3.16 \(\int \text {csch}^4(c+d x) (a+b \text {sech}^2(c+d x))^2 \, dx\)

Optimal. Leaf size=75 \[ \frac {b (2 a+3 b) \tanh (c+d x)}{d}-\frac {(a+b)^2 \coth ^3(c+d x)}{3 d}+\frac {(a+b) (a+3 b) \coth (c+d x)}{d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4132, 448} \[ \frac {b (2 a+3 b) \tanh (c+d x)}{d}-\frac {(a+b)^2 \coth ^3(c+d x)}{3 d}+\frac {(a+b) (a+3 b) \coth (c+d x)}{d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 448

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

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

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

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Mathematica [B]  time = 1.32, size = 151, normalized size = 2.01 \[ -\frac {\text {csch}(2 c) \text {csch}^3(2 (c+d x)) \left (-3 a^2 \sinh (2 (c+d x))+a^2 \sinh (6 (c+d x))+3 a^2 \sinh (4 c+2 d x)+a^2 \sinh (4 c+6 d x)-6 a b \sinh (2 (c+d x))+2 a b \sinh (6 (c+d x))+8 a b \sinh (4 c+6 d x)+8 a (a+2 b) \sinh (2 c)-6 (a+2 b)^2 \sinh (2 d x)+8 b^2 \sinh (4 c+6 d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(Csch[2*c]*Csch[2*(c + d*x)]^3*(8*a*(a + 2*b)*Sinh[2*c] - 6*(a + 2*b)^2*Sinh[2*d*x] - 3*a^2*Sinh[2*(c + d
*x)] - 6*a*b*Sinh[2*(c + d*x)] + a^2*Sinh[6*(c + d*x)] + 2*a*b*Sinh[6*(c + d*x)] + 3*a^2*Sinh[4*c + 2*d*x] + a
^2*Sinh[4*c + 6*d*x] + 8*a*b*Sinh[4*c + 6*d*x] + 8*b^2*Sinh[4*c + 6*d*x]))/d

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fricas [B]  time = 0.40, size = 408, normalized size = 5.44 \[ -\frac {8 \, {\left ({\left (a^{2} - 4 \, a b - 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 8 \, {\left (a^{2} + 2 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{2} - 4 \, a b - 4 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{2} - 4 \, a b - 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, a^{2} + 4 \, a b\right )} \sinh \left (d x + c\right )^{2} + 3 \, a^{2} + 12 \, a b + 12 \, b^{2} + 8 \, {\left ({\left (a^{2} + 2 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{8} + 56 \, d \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right )^{5} + 28 \, d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{6} + 8 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + d \sinh \left (d x + c\right )^{8} - 4 \, d \cosh \left (d x + c\right )^{4} + 2 \, {\left (35 \, d \cosh \left (d x + c\right )^{4} - 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 8 \, {\left (7 \, d \cosh \left (d x + c\right )^{5} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 4 \, {\left (7 \, d \cosh \left (d x + c\right )^{6} - 6 \, d \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} + 8 \, {\left (d \cosh \left (d x + c\right )^{7} - d \cosh \left (d x + c\right )^{3}\right )} \sinh \left (d x + c\right ) + 3 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3*((a^2 - 4*a*b - 4*b^2)*cosh(d*x + c)^4 + 8*(a^2 + 2*a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 - 4
*a*b - 4*b^2)*sinh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(a^2 - 4*a*b - 4*b^2)*cosh(d*x + c)^2 +
2*a^2 + 4*a*b)*sinh(d*x + c)^2 + 3*a^2 + 12*a*b + 12*b^2 + 8*((a^2 + 2*a*b + 2*b^2)*cosh(d*x + c)^3 + (a^2 +
2*a*b)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^8 + 56*d*cosh(d*x + c)^3*sinh(d*x + c)^5 + 28*d*cosh(d*x
+ c)^2*sinh(d*x + c)^6 + 8*d*cosh(d*x + c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^8 - 4*d*cosh(d*x + c)^4 + 2*(35*
d*cosh(d*x + c)^4 - 2*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 - d*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(7*d*
cosh(d*x + c)^6 - 6*d*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 8*(d*cosh(d*x + c)^7 - d*cosh(d*x + c)^3)*sinh(d*x +
c) + 3*d)

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giac [A]  time = 0.18, size = 115, normalized size = 1.53 \[ -\frac {4 \, {\left (3 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 8 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 16 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 6 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 24 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - a^{2} - 8 \, a b - 8 \, b^{2}\right )}}{3 \, d {\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(3*a^2*e^(8*d*x + 8*c) + 8*a^2*e^(6*d*x + 6*c) + 16*a*b*e^(6*d*x + 6*c) + 6*a^2*e^(4*d*x + 4*c) + 24*a*b*
e^(4*d*x + 4*c) + 24*b^2*e^(4*d*x + 4*c) - a^2 - 8*a*b - 8*b^2)/(d*(e^(4*d*x + 4*c) - 1)^3)

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maple [A]  time = 0.55, size = 138, normalized size = 1.84 \[ \frac {a^{2} \left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+2 a b \left (-\frac {1}{3 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )}+\frac {4}{3 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )}+\frac {8 \tanh \left (d x +c \right )}{3}\right )+b^{2} \left (-\frac {1}{3 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )^{3}}+\frac {2}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}+8 \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+2*a*b*(-1/3/sinh(d*x+c)^3/cosh(d*x+c)+4/3/sinh(d*x+c)/cosh(d*x+c)
+8/3*tanh(d*x+c))+b^2*(-1/3/sinh(d*x+c)^3/cosh(d*x+c)^3+2/sinh(d*x+c)/cosh(d*x+c)^3+8*(2/3+1/3*sech(d*x+c)^2)*
tanh(d*x+c)))

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maxima [B]  time = 0.34, size = 285, normalized size = 3.80 \[ \frac {4}{3} \, a^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {32}{3} \, a b {\left (\frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} - \frac {1}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + \frac {32}{3} \, b^{2} {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 3 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 3 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*a^2*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1)) - 1/(d*(3*e^(
-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1))) + 32/3*a*b*(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))) + 32/3*b^2*(3*e^(-4*d*x - 4*c)/(d*(3*e^(-4*d*x - 4*c) - 3*e^(-8*d*x - 8*c) + e^(-12*d*x - 12*c)
- 1)) - 1/(d*(3*e^(-4*d*x - 4*c) - 3*e^(-8*d*x - 8*c) + e^(-12*d*x - 12*c) - 1)))

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mupad [B]  time = 0.21, size = 115, normalized size = 1.53 \[ -\frac {4\,\left (6\,a^2\,{\mathrm {e}}^{4\,c+4\,d\,x}-a^2-8\,b^2-8\,a\,b+8\,a^2\,{\mathrm {e}}^{6\,c+6\,d\,x}+3\,a^2\,{\mathrm {e}}^{8\,c+8\,d\,x}+24\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}+24\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}+16\,a\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}\right )}{3\,d\,{\left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cosh(c + d*x)^2)^2/sinh(c + d*x)^4,x)

[Out]

-(4*(6*a^2*exp(4*c + 4*d*x) - a^2 - 8*b^2 - 8*a*b + 8*a^2*exp(6*c + 6*d*x) + 3*a^2*exp(8*c + 8*d*x) + 24*b^2*e
xp(4*c + 4*d*x) + 24*a*b*exp(4*c + 4*d*x) + 16*a*b*exp(6*c + 6*d*x)))/(3*d*(exp(4*c + 4*d*x) - 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2} \operatorname {csch}^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**4*(a+b*sech(d*x+c)**2)**2,x)

[Out]

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

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