### 3.93 $$\int \text{sech}(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx$$

Optimal. Leaf size=91 $\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}-\frac{3 b (2 a+b) \tanh (c+d x) \text{sech}(c+d x)}{8 d}-\frac{b \tanh (c+d x) \text{sech}^3(c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{4 d}$

[Out]

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

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Rubi [A]  time = 0.085125, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.19, Rules used = {3676, 413, 385, 203} $\frac{\left (8 a^2+8 a b+3 b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}-\frac{3 b (2 a+b) \tanh (c+d x) \text{sech}(c+d x)}{8 d}-\frac{b \tanh (c+d x) \text{sech}^3(c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{4 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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]

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 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 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])

Rubi steps

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

Mathematica [C]  time = 7.22038, size = 427, normalized size = 4.69 $-\frac{\text{csch}^3(c+d x) \left (128 \sinh ^6(c+d x) \left (a^2 \left (5 \sinh ^4(c+d x)+12 \sinh ^2(c+d x)+7\right )+2 a b \left (5 \sinh ^2(c+d x)+6\right ) \sinh ^2(c+d x)+5 b^2 \sinh ^4(c+d x)\right ) \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2\right \},\left \{1,1,\frac{9}{2}\right \},-\sinh ^2(c+d x)\right )+128 \sinh ^6(c+d x) \left (a \sinh ^2(c+d x)+a+b \sinh ^2(c+d x)\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2\right \},\left \{1,1,1,\frac{9}{2}\right \},-\sinh ^2(c+d x)\right )+35 \left (a^2 \left (485 \sinh ^6(c+d x)+3161 \sinh ^4(c+d x)+5907 \sinh ^2(c+d x)+3375\right )+2 a b \left (485 \sinh ^4(c+d x)+2554 \sinh ^2(c+d x)+2625\right ) \sinh ^2(c+d x)+b^2 \left (485 \sinh ^2(c+d x)+1947\right ) \sinh ^4(c+d x)\right )-\frac{105 \tanh ^{-1}\left (\sqrt{-\sinh ^2(c+d x)}\right ) \left (a^2 \left (9 \sinh ^8(c+d x)+400 \sinh ^6(c+d x)+1674 \sinh ^4(c+d x)+2344 \sinh ^2(c+d x)+1125\right )+2 a b \left (9 \sinh ^6(c+d x)+389 \sinh ^4(c+d x)+1143 \sinh ^2(c+d x)+875\right ) \sinh ^2(c+d x)+b^2 \left (9 \sinh ^4(c+d x)+378 \sinh ^2(c+d x)+649\right ) \sinh ^4(c+d x)\right )}{\sqrt{-\sinh ^2(c+d x)}}\right )}{6720 d}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Csch[c + d*x]^3*(128*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^6*
(a + a*Sinh[c + d*x]^2 + b*Sinh[c + d*x]^2)^2 + 128*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, -Sinh[c + d
*x]^2]*Sinh[c + d*x]^6*(5*b^2*Sinh[c + d*x]^4 + 2*a*b*Sinh[c + d*x]^2*(6 + 5*Sinh[c + d*x]^2) + a^2*(7 + 12*Si
nh[c + d*x]^2 + 5*Sinh[c + d*x]^4)) + 35*(b^2*Sinh[c + d*x]^4*(1947 + 485*Sinh[c + d*x]^2) + 2*a*b*Sinh[c + d*
x]^2*(2625 + 2554*Sinh[c + d*x]^2 + 485*Sinh[c + d*x]^4) + a^2*(3375 + 5907*Sinh[c + d*x]^2 + 3161*Sinh[c + d*
x]^4 + 485*Sinh[c + d*x]^6)) - (105*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^2*Sinh[c + d*x]^4*(649 + 378*Sinh[c + d
*x]^2 + 9*Sinh[c + d*x]^4) + 2*a*b*Sinh[c + d*x]^2*(875 + 1143*Sinh[c + d*x]^2 + 389*Sinh[c + d*x]^4 + 9*Sinh[
c + d*x]^6) + a^2*(1125 + 2344*Sinh[c + d*x]^2 + 1674*Sinh[c + d*x]^4 + 400*Sinh[c + d*x]^6 + 9*Sinh[c + d*x]^
8)))/Sqrt[-Sinh[c + d*x]^2]))/(6720*d)

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Maple [B]  time = 0.036, size = 173, normalized size = 1.9 \begin{align*} 2\,{\frac{{a}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}-2\,{\frac{ab\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{d}}+2\,{\frac{ab\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}-{\frac{{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{2}\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{2}\tanh \left ( dx+c \right ) \left ({\rm sech} \left (dx+c\right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{4\,d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a^2*arctan(exp(d*x+c))-2/d*a*b*sinh(d*x+c)/cosh(d*x+c)^2+1/d*a*b*sech(d*x+c)*tanh(d*x+c)+2/d*a*b*arctan(ex
p(d*x+c))-1/d*b^2*sinh(d*x+c)^3/cosh(d*x+c)^4-1/d*b^2*sinh(d*x+c)/cosh(d*x+c)^4+1/4/d*b^2*tanh(d*x+c)*sech(d*x
+c)^3+3/8/d*b^2*sech(d*x+c)*tanh(d*x+c)+3/4/d*b^2*arctan(exp(d*x+c))

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Maxima [B]  time = 1.61505, size = 269, normalized size = 2.96 \begin{align*} -\frac{1}{4} \, b^{2}{\left (\frac{3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{5 \, e^{\left (-d x - c\right )} - 3 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} - 5 \, 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 )} - 2 \, a b{\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 )} + \frac{a^{2} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b^2*(3*arctan(e^(-d*x - c))/d + (5*e^(-d*x - c) - 3*e^(-3*d*x - 3*c) + 3*e^(-5*d*x - 5*c) - 5*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))) - 2*a*b*(arc
tan(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))) + a^2
*arctan(sinh(d*x + c))/d

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Fricas [B]  time = 2.19051, size = 3433, normalized size = 37.73 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((8*a*b + 5*b^2)*cosh(d*x + c)^7 + 7*(8*a*b + 5*b^2)*cosh(d*x + c)*sinh(d*x + c)^6 + (8*a*b + 5*b^2)*sinh
(d*x + c)^7 + (8*a*b - 3*b^2)*cosh(d*x + c)^5 + (21*(8*a*b + 5*b^2)*cosh(d*x + c)^2 + 8*a*b - 3*b^2)*sinh(d*x
+ c)^5 + 5*(7*(8*a*b + 5*b^2)*cosh(d*x + c)^3 + (8*a*b - 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 - (8*a*b - 3*b^
2)*cosh(d*x + c)^3 + (35*(8*a*b + 5*b^2)*cosh(d*x + c)^4 + 10*(8*a*b - 3*b^2)*cosh(d*x + c)^2 - 8*a*b + 3*b^2)
*sinh(d*x + c)^3 + (21*(8*a*b + 5*b^2)*cosh(d*x + c)^5 + 10*(8*a*b - 3*b^2)*cosh(d*x + c)^3 - 3*(8*a*b - 3*b^2
)*cosh(d*x + c))*sinh(d*x + c)^2 - ((8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^8 + 8*(8*a^2 + 8*a*b + 3*b^2)*cosh(d
*x + c)*sinh(d*x + c)^7 + (8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^8 + 4*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^6
+ 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^6 + 8*(7*(8*a^2 + 8*a*b
+ 3*b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 6*(8*a^2 + 8*a*b + 3*b^2
)*cosh(d*x + c)^4 + 2*(35*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^4 + 30*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2
+ 24*a^2 + 24*a*b + 9*b^2)*sinh(d*x + c)^4 + 8*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^5 + 10*(8*a^2 + 8*a*b
+ 3*b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(8*a^2 + 8*a*b + 3*b^
2)*cosh(d*x + c)^2 + 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^6 + 15*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^4
+ 9*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^2 + 8*a^2 + 8*a*b + 3*b^2
+ 8*((8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^7 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^5 + 3*(8*a^2 + 8*a*b +
3*b^2)*cosh(d*x + c)^3 + (8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x
+ c)) - (8*a*b + 5*b^2)*cosh(d*x + c) + (7*(8*a*b + 5*b^2)*cosh(d*x + c)^6 + 5*(8*a*b - 3*b^2)*cosh(d*x + c)^
4 - 3*(8*a*b - 3*b^2)*cosh(d*x + c)^2 - 8*a*b - 5*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^8 + 8*d*cosh(d*x + c)*s
inh(d*x + c)^7 + d*sinh(d*x + c)^8 + 4*d*cosh(d*x + c)^6 + 4*(7*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^6 + 8*(7*
d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(d*x + c)^4 + 2*(35*d*cosh(d*x + c)^4 + 30*d*
cosh(d*x + c)^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*si
nh(d*x + c)^3 + 4*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)^6 + 15*d*cosh(d*x + c)^4 + 9*d*cosh(d*x + c)^2 + d)
*sinh(d*x + c)^2 + 8*(d*cosh(d*x + c)^7 + 3*d*cosh(d*x + c)^5 + 3*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*
x + c) + d)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39557, size = 212, normalized size = 2.33 \begin{align*} \frac{{\left (8 \, a^{2} e^{c} + 8 \, a b e^{c} + 3 \, b^{2} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )} - \frac{8 \, a b e^{\left (7 \, d x + 7 \, c\right )} + 5 \, b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 8 \, a b e^{\left (5 \, d x + 5 \, c\right )} - 3 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} - 8 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 3 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 8 \, a b e^{\left (d x + c\right )} - 5 \, b^{2} e^{\left (d x + c\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}}{4 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((8*a^2*e^c + 8*a*b*e^c + 3*b^2*e^c)*arctan(e^(d*x + c))*e^(-c) - (8*a*b*e^(7*d*x + 7*c) + 5*b^2*e^(7*d*x
+ 7*c) + 8*a*b*e^(5*d*x + 5*c) - 3*b^2*e^(5*d*x + 5*c) - 8*a*b*e^(3*d*x + 3*c) + 3*b^2*e^(3*d*x + 3*c) - 8*a*b
*e^(d*x + c) - 5*b^2*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^4)/d