3.130 \(\int \coth ^2(c+d x) (a+b \text {sech}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=61 \[ a^3 x-\frac {b^2 (3 a+2 b) \tanh (c+d x)}{d}-\frac {(a+b)^3 \coth (c+d x)}{d}+\frac {b^3 \tanh ^3(c+d x)}{3 d} \]

[Out]

a^3*x-(a+b)^3*coth(d*x+c)/d-b^2*(3*a+2*b)*tanh(d*x+c)/d+1/3*b^3*tanh(d*x+c)^3/d

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Rubi [A]  time = 0.10, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4141, 1802, 207} \[ a^3 x-\frac {b^2 (3 a+2 b) \tanh (c+d x)}{d}-\frac {(a+b)^3 \coth (c+d x)}{d}+\frac {b^3 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^2*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

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

Rule 207

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

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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])

Rubi steps

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

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Mathematica [B]  time = 1.80, size = 126, normalized size = 2.07 \[ \frac {8 (a \cosh (c+d x)+b \text {sech}(c+d x))^3 \left (3 a^3 d x \cosh ^3(c+d x)+\sinh (d x) \cosh ^2(c+d x) \left (3 (a+b)^3 \text {csch}(c) \coth (c+d x)-b^2 (9 a+5 b) \text {sech}(c)\right )-b^3 \tanh (c) \cosh (c+d x)+b^3 (-\text {sech}(c)) \sinh (d x)\right )}{3 d (a \cosh (2 (c+d x))+a+2 b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^2*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

(8*(a*Cosh[c + d*x] + b*Sech[c + d*x])^3*(3*a^3*d*x*Cosh[c + d*x]^3 - b^3*Sech[c]*Sinh[d*x] + Cosh[c + d*x]^2*
(3*(a + b)^3*Coth[c + d*x]*Csch[c] - b^2*(9*a + 5*b)*Sech[c])*Sinh[d*x] - b^3*Cosh[c + d*x]*Tanh[c]))/(3*d*(a
+ 2*b + a*Cosh[2*(c + d*x)])^3)

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fricas [B]  time = 0.41, size = 359, normalized size = 5.89 \[ -\frac {{\left (3 \, a^{3} + 9 \, a^{2} b + 18 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} - 4 \, {\left (3 \, a^{3} d x + 3 \, a^{3} + 9 \, a^{2} b + 18 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (3 \, a^{3} + 9 \, a^{2} b + 18 \, a b^{2} + 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} + 9 \, a^{3} + 27 \, a^{2} b + 18 \, a b^{2} + 4 \, {\left (3 \, a^{3} + 9 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{3} + 18 \, a^{2} b + 18 \, a b^{2} + 8 \, b^{3} + 3 \, {\left (3 \, a^{3} + 9 \, a^{2} b + 18 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} - 4 \, {\left ({\left (3 \, a^{3} d x + 3 \, a^{3} + 9 \, a^{2} b + 18 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + {\left (3 \, a^{3} d x + 3 \, a^{3} + 9 \, a^{2} b + 18 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{12 \, {\left (d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/12*((3*a^3 + 9*a^2*b + 18*a*b^2 + 8*b^3)*cosh(d*x + c)^4 - 4*(3*a^3*d*x + 3*a^3 + 9*a^2*b + 18*a*b^2 + 8*b^
3)*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a^3 + 9*a^2*b + 18*a*b^2 + 8*b^3)*sinh(d*x + c)^4 + 9*a^3 + 27*a^2*b + 1
8*a*b^2 + 4*(3*a^3 + 9*a^2*b + 9*a*b^2 + 4*b^3)*cosh(d*x + c)^2 + 2*(6*a^3 + 18*a^2*b + 18*a*b^2 + 8*b^3 + 3*(
3*a^3 + 9*a^2*b + 18*a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 4*((3*a^3*d*x + 3*a^3 + 9*a^2*b + 18*a*
b^2 + 8*b^3)*cosh(d*x + c)^3 + (3*a^3*d*x + 3*a^3 + 9*a^2*b + 18*a*b^2 + 8*b^3)*cosh(d*x + c))*sinh(d*x + c))/
(d*cosh(d*x + c)*sinh(d*x + c)^3 + (d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c))

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giac [B]  time = 0.21, size = 132, normalized size = 2.16 \[ \frac {3 \, a^{3} d x - \frac {6 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + \frac {2 \, {\left (9 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a b^{2} + 5 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/3*(3*a^3*d*x - 6*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)/(e^(2*d*x + 2*c) - 1) + 2*(9*a*b^2*e^(4*d*x + 4*c) + 3*b^3*
e^(4*d*x + 4*c) + 18*a*b^2*e^(2*d*x + 2*c) + 12*b^3*e^(2*d*x + 2*c) + 9*a*b^2 + 5*b^3)/(e^(2*d*x + 2*c) + 1)^3
)/d

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maple [A]  time = 0.57, size = 111, normalized size = 1.82 \[ \frac {a^{3} \left (d x +c -\coth \left (d x +c \right )\right )-3 a^{2} b \coth \left (d x +c \right )+3 a \,b^{2} \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}-2 \tanh \left (d x +c \right )\right )+b^{3} \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}-4 \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(d*x+c-coth(d*x+c))-3*a^2*b*coth(d*x+c)+3*a*b^2*(-1/sinh(d*x+c)/cosh(d*x+c)-2*tanh(d*x+c))+b^3*(-1/si
nh(d*x+c)/cosh(d*x+c)^3-4*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)))

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maxima [B]  time = 0.41, size = 172, normalized size = 2.82 \[ a^{3} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} - \frac {16}{3} \, b^{3} {\left (\frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}} + \frac {1}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + \frac {6 \, a^{2} b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} + \frac {12 \, a b^{2}}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

a^3*(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1))) - 16/3*b^3*(2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - 2*e^(-6*d
*x - 6*c) - e^(-8*d*x - 8*c) + 1)) + 1/(d*(2*e^(-2*d*x - 2*c) - 2*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) + 1))) +
6*a^2*b/(d*(e^(-2*d*x - 2*c) - 1)) + 12*a*b^2/(d*(e^(-4*d*x - 4*c) - 1))

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mupad [B]  time = 0.14, size = 234, normalized size = 3.84 \[ \frac {\frac {2\,\left (b^3+3\,a\,b^2\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^3+a\,b^2\right )}{d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^3+3\,a\,b^2\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+a^3\,x+\frac {\frac {2\,\left (b^3+a\,b^2\right )}{d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^3+3\,a\,b^2\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {2\,\left (b^3+3\,a\,b^2\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^2*(a + b/cosh(c + d*x)^2)^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3} \coth ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**2*(a+b*sech(d*x+c)**2)**3,x)

[Out]

Integral((a + b*sech(c + d*x)**2)**3*coth(c + d*x)**2, x)

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