3.503 \(\int x \text {csch}^2(a+b x) \text {sech}^3(a+b x) \, dx\)
Optimal. Leaf size=120 \[ \frac {3 i \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}-\frac {3 i \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}-\frac {\text {sech}(a+b x)}{2 b^2}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {3 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 x \text {csch}(a+b x)}{2 b}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b} \]
[Out]
-3*x*arctan(exp(b*x+a))/b-arctanh(cosh(b*x+a))/b^2-3/2*x*csch(b*x+a)/b+3/2*I*polylog(2,-I*exp(b*x+a))/b^2-3/2*
I*polylog(2,I*exp(b*x+a))/b^2-1/2*sech(b*x+a)/b^2+1/2*x*csch(b*x+a)*sech(b*x+a)^2/b
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Rubi [A] time = 0.17, antiderivative size = 120, normalized size of antiderivative =
1.00, number of steps used = 13, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) =
0.667, Rules used = {2621, 288, 321, 207, 5462, 5203, 12, 4180, 2279, 2391, 3770, 2622}
\[ \frac {3 i \text {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}-\frac {3 i \text {PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}-\frac {\text {sech}(a+b x)}{2 b^2}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {3 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 x \text {csch}(a+b x)}{2 b}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[x*Csch[a + b*x]^2*Sech[a + b*x]^3,x]
[Out]
(-3*x*ArcTan[E^(a + b*x)])/b - ArcTanh[Cosh[a + b*x]]/b^2 - (3*x*Csch[a + b*x])/(2*b) + (((3*I)/2)*PolyLog[2,
(-I)*E^(a + b*x)])/b^2 - (((3*I)/2)*PolyLog[2, I*E^(a + b*x)])/b^2 - Sech[a + b*x]/(2*b^2) + (x*Csch[a + b*x]*
Sech[a + b*x]^2)/(2*b)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rule 288
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 2279
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Rule 2391
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]
Rule 2621
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Rule 2622
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 4180
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Rule 5203
Int[ArcTan[u_], x_Symbol] :> Simp[x*ArcTan[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 + u^2), x], x] /; Inv
erseFunctionFreeQ[u, x]
Rule 5462
Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Wit
h[{u = IntHide[Csch[a + b*x]^n*Sech[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[(c + d*x)^(m - 1)
*u, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]
Rubi steps
\begin {align*} \int x \text {csch}^2(a+b x) \text {sech}^3(a+b x) \, dx &=-\frac {3 x \tan ^{-1}(\sinh (a+b x))}{2 b}-\frac {3 x \text {csch}(a+b x)}{2 b}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}-\int \left (-\frac {3 \tan ^{-1}(\sinh (a+b x))}{2 b}-\frac {3 \text {csch}(a+b x)}{2 b}+\frac {\text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}\right ) \, dx\\ &=-\frac {3 x \tan ^{-1}(\sinh (a+b x))}{2 b}-\frac {3 x \text {csch}(a+b x)}{2 b}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}-\frac {\int \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx}{2 b}+\frac {3 \int \tan ^{-1}(\sinh (a+b x)) \, dx}{2 b}+\frac {3 \int \text {csch}(a+b x) \, dx}{2 b}\\ &=-\frac {3 \tanh ^{-1}(\cosh (a+b x))}{2 b^2}-\frac {3 x \text {csch}(a+b x)}{2 b}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}-\frac {\operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\text {sech}(a+b x)\right )}{2 b^2}-\frac {3 \int b x \text {sech}(a+b x) \, dx}{2 b}\\ &=-\frac {3 \tanh ^{-1}(\cosh (a+b x))}{2 b^2}-\frac {3 x \text {csch}(a+b x)}{2 b}-\frac {\text {sech}(a+b x)}{2 b^2}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}-\frac {3}{2} \int x \text {sech}(a+b x) \, dx-\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\text {sech}(a+b x)\right )}{2 b^2}\\ &=-\frac {3 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {3 x \text {csch}(a+b x)}{2 b}-\frac {\text {sech}(a+b x)}{2 b^2}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}+\frac {(3 i) \int \log \left (1-i e^{a+b x}\right ) \, dx}{2 b}-\frac {(3 i) \int \log \left (1+i e^{a+b x}\right ) \, dx}{2 b}\\ &=-\frac {3 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {3 x \text {csch}(a+b x)}{2 b}-\frac {\text {sech}(a+b x)}{2 b^2}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}\\ &=-\frac {3 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {3 x \text {csch}(a+b x)}{2 b}+\frac {3 i \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}-\frac {3 i \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}-\frac {\text {sech}(a+b x)}{2 b^2}+\frac {x \text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b}\\ \end {align*}
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Mathematica [A] time = 2.49, size = 209, normalized size = 1.74 \[ -\frac {-3 i \text {Li}_2\left (-i e^{-a-b x}\right )+3 i \text {Li}_2\left (i e^{-a-b x}\right )-3 i a \log \left (1-i e^{-a-b x}\right )-3 i b x \log \left (1-i e^{-a-b x}\right )+3 i a \log \left (1+i e^{-a-b x}\right )+3 i b x \log \left (1+i e^{-a-b x}\right )-b x \tanh \left (\frac {1}{2} (a+b x)\right )+b x \coth \left (\frac {1}{2} (a+b x)\right )+\text {sech}(a+b x)-2 \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )-3 a \tan ^{-1}(\sinh (a+b x))+b x \tanh (a+b x) \text {sech}(a+b x)}{2 b^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x*Csch[a + b*x]^2*Sech[a + b*x]^3,x]
[Out]
-1/2*(-3*a*ArcTan[Sinh[a + b*x]] + b*x*Coth[(a + b*x)/2] - (3*I)*a*Log[1 - I*E^(-a - b*x)] - (3*I)*b*x*Log[1 -
I*E^(-a - b*x)] + (3*I)*a*Log[1 + I*E^(-a - b*x)] + (3*I)*b*x*Log[1 + I*E^(-a - b*x)] - 2*Log[Tanh[(a + b*x)/
2]] - (3*I)*PolyLog[2, (-I)*E^(-a - b*x)] + (3*I)*PolyLog[2, I*E^(-a - b*x)] + Sech[a + b*x] - b*x*Tanh[(a + b
*x)/2] + b*x*Sech[a + b*x]*Tanh[a + b*x])/b^2
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fricas [B] time = 0.50, size = 2201, normalized size = 18.34 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*csch(b*x+a)^2*sech(b*x+a)^3,x, algorithm="fricas")
[Out]
-1/2*(2*(3*b*x + 1)*cosh(b*x + a)^5 + 10*(3*b*x + 1)*cosh(b*x + a)*sinh(b*x + a)^4 + 2*(3*b*x + 1)*sinh(b*x +
a)^5 + 4*b*x*cosh(b*x + a)^3 + 4*(5*(3*b*x + 1)*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^3 + 4*(5*(3*b*x + 1)*cosh
(b*x + a)^3 + 3*b*x*cosh(b*x + a))*sinh(b*x + a)^2 + 2*(3*b*x - 1)*cosh(b*x + a) - (-3*I*cosh(b*x + a)^6 - 18*
I*cosh(b*x + a)*sinh(b*x + a)^5 - 3*I*sinh(b*x + a)^6 + (-45*I*cosh(b*x + a)^2 - 3*I)*sinh(b*x + a)^4 - 3*I*co
sh(b*x + a)^4 + (-60*I*cosh(b*x + a)^3 - 12*I*cosh(b*x + a))*sinh(b*x + a)^3 + (-45*I*cosh(b*x + a)^4 - 18*I*c
osh(b*x + a)^2 + 3*I)*sinh(b*x + a)^2 + 3*I*cosh(b*x + a)^2 + (-18*I*cosh(b*x + a)^5 - 12*I*cosh(b*x + a)^3 +
6*I*cosh(b*x + a))*sinh(b*x + a) + 3*I)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (3*I*cosh(b*x + a)^6 + 18*I
*cosh(b*x + a)*sinh(b*x + a)^5 + 3*I*sinh(b*x + a)^6 + (45*I*cosh(b*x + a)^2 + 3*I)*sinh(b*x + a)^4 + 3*I*cosh
(b*x + a)^4 + (60*I*cosh(b*x + a)^3 + 12*I*cosh(b*x + a))*sinh(b*x + a)^3 + (45*I*cosh(b*x + a)^4 + 18*I*cosh(
b*x + a)^2 - 3*I)*sinh(b*x + a)^2 - 3*I*cosh(b*x + a)^2 + (18*I*cosh(b*x + a)^5 + 12*I*cosh(b*x + a)^3 - 6*I*c
osh(b*x + a))*sinh(b*x + a) - 3*I)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 2*(cosh(b*x + a)^6 + 6*cosh(b*x
+ a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + (15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + cosh(b*x + a)^4 + 4*(5*co
sh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 + 6*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2
- cosh(b*x + a)^2 + 2*(3*cosh(b*x + a)^5 + 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) - 1)*log(cosh(b*x
+ a) + sinh(b*x + a) + 1) - (3*I*a*cosh(b*x + a)^6 + 18*I*a*cosh(b*x + a)*sinh(b*x + a)^5 + 3*I*a*sinh(b*x + a
)^6 + 3*I*a*cosh(b*x + a)^4 + (45*I*a*cosh(b*x + a)^2 + 3*I*a)*sinh(b*x + a)^4 + (60*I*a*cosh(b*x + a)^3 + 12*
I*a*cosh(b*x + a))*sinh(b*x + a)^3 - 3*I*a*cosh(b*x + a)^2 + (45*I*a*cosh(b*x + a)^4 + 18*I*a*cosh(b*x + a)^2
- 3*I*a)*sinh(b*x + a)^2 + (18*I*a*cosh(b*x + a)^5 + 12*I*a*cosh(b*x + a)^3 - 6*I*a*cosh(b*x + a))*sinh(b*x +
a) - 3*I*a)*log(cosh(b*x + a) + sinh(b*x + a) + I) - (-3*I*a*cosh(b*x + a)^6 - 18*I*a*cosh(b*x + a)*sinh(b*x +
a)^5 - 3*I*a*sinh(b*x + a)^6 - 3*I*a*cosh(b*x + a)^4 + (-45*I*a*cosh(b*x + a)^2 - 3*I*a)*sinh(b*x + a)^4 + (-
60*I*a*cosh(b*x + a)^3 - 12*I*a*cosh(b*x + a))*sinh(b*x + a)^3 + 3*I*a*cosh(b*x + a)^2 + (-45*I*a*cosh(b*x + a
)^4 - 18*I*a*cosh(b*x + a)^2 + 3*I*a)*sinh(b*x + a)^2 + (-18*I*a*cosh(b*x + a)^5 - 12*I*a*cosh(b*x + a)^3 + 6*
I*a*cosh(b*x + a))*sinh(b*x + a) + 3*I*a)*log(cosh(b*x + a) + sinh(b*x + a) - I) - 2*(cosh(b*x + a)^6 + 6*cosh
(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + (15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + cosh(b*x + a)^4 + 4*(
5*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 + 6*cosh(b*x + a)^2 - 1)*sinh(b*x + a
)^2 - cosh(b*x + a)^2 + 2*(3*cosh(b*x + a)^5 + 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) - 1)*log(cosh(
b*x + a) + sinh(b*x + a) - 1) - ((3*I*b*x + 3*I*a)*cosh(b*x + a)^6 + (18*I*b*x + 18*I*a)*cosh(b*x + a)*sinh(b*
x + a)^5 + (3*I*b*x + 3*I*a)*sinh(b*x + a)^6 + (3*I*b*x + 3*I*a)*cosh(b*x + a)^4 + ((45*I*b*x + 45*I*a)*cosh(b
*x + a)^2 + 3*I*b*x + 3*I*a)*sinh(b*x + a)^4 + ((60*I*b*x + 60*I*a)*cosh(b*x + a)^3 + (12*I*b*x + 12*I*a)*cosh
(b*x + a))*sinh(b*x + a)^3 + (-3*I*b*x - 3*I*a)*cosh(b*x + a)^2 + ((45*I*b*x + 45*I*a)*cosh(b*x + a)^4 + (18*I
*b*x + 18*I*a)*cosh(b*x + a)^2 - 3*I*b*x - 3*I*a)*sinh(b*x + a)^2 - 3*I*b*x + ((18*I*b*x + 18*I*a)*cosh(b*x +
a)^5 + (12*I*b*x + 12*I*a)*cosh(b*x + a)^3 + (-6*I*b*x - 6*I*a)*cosh(b*x + a))*sinh(b*x + a) - 3*I*a)*log(I*co
sh(b*x + a) + I*sinh(b*x + a) + 1) - ((-3*I*b*x - 3*I*a)*cosh(b*x + a)^6 + (-18*I*b*x - 18*I*a)*cosh(b*x + a)*
sinh(b*x + a)^5 + (-3*I*b*x - 3*I*a)*sinh(b*x + a)^6 + (-3*I*b*x - 3*I*a)*cosh(b*x + a)^4 + ((-45*I*b*x - 45*I
*a)*cosh(b*x + a)^2 - 3*I*b*x - 3*I*a)*sinh(b*x + a)^4 + ((-60*I*b*x - 60*I*a)*cosh(b*x + a)^3 + (-12*I*b*x -
12*I*a)*cosh(b*x + a))*sinh(b*x + a)^3 + (3*I*b*x + 3*I*a)*cosh(b*x + a)^2 + ((-45*I*b*x - 45*I*a)*cosh(b*x +
a)^4 + (-18*I*b*x - 18*I*a)*cosh(b*x + a)^2 + 3*I*b*x + 3*I*a)*sinh(b*x + a)^2 + 3*I*b*x + ((-18*I*b*x - 18*I*
a)*cosh(b*x + a)^5 + (-12*I*b*x - 12*I*a)*cosh(b*x + a)^3 + (6*I*b*x + 6*I*a)*cosh(b*x + a))*sinh(b*x + a) + 3
*I*a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + 2*(5*(3*b*x + 1)*cosh(b*x + a)^4 + 6*b*x*cosh(b*x + a)^2 +
3*b*x - 1)*sinh(b*x + a))/(b^2*cosh(b*x + a)^6 + 6*b^2*cosh(b*x + a)*sinh(b*x + a)^5 + b^2*sinh(b*x + a)^6 +
b^2*cosh(b*x + a)^4 + (15*b^2*cosh(b*x + a)^2 + b^2)*sinh(b*x + a)^4 - b^2*cosh(b*x + a)^2 + 4*(5*b^2*cosh(b*x
+ a)^3 + b^2*cosh(b*x + a))*sinh(b*x + a)^3 + (15*b^2*cosh(b*x + a)^4 + 6*b^2*cosh(b*x + a)^2 - b^2)*sinh(b*x
+ a)^2 - b^2 + 2*(3*b^2*cosh(b*x + a)^5 + 2*b^2*cosh(b*x + a)^3 - b^2*cosh(b*x + a))*sinh(b*x + a))
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {csch}\left (b x + a\right )^{2} \operatorname {sech}\left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*csch(b*x+a)^2*sech(b*x+a)^3,x, algorithm="giac")
[Out]
integrate(x*csch(b*x + a)^2*sech(b*x + a)^3, x)
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maple [B] time = 0.71, size = 232, normalized size = 1.93 \[ -\frac {{\mathrm e}^{b x +a} \left (3 b x \,{\mathrm e}^{4 b x +4 a}+2 b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{4 b x +4 a}+3 b x -1\right )}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{2}}-\frac {\ln \left (1+{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {3 a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {3 i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{2 b}+\frac {3 i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {3 i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {3 i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {3 i \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {3 i \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*csch(b*x+a)^2*sech(b*x+a)^3,x)
[Out]
-exp(b*x+a)*(3*b*x*exp(4*b*x+4*a)+2*b*x*exp(2*b*x+2*a)+exp(4*b*x+4*a)+3*b*x-1)/b^2/(1+exp(2*b*x+2*a))^2/(exp(2
*b*x+2*a)-1)+1/b^2*ln(exp(b*x+a)-1)-1/b^2*ln(1+exp(b*x+a))+3/b^2*a*arctan(exp(b*x+a))+3/2*I/b*ln(1+I*exp(b*x+a
))*x+3/2*I/b^2*ln(1+I*exp(b*x+a))*a-3/2*I/b*ln(1-I*exp(b*x+a))*x-3/2*I/b^2*ln(1-I*exp(b*x+a))*a-3/2*I/b^2*dilo
g(1-I*exp(b*x+a))+3/2*I/b^2*dilog(1+I*exp(b*x+a))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, b x e^{\left (3 \, b x + 3 \, a\right )} + {\left (3 \, b x e^{\left (5 \, a\right )} + e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )} + {\left (3 \, b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (6 \, b x + 6 \, a\right )} + b^{2} e^{\left (4 \, b x + 4 \, a\right )} - b^{2} e^{\left (2 \, b x + 2 \, a\right )} - b^{2}} - \frac {\log \left ({\left (e^{\left (b x + a\right )} + 1\right )} e^{\left (-a\right )}\right )}{b^{2}} + \frac {\log \left ({\left (e^{\left (b x + a\right )} - 1\right )} e^{\left (-a\right )}\right )}{b^{2}} - 96 \, \int \frac {x e^{\left (b x + a\right )}}{32 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*csch(b*x+a)^2*sech(b*x+a)^3,x, algorithm="maxima")
[Out]
-(2*b*x*e^(3*b*x + 3*a) + (3*b*x*e^(5*a) + e^(5*a))*e^(5*b*x) + (3*b*x*e^a - e^a)*e^(b*x))/(b^2*e^(6*b*x + 6*a
) + b^2*e^(4*b*x + 4*a) - b^2*e^(2*b*x + 2*a) - b^2) - log((e^(b*x + a) + 1)*e^(-a))/b^2 + log((e^(b*x + a) -
1)*e^(-a))/b^2 - 96*integrate(1/32*x*e^(b*x + a)/(e^(2*b*x + 2*a) + 1), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/(cosh(a + b*x)^3*sinh(a + b*x)^2),x)
[Out]
int(x/(cosh(a + b*x)^3*sinh(a + b*x)^2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {csch}^{2}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*csch(b*x+a)**2*sech(b*x+a)**3,x)
[Out]
Integral(x*csch(a + b*x)**2*sech(a + b*x)**3, x)
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