3.614 \(\int (a \text {sech}(x)+b \tanh (x))^5 \, dx\)

Optimal. Leaf size=124 \[ -\frac {1}{8} a b^2 \left (3 a^2+7 b^2\right ) \sinh (x)-\frac {1}{8} \text {sech}^2(x) (a+b \sinh (x))^2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \sinh (x)\right )+\frac {1}{8} a \left (3 a^4+10 a^2 b^2+15 b^4\right ) \tan ^{-1}(\sinh (x))-\frac {1}{4} \text {sech}^4(x) (b-a \sinh (x)) (a+b \sinh (x))^4+b^5 \log (\cosh (x)) \]

[Out]

1/8*a*(3*a^4+10*a^2*b^2+15*b^4)*arctan(sinh(x))+b^5*ln(cosh(x))-1/8*a*b^2*(3*a^2+7*b^2)*sinh(x)-1/4*sech(x)^4*
(b-a*sinh(x))*(a+b*sinh(x))^4-1/8*sech(x)^2*(a+b*sinh(x))^2*(2*b*(a^2+2*b^2)-a*(3*a^2+5*b^2)*sinh(x))

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Rubi [A]  time = 0.19, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {4391, 2668, 739, 819, 774, 635, 204, 260} \[ -\frac {1}{8} a b^2 \left (3 a^2+7 b^2\right ) \sinh (x)+\frac {1}{8} a \left (10 a^2 b^2+3 a^4+15 b^4\right ) \tan ^{-1}(\sinh (x))-\frac {1}{8} \text {sech}^2(x) (a+b \sinh (x))^2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \sinh (x)\right )-\frac {1}{4} \text {sech}^4(x) (b-a \sinh (x)) (a+b \sinh (x))^4+b^5 \log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(a*Sech[x] + b*Tanh[x])^5,x]

[Out]

(a*(3*a^4 + 10*a^2*b^2 + 15*b^4)*ArcTan[Sinh[x]])/8 + b^5*Log[Cosh[x]] - (a*b^2*(3*a^2 + 7*b^2)*Sinh[x])/8 - (
Sech[x]^4*(b - a*Sinh[x])*(a + b*Sinh[x])^4)/4 - (Sech[x]^2*(a + b*Sinh[x])^2*(2*b*(a^2 + 2*b^2) - a*(3*a^2 +
5*b^2)*Sinh[x]))/8

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 739

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*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 774

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

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int (a \text {sech}(x)+b \tanh (x))^5 \, dx &=\int \text {sech}^5(x) (a+b \sinh (x))^5 \, dx\\ &=-\left (b^5 \operatorname {Subst}\left (\int \frac {(a+x)^5}{\left (-b^2-x^2\right )^3} \, dx,x,b \sinh (x)\right )\right )\\ &=-\frac {1}{4} \text {sech}^4(x) (b-a \sinh (x)) (a+b \sinh (x))^4-\frac {1}{4} b^3 \operatorname {Subst}\left (\int \frac {(a+x)^3 \left (-3 a^2-4 b^2+a x\right )}{\left (-b^2-x^2\right )^2} \, dx,x,b \sinh (x)\right )\\ &=-\frac {1}{4} \text {sech}^4(x) (b-a \sinh (x)) (a+b \sinh (x))^4-\frac {1}{8} \text {sech}^2(x) (a+b \sinh (x))^2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \sinh (x)\right )-\frac {1}{8} b \operatorname {Subst}\left (\int \frac {(a+x) \left (3 a^4+7 a^2 b^2+8 b^4-a \left (3 a^2+7 b^2\right ) x\right )}{-b^2-x^2} \, dx,x,b \sinh (x)\right )\\ &=-\frac {1}{8} a b^2 \left (3 a^2+7 b^2\right ) \sinh (x)-\frac {1}{4} \text {sech}^4(x) (b-a \sinh (x)) (a+b \sinh (x))^4-\frac {1}{8} \text {sech}^2(x) (a+b \sinh (x))^2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \sinh (x)\right )+\frac {1}{8} b \operatorname {Subst}\left (\int \frac {-a b^2 \left (3 a^2+7 b^2\right )-a \left (3 a^4+7 a^2 b^2+8 b^4\right )-\left (3 a^4+7 a^2 b^2+8 b^4-a^2 \left (3 a^2+7 b^2\right )\right ) x}{-b^2-x^2} \, dx,x,b \sinh (x)\right )\\ &=-\frac {1}{8} a b^2 \left (3 a^2+7 b^2\right ) \sinh (x)-\frac {1}{4} \text {sech}^4(x) (b-a \sinh (x)) (a+b \sinh (x))^4-\frac {1}{8} \text {sech}^2(x) (a+b \sinh (x))^2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \sinh (x)\right )-b^5 \operatorname {Subst}\left (\int \frac {x}{-b^2-x^2} \, dx,x,b \sinh (x)\right )-\frac {1}{8} \left (a b \left (3 a^4+10 a^2 b^2+15 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-b^2-x^2} \, dx,x,b \sinh (x)\right )\\ &=\frac {1}{8} a \left (3 a^4+10 a^2 b^2+15 b^4\right ) \tan ^{-1}(\sinh (x))+b^5 \log (\cosh (x))-\frac {1}{8} a b^2 \left (3 a^2+7 b^2\right ) \sinh (x)-\frac {1}{4} \text {sech}^4(x) (b-a \sinh (x)) (a+b \sinh (x))^4-\frac {1}{8} \text {sech}^2(x) (a+b \sinh (x))^2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \sinh (x)\right )\\ \end {align*}

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Mathematica [B]  time = 1.96, size = 355, normalized size = 2.86 \[ \frac {\frac {2 \text {sech}^2(x) \left (a \left (3 a^2-5 b^2\right ) \sinh (x)+6 a^2 b-2 b^3\right ) (a+b \sinh (x))^6}{a^2+b^2}+\frac {b \left (2 a b^5 \left (5 b^2-3 a^2\right ) \sinh ^5(x)+4 b^4 \left (-9 a^4+12 a^2 b^2+b^4\right ) \sinh ^4(x)+10 a b^3 \left (-9 a^4+8 a^2 b^2+b^4\right ) \sinh ^3(x)-8 b^2 \left (15 a^6-4 a^4 b^2+2 a^2 b^4+b^6\right ) \sinh ^2(x)-10 a b \left (9 a^6+6 a^4 b^2+8 a^2 b^4+3 b^6\right ) \sinh (x)+\frac {\left (a^2+b^2\right )^2 \left (\left (3 a^5+10 a^3 b^2+15 a b^4+8 b^4 \sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \sinh (x)\right )+\left (-3 a^5-10 a^3 b^2-15 a b^4+8 \left (-b^2\right )^{5/2}\right ) \log \left (\sqrt {-b^2}+b \sinh (x)\right )\right )}{\sqrt {-b^2}}\right )}{a^2+b^2}+4 \text {sech}^4(x) (a \sinh (x)+b) (a+b \sinh (x))^6}{16 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sech[x] + b*Tanh[x])^5,x]

[Out]

(4*Sech[x]^4*(b + a*Sinh[x])*(a + b*Sinh[x])^6 + (2*Sech[x]^2*(a + b*Sinh[x])^6*(6*a^2*b - 2*b^3 + a*(3*a^2 -
5*b^2)*Sinh[x]))/(a^2 + b^2) + (b*(((a^2 + b^2)^2*((3*a^5 + 10*a^3*b^2 + 15*a*b^4 + 8*b^4*Sqrt[-b^2])*Log[Sqrt
[-b^2] - b*Sinh[x]] + (-3*a^5 - 10*a^3*b^2 - 15*a*b^4 + 8*(-b^2)^(5/2))*Log[Sqrt[-b^2] + b*Sinh[x]]))/Sqrt[-b^
2] - 10*a*b*(9*a^6 + 6*a^4*b^2 + 8*a^2*b^4 + 3*b^6)*Sinh[x] - 8*b^2*(15*a^6 - 4*a^4*b^2 + 2*a^2*b^4 + b^6)*Sin
h[x]^2 + 10*a*b^3*(-9*a^4 + 8*a^2*b^2 + b^4)*Sinh[x]^3 + 4*b^4*(-9*a^4 + 12*a^2*b^2 + b^4)*Sinh[x]^4 + 2*a*b^5
*(-3*a^2 + 5*b^2)*Sinh[x]^5))/(a^2 + b^2))/(16*(a^2 + b^2))

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fricas [B]  time = 0.46, size = 2040, normalized size = 16.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)+b*tanh(x))^5,x, algorithm="fricas")

[Out]

-1/4*(4*b^5*x*cosh(x)^8 + 4*b^5*x*sinh(x)^8 - (3*a^5 + 10*a^3*b^2 - 25*a*b^4)*cosh(x)^7 + (32*b^5*x*cosh(x) -
3*a^5 - 10*a^3*b^2 + 25*a*b^4)*sinh(x)^7 + 16*(b^5*x + 5*a^2*b^3 - b^5)*cosh(x)^6 + (112*b^5*x*cosh(x)^2 + 16*
b^5*x + 80*a^2*b^3 - 16*b^5 - 7*(3*a^5 + 10*a^3*b^2 - 25*a*b^4)*cosh(x))*sinh(x)^6 + 4*b^5*x - (11*a^5 - 70*a^
3*b^2 + 15*a*b^4)*cosh(x)^5 + (224*b^5*x*cosh(x)^3 - 11*a^5 + 70*a^3*b^2 - 15*a*b^4 - 21*(3*a^5 + 10*a^3*b^2 -
25*a*b^4)*cosh(x)^2 + 96*(b^5*x + 5*a^2*b^3 - b^5)*cosh(x))*sinh(x)^5 + 8*(3*b^5*x + 10*a^4*b - 2*b^5)*cosh(x
)^4 + (280*b^5*x*cosh(x)^4 + 24*b^5*x + 80*a^4*b - 16*b^5 - 35*(3*a^5 + 10*a^3*b^2 - 25*a*b^4)*cosh(x)^3 + 240
*(b^5*x + 5*a^2*b^3 - b^5)*cosh(x)^2 - 5*(11*a^5 - 70*a^3*b^2 + 15*a*b^4)*cosh(x))*sinh(x)^4 + (11*a^5 - 70*a^
3*b^2 + 15*a*b^4)*cosh(x)^3 + (224*b^5*x*cosh(x)^5 + 11*a^5 - 70*a^3*b^2 + 15*a*b^4 - 35*(3*a^5 + 10*a^3*b^2 -
25*a*b^4)*cosh(x)^4 + 320*(b^5*x + 5*a^2*b^3 - b^5)*cosh(x)^3 - 10*(11*a^5 - 70*a^3*b^2 + 15*a*b^4)*cosh(x)^2
+ 32*(3*b^5*x + 10*a^4*b - 2*b^5)*cosh(x))*sinh(x)^3 + 16*(b^5*x + 5*a^2*b^3 - b^5)*cosh(x)^2 + (112*b^5*x*co
sh(x)^6 + 16*b^5*x - 21*(3*a^5 + 10*a^3*b^2 - 25*a*b^4)*cosh(x)^5 + 80*a^2*b^3 - 16*b^5 + 240*(b^5*x + 5*a^2*b
^3 - b^5)*cosh(x)^4 - 10*(11*a^5 - 70*a^3*b^2 + 15*a*b^4)*cosh(x)^3 + 48*(3*b^5*x + 10*a^4*b - 2*b^5)*cosh(x)^
2 + 3*(11*a^5 - 70*a^3*b^2 + 15*a*b^4)*cosh(x))*sinh(x)^2 - ((3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^8 + 8*(3*
a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)*sinh(x)^7 + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*sinh(x)^8 + 4*(3*a^5 + 10*a^3
*b^2 + 15*a*b^4)*cosh(x)^6 + 4*(3*a^5 + 10*a^3*b^2 + 15*a*b^4 + 7*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^2)*s
inh(x)^6 + 8*(7*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^3 + 3*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x))*sinh(x)
^5 + 3*a^5 + 10*a^3*b^2 + 15*a*b^4 + 6*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^4 + 2*(9*a^5 + 30*a^3*b^2 + 45*
a*b^4 + 35*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^4 + 30*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^2)*sinh(x)^4
+ 8*(7*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^5 + 10*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^3 + 3*(3*a^5 +
10*a^3*b^2 + 15*a*b^4)*cosh(x))*sinh(x)^3 + 4*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^2 + 4*(7*(3*a^5 + 10*a^3
*b^2 + 15*a*b^4)*cosh(x)^6 + 3*a^5 + 10*a^3*b^2 + 15*a*b^4 + 15*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^4 + 9*
(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^2)*sinh(x)^2 + 8*((3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^7 + 3*(3*a^5
+ 10*a^3*b^2 + 15*a*b^4)*cosh(x)^5 + 3*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*cosh(x)^3 + (3*a^5 + 10*a^3*b^2 + 15*a
*b^4)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) + (3*a^5 + 10*a^3*b^2 - 25*a*b^4)*cosh(x) - 4*(b^5*cosh(x)^8
+ 8*b^5*cosh(x)*sinh(x)^7 + b^5*sinh(x)^8 + 4*b^5*cosh(x)^6 + 6*b^5*cosh(x)^4 + 4*b^5*cosh(x)^2 + 4*(7*b^5*co
sh(x)^2 + b^5)*sinh(x)^6 + 8*(7*b^5*cosh(x)^3 + 3*b^5*cosh(x))*sinh(x)^5 + b^5 + 2*(35*b^5*cosh(x)^4 + 30*b^5*
cosh(x)^2 + 3*b^5)*sinh(x)^4 + 8*(7*b^5*cosh(x)^5 + 10*b^5*cosh(x)^3 + 3*b^5*cosh(x))*sinh(x)^3 + 4*(7*b^5*cos
h(x)^6 + 15*b^5*cosh(x)^4 + 9*b^5*cosh(x)^2 + b^5)*sinh(x)^2 + 8*(b^5*cosh(x)^7 + 3*b^5*cosh(x)^5 + 3*b^5*cosh
(x)^3 + b^5*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + (32*b^5*x*cosh(x)^7 - 7*(3*a^5 + 10*a^3*b^2
- 25*a*b^4)*cosh(x)^6 + 96*(b^5*x + 5*a^2*b^3 - b^5)*cosh(x)^5 + 3*a^5 + 10*a^3*b^2 - 25*a*b^4 - 5*(11*a^5 -
70*a^3*b^2 + 15*a*b^4)*cosh(x)^4 + 32*(3*b^5*x + 10*a^4*b - 2*b^5)*cosh(x)^3 + 3*(11*a^5 - 70*a^3*b^2 + 15*a*b
^4)*cosh(x)^2 + 32*(b^5*x + 5*a^2*b^3 - b^5)*cosh(x))*sinh(x))/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 +
4*(7*cosh(x)^2 + 1)*sinh(x)^6 + 4*cosh(x)^6 + 8*(7*cosh(x)^3 + 3*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 + 30*cos
h(x)^2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 + 10*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 +
15*cosh(x)^4 + 9*cosh(x)^2 + 1)*sinh(x)^2 + 4*cosh(x)^2 + 8*(cosh(x)^7 + 3*cosh(x)^5 + 3*cosh(x)^3 + cosh(x))*
sinh(x) + 1)

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giac [B]  time = 0.14, size = 240, normalized size = 1.94 \[ \frac {1}{2} \, b^{5} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right ) + \frac {1}{16} \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} - \frac {3 \, b^{5} {\left (e^{\left (-x\right )} - e^{x}\right )}^{4} + 3 \, a^{5} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 10 \, a^{3} b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - 25 \, a b^{4} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 80 \, a^{2} b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 8 \, b^{5} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 20 \, a^{5} {\left (e^{\left (-x\right )} - e^{x}\right )} - 40 \, a^{3} b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )} - 60 \, a b^{4} {\left (e^{\left (-x\right )} - e^{x}\right )} + 80 \, a^{4} b + 160 \, a^{2} b^{3}}{4 \, {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)+b*tanh(x))^5,x, algorithm="giac")

[Out]

1/2*b^5*log((e^(-x) - e^x)^2 + 4) + 1/16*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*(3*a^5 + 10*a^3*b^2 + 15*a*
b^4) - 1/4*(3*b^5*(e^(-x) - e^x)^4 + 3*a^5*(e^(-x) - e^x)^3 + 10*a^3*b^2*(e^(-x) - e^x)^3 - 25*a*b^4*(e^(-x) -
e^x)^3 + 80*a^2*b^3*(e^(-x) - e^x)^2 + 8*b^5*(e^(-x) - e^x)^2 + 20*a^5*(e^(-x) - e^x) - 40*a^3*b^2*(e^(-x) -
e^x) - 60*a*b^4*(e^(-x) - e^x) + 80*a^4*b + 160*a^2*b^3)/((e^(-x) - e^x)^2 + 4)^2

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maple [A]  time = 0.34, size = 201, normalized size = 1.62 \[ \frac {a^{5} \tanh \relax (x ) \mathrm {sech}\relax (x )^{3}}{4}+\frac {3 a^{5} \mathrm {sech}\relax (x ) \tanh \relax (x )}{8}+\frac {3 a^{5} \arctan \left ({\mathrm e}^{x}\right )}{4}-\frac {5 a^{4} b}{4 \cosh \relax (x )^{4}}-\frac {10 a^{3} b^{2} \sinh \relax (x )}{3 \cosh \relax (x )^{4}}+\frac {5 a^{3} b^{2} \tanh \relax (x ) \mathrm {sech}\relax (x )^{3}}{6}+\frac {5 a^{3} b^{2} \mathrm {sech}\relax (x ) \tanh \relax (x )}{4}+\frac {5 a^{3} b^{2} \arctan \left ({\mathrm e}^{x}\right )}{2}-\frac {5 a^{2} b^{3} \left (\sinh ^{2}\relax (x )\right )}{\cosh \relax (x )^{4}}-\frac {5 a^{2} b^{3}}{2 \cosh \relax (x )^{4}}-\frac {5 a \,b^{4} \left (\sinh ^{3}\relax (x )\right )}{\cosh \relax (x )^{4}}-\frac {5 a \,b^{4} \sinh \relax (x )}{\cosh \relax (x )^{4}}+\frac {5 a \,b^{4} \tanh \relax (x ) \mathrm {sech}\relax (x )^{3}}{4}+\frac {15 a \,b^{4} \mathrm {sech}\relax (x ) \tanh \relax (x )}{8}+\frac {15 a \,b^{4} \arctan \left ({\mathrm e}^{x}\right )}{4}+b^{5} \ln \left (\cosh \relax (x )\right )-\frac {b^{5} \left (\tanh ^{2}\relax (x )\right )}{2}-\frac {b^{5} \left (\tanh ^{4}\relax (x )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sech(x)+b*tanh(x))^5,x)

[Out]

1/4*a^5*tanh(x)*sech(x)^3+3/8*a^5*sech(x)*tanh(x)+3/4*a^5*arctan(exp(x))-5/4*a^4*b/cosh(x)^4-10/3*a^3*b^2*sinh
(x)/cosh(x)^4+5/6*a^3*b^2*tanh(x)*sech(x)^3+5/4*a^3*b^2*sech(x)*tanh(x)+5/2*a^3*b^2*arctan(exp(x))-5*a^2*b^3*s
inh(x)^2/cosh(x)^4-5/2*a^2*b^3/cosh(x)^4-5*a*b^4*sinh(x)^3/cosh(x)^4-5*a*b^4*sinh(x)/cosh(x)^4+5/4*a*b^4*tanh(
x)*sech(x)^3+15/8*a*b^4*sech(x)*tanh(x)+15/4*a*b^4*arctan(exp(x))+b^5*ln(cosh(x))-1/2*b^5*tanh(x)^2-1/4*b^5*ta
nh(x)^4

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maxima [B]  time = 0.54, size = 279, normalized size = 2.25 \[ \frac {5}{2} \, a^{2} b^{3} \tanh \relax (x)^{4} + b^{5} {\left (x + \frac {4 \, {\left (e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1} + \log \left (e^{\left (-2 \, x\right )} + 1\right )\right )} - \frac {5}{4} \, a b^{4} {\left (\frac {5 \, e^{\left (-x\right )} - 3 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-5 \, x\right )} - 5 \, e^{\left (-7 \, x\right )}}{4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1} + 3 \, \arctan \left (e^{\left (-x\right )}\right )\right )} + \frac {1}{4} \, a^{5} {\left (\frac {3 \, e^{\left (-x\right )} + 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} - 3 \, e^{\left (-7 \, x\right )}}{4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1} - 3 \, \arctan \left (e^{\left (-x\right )}\right )\right )} + \frac {5}{2} \, a^{3} b^{2} {\left (\frac {e^{\left (-x\right )} - 7 \, e^{\left (-3 \, x\right )} + 7 \, e^{\left (-5 \, x\right )} - e^{\left (-7 \, x\right )}}{4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1} - \arctan \left (e^{\left (-x\right )}\right )\right )} - \frac {20 \, a^{4} b}{{\left (e^{\left (-x\right )} + e^{x}\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)+b*tanh(x))^5,x, algorithm="maxima")

[Out]

5/2*a^2*b^3*tanh(x)^4 + b^5*(x + 4*(e^(-2*x) + e^(-4*x) + e^(-6*x))/(4*e^(-2*x) + 6*e^(-4*x) + 4*e^(-6*x) + e^
(-8*x) + 1) + log(e^(-2*x) + 1)) - 5/4*a*b^4*((5*e^(-x) - 3*e^(-3*x) + 3*e^(-5*x) - 5*e^(-7*x))/(4*e^(-2*x) +
6*e^(-4*x) + 4*e^(-6*x) + e^(-8*x) + 1) + 3*arctan(e^(-x))) + 1/4*a^5*((3*e^(-x) + 11*e^(-3*x) - 11*e^(-5*x) -
3*e^(-7*x))/(4*e^(-2*x) + 6*e^(-4*x) + 4*e^(-6*x) + e^(-8*x) + 1) - 3*arctan(e^(-x))) + 5/2*a^3*b^2*((e^(-x)
- 7*e^(-3*x) + 7*e^(-5*x) - e^(-7*x))/(4*e^(-2*x) + 6*e^(-4*x) + 4*e^(-6*x) + e^(-8*x) + 1) - arctan(e^(-x)))
- 20*a^4*b/(e^(-x) + e^x)^4

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mupad [B]  time = 1.92, size = 495, normalized size = 3.99 \[ \frac {{\mathrm {e}}^x\,\left (4\,a^5-40\,a^3\,b^2+20\,a\,b^4\right )-20\,a^4\,b-4\,b^5+40\,a^2\,b^3}{4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1}+b^5\,\ln \left (\left (\frac {3\,a^5\,{\mathrm {e}}^x}{4}-2\,\sqrt {-\frac {9\,a^{10}}{64}-\frac {15\,a^8\,b^2}{16}-\frac {95\,a^6\,b^4}{32}-\frac {75\,a^4\,b^6}{16}-\frac {225\,a^2\,b^8}{64}}+\frac {15\,a\,b^4\,{\mathrm {e}}^x}{4}+\frac {5\,a^3\,b^2\,{\mathrm {e}}^x}{2}\right )\,\left (2\,\sqrt {-\frac {9\,a^{10}}{64}-\frac {15\,a^8\,b^2}{16}-\frac {95\,a^6\,b^4}{32}-\frac {75\,a^4\,b^6}{16}-\frac {225\,a^2\,b^8}{64}}+\frac {3\,a^5\,{\mathrm {e}}^x}{4}+\frac {15\,a\,b^4\,{\mathrm {e}}^x}{4}+\frac {5\,a^3\,b^2\,{\mathrm {e}}^x}{2}\right )\right )-\frac {{\mathrm {e}}^x\,\left (6\,a^5-60\,a^3\,b^2+30\,a\,b^4\right )-40\,a^4\,b-8\,b^5+80\,a^2\,b^3}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-b^5\,x+\frac {\mathrm {atan}\left (\frac {4\,{\mathrm {e}}^x\,\left (\frac {3\,a^5}{4}+\frac {5\,a^3\,b^2}{2}+\frac {15\,a\,b^4}{4}\right )}{\sqrt {9\,a^{10}+60\,a^8\,b^2+190\,a^6\,b^4+300\,a^4\,b^6+225\,a^2\,b^8}}\right )\,\sqrt {9\,a^{10}+60\,a^8\,b^2+190\,a^6\,b^4+300\,a^4\,b^6+225\,a^2\,b^8}}{4}+\frac {{\mathrm {e}}^x\,\left (\frac {3\,a^5}{4}+\frac {5\,a^3\,b^2}{2}-\frac {25\,a\,b^4}{4}\right )+4\,b^5-20\,a^2\,b^3}{{\mathrm {e}}^{2\,x}+1}+\frac {{\mathrm {e}}^x\,\left (\frac {a^5}{2}-25\,a^3\,b^2+\frac {45\,a\,b^4}{2}\right )-20\,a^4\,b-8\,b^5+60\,a^2\,b^3}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tanh(x) + a/cosh(x))^5,x)

[Out]

(exp(x)*(20*a*b^4 + 4*a^5 - 40*a^3*b^2) - 20*a^4*b - 4*b^5 + 40*a^2*b^3)/(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x)
+ exp(8*x) + 1) + b^5*log(((3*a^5*exp(x))/4 - 2*(- (9*a^10)/64 - (225*a^2*b^8)/64 - (75*a^4*b^6)/16 - (95*a^6
*b^4)/32 - (15*a^8*b^2)/16)^(1/2) + (15*a*b^4*exp(x))/4 + (5*a^3*b^2*exp(x))/2)*(2*(- (9*a^10)/64 - (225*a^2*b
^8)/64 - (75*a^4*b^6)/16 - (95*a^6*b^4)/32 - (15*a^8*b^2)/16)^(1/2) + (3*a^5*exp(x))/4 + (15*a*b^4*exp(x))/4 +
(5*a^3*b^2*exp(x))/2)) - (exp(x)*(30*a*b^4 + 6*a^5 - 60*a^3*b^2) - 40*a^4*b - 8*b^5 + 80*a^2*b^3)/(3*exp(2*x)
+ 3*exp(4*x) + exp(6*x) + 1) - b^5*x + (atan((4*exp(x)*((15*a*b^4)/4 + (3*a^5)/4 + (5*a^3*b^2)/2))/(9*a^10 +
225*a^2*b^8 + 300*a^4*b^6 + 190*a^6*b^4 + 60*a^8*b^2)^(1/2))*(9*a^10 + 225*a^2*b^8 + 300*a^4*b^6 + 190*a^6*b^4
+ 60*a^8*b^2)^(1/2))/4 + (exp(x)*((3*a^5)/4 - (25*a*b^4)/4 + (5*a^3*b^2)/2) + 4*b^5 - 20*a^2*b^3)/(exp(2*x) +
1) + (exp(x)*((45*a*b^4)/2 + a^5/2 - 25*a^3*b^2) - 20*a^4*b - 8*b^5 + 60*a^2*b^3)/(2*exp(2*x) + exp(4*x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \operatorname {sech}{\relax (x )} + b \tanh {\relax (x )}\right )^{5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)+b*tanh(x))**5,x)

[Out]

Integral((a*sech(x) + b*tanh(x))**5, x)

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