*d) + (b*Coth[c + d*x])/(6*a*(a + b)*d*(a + b*Coth[c + d*x]^2)^3) + (b*(11*a + 5*b)*Coth[c + d*x
])/(24*a^2*(a + b)^2*d*(a + b*Coth[c + d*x]^2)^2) + (b*(19*a^2 + 16*a*b + 5*b^2)*Coth[c + d*x])/(16*a^3*(a + b
)^3*d*(a + b*Coth[c + d*x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.278509, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3661, 414, 527, 522, 206, 205} \[ \frac{b \left (19 a^2+16 a b+5 b^2\right ) \coth (c+d x)}{16 a^3 d (a+b)^3 \left (a+b \coth ^2(c+d x)\right )}-\frac{\sqrt{b} \left (35 a^2 b+35 a^3+21 a b^2+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{b}}\right )}{16 a^{7/2} d (a+b)^4}+\frac{b (11 a+5 b) \coth (c+d x)}{24 a^2 d (a+b)^2 \left (a+b \coth ^2(c+d x)\right )^2}+\frac{b \coth (c+d x)}{6 a d (a+b) \left (a+b \coth ^2(c+d x)\right )^3}+\frac{x}{(a+b)^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x]^2)^(-4),x]

[Out]

x/(a + b)^4 - (Sqrt[b]*(35*a^3 + 35*a^2*b + 21*a*b^2 + 5*b^3)*ArcTan[(Sqrt[a]*Tanh[c + d*x])/Sqrt[b]])/(16*a^(
7/2)*(a + b)^4*d) + (b*Coth[c + d*x])/(6*a*(a + b)*d*(a + b*Coth[c + d*x]^2)^3) + (b*(11*a + 5*b)*Coth[c + d*x
])/(24*a^2*(a + b)^2*d*(a + b*Coth[c + d*x]^2)^2) + (b*(19*a^2 + 16*a*b + 5*b^2)*Coth[c + d*x])/(16*a^3*(a + b
)^3*d*(a + b*Coth[c + d*x]^2))

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
 n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -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 205

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \coth ^2(c+d x)\right )^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \left (a+b x^2\right )^4} \, dx,x,\coth (c+d x)\right )}{d}\\ &=\frac{b \coth (c+d x)}{6 a (a+b) d \left (a+b \coth ^2(c+d x)\right )^3}-\frac{\operatorname{Subst}\left (\int \frac{b-6 (a+b)+5 b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\coth (c+d x)\right )}{6 a (a+b) d}\\ &=\frac{b \coth (c+d x)}{6 a (a+b) d \left (a+b \coth ^2(c+d x)\right )^3}+\frac{b (11 a+5 b) \coth (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \coth ^2(c+d x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{3 \left (8 a^2+11 a b+5 b^2\right )-3 b (11 a+5 b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\coth (c+d x)\right )}{24 a^2 (a+b)^2 d}\\ &=\frac{b \coth (c+d x)}{6 a (a+b) d \left (a+b \coth ^2(c+d x)\right )^3}+\frac{b (11 a+5 b) \coth (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \coth ^2(c+d x)\right )^2}+\frac{b \left (19 a^2+16 a b+5 b^2\right ) \coth (c+d x)}{16 a^3 (a+b)^3 d \left (a+b \coth ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 \left (16 a^3+19 a^2 b+16 a b^2+5 b^3\right )+3 b \left (19 a^2+16 a b+5 b^2\right ) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\coth (c+d x)\right )}{48 a^3 (a+b)^3 d}\\ &=\frac{b \coth (c+d x)}{6 a (a+b) d \left (a+b \coth ^2(c+d x)\right )^3}+\frac{b (11 a+5 b) \coth (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \coth ^2(c+d x)\right )^2}+\frac{b \left (19 a^2+16 a b+5 b^2\right ) \coth (c+d x)}{16 a^3 (a+b)^3 d \left (a+b \coth ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\coth (c+d x)\right )}{(a+b)^4 d}+\frac{\left (b \left (35 a^3+35 a^2 b+21 a b^2+5 b^3\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\coth (c+d x)\right )}{16 a^3 (a+b)^4 d}\\ &=\frac{x}{(a+b)^4}-\frac{\sqrt{b} \left (35 a^3+35 a^2 b+21 a b^2+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{b}}\right )}{16 a^{7/2} (a+b)^4 d}+\frac{b \coth (c+d x)}{6 a (a+b) d \left (a+b \coth ^2(c+d x)\right )^3}+\frac{b (11 a+5 b) \coth (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \coth ^2(c+d x)\right )^2}+\frac{b \left (19 a^2+16 a b+5 b^2\right ) \coth (c+d x)}{16 a^3 (a+b)^3 d \left (a+b \coth ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.57054, size = 203, normalized size = 1.01 \[ \frac{\frac{3 b \left (19 a^2+16 a b+5 b^2\right ) (a+b) \coth (c+d x)}{a^3 \left (a+b \coth ^2(c+d x)\right )}+\frac{3 \sqrt{b} \left (35 a^2 b+35 a^3+21 a b^2+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} \coth (c+d x)}{\sqrt{a}}\right )}{a^{7/2}}+\frac{2 b (11 a+5 b) (a+b)^2 \coth (c+d x)}{a^2 \left (a+b \coth ^2(c+d x)\right )^2}+\frac{8 b (a+b)^3 \coth (c+d x)}{a \left (a+b \coth ^2(c+d x)\right )^3}-24 \log (1-\coth (c+d x))+24 \log (\coth (c+d x)+1)}{48 d (a+b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*Sqrt[b]*(35*a^3 + 35*a^2*b + 21*a*b^2 + 5*b^3)*ArcTan[(Sqrt[b]*Coth[c + d*x])/Sqrt[a]])/a^(7/2) + (8*b*(a
+ b)^3*Coth[c + d*x])/(a*(a + b*Coth[c + d*x]^2)^3) + (2*b*(a + b)^2*(11*a + 5*b)*Coth[c + d*x])/(a^2*(a + b*C
oth[c + d*x]^2)^2) + (3*b*(a + b)*(19*a^2 + 16*a*b + 5*b^2)*Coth[c + d*x])/(a^3*(a + b*Coth[c + d*x]^2)) - 24*
Log[1 - Coth[c + d*x]] + 24*Log[1 + Coth[c + d*x]])/(48*(a + b)^4*d)

________________________________________________________________________________________

Maple [B]  time = 0.03, size = 608, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*coth(d*x+c)^2)^4,x)

[Out]

1/2/d/(a+b)^4*ln(coth(d*x+c)+1)+19/16/d/(a+b)^4*b^3/(a+b*coth(d*x+c)^2)^3*coth(d*x+c)^5+35/16/d/(a+b)^4*b^4/(a
+b*coth(d*x+c)^2)^3/a*coth(d*x+c)^5+21/16/d/(a+b)^4*b^5/(a+b*coth(d*x+c)^2)^3/a^2*coth(d*x+c)^5+5/16/d/(a+b)^4
*b^6/(a+b*coth(d*x+c)^2)^3/a^3*coth(d*x+c)^5+17/6/d/(a+b)^4*b^2/(a+b*coth(d*x+c)^2)^3*a*coth(d*x+c)^3+11/2/d/(
a+b)^4*b^3/(a+b*coth(d*x+c)^2)^3*coth(d*x+c)^3+7/2/d/(a+b)^4*b^4/(a+b*coth(d*x+c)^2)^3/a*coth(d*x+c)^3+5/6/d/(
a+b)^4*b^5/(a+b*coth(d*x+c)^2)^3/a^2*coth(d*x+c)^3+29/16/d/(a+b)^4*b/(a+b*coth(d*x+c)^2)^3*a^2*coth(d*x+c)+61/
16/d/(a+b)^4*b^2/(a+b*coth(d*x+c)^2)^3*a*coth(d*x+c)+43/16/d/(a+b)^4*b^3/(a+b*coth(d*x+c)^2)^3*coth(d*x+c)+11/
16/d/(a+b)^4*b^4/(a+b*coth(d*x+c)^2)^3/a*coth(d*x+c)+35/16/d/(a+b)^4*b/(a*b)^(1/2)*arctan(coth(d*x+c)*b/(a*b)^
(1/2))+35/16/d/(a+b)^4*b^2/a/(a*b)^(1/2)*arctan(coth(d*x+c)*b/(a*b)^(1/2))+21/16/d/(a+b)^4*b^3/a^2/(a*b)^(1/2)
*arctan(coth(d*x+c)*b/(a*b)^(1/2))+5/16/d/(a+b)^4*b^4/a^3/(a*b)^(1/2)*arctan(coth(d*x+c)*b/(a*b)^(1/2))-1/2/d/
(a+b)^4*ln(coth(d*x+c)-1)

________________________________________________________________________________________

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(1/(a+b*coth(d*x+c)^2)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c)^2)^4,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c)**2)**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.29303, size = 1030, normalized size = 5.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c)^2)^4,x, algorithm="giac")

[Out]

-1/16*(35*a^3*b + 35*a^2*b^2 + 21*a*b^3 + 5*b^4)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) - a + b)/sq
rt(a*b))/((a^7*d + 4*a^6*b*d + 6*a^5*b^2*d + 4*a^4*b^3*d + a^3*b^4*d)*sqrt(a*b)) + (d*x + c)/(a^4*d + 4*a^3*b*
d + 6*a^2*b^2*d + 4*a*b^3*d + b^4*d) + 1/24*(87*a^5*b*e^(10*d*x + 10*c) + 69*a^4*b^2*e^(10*d*x + 10*c) - 186*a
^3*b^3*e^(10*d*x + 10*c) - 246*a^2*b^4*e^(10*d*x + 10*c) - 93*a*b^5*e^(10*d*x + 10*c) - 15*b^6*e^(10*d*x + 10*
c) - 435*a^5*b*e^(8*d*x + 8*c) - 51*a^4*b^2*e^(8*d*x + 8*c) + 174*a^3*b^3*e^(8*d*x + 8*c) - 450*a^2*b^4*e^(8*d
*x + 8*c) - 315*a*b^5*e^(8*d*x + 8*c) - 75*b^6*e^(8*d*x + 8*c) + 870*a^5*b*e^(6*d*x + 6*c) + 58*a^4*b^2*e^(6*d
*x + 6*c) + 324*a^3*b^3*e^(6*d*x + 6*c) - 612*a^2*b^4*e^(6*d*x + 6*c) - 490*a*b^5*e^(6*d*x + 6*c) - 150*b^6*e^
(6*d*x + 6*c) - 870*a^5*b*e^(4*d*x + 4*c) - 558*a^4*b^2*e^(4*d*x + 4*c) + 36*a^3*b^3*e^(4*d*x + 4*c) - 636*a^2
*b^4*e^(4*d*x + 4*c) - 510*a*b^5*e^(4*d*x + 4*c) - 150*b^6*e^(4*d*x + 4*c) + 435*a^5*b*e^(2*d*x + 2*c) + 801*a
^4*b^2*e^(2*d*x + 2*c) + 102*a^3*b^3*e^(2*d*x + 2*c) - 534*a^2*b^4*e^(2*d*x + 2*c) - 345*a*b^5*e^(2*d*x + 2*c)
 - 75*b^6*e^(2*d*x + 2*c) - 87*a^5*b - 319*a^4*b^2 - 450*a^3*b^3 - 306*a^2*b^4 - 103*a*b^5 - 15*b^6)/((a^7*d +
 4*a^6*b*d + 6*a^5*b^2*d + 4*a^4*b^3*d + a^3*b^4*d)*(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) - 2*a*e^(2*d*x + 2*
c) + 2*b*e^(2*d*x + 2*c) + a + b)^3)