3.389 \(\int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=121 \[ \frac {a^2 b \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a \left (a^2-b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}-\frac {a^2 b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {\text {sech}^2(c+d x) (a \sinh (c+d x)+b)}{2 d \left (a^2+b^2\right )} \]

[Out]

1/2*a*(a^2-b^2)*arctan(sinh(d*x+c))/(a^2+b^2)^2/d-a^2*b*ln(cosh(d*x+c))/(a^2+b^2)^2/d+a^2*b*ln(a+b*sinh(d*x+c)
)/(a^2+b^2)^2/d-1/2*sech(d*x+c)^2*(b+a*sinh(d*x+c))/(a^2+b^2)/d

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Rubi [A]  time = 0.23, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2837, 12, 1647, 801, 635, 203, 260} \[ \frac {a^2 b \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a \left (a^2-b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}-\frac {a^2 b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {\text {sech}^2(c+d x) (a \sinh (c+d x)+b)}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[c + d*x]*Tanh[c + d*x]^2)/(a + b*Sinh[c + d*x]),x]

[Out]

(a*(a^2 - b^2)*ArcTan[Sinh[c + d*x]])/(2*(a^2 + b^2)^2*d) - (a^2*b*Log[Cosh[c + d*x]])/((a^2 + b^2)^2*d) + (a^
2*b*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [C]  time = 0.33, size = 130, normalized size = 1.07 \[ -\frac {b \left (a^2+b^2\right ) \text {sech}^2(c+d x)+a \left (a^2+b^2\right ) \tanh (c+d x) \text {sech}(c+d x)+a \left (\left (a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))+a ((b+i a) \log (-\sinh (c+d x)+i)+(b-i a) \log (\sinh (c+d x)+i)-2 b \log (a+b \sinh (c+d x)))\right )}{2 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[c + d*x]*Tanh[c + d*x]^2)/(a + b*Sinh[c + d*x]),x]

[Out]

-1/2*(a*((a^2 + b^2)*ArcTan[Sinh[c + d*x]] + a*((I*a + b)*Log[I - Sinh[c + d*x]] + ((-I)*a + b)*Log[I + Sinh[c
+ d*x]] - 2*b*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])/((a^2 + b^2)^2*d)

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fricas [B]  time = 0.51, size = 917, normalized size = 7.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)*tanh(d*x+c)^2/(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 - a*b^2)*cosh(d*x + c)^4 + 4*(a^3 - a*b^2)*cos
h(d*x + c)*sinh(d*x + c)^3 + (a^3 - a*b^2)*sinh(d*x + c)^4 + a^3 - a*b^2 + 2*(a^3 - a*b^2)*cosh(d*x + c)^2 + 2
*(a^3 - a*b^2 + 3*(a^3 - a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 - a*b^2)*cosh(d*x + c)^3 + (a^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)*cosh(d*x + c) - (a^2
*b*cosh(d*x + c)^4 + 4*a^2*b*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*b*sinh(d*x + c)^4 + 2*a^2*b*cosh(d*x + c)^2 +
a^2*b + 2*(3*a^2*b*cosh(d*x + c)^2 + a^2*b)*sinh(d*x + c)^2 + 4*(a^2*b*cosh(d*x + c)^3 + a^2*b*cosh(d*x + c))
*sinh(d*x + c))*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + (a^2*b*cosh(d*x + c)^4 + 4*a^2*
b*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*b*sinh(d*x + c)^4 + 2*a^2*b*cosh(d*x + c)^2 + a^2*b + 2*(3*a^2*b*cosh(d*
x + c)^2 + a^2*b)*sinh(d*x + c)^2 + 4*(a^2*b*cosh(d*x + c)^3 + a^2*b*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)*c
osh(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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giac [A]  time = 1.13, size = 226, normalized size = 1.87 \[ -\frac {\frac {a^{2} b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a^{2} b \log \left ({\left | -b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a e^{\left (d x + c\right )} + b \right |}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a^{3} e^{c} - a b^{2} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} e^{\left (3 \, c\right )} + a b^{2} e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + 2 \, {\left (a^{2} b e^{\left (2 \, c\right )} + b^{3} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - {\left (a^{3} e^{c} + a b^{2} e^{c}\right )} e^{\left (d x\right )}}{{\left (a^{2} + b^{2}\right )}^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)*tanh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.00, size = 475, normalized size = 3.93 \[ \frac {4 a^{2} b \ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d \left (4 a^{4}+8 a^{2} b^{2}+4 b^{4}\right )}+\frac {\left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a \,b^{2}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\ln \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {\arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {\arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)*tanh(d*x+c)^2/(a+b*sinh(d*x+c)),x)

[Out]

4/d*a^2*b/(4*a^4+8*a^2*b^2+4*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
)/(tanh(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*ta
nh(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)*ln(tanh(1/2*d*x+1/2*c)^2+1)*a^2*b+1/d/(a^4+2*a^2*b^2+b^4)*arctan(tanh(1/2*d*x+
1/2*c))*a^3-1/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 = 0.74, size = 219, normalized size = 1.81 \[ \frac {a^{2} b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {a^{2} b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {{\left (a^{3} - a b^{2}\right )} \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {a e^{\left (-d x - c\right )} + 2 \, b e^{\left (-2 \, d x - 2 \, c\right )} - a e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)*tanh(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

a^2*b*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) - a^2*b*log(e^(-2*d*x - 2*c)
+ 1)/((a^4 + 2*a^2*b^2 + b^4)*d) - (a^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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mupad [B]  time = 2.02, size = 339, normalized size = 2.80 \[ \frac {\frac {2\,b}{d\,\left (a^2+b^2\right )}+\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (a^2\,b+b^3\right )}{d\,{\left (a^2+b^2\right )}^2}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (a^3+a\,b^2\right )}{d\,{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {a\,\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}{2\,\left (-1{}\mathrm {i}\,d\,a^2+2\,d\,a\,b+1{}\mathrm {i}\,d\,b^2\right )}+\frac {a^2\,b\,\ln \left (2\,a^7\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a^2\,b^5-14\,a^4\,b^3-a^6\,b+a^6\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a^3\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+28\,a^5\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+a^2\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+14\,a^4\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )}{d\,a^4+2\,d\,a^2\,b^2+d\,b^4}-\frac {a\,\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,\left (-d\,a^2+2{}\mathrm {i}\,d\,a\,b+d\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*b)/(d*(a^2 + b^2)) + (2*a*exp(c + d*x))/(d*(a^2 + b^2)))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2
*(a^2*b + b^3))/(d*(a^2 + b^2)^2) + (exp(c + d*x)*(a*b^2 + a^3))/(d*(a^2 + b^2)^2))/(exp(2*c + 2*d*x) + 1) - (
a*log(exp(c + d*x) + 1i)*1i)/(2*(b^2*d - a^2*d + a*b*d*2i)) - (a*log(exp(c + d*x)*1i + 1))/(2*(b^2*d*1i - a^2*
d*1i + 2*a*b*d)) + (a^2*b*log(2*a^7*exp(d*x)*exp(c) - a^2*b^5 - 14*a^4*b^3 - a^6*b + a^6*b*exp(2*c)*exp(2*d*x)
+ 2*a^3*b^4*exp(d*x)*exp(c) + 28*a^5*b^2*exp(d*x)*exp(c) + a^2*b^5*exp(2*c)*exp(2*d*x) + 14*a^4*b^3*exp(2*c)*
exp(2*d*x)))/(a^4*d + b^4*d + 2*a^2*b^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)*tanh(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Integral(tanh(c + d*x)**2*sech(c + d*x)/(a + b*sinh(c + d*x)), x)

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