3.95 \(\int \text {sech}^3(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=125 \[ \frac {\left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{16 d}+\frac {\left (8 a^2+4 a b+b^2\right ) \tanh (c+d x) \text {sech}(c+d x)}{16 d}-\frac {b (8 a+3 b) \tanh (c+d x) \text {sech}^3(c+d x)}{24 d}-\frac {b \tanh (c+d x) \text {sech}^5(c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{6 d} \]

[Out]

1/16*(8*a^2+4*a*b+b^2)*arctan(sinh(d*x+c))/d+1/16*(8*a^2+4*a*b+b^2)*sech(d*x+c)*tanh(d*x+c)/d-1/24*b*(8*a+3*b)
*sech(d*x+c)^3*tanh(d*x+c)/d-1/6*b*sech(d*x+c)^5*(a+(a+b)*sinh(d*x+c)^2)*tanh(d*x+c)/d

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Rubi [A]  time = 0.15, antiderivative size = 125, 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 = {3676, 413, 385, 199, 203} \[ \frac {\left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{16 d}+\frac {\left (8 a^2+4 a b+b^2\right ) \tanh (c+d x) \text {sech}(c+d x)}{16 d}-\frac {b (8 a+3 b) \tanh (c+d x) \text {sech}^3(c+d x)}{24 d}-\frac {b \tanh (c+d x) \text {sech}^5(c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

((8*a^2 + 4*a*b + b^2)*ArcTan[Sinh[c + d*x]])/(16*d) + ((8*a^2 + 4*a*b + b^2)*Sech[c + d*x]*Tanh[c + d*x])/(16
*d) - (b*(8*a + 3*b)*Sech[c + d*x]^3*Tanh[c + d*x])/(24*d) - (b*Sech[c + d*x]^5*(a + (a + b)*Sinh[c + d*x]^2)*
Tanh[c + d*x])/(6*d)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 385

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

Rule 413

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

Rule 3676

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 \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+(a+b) x^2\right )^2}{\left (1+x^2\right )^4} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=-\frac {b \text {sech}^5(c+d x) \left (a+(a+b) \sinh ^2(c+d x)\right ) \tanh (c+d x)}{6 d}+\frac {\operatorname {Subst}\left (\int \frac {a (6 a+b)+3 (a+b) (2 a+b) x^2}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{6 d}\\ &=-\frac {b (8 a+3 b) \text {sech}^3(c+d x) \tanh (c+d x)}{24 d}-\frac {b \text {sech}^5(c+d x) \left (a+(a+b) \sinh ^2(c+d x)\right ) \tanh (c+d x)}{6 d}+\frac {\left (8 a^2+4 a b+b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{8 d}\\ &=\frac {\left (8 a^2+4 a b+b^2\right ) \text {sech}(c+d x) \tanh (c+d x)}{16 d}-\frac {b (8 a+3 b) \text {sech}^3(c+d x) \tanh (c+d x)}{24 d}-\frac {b \text {sech}^5(c+d x) \left (a+(a+b) \sinh ^2(c+d x)\right ) \tanh (c+d x)}{6 d}+\frac {\left (8 a^2+4 a b+b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{16 d}\\ &=\frac {\left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{16 d}+\frac {\left (8 a^2+4 a b+b^2\right ) \text {sech}(c+d x) \tanh (c+d x)}{16 d}-\frac {b (8 a+3 b) \text {sech}^3(c+d x) \tanh (c+d x)}{24 d}-\frac {b \text {sech}^5(c+d x) \left (a+(a+b) \sinh ^2(c+d x)\right ) \tanh (c+d x)}{6 d}\\ \end {align*}

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Mathematica [C]  time = 8.76, size = 792, normalized size = 6.34 \[ \frac {a^2 \sinh (c+d x) \left (-\frac {380 (a+b)^2 \sinh ^6(c+d x) \, _4F_3\left (\frac {3}{2},2,2,2;1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )}{a^2}-\frac {128 (a+b)^2 \sinh ^6(c+d x) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )}{a^2}-\frac {16 (a+b)^2 \sinh ^6(c+d x) \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )}{a^2}-\frac {968 (a+b) \sinh ^4(c+d x) \, _4F_3\left (\frac {3}{2},2,2,2;1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )}{a}-\frac {288 (a+b) \sinh ^4(c+d x) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )}{a}-\frac {32 (a+b) \sinh ^4(c+d x) \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )}{a}-620 \sinh ^2(c+d x) \, _4F_3\left (\frac {3}{2},2,2,2;1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )-160 \sinh ^2(c+d x) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )-16 \sinh ^2(c+d x) \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {9}{2};-\sinh ^2(c+d x)\right )-\frac {8855 (a+b)^2 \sinh ^2(c+d x)}{a^2}-\frac {19845 (a+b)^2 \sqrt {-\sinh ^2(c+d x)} \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{a^2}+\frac {32970 (a+b)^2 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{a^2 \sqrt {-\sinh ^2(c+d x)}}+\frac {525 (a+b)^2 \sinh ^4(c+d x) \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{a^2 \sqrt {-\sinh ^2(c+d x)}}-\frac {32970 (a+b)^2}{a^2}-\frac {91875 (a+b) \text {csch}^2(c+d x)}{a}-\frac {1365 (a+b) \sqrt {-\sinh ^2(c+d x)} \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{a}+\frac {54180 (a+b) \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{a \sqrt {-\sinh ^2(c+d x)}}-\frac {91875 (a+b) \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{a \left (-\sinh ^2(c+d x)\right )^{3/2}}-\frac {23555 (a+b)}{a}-65625 \text {csch}^4(c+d x)-14980 \text {csch}^2(c+d x)-\frac {36855 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{\left (-\sinh ^2(c+d x)\right )^{3/2}}+\frac {65625 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{\left (-\sinh ^2(c+d x)\right )^{5/2}}+\frac {1680 \sinh ^4(c+d x) \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{\left (-\sinh ^2(c+d x)\right )^{5/2}}\right )}{2520 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sech[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

(a^2*Sinh[c + d*x]*((-23555*(a + b))/a - (32970*(a + b)^2)/a^2 - 14980*Csch[c + d*x]^2 - (91875*(a + b)*Csch[c
+ d*x]^2)/a - 65625*Csch[c + d*x]^4 - (8855*(a + b)^2*Sinh[c + d*x]^2)/a^2 - 620*HypergeometricPFQ[{3/2, 2, 2
, 2}, {1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^2 - 160*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}
, -Sinh[c + d*x]^2]*Sinh[c + d*x]^2 - 16*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, -Sinh[c +
d*x]^2]*Sinh[c + d*x]^2 - (968*(a + b)*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c
+ d*x]^4)/a - (288*(a + b)*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*
x]^4)/a - (32*(a + b)*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*
x]^4)/a - (380*(a + b)^2*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^6)/a^2
- (128*(a + b)^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^6)/a^2
- (16*(a + b)^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^6)/
a^2 + (65625*ArcTanh[Sqrt[-Sinh[c + d*x]^2]])/(-Sinh[c + d*x]^2)^(5/2) + (1680*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]
*Sinh[c + d*x]^4)/(-Sinh[c + d*x]^2)^(5/2) - (36855*ArcTanh[Sqrt[-Sinh[c + d*x]^2]])/(-Sinh[c + d*x]^2)^(3/2)
- (91875*(a + b)*ArcTanh[Sqrt[-Sinh[c + d*x]^2]])/(a*(-Sinh[c + d*x]^2)^(3/2)) + (54180*(a + b)*ArcTanh[Sqrt[-
Sinh[c + d*x]^2]])/(a*Sqrt[-Sinh[c + d*x]^2]) + (32970*(a + b)^2*ArcTanh[Sqrt[-Sinh[c + d*x]^2]])/(a^2*Sqrt[-S
inh[c + d*x]^2]) + (525*(a + b)^2*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*Sinh[c + d*x]^4)/(a^2*Sqrt[-Sinh[c + d*x]^2]
) - (1365*(a + b)*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*Sqrt[-Sinh[c + d*x]^2])/a - (19845*(a + b)^2*ArcTanh[Sqrt[-S
inh[c + d*x]^2]]*Sqrt[-Sinh[c + d*x]^2])/a^2))/(2520*d)

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fricas [B]  time = 0.47, size = 2824, normalized size = 22.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/24*(3*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^11 + 33*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^10 + 3*(
8*a^2 + 4*a*b + b^2)*sinh(d*x + c)^11 + (72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^9 + (165*(8*a^2 + 4*a*b + b^2
)*cosh(d*x + c)^2 + 72*a^2 - 60*a*b - 47*b^2)*sinh(d*x + c)^9 + 9*(55*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^3 +
(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c))*sinh(d*x + c)^8 + 6*(8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c)^7 + 6*(1
65*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 + 6*(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^2 + 8*a^2 - 12*a*b + 13*
b^2)*sinh(d*x + c)^7 + 42*(33*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^5 + 2*(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x +
c)^3 + (8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c))*sinh(d*x + c)^6 - 6*(8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c)^5
+ 6*(231*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^6 + 21*(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^4 + 21*(8*a^2 - 1
2*a*b + 13*b^2)*cosh(d*x + c)^2 - 8*a^2 + 12*a*b - 13*b^2)*sinh(d*x + c)^5 + 6*(165*(8*a^2 + 4*a*b + b^2)*cosh
(d*x + c)^7 + 21*(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^5 + 35*(8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c)^3 - 5
*(8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 - (72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^3 + (495*
(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^8 + 84*(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^6 + 210*(8*a^2 - 12*a*b +
13*b^2)*cosh(d*x + c)^4 - 60*(8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c)^2 - 72*a^2 + 60*a*b + 47*b^2)*sinh(d*x +
c)^3 + 3*(55*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^9 + 12*(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^7 + 42*(8*a^2
- 12*a*b + 13*b^2)*cosh(d*x + c)^5 - 20*(8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c)^3 - (72*a^2 - 60*a*b - 47*b^2
)*cosh(d*x + c))*sinh(d*x + c)^2 + 3*((8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^12 + 12*(8*a^2 + 4*a*b + b^2)*cosh(d
*x + c)*sinh(d*x + c)^11 + (8*a^2 + 4*a*b + b^2)*sinh(d*x + c)^12 + 6*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^10 +
6*(11*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^2 + 8*a^2 + 4*a*b + b^2)*sinh(d*x + c)^10 + 20*(11*(8*a^2 + 4*a*b +
b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^9 + 15*(8*a^2 + 4*a*b + b^2)*cosh
(d*x + c)^8 + 15*(33*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 + 18*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^2 + 8*a^2
+ 4*a*b + b^2)*sinh(d*x + c)^8 + 24*(33*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^5 + 30*(8*a^2 + 4*a*b + b^2)*cosh(
d*x + c)^3 + 5*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^7 + 20*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^6
+ 4*(231*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^6 + 315*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 + 105*(8*a^2 + 4*a
*b + b^2)*cosh(d*x + c)^2 + 40*a^2 + 20*a*b + 5*b^2)*sinh(d*x + c)^6 + 24*(33*(8*a^2 + 4*a*b + b^2)*cosh(d*x +
c)^7 + 63*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^5 + 35*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^3 + 5*(8*a^2 + 4*a*b
+ b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 15*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 + 15*(33*(8*a^2 + 4*a*b + b^
2)*cosh(d*x + c)^8 + 84*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^6 + 70*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 + 20*
(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^2 + 8*a^2 + 4*a*b + b^2)*sinh(d*x + c)^4 + 20*(11*(8*a^2 + 4*a*b + b^2)*co
sh(d*x + c)^9 + 36*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^7 + 42*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^5 + 20*(8*a^
2 + 4*a*b + b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 6*(8*a^2 + 4*a*b +
b^2)*cosh(d*x + c)^2 + 6*(11*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^10 + 45*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^
8 + 70*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^6 + 50*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 + 15*(8*a^2 + 4*a*b +
b^2)*cosh(d*x + c)^2 + 8*a^2 + 4*a*b + b^2)*sinh(d*x + c)^2 + 8*a^2 + 4*a*b + b^2 + 12*((8*a^2 + 4*a*b + b^2)*
cosh(d*x + c)^11 + 5*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^9 + 10*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^7 + 10*(8*
a^2 + 4*a*b + b^2)*cosh(d*x + c)^5 + 5*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c)^3 + (8*a^2 + 4*a*b + b^2)*cosh(d*x
+ c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - 3*(8*a^2 + 4*a*b + b^2)*cosh(d*x + c) + 3*(11*(8*
a^2 + 4*a*b + b^2)*cosh(d*x + c)^10 + 3*(72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^8 + 14*(8*a^2 - 12*a*b + 13*b
^2)*cosh(d*x + c)^6 - 10*(8*a^2 - 12*a*b + 13*b^2)*cosh(d*x + c)^4 - (72*a^2 - 60*a*b - 47*b^2)*cosh(d*x + c)^
2 - 8*a^2 - 4*a*b - b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^12 + 12*d*cosh(d*x + c)*sinh(d*x + c)^11 + d*sinh(d*x
+ c)^12 + 6*d*cosh(d*x + c)^10 + 6*(11*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^10 + 20*(11*d*cosh(d*x + c)^3 + 3
*d*cosh(d*x + c))*sinh(d*x + c)^9 + 15*d*cosh(d*x + c)^8 + 15*(33*d*cosh(d*x + c)^4 + 18*d*cosh(d*x + c)^2 + d
)*sinh(d*x + c)^8 + 24*(33*d*cosh(d*x + c)^5 + 30*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c)^7 + 20*
d*cosh(d*x + c)^6 + 4*(231*d*cosh(d*x + c)^6 + 315*d*cosh(d*x + c)^4 + 105*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x +
c)^6 + 24*(33*d*cosh(d*x + c)^7 + 63*d*cosh(d*x + c)^5 + 35*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x +
c)^5 + 15*d*cosh(d*x + c)^4 + 15*(33*d*cosh(d*x + c)^8 + 84*d*cosh(d*x + c)^6 + 70*d*cosh(d*x + c)^4 + 20*d*c
osh(d*x + c)^2 + d)*sinh(d*x + c)^4 + 20*(11*d*cosh(d*x + c)^9 + 36*d*cosh(d*x + c)^7 + 42*d*cosh(d*x + c)^5 +
20*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^3 + 6*d*cosh(d*x + c)^2 + 6*(11*d*cosh(d*x + c)^10 +
45*d*cosh(d*x + c)^8 + 70*d*cosh(d*x + c)^6 + 50*d*cosh(d*x + c)^4 + 15*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2
+ 12*(d*cosh(d*x + c)^11 + 5*d*cosh(d*x + c)^9 + 10*d*cosh(d*x + c)^7 + 10*d*cosh(d*x + c)^5 + 5*d*cosh(d*x +
c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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giac [B]  time = 0.22, size = 291, normalized size = 2.33 \[ \frac {3 \, {\left (8 \, a^{2} e^{c} + 4 \, a b e^{c} + b^{2} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )} + \frac {24 \, a^{2} e^{\left (11 \, d x + 11 \, c\right )} + 12 \, a b e^{\left (11 \, d x + 11 \, c\right )} + 3 \, b^{2} e^{\left (11 \, d x + 11 \, c\right )} + 72 \, a^{2} e^{\left (9 \, d x + 9 \, c\right )} - 60 \, a b e^{\left (9 \, d x + 9 \, c\right )} - 47 \, b^{2} e^{\left (9 \, d x + 9 \, c\right )} + 48 \, a^{2} e^{\left (7 \, d x + 7 \, c\right )} - 72 \, a b e^{\left (7 \, d x + 7 \, c\right )} + 78 \, b^{2} e^{\left (7 \, d x + 7 \, c\right )} - 48 \, a^{2} e^{\left (5 \, d x + 5 \, c\right )} + 72 \, a b e^{\left (5 \, d x + 5 \, c\right )} - 78 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} - 72 \, a^{2} e^{\left (3 \, d x + 3 \, c\right )} + 60 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 47 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 24 \, a^{2} e^{\left (d x + c\right )} - 12 \, a b e^{\left (d x + c\right )} - 3 \, b^{2} e^{\left (d x + c\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{6}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/24*(3*(8*a^2*e^c + 4*a*b*e^c + b^2*e^c)*arctan(e^(d*x + c))*e^(-c) + (24*a^2*e^(11*d*x + 11*c) + 12*a*b*e^(1
1*d*x + 11*c) + 3*b^2*e^(11*d*x + 11*c) + 72*a^2*e^(9*d*x + 9*c) - 60*a*b*e^(9*d*x + 9*c) - 47*b^2*e^(9*d*x +
9*c) + 48*a^2*e^(7*d*x + 7*c) - 72*a*b*e^(7*d*x + 7*c) + 78*b^2*e^(7*d*x + 7*c) - 48*a^2*e^(5*d*x + 5*c) + 72*
a*b*e^(5*d*x + 5*c) - 78*b^2*e^(5*d*x + 5*c) - 72*a^2*e^(3*d*x + 3*c) + 60*a*b*e^(3*d*x + 3*c) + 47*b^2*e^(3*d
*x + 3*c) - 24*a^2*e^(d*x + c) - 12*a*b*e^(d*x + c) - 3*b^2*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^6)/d

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maple [B]  time = 0.51, size = 236, normalized size = 1.89 \[ \frac {a^{2} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2 d}+\frac {a^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{d}-\frac {2 a b \sinh \left (d x +c \right )}{3 d \cosh \left (d x +c \right )^{4}}+\frac {a b \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{6 d}+\frac {a b \,\mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{4 d}+\frac {a b \arctan \left ({\mathrm e}^{d x +c}\right )}{2 d}-\frac {b^{2} \left (\sinh ^{3}\left (d x +c \right )\right )}{3 d \cosh \left (d x +c \right )^{6}}-\frac {b^{2} \sinh \left (d x +c \right )}{5 d \cosh \left (d x +c \right )^{6}}+\frac {b^{2} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{5}}{30 d}+\frac {b^{2} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{24 d}+\frac {b^{2} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{16 d}+\frac {b^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x)

[Out]

1/2/d*a^2*sech(d*x+c)*tanh(d*x+c)+1/d*a^2*arctan(exp(d*x+c))-2/3/d*a*b*sinh(d*x+c)/cosh(d*x+c)^4+1/6/d*a*b*tan
h(d*x+c)*sech(d*x+c)^3+1/4/d*a*b*sech(d*x+c)*tanh(d*x+c)+1/2/d*a*b*arctan(exp(d*x+c))-1/3/d*b^2*sinh(d*x+c)^3/
cosh(d*x+c)^6-1/5/d*b^2*sinh(d*x+c)/cosh(d*x+c)^6+1/30/d*b^2*tanh(d*x+c)*sech(d*x+c)^5+1/24/d*b^2*tanh(d*x+c)*
sech(d*x+c)^3+1/16/d*b^2*sech(d*x+c)*tanh(d*x+c)+1/8/d*b^2*arctan(exp(d*x+c))

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maxima [B]  time = 0.43, size = 345, normalized size = 2.76 \[ -\frac {1}{24} \, b^{2} {\left (\frac {3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {3 \, e^{\left (-d x - c\right )} - 47 \, e^{\left (-3 \, d x - 3 \, c\right )} + 78 \, e^{\left (-5 \, d x - 5 \, c\right )} - 78 \, e^{\left (-7 \, d x - 7 \, c\right )} + 47 \, e^{\left (-9 \, d x - 9 \, c\right )} - 3 \, e^{\left (-11 \, d x - 11 \, c\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} + 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}}\right )} - \frac {1}{2} \, a b {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - 7 \, e^{\left (-3 \, d x - 3 \, c\right )} + 7 \, e^{\left (-5 \, d x - 5 \, c\right )} - e^{\left (-7 \, d x - 7 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} - a^{2} {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

-1/24*b^2*(3*arctan(e^(-d*x - c))/d - (3*e^(-d*x - c) - 47*e^(-3*d*x - 3*c) + 78*e^(-5*d*x - 5*c) - 78*e^(-7*d
*x - 7*c) + 47*e^(-9*d*x - 9*c) - 3*e^(-11*d*x - 11*c))/(d*(6*e^(-2*d*x - 2*c) + 15*e^(-4*d*x - 4*c) + 20*e^(-
6*d*x - 6*c) + 15*e^(-8*d*x - 8*c) + 6*e^(-10*d*x - 10*c) + e^(-12*d*x - 12*c) + 1))) - 1/2*a*b*(arctan(e^(-d*
x - c))/d - (e^(-d*x - c) - 7*e^(-3*d*x - 3*c) + 7*e^(-5*d*x - 5*c) - e^(-7*d*x - 7*c))/(d*(4*e^(-2*d*x - 2*c)
+ 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) - a^2*(arctan(e^(-d*x - c))/d - (e^(-d*x
- c) - e^(-3*d*x - 3*c))/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1)))

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mupad [B]  time = 0.18, size = 572, normalized size = 4.58 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (8\,a^2\,\sqrt {d^2}+b^2\,\sqrt {d^2}+4\,a\,b\,\sqrt {d^2}\right )}{d\,\sqrt {64\,a^4+64\,a^3\,b+32\,a^2\,b^2+8\,a\,b^3+b^4}}\right )\,\sqrt {64\,a^4+64\,a^3\,b+32\,a^2\,b^2+8\,a\,b^3+b^4}}{8\,\sqrt {d^2}}-\frac {\frac {2\,{\mathrm {e}}^{c+d\,x}\,{\left (a+b\right )}^2}{3\,d}+\frac {8\,{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (a^2-b^2\right )}{3\,d}+\frac {8\,{\mathrm {e}}^{7\,c+7\,d\,x}\,\left (a^2-b^2\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{9\,c+9\,d\,x}\,{\left (a+b\right )}^2}{3\,d}+\frac {4\,{\mathrm {e}}^{5\,c+5\,d\,x}\,\left (3\,a^2-2\,a\,b+3\,b^2\right )}{3\,d}}{6\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,{\mathrm {e}}^{4\,c+4\,d\,x}+20\,{\mathrm {e}}^{6\,c+6\,d\,x}+15\,{\mathrm {e}}^{8\,c+8\,d\,x}+6\,{\mathrm {e}}^{10\,c+10\,d\,x}+{\mathrm {e}}^{12\,c+12\,d\,x}+1}-\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (15\,b^2+4\,a\,b\right )}{3\,d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}+\frac {16\,b^2\,{\mathrm {e}}^{c+d\,x}}{3\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (8\,a^2+4\,a\,b+b^2\right )}{8\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (16\,a^2+44\,a\,b+23\,b^2\right )}{12\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (21\,b^2+20\,a\,b\right )}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(c + d*x)^2)^2/cosh(c + d*x)^3,x)

[Out]

(atan((exp(d*x)*exp(c)*(8*a^2*(d^2)^(1/2) + b^2*(d^2)^(1/2) + 4*a*b*(d^2)^(1/2)))/(d*(8*a*b^3 + 64*a^3*b + 64*
a^4 + b^4 + 32*a^2*b^2)^(1/2)))*(8*a*b^3 + 64*a^3*b + 64*a^4 + b^4 + 32*a^2*b^2)^(1/2))/(8*(d^2)^(1/2)) - ((2*
exp(c + d*x)*(a + b)^2)/(3*d) + (8*exp(3*c + 3*d*x)*(a^2 - b^2))/(3*d) + (8*exp(7*c + 7*d*x)*(a^2 - b^2))/(3*d
) + (2*exp(9*c + 9*d*x)*(a + b)^2)/(3*d) + (4*exp(5*c + 5*d*x)*(3*a^2 - 2*a*b + 3*b^2))/(3*d))/(6*exp(2*c + 2*
d*x) + 15*exp(4*c + 4*d*x) + 20*exp(6*c + 6*d*x) + 15*exp(8*c + 8*d*x) + 6*exp(10*c + 10*d*x) + exp(12*c + 12*
d*x) + 1) - (2*exp(c + d*x)*(4*a*b + 15*b^2))/(3*d*(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*
x) + exp(8*c + 8*d*x) + 1)) + (16*b^2*exp(c + d*x))/(3*d*(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x) + 10*exp(6*
c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1)) + (exp(c + d*x)*(4*a*b + 8*a^2 + b^2))/(8*d*(exp(2*
c + 2*d*x) + 1)) - (exp(c + d*x)*(44*a*b + 16*a^2 + 23*b^2))/(12*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)
) + (exp(c + d*x)*(20*a*b + 21*b^2))/(3*d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \operatorname {sech}^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**3*(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**2*sech(c + d*x)**3, x)

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