3.418 \(\int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=120 \[ \frac {b \left (3 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}+\frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 d \left (a^2+b^2\right )}-\frac {a^3 \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a^3 \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2} \]

[Out]

1/2*b*(3*a^2+b^2)*arctan(sinh(d*x+c))/(a^2+b^2)^2/d+a^3*ln(cosh(d*x+c))/(a^2+b^2)^2/d-a^3*ln(a+b*sinh(d*x+c))/
(a^2+b^2)^2/d+1/2*sech(d*x+c)^2*(a-b*sinh(d*x+c))/(a^2+b^2)/d

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Rubi [A]  time = 0.20, antiderivative size = 120, normalized size of antiderivative = 1.00, 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 = {2721, 1647, 801, 635, 203, 260} \[ -\frac {a^3 \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {b \left (3 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}+\frac {a^3 \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [C]  time = 0.39, size = 152, normalized size = 1.27 \[ -\frac {2 a^3 \log (a+b \sinh (c+d x))-a \left (a^2+b^2\right ) \text {sech}^2(c+d x)+b \left (a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))+b \left (a^2+b^2\right ) \tanh (c+d x) \text {sech}(c+d x)-\left (a^3-i \left (2 a^2 b+b^3\right )\right ) \log (-\sinh (c+d x)+i)-\left (a^3+i \left (2 a^2 b+b^3\right )\right ) \log (\sinh (c+d x)+i)}{2 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*(a^2 + b^2)*ArcTan[Sinh[c + d*x]] - (a^3 - I*(2*a^2*b + b^3))*Log[I - Sinh[c + d*x]] - (a^3 + I*(2*a^2
*b + b^3))*Log[I + Sinh[c + d*x]] + 2*a^3*Log[a + b*Sinh[c + d*x]] - a*(a^2 + b^2)*Sech[c + d*x]^2 + b*(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 = 0.49, size = 896, normalized size = 7.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2*b + b^3)*cosh(d*x + c)^3 + (a^2*b + b^3)*sinh(d*x + c)^3 - 2*(a^3 + a*b^2)*cosh(d*x + c)^2 - (2*a^3 + 2
*a*b^2 - 3*(a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - ((3*a^2*b + b^3)*cosh(d*x + c)^4 + 4*(3*a^2*b + b^3)
*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a^2*b + b^3)*sinh(d*x + c)^4 + 3*a^2*b + b^3 + 2*(3*a^2*b + b^3)*cosh(d*x
+ c)^2 + 2*(3*a^2*b + b^3 + 3*(3*a^2*b + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((3*a^2*b + b^3)*cosh(d*x +
c)^3 + (3*a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - (a^2*b + b^3)*co
sh(d*x + c) + (a^3*cosh(d*x + c)^4 + 4*a^3*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*sinh(d*x + c)^4 + 2*a^3*cosh(d*
x + c)^2 + a^3 + 2*(3*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^2 + 4*(a^3*cosh(d*x + c)^3 + a^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))) - (a^3*cosh(d*x + c)^4 + 4*a^3*co
sh(d*x + c)*sinh(d*x + c)^3 + a^3*sinh(d*x + c)^4 + 2*a^3*cosh(d*x + c)^2 + a^3 + 2*(3*a^3*cosh(d*x + c)^2 + a
^3)*sinh(d*x + c)^2 + 4*(a^3*cosh(d*x + c)^3 + a^3*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x
+ c) - sinh(d*x + c))) - (a^2*b + b^3 - 3*(a^2*b + b^3)*cosh(d*x + c)^2 + 4*(a^3 + a*b^2)*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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giac [A]  time = 3.78, size = 223, normalized size = 1.86 \[ \frac {\frac {a^{3} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a^{3} \log \left ({\left | -b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a e^{\left (d x + c\right )} + b \right |}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (3 \, a^{2} b e^{c} + b^{3} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a^{2} b e^{\left (3 \, c\right )} + b^{3} e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} - 2 \, {\left (a^{3} e^{\left (2 \, c\right )} + a b^{2} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - {\left (a^{2} b e^{c} + b^{3} e^{c}\right )} e^{\left (d x\right )}}{{\left (a^{2} + b^{2}\right )}^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*log(e^(2*d*x + 2*c) + 1)/(a^4 + 2*a^2*b^2 + b^4) - a^3*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) + (3*a^2*b*e^c + b^3*e^c)*arctan(e^(d*x + c))*e^(-c)/(a^4 + 2*a^2*b^2 + b^4) - ((a^2*
b*e^(3*c) + b^3*e^(3*c))*e^(3*d*x) - 2*(a^3*e^(2*c) + a*b^2*e^(2*c))*e^(2*d*x) - (a^2*b*e^c + b^3*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 = 472, normalized size = 3.93 \[ -\frac {8 a^{3} \ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d \left (8 a^{4}+16 a^{2} b^{2}+8 b^{4}\right )}+\frac {\left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a^{3} \ln \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {3 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {\arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/d*a^3/(8*a^4+16*a^2*b^2+8*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^2*b+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*b^3-2/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^3-2/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*b^2-1/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^2*b-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)*b^3+1/d/(a^4+2*a^2*b^2+b^4)*a^3*ln(tanh(1/2*d*x+1/2*c)^2+1)+3/d/(a^4+2*a^2*b^2+b^4)*arctan(tanh(1/2*d*x+1/
2*c))*a^2*b+1/d/(a^4+2*a^2*b^2+b^4)*arctan(tanh(1/2*d*x+1/2*c))*b^3

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maxima [A]  time = 0.41, size = 217, normalized size = 1.81 \[ -\frac {a^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} + \frac {a^{3} \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {b e^{\left (-d x - c\right )} - 2 \, a e^{\left (-2 \, d x - 2 \, c\right )} - b e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^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) + a^3*log(e^(-2*d*x - 2*c) +
1)/((a^4 + 2*a^2*b^2 + b^4)*d) - (3*a^2*b + b^3)*arctan(e^(-d*x - c))/((a^4 + 2*a^2*b^2 + b^4)*d) - (b*e^(-d*x
- c) - 2*a*e^(-2*d*x - 2*c) - b*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 = 2.37, size = 381, normalized size = 3.18 \[ \frac {\frac {2\,\left (a^3+a\,b^2\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (a^2\,b+b^3\right )}{d\,{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,a}{d\,\left (a^2+b^2\right )}-\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,\left (2\,a+b\,1{}\mathrm {i}\right )}{2\,\left (d\,a^2+2{}\mathrm {i}\,d\,a\,b-d\,b^2\right )}+\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )\,\left (b+a\,2{}\mathrm {i}\right )}{2\,\left (1{}\mathrm {i}\,d\,a^2+2\,d\,a\,b-1{}\mathrm {i}\,d\,b^2\right )}-\frac {a^3\,\ln \left (32\,a^7\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-b^7-6\,a^2\,b^5-9\,a^4\,b^3-16\,a^6\,b+b^7\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+16\,a^6\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+12\,a^3\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+18\,a^5\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+6\,a^2\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+9\,a^4\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^4+2\,d\,a^2\,b^2+d\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*(a*b^2 + a^3))/(d*(a^2 + b^2)^2) - (exp(c + d*x)*(a^2*b + b^3))/(d*(a^2 + b^2)^2))/(exp(2*c + 2*d*x) + 1)
- ((2*a)/(d*(a^2 + b^2)) - (2*b*exp(c + d*x))/(d*(a^2 + b^2)))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) + (
log(exp(c + d*x)*1i + 1)*(2*a + b*1i))/(2*(a^2*d - b^2*d + a*b*d*2i)) + (log(exp(c + d*x) + 1i)*(a*2i + b))/(2
*(a^2*d*1i - b^2*d*1i + 2*a*b*d)) - (a^3*log(32*a^7*exp(d*x)*exp(c) - b^7 - 6*a^2*b^5 - 9*a^4*b^3 - 16*a^6*b +
b^7*exp(2*c)*exp(2*d*x) + 16*a^6*b*exp(2*c)*exp(2*d*x) + 12*a^3*b^4*exp(d*x)*exp(c) + 18*a^5*b^2*exp(d*x)*exp
(c) + 6*a^2*b^5*exp(2*c)*exp(2*d*x) + 9*a^4*b^3*exp(2*c)*exp(2*d*x) + 2*a*b^6*exp(d*x)*exp(c)))/(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 \[ \int \frac {\tanh ^{3}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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