3.2.22 \(\int \frac {\text {sech}^5(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [122]

Optimal. Leaf size=102 \[ \frac {\text {ArcTan}(\sinh (c+d x))}{b^2 d}-\frac {(2 a-b) \sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^2 d}+\frac {(a+b) \sinh (c+d x)}{2 a b d \left (a+(a+b) \sinh ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3757, 425, 536, 209, 211} \begin {gather*} -\frac {(2 a-b) \sqrt {a+b} \text {ArcTan}\left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^2 d}+\frac {(a+b) \sinh (c+d x)}{2 a b d \left ((a+b) \sinh ^2(c+d x)+a\right )}+\frac {\text {ArcTan}(\sinh (c+d x))}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

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

Rule 209

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

Rule 211

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

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

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

Rule 3757

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {(a+b) \sinh (c+d x)}{2 a b d \left (a+(a+b) \sinh ^2(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {a-b+(-a-b) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\sinh (c+d x)\right )}{2 a b d}\\ &=\frac {(a+b) \sinh (c+d x)}{2 a b d \left (a+(a+b) \sinh ^2(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{b^2 d}-\frac {((2 a-b) (a+b)) \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\sinh (c+d x)\right )}{2 a b^2 d}\\ &=\frac {\tan ^{-1}(\sinh (c+d x))}{b^2 d}-\frac {(2 a-b) \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^2 d}+\frac {(a+b) \sinh (c+d x)}{2 a b d \left (a+(a+b) \sinh ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 203, normalized size = 1.99 \begin {gather*} \frac {(a-b) \left (\left (2 a^2+a b-b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )+4 a^{3/2} \sqrt {a+b} \text {ArcTan}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )\right )+(a+b) \left (\left (2 a^2+a b-b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )+4 a^{3/2} \sqrt {a+b} \text {ArcTan}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )\right ) \cosh (2 (c+d x))+2 \sqrt {a} b (a+b)^{3/2} \sinh (c+d x)}{2 a^{3/2} b^2 \sqrt {a+b} d (a-b+(a+b) \cosh (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

((a - b)*((2*a^2 + a*b - b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[a + b]] + 4*a^(3/2)*Sqrt[a + b]*ArcTan[Tanh[
(c + d*x)/2]]) + (a + b)*((2*a^2 + a*b - b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[a + b]] + 4*a^(3/2)*Sqrt[a +
b]*ArcTan[Tanh[(c + d*x)/2]])*Cosh[2*(c + d*x)] + 2*Sqrt[a]*b*(a + b)^(3/2)*Sinh[c + d*x])/(2*a^(3/2)*b^2*Sqr
t[a + b]*d*(a - b + (a + b)*Cosh[2*(c + d*x)]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(272\) vs. \(2(90)=180\).
time = 2.45, size = 273, normalized size = 2.68

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 \left (\frac {\frac {b \left (a +b \right ) \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {b \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}}{a \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a}+\frac {\left (2 a^{2}+a b -b^{2}\right ) \left (-\frac {\left (\sqrt {b \left (a +b \right )}-b \right ) \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{2 a \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}+\frac {\left (\sqrt {b \left (a +b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{2 a \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{2}\right )}{b^{2}}}{d}\)	\(273\)
default	\(\frac {\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 \left (\frac {\frac {b \left (a +b \right ) \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {b \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}}{a \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a}+\frac {\left (2 a^{2}+a b -b^{2}\right ) \left (-\frac {\left (\sqrt {b \left (a +b \right )}-b \right ) \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{2 a \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}+\frac {\left (\sqrt {b \left (a +b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{2 a \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{2}\right )}{b^{2}}}{d}\)	\(273\)
risch	\(\frac {{\mathrm e}^{d x +c} \left (a +b \right ) \left ({\mathrm e}^{2 d x +2 c}-1\right )}{b d a \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right )}{d \,b^{2}}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right )}{d \,b^{2}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a \left (a +b \right )}\, {\mathrm e}^{d x +c}}{a +b}-1\right )}{4 a^{2} d b}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a \left (a +b \right )}\, {\mathrm e}^{d x +c}}{a +b}-1\right )}{2 a d \,b^{2}}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a \left (a +b \right )}\, {\mathrm e}^{d x +c}}{a +b}-1\right )}{4 a^{2} d b}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a \left (a +b \right )}\, {\mathrm e}^{d x +c}}{a +b}-1\right )}{2 a d \,b^{2}}\)	\(329\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

((a*e^(3*c) + b*e^(3*c))*e^(3*d*x) - (a*e^c + b*e^c)*e^(d*x))/(a^2*b*d + a*b^2*d + (a^2*b*d*e^(4*c) + a*b^2*d*
e^(4*c))*e^(4*d*x) + 2*(a^2*b*d*e^(2*c) - a*b^2*d*e^(2*c))*e^(2*d*x)) + 2*arctan(e^(d*x + c))/(b^2*d) - 32*int
egrate(1/32*((2*a^2*e^(3*c) + a*b*e^(3*c) - b^2*e^(3*c))*e^(3*d*x) + (2*a^2*e^c + a*b*e^c - b^2*e^c)*e^(d*x))/
(a^2*b^2 + a*b^3 + (a^2*b^2*e^(4*c) + a*b^3*e^(4*c))*e^(4*d*x) + 2*(a^2*b^2*e^(2*c) - a*b^3*e^(2*c))*e^(2*d*x)
), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1078 vs. \(2 (90) = 180\).
time = 0.39, size = 2140, normalized size = 20.98 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(4*(a*b + b^2)*cosh(d*x + c)^3 + 12*(a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + 4*(a*b + b^2)*sinh(d*x +
c)^3 - ((2*a^2 + a*b - b^2)*cosh(d*x + c)^4 + 4*(2*a^2 + a*b - b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a^2 + a
*b - b^2)*sinh(d*x + c)^4 + 2*(2*a^2 - 3*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(2*a^2 + a*b - b^2)*cosh(d*x + c)^2
+ 2*a^2 - 3*a*b + b^2)*sinh(d*x + c)^2 + 2*a^2 + a*b - b^2 + 4*((2*a^2 + a*b - b^2)*cosh(d*x + c)^3 + (2*a^2
- 3*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-(a + b)/a)*log(((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*
x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 - 2*(3*a + b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2
- 3*a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 - (3*a + b)*cosh(d*x + c))*sinh(d*x + c) + 4*(a*cosh(d
*x + c)^3 + 3*a*cosh(d*x + c)*sinh(d*x + c)^2 + a*sinh(d*x + c)^3 - a*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 - a
)*sinh(d*x + c))*sqrt(-(a + b)/a) + a + b)/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3
+ (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2
+ 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) + 8*((a^2 + a*b)*cosh(d*x + c)^4
+ 4*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + a*b)*sinh(d*x + c)^4 + 2*(a^2 - a*b)*cosh(d*x + c)^2 +
2*(3*(a^2 + a*b)*cosh(d*x + c)^2 + a^2 - a*b)*sinh(d*x + c)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(d*x + c)^3 +
(a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - 4*(a*b + b^2)*cosh(d*x + c)
+ 4*(3*(a*b + b^2)*cosh(d*x + c)^2 - a*b - b^2)*sinh(d*x + c))/((a^2*b^2 + a*b^3)*d*cosh(d*x + c)^4 + 4*(a^2*b
^2 + a*b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2*b^2 + a*b^3)*d*sinh(d*x + c)^4 + 2*(a^2*b^2 - a*b^3)*d*cosh
(d*x + c)^2 + 2*(3*(a^2*b^2 + a*b^3)*d*cosh(d*x + c)^2 + (a^2*b^2 - a*b^3)*d)*sinh(d*x + c)^2 + (a^2*b^2 + a*b
^3)*d + 4*((a^2*b^2 + a*b^3)*d*cosh(d*x + c)^3 + (a^2*b^2 - a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*(a*
b + b^2)*cosh(d*x + c)^3 + 6*(a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + 2*(a*b + b^2)*sinh(d*x + c)^3 - ((2*a
^2 + a*b - b^2)*cosh(d*x + c)^4 + 4*(2*a^2 + a*b - b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a^2 + a*b - b^2)*si
nh(d*x + c)^4 + 2*(2*a^2 - 3*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(2*a^2 + a*b - b^2)*cosh(d*x + c)^2 + 2*a^2 - 3
*a*b + b^2)*sinh(d*x + c)^2 + 2*a^2 + a*b - b^2 + 4*((2*a^2 + a*b - b^2)*cosh(d*x + c)^3 + (2*a^2 - 3*a*b + b^
2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a + b)/a)*arctan(1/2*sqrt((a + b)/a)*(cosh(d*x + c) + sinh(d*x + c))) -
((2*a^2 + a*b - b^2)*cosh(d*x + c)^4 + 4*(2*a^2 + a*b - b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a^2 + a*b - b
^2)*sinh(d*x + c)^4 + 2*(2*a^2 - 3*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(2*a^2 + a*b - b^2)*cosh(d*x + c)^2 + 2*a
^2 - 3*a*b + b^2)*sinh(d*x + c)^2 + 2*a^2 + a*b - b^2 + 4*((2*a^2 + a*b - b^2)*cosh(d*x + c)^3 + (2*a^2 - 3*a*
b + b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a + b)/a)*arctan(1/2*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*
x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3 + (3*a - b)*cosh(d*x + c) + (3*(a + b)*cosh(d*x + c)^2 + 3*a
- b)*sinh(d*x + c))*sqrt((a + b)/a)/(a + b)) + 4*((a^2 + a*b)*cosh(d*x + c)^4 + 4*(a^2 + a*b)*cosh(d*x + c)*si
nh(d*x + c)^3 + (a^2 + a*b)*sinh(d*x + c)^4 + 2*(a^2 - a*b)*cosh(d*x + c)^2 + 2*(3*(a^2 + a*b)*cosh(d*x + c)^2
+ a^2 - a*b)*sinh(d*x + c)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(d*x + c)^3 + (a^2 - a*b)*cosh(d*x + c))*sinh(d
*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - 2*(a*b + b^2)*cosh(d*x + c) + 2*(3*(a*b + b^2)*cosh(d*x + c)^
2 - a*b - b^2)*sinh(d*x + c))/((a^2*b^2 + a*b^3)*d*cosh(d*x + c)^4 + 4*(a^2*b^2 + a*b^3)*d*cosh(d*x + c)*sinh(
d*x + c)^3 + (a^2*b^2 + a*b^3)*d*sinh(d*x + c)^4 + 2*(a^2*b^2 - a*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^2*b^2 + a*b
^3)*d*cosh(d*x + c)^2 + (a^2*b^2 - a*b^3)*d)*sinh(d*x + c)^2 + (a^2*b^2 + a*b^3)*d + 4*((a^2*b^2 + a*b^3)*d*co
sh(d*x + c)^3 + (a^2*b^2 - a*b^3)*d*cosh(d*x + c))*sinh(d*x + c))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {sech}^{5}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**5/(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Integral(sech(c + d*x)**5/(a + b*tanh(c + d*x)**2)**2, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^5\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)^5*(a + b*tanh(c + d*x)^2)^2),x)

[Out]

int(1/(cosh(c + d*x)^5*(a + b*tanh(c + d*x)^2)^2), x)

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