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

Optimal. Leaf size=54 \[ \frac{(a+3 b) \tanh ^{-1}(\cosh (c+d x))}{2 d}-\frac{(a+b) \coth (c+d x) \text{csch}(c+d x)}{2 d}-\frac{b \text{sech}(c+d x)}{d} \]

[Out]

((a + 3*b)*ArcTanh[Cosh[c + d*x]])/(2*d) - ((a + b)*Coth[c + d*x]*Csch[c + d*x])/(2*d) - (b*Sech[c + d*x])/d

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Rubi [A]  time = 0.0713326, antiderivative size = 54, 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 = {4133, 456, 453, 206} \[ \frac{(a+3 b) \tanh ^{-1}(\cosh (c+d x))}{2 d}-\frac{(a+b) \coth (c+d x) \text{csch}(c+d x)}{2 d}-\frac{b \text{sech}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 3*b)*ArcTanh[Cosh[c + d*x]])/(2*d) - ((a + b)*Coth[c + d*x]*Csch[c + d*x])/(2*d) - (b*Sech[c + d*x])/d

Rule 4133

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

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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

Mathematica [B]  time = 0.0468217, size = 131, normalized size = 2.43 \[ -\frac{a \text{csch}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a \text{sech}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{b \text{csch}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{b \text{sech}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{b \text{sech}(c+d x)}{d}-\frac{3 b \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Csch[(c + d*x)/2]^2)/(8*d) - (b*Csch[(c + d*x)/2]^2)/(8*d) - (a*Log[Tanh[(c + d*x)/2]])/(2*d) - (3*b*Log[T
anh[(c + d*x)/2]])/(2*d) - (a*Sech[(c + d*x)/2]^2)/(8*d) - (b*Sech[(c + d*x)/2]^2)/(8*d) - (b*Sech[c + d*x])/d

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Maple [A]  time = 0.036, size = 70, normalized size = 1.3 \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 \left ( -{\frac{1}{2\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}\cosh \left ( dx+c \right ) }}-{\frac{3}{2\,\cosh \left ( dx+c \right ) }}+3\,{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-1/2*csch(d*x+c)*coth(d*x+c)+arctanh(exp(d*x+c)))+b*(-1/2/sinh(d*x+c)^2/cosh(d*x+c)-3/2/cosh(d*x+c)+3*
arctanh(exp(d*x+c))))

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Maxima [B]  time = 1.02697, size = 267, normalized size = 4.94 \begin{align*} \frac{1}{2} \, b{\left (\frac{3 \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac{3 \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac{2 \,{\left (3 \, e^{\left (-d x - c\right )} - 2 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )}\right )}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} - e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\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*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2*b*(3*log(e^(-d*x - c) + 1)/d - 3*log(e^(-d*x - c) - 1)/d + 2*(3*e^(-d*x - c) - 2*e^(-3*d*x - 3*c) + 3*e^(-
5*d*x - 5*c))/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) - e^(-6*d*x - 6*c) - 1))) + 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 = 2.71236, size = 2469, normalized size = 45.72 \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*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(2*(a + 3*b)*cosh(d*x + c)^5 + 10*(a + 3*b)*cosh(d*x + c)*sinh(d*x + c)^4 + 2*(a + 3*b)*sinh(d*x + c)^5 +
4*(a - b)*cosh(d*x + c)^3 + 4*(5*(a + 3*b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^3 + 4*(5*(a + 3*b)*cosh(d*x
+ c)^3 + 3*(a - b)*cosh(d*x + c))*sinh(d*x + c)^2 + 2*(a + 3*b)*cosh(d*x + c) - ((a + 3*b)*cosh(d*x + c)^6 +
6*(a + 3*b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a + 3*b)*sinh(d*x + c)^6 - (a + 3*b)*cosh(d*x + c)^4 + (15*(a + 3
*b)*cosh(d*x + c)^2 - a - 3*b)*sinh(d*x + c)^4 + 4*(5*(a + 3*b)*cosh(d*x + c)^3 - (a + 3*b)*cosh(d*x + c))*sin
h(d*x + c)^3 - (a + 3*b)*cosh(d*x + c)^2 + (15*(a + 3*b)*cosh(d*x + c)^4 - 6*(a + 3*b)*cosh(d*x + c)^2 - a - 3
*b)*sinh(d*x + c)^2 + 2*(3*(a + 3*b)*cosh(d*x + c)^5 - 2*(a + 3*b)*cosh(d*x + c)^3 - (a + 3*b)*cosh(d*x + c))*
sinh(d*x + c) + a + 3*b)*log(cosh(d*x + c) + sinh(d*x + c) + 1) + ((a + 3*b)*cosh(d*x + c)^6 + 6*(a + 3*b)*cos
h(d*x + c)*sinh(d*x + c)^5 + (a + 3*b)*sinh(d*x + c)^6 - (a + 3*b)*cosh(d*x + c)^4 + (15*(a + 3*b)*cosh(d*x +
c)^2 - a - 3*b)*sinh(d*x + c)^4 + 4*(5*(a + 3*b)*cosh(d*x + c)^3 - (a + 3*b)*cosh(d*x + c))*sinh(d*x + c)^3 -
(a + 3*b)*cosh(d*x + c)^2 + (15*(a + 3*b)*cosh(d*x + c)^4 - 6*(a + 3*b)*cosh(d*x + c)^2 - a - 3*b)*sinh(d*x +
c)^2 + 2*(3*(a + 3*b)*cosh(d*x + c)^5 - 2*(a + 3*b)*cosh(d*x + c)^3 - (a + 3*b)*cosh(d*x + c))*sinh(d*x + c) +
a + 3*b)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(5*(a + 3*b)*cosh(d*x + c)^4 + 6*(a - b)*cosh(d*x + c)^2
+ a + 3*b)*sinh(d*x + c))/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 - d*cosh(
d*x + c)^4 + (15*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x +
c)^3 - d*cosh(d*x + c)^2 + (15*d*cosh(d*x + c)^4 - 6*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x
+ c)^5 - 2*d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right ) \operatorname{csch}^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.16261, size = 198, normalized size = 3.67 \begin{align*} \frac{{\left (a + 3 \, b\right )} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} + 2\right )}{4 \, d} - \frac{{\left (a + 3 \, b\right )} \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 )}^{2} + 3 \, b{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} - 8 \, b}{{\left ({\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{3} - 4 \, e^{\left (d x + c\right )} - 4 \, e^{\left (-d x - c\right )}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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