3.316 \(\int \frac{\text{sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=119 \[ \frac{b^3 \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{a \left (a^2+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}-\frac{b^3 \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{\text{sech}^2(c+d x) (a \sinh (c+d x)+b)}{2 d \left (a^2+b^2\right )} \]

[Out]

(a*(a^2 + 3*b^2)*ArcTan[Sinh[c + d*x]])/(2*(a^2 + b^2)^2*d) - (b^3*Log[Cosh[c + d*x]])/((a^2 + b^2)^2*d) + (b^
3*Log[a + b*Sinh[c + d*x]])/((a^2 + b^2)^2*d) + (Sech[c + d*x]^2*(b + a*Sinh[c + d*x]))/(2*(a^2 + b^2)*d)

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Rubi [A]  time = 0.143604, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2668, 741, 801, 635, 203, 260} \[ \frac{b^3 \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{a \left (a^2+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}-\frac{b^3 \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{\text{sech}^2(c+d x) (a \sinh (c+d x)+b)}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2 + 3*b^2)*ArcTan[Sinh[c + d*x]])/(2*(a^2 + b^2)^2*d) - (b^3*Log[Cosh[c + d*x]])/((a^2 + b^2)^2*d) + (b^
3*Log[a + b*Sinh[c + d*x]])/((a^2 + b^2)^2*d) + (Sech[c + d*x]^2*(b + a*Sinh[c + d*x]))/(2*(a^2 + b^2)*d)

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.188339, size = 104, normalized size = 0.87 \[ \frac{b \left (a^2+b^2\right ) \text{sech}^2(c+d x)+2 a \left (a^2+3 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+a \left (a^2+b^2\right ) \tanh (c+d x) \text{sech}(c+d x)+2 b^3 (\log (a+b \sinh (c+d x))-\log (\cosh (c+d x)))}{2 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a^2 + 3*b^2)*ArcTan[Tanh[(c + d*x)/2]] + 2*b^3*(-Log[Cosh[c + d*x]] + Log[a + b*Sinh[c + d*x]]) + b*(a^2
+ b^2)*Sech[c + d*x]^2 + a*(a^2 + b^2)*Sech[c + d*x]*Tanh[c + d*x])/(2*(a^2 + b^2)^2*d)

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Maple [B]  time = 0.003, size = 468, normalized size = 3.9 \begin{align*}{\frac{{b}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }-{\frac{{a}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-{\frac{a{b}^{2}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-2\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}{a}^{2}b}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}{b}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{{a}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{a{b}^{2}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-{\frac{{b}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) }+{\frac{{a}^{3}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\arctan \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+3\,{\frac{\arctan \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) a{b}^{2}}{d \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^3/(a+b*sinh(d*x+c)),x)

[Out]

1/d*b^3/(a^4+2*a^2*b^2+b^4)*ln(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)-1/d/(a^4+2*a^2*b^2+b^4)/(tan
h(1/2*d*x+1/2*c)^2+1)^2*tanh(1/2*d*x+1/2*c)^3*a^3-1/d/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*d*x+1/2*c)^2+1)^2*tanh(1/2
*d*x+1/2*c)^3*a*b^2-2/d/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*d*x+1/2*c)^2+1)^2*tanh(1/2*d*x+1/2*c)^2*a^2*b-2/d/(a^4+2
*a^2*b^2+b^4)/(tanh(1/2*d*x+1/2*c)^2+1)^2*tanh(1/2*d*x+1/2*c)^2*b^3+1/d/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*d*x+1/2*
c)^2+1)^2*tanh(1/2*d*x+1/2*c)*a^3+1/d/(a^4+2*a^2*b^2+b^4)/(tanh(1/2*d*x+1/2*c)^2+1)^2*tanh(1/2*d*x+1/2*c)*a*b^
2-1/d/(a^4+2*a^2*b^2+b^4)*b^3*ln(tanh(1/2*d*x+1/2*c)^2+1)+1/d/(a^4+2*a^2*b^2+b^4)*arctan(tanh(1/2*d*x+1/2*c))*
a^3+3/d/(a^4+2*a^2*b^2+b^4)*arctan(tanh(1/2*d*x+1/2*c))*a*b^2

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Maxima [A]  time = 1.96289, size = 292, normalized size = 2.45 \begin{align*} \frac{b^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac{b^{3} \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac{{\left (a^{3} + 3 \, a b^{2}\right )} \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} + \frac{a e^{\left (-d x - c\right )} + 2 \, b e^{\left (-2 \, d x - 2 \, c\right )} - a e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{2} + b^{2} + 2 \,{\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a^{2} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.43699, size = 2215, normalized size = 18.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**3/(a+b*sinh(d*x+c)),x)

[Out]

Integral(sech(c + d*x)**3/(a + b*sinh(c + d*x)), x)

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Giac [B]  time = 1.36663, size = 400, normalized size = 3.36 \begin{align*} \frac{b^{4} \log \left ({\left | b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{4} b d + 2 \, a^{2} b^{3} d + b^{5} d} - \frac{b^{3} \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{2 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )}{\left (a^{3} + 3 \, a b^{2}\right )}}{4 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} + \frac{b^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 2 \, a^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 4 \, a^{2} b + 8 \, b^{3}}{2 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^4*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^4*b*d + 2*a^2*b^3*d + b^5*d) - 1/2*b^3*log((e^(d*x + c)
- e^(-d*x - c))^2 + 4)/(a^4*d + 2*a^2*b^2*d + b^4*d) + 1/4*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x -
c)))*(a^3 + 3*a*b^2)/(a^4*d + 2*a^2*b^2*d + b^4*d) + 1/2*(b^3*(e^(d*x + c) - e^(-d*x - c))^2 + 2*a^3*(e^(d*x +
c) - e^(-d*x - c)) + 2*a*b^2*(e^(d*x + c) - e^(-d*x - c)) + 4*a^2*b + 8*b^3)/((a^4*d + 2*a^2*b^2*d + b^4*d)*(
(e^(d*x + c) - e^(-d*x - c))^2 + 4))