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

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

[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

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Rubi [A]  time = 0.185464, antiderivative size = 124, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 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]

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*}

Mathematica [B]  time = 1.87455, size = 355, normalized size = 2.86 $\frac{\frac{b \left (2 a b^5 \left (5 b^2-3 a^2\right ) \sinh ^5(x)+4 b^4 \left (12 a^2 b^2-9 a^4+b^4\right ) \sinh ^4(x)+10 a b^3 \left (8 a^2 b^2-9 a^4+b^4\right ) \sinh ^3(x)-8 b^2 \left (-4 a^4 b^2+2 a^2 b^4+15 a^6+b^6\right ) \sinh ^2(x)-10 a b \left (6 a^4 b^2+8 a^2 b^4+9 a^6+3 b^6\right ) \sinh (x)+\frac{\left (a^2+b^2\right )^2 \left (\left (10 a^3 b^2+3 a^5+15 a b^4+8 b^4 \sqrt{-b^2}\right ) \log \left (\sqrt{-b^2}-b \sinh (x)\right )+\left (-10 a^3 b^2-3 a^5-15 a b^4+8 \left (-b^2\right )^{5/2}\right ) \log \left (\sqrt{-b^2}+b \sinh (x)\right )\right )}{\sqrt{-b^2}}\right )}{a^2+b^2}+\frac{2 \text{sech}^2(x) \left (a \left (3 a^2-5 b^2\right ) \sinh (x)+6 a^2 b-2 b^3\right ) (a+b \sinh (x))^6}{a^2+b^2}+4 \text{sech}^4(x) (a \sinh (x)+b) (a+b \sinh (x))^6}{16 \left (a^2+b^2\right )}$

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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Maple [A]  time = 0.046, size = 223, normalized size = 1.8 \begin{align*}{\frac{{a}^{5}\tanh \left ( x \right ) \left ({\rm sech} \left (x\right ) \right ) ^{3}}{4}}+{\frac{3\,{a}^{5}{\rm sech} \left (x\right )\tanh \left ( x \right ) }{8}}+{\frac{3\,{a}^{5}\arctan \left ({{\rm e}^{x}} \right ) }{4}}+{\frac{5\,{a}^{4}b \left ( \sinh \left ( x \right ) \right ) ^{2}}{4\, \left ( \cosh \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{a}^{4}b \left ( \sinh \left ( x \right ) \right ) ^{2}}{4\, \left ( \cosh \left ( x \right ) \right ) ^{2}}}-{\frac{10\,{a}^{3}{b}^{2}\sinh \left ( x \right ) }{3\, \left ( \cosh \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{a}^{3}{b}^{2}\tanh \left ( x \right ) \left ({\rm sech} \left (x\right ) \right ) ^{3}}{6}}+{\frac{5\,{a}^{3}{b}^{2}{\rm sech} \left (x\right )\tanh \left ( x \right ) }{4}}+{\frac{5\,{a}^{3}{b}^{2}\arctan \left ({{\rm e}^{x}} \right ) }{2}}-{\frac{5\,{a}^{2}{b}^{3} \left ( \sinh \left ( x \right ) \right ) ^{2}}{2\, \left ( \cosh \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{a}^{2}{b}^{3} \left ( \sinh \left ( x \right ) \right ) ^{2}}{2\, \left ( \cosh \left ( x \right ) \right ) ^{2}}}-5\,{\frac{{b}^{4}a \left ( \sinh \left ( x \right ) \right ) ^{3}}{ \left ( \cosh \left ( x \right ) \right ) ^{4}}}-5\,{\frac{{b}^{4}a\sinh \left ( x \right ) }{ \left ( \cosh \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{b}^{4}a\tanh \left ( x \right ) \left ({\rm sech} \left (x\right ) \right ) ^{3}}{4}}+{\frac{15\,{b}^{4}a{\rm sech} \left (x\right )\tanh \left ( x \right ) }{8}}+{\frac{15\,{b}^{4}a\arctan \left ({{\rm e}^{x}} \right ) }{4}}+{b}^{5}\ln \left ( \cosh \left ( x \right ) \right ) -{\frac{{b}^{5} \left ( \tanh \left ( x \right ) \right ) ^{2}}{2}}-{\frac{{b}^{5} \left ( \tanh \left ( x \right ) \right ) ^{4}}{4}} \end{align*}

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

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Maxima [B]  time = 1.6891, size = 377, normalized size = 3.04 \begin{align*} \frac{5}{2} \, a^{2} b^{3} \tanh \left (x\right )^{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}} \end{align*}

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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Fricas [B]  time = 2.81096, size = 5216, normalized size = 42.06 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \operatorname{sech}{\left (x \right )} + b \tanh{\left (x \right )}\right )^{5}\, dx \end{align*}

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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Giac [B]  time = 1.13656, size = 324, normalized size = 2.61 \begin{align*} \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}} \end{align*}

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