3.191 \(\int \text{csch}^3(c+d x) (a+b \sinh ^4(c+d x)) \, dx\)

Optimal. Leaf size=47 \[ \frac{a \tanh ^{-1}(\cosh (c+d x))}{2 d}-\frac{a \coth (c+d x) \text{csch}(c+d x)}{2 d}+\frac{b \cosh (c+d x)}{d} \]

[Out]

(a*ArcTanh[Cosh[c + d*x]])/(2*d) + (b*Cosh[c + d*x])/d - (a*Coth[c + d*x]*Csch[c + d*x])/(2*d)

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Rubi [A]  time = 0.0539296, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3215, 1157, 388, 206} \[ \frac{a \tanh ^{-1}(\cosh (c+d x))}{2 d}-\frac{a \coth (c+d x) \text{csch}(c+d x)}{2 d}+\frac{b \cosh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^3*(a + b*Sinh[c + d*x]^4),x]

[Out]

(a*ArcTanh[Cosh[c + d*x]])/(2*d) + (b*Cosh[c + d*x])/d - (a*Coth[c + d*x]*Csch[c + d*x])/(2*d)

Rule 3215

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0260933, size = 82, normalized size = 1.74 \[ -\frac{a \text{csch}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a \text{sech}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{b \sinh (c) \sinh (d x)}{d}+\frac{b \cosh (c) \cosh (d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^3*(a + b*Sinh[c + d*x]^4),x]

[Out]

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

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Maple [A]  time = 0.04, size = 38, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{{\rm csch} \left (dx+c\right ){\rm coth} \left (dx+c\right )}{2}}+{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) \right ) +b\cosh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^3*(a+b*sinh(d*x+c)^4),x)

[Out]

1/d*(a*(-1/2*csch(d*x+c)*coth(d*x+c)+arctanh(exp(d*x+c)))+b*cosh(d*x+c))

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Maxima [B]  time = 1.23746, size = 155, normalized size = 3.3 \begin{align*} \frac{1}{2} \, b{\left (\frac{e^{\left (d x + c\right )}}{d} + \frac{e^{\left (-d x - c\right )}}{d}\right )} + \frac{1}{2} \, a{\left (\frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac{2 \,{\left (e^{\left (-d x - c\right )} + e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3*(a+b*sinh(d*x+c)^4),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.7422, size = 1867, normalized size = 39.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3*(a+b*sinh(d*x+c)^4),x, algorithm="fricas")

[Out]

1/2*(b*cosh(d*x + c)^6 + 6*b*cosh(d*x + c)*sinh(d*x + c)^5 + b*sinh(d*x + c)^6 - (2*a + b)*cosh(d*x + c)^4 + (
15*b*cosh(d*x + c)^2 - 2*a - b)*sinh(d*x + c)^4 + 4*(5*b*cosh(d*x + c)^3 - (2*a + b)*cosh(d*x + c))*sinh(d*x +
c)^3 - (2*a + b)*cosh(d*x + c)^2 + (15*b*cosh(d*x + c)^4 - 6*(2*a + b)*cosh(d*x + c)^2 - 2*a - b)*sinh(d*x +
c)^2 + (a*cosh(d*x + c)^5 + 5*a*cosh(d*x + c)*sinh(d*x + c)^4 + a*sinh(d*x + c)^5 - 2*a*cosh(d*x + c)^3 + 2*(5
*a*cosh(d*x + c)^2 - a)*sinh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^3 - 3*a*cosh(d*x + c))*sinh(d*x + c)^2 + a*cosh
(d*x + c) + (5*a*cosh(d*x + c)^4 - 6*a*cosh(d*x + c)^2 + a)*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) +
1) - (a*cosh(d*x + c)^5 + 5*a*cosh(d*x + c)*sinh(d*x + c)^4 + a*sinh(d*x + c)^5 - 2*a*cosh(d*x + c)^3 + 2*(5*
a*cosh(d*x + c)^2 - a)*sinh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^3 - 3*a*cosh(d*x + c))*sinh(d*x + c)^2 + a*cosh(
d*x + c) + (5*a*cosh(d*x + c)^4 - 6*a*cosh(d*x + c)^2 + a)*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) -
1) + 2*(3*b*cosh(d*x + c)^5 - 2*(2*a + b)*cosh(d*x + c)^3 - (2*a + b)*cosh(d*x + c))*sinh(d*x + c) + b)/(d*cos
h(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5 - 2*d*cosh(d*x + c)^3 + 2*(5*d*cosh(d*x +
c)^2 - d)*sinh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^3 - 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + d*cosh(d*x + c) + (
5*d*cosh(d*x + c)^4 - 6*d*cosh(d*x + c)^2 + d)*sinh(d*x + c))

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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(csch(d*x+c)**3*(a+b*sinh(d*x+c)**4),x)

[Out]

Timed out

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Giac [B]  time = 1.19865, size = 155, normalized size = 3.3 \begin{align*} \frac{b{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{2 \, d} + \frac{a \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} + 2\right )}{4 \, d} - \frac{a \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} - 2\right )}{4 \, d} - \frac{a{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} - 4\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3*(a+b*sinh(d*x+c)^4),x, algorithm="giac")

[Out]

1/2*b*(e^(d*x + c) + e^(-d*x - c))/d + 1/4*a*log(e^(d*x + c) + e^(-d*x - c) + 2)/d - 1/4*a*log(e^(d*x + c) + e
^(-d*x - c) - 2)/d - a*(e^(d*x + c) + e^(-d*x - c))/(((e^(d*x + c) + e^(-d*x - c))^2 - 4)*d)