3.35 \(\int \frac {\sinh ^2(c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\)

Optimal. Leaf size=131 \[ \frac {\sqrt {b} (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^3 d \sqrt {a+b}}-\frac {x (a+4 b)}{2 a^3}+\frac {b \tanh (c+d x)}{a^2 d \left (a-b \tanh ^2(c+d x)+b\right )}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 a d \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4132, 471, 527, 522, 206, 208} \[ \frac {\sqrt {b} (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^3 d \sqrt {a+b}}+\frac {b \tanh (c+d x)}{a^2 d \left (a-b \tanh ^2(c+d x)+b\right )}-\frac {x (a+4 b)}{2 a^3}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 a d \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 208

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

Rule 471

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

Rule 522

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 527

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

Rule 4132

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

Rubi steps

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

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Mathematica [B]  time = 11.33, size = 791, normalized size = 6.04 \[ \frac {\text {sech}^4(c+d x) (a \cosh (2 c+2 d x)+a+2 b)^2 \left (\frac {\left (a^2+8 a b+8 b^2\right ) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}+\frac {\left (a^3-6 a^2 b-24 a b^2-16 b^3\right ) (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )}{b d (a+b)^{3/2} \sqrt {b (\cosh (c)-\sinh (c))^4}}+16 x\right )}{128 a^2 \left (a+b \text {sech}^2(c+d x)\right )^2}+\frac {\text {sech}^4(c+d x) (a \cosh (2 c+2 d x)+a+2 b)^2 \left (-\frac {\left (a^3+18 a^2 b+48 a b^2+32 b^3\right ) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}+\frac {\left (-a^4+16 a^3 b+144 a^2 b^2+256 a b^3+128 b^4\right ) (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )}{b d (a+b)^{3/2} \sqrt {b (\cosh (c)-\sinh (c))^4}}-64 x (a+2 b)+\frac {16 a \sinh (2 c) \cosh (2 d x)}{d}+\frac {16 a \cosh (2 c) \sinh (2 d x)}{d}\right )}{256 a^3 \left (a+b \text {sech}^2(c+d x)\right )^2}-\frac {\text {sech}^4(c+d x) (a \cosh (2 c+2 d x)+a+2 b)^2 \left (\frac {\sqrt {b} (a+2 b) \sinh (2 (c+d x))}{(a+b) (a \cosh (2 (c+d x))+a+2 b)}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}\right )}{256 b^{3/2} d \left (a+b \text {sech}^2(c+d x)\right )^2}+\frac {\text {sech}^4(c+d x) (a \cosh (2 c+2 d x)+a+2 b)^2 \left (\frac {a \sinh (2 (c+d x))}{8 b d (a+b) (a \cosh (2 (c+d x))+a+2 b)}-\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{8 b^{3/2} d (a+b)^{3/2}}\right )}{16 \left (a+b \text {sech}^2(c+d x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*c + 2*d*x])^2*Sech[c + d*x]^4*(16*x + ((a^3 - 6*a^2*b - 24*a*b^2 - 16*b^3)*ArcTanh[(Sech[
d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[
c])^4])]*(Cosh[2*c] - Sinh[2*c]))/(b*(a + b)^(3/2)*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + ((a^2 + 8*a*b + 8*b^2)*S
ech[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/(b*(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)]))))/(128*a^2*(a +
b*Sech[c + d*x]^2)^2) + ((a + 2*b + a*Cosh[2*c + 2*d*x])^2*Sech[c + d*x]^4*(-64*(a + 2*b)*x + ((-a^4 + 16*a^3*
b + 144*a^2*b^2 + 256*a*b^3 + 128*b^4)*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sin
h[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(Cosh[2*c] - Sinh[2*c]))/(b*(a + b)^(3/2)*d*Sqrt
[b*(Cosh[c] - Sinh[c])^4]) + (16*a*Cosh[2*d*x]*Sinh[2*c])/d + (16*a*Cosh[2*c]*Sinh[2*d*x])/d - ((a^3 + 18*a^2*
b + 48*a*b^2 + 32*b^3)*Sech[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/(b*(a + b)*d*(a + 2*b + a*Cosh[2*(c +
d*x)]))))/(256*a^3*(a + b*Sech[c + d*x]^2)^2) - ((a + 2*b + a*Cosh[2*c + 2*d*x])^2*Sech[c + d*x]^4*(-((a*ArcTa
nh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a + b)^(3/2)) + (Sqrt[b]*(a + 2*b)*Sinh[2*(c + d*x)])/((a + b)*(a +
2*b + a*Cosh[2*(c + d*x)]))))/(256*b^(3/2)*d*(a + b*Sech[c + d*x]^2)^2) + ((a + 2*b + a*Cosh[2*c + 2*d*x])^2*S
ech[c + d*x]^4*(-1/8*((a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(b^(3/2)*(a + b)^(3/2)*d) + (a*S
inh[2*(c + d*x)])/(8*b*(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)]))))/(16*(a + b*Sech[c + d*x]^2)^2)

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fricas [B]  time = 0.50, size = 2925, normalized size = 22.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(a^2*cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 - 2*(2*(a^2 + 4*a*b)*d*x
- a^2 - 2*a*b)*cosh(d*x + c)^6 + 2*(14*a^2*cosh(d*x + c)^2 - 2*(a^2 + 4*a*b)*d*x + a^2 + 2*a*b)*sinh(d*x + c)
^6 + 4*(14*a^2*cosh(d*x + c)^3 - 3*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c))*sinh(d*x + c)^5 - 8*((a^
2 + 6*a*b + 8*b^2)*d*x + a*b + 2*b^2)*cosh(d*x + c)^4 + 2*(35*a^2*cosh(d*x + c)^4 - 4*(a^2 + 6*a*b + 8*b^2)*d*
x - 15*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^2 - 4*a*b - 8*b^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*
x + c)^5 - 5*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^3 - 4*((a^2 + 6*a*b + 8*b^2)*d*x + a*b + 2*b^2)
*cosh(d*x + c))*sinh(d*x + c)^3 - 2*(2*(a^2 + 4*a*b)*d*x + a^2 + 6*a*b)*cosh(d*x + c)^2 + 2*(14*a^2*cosh(d*x +
c)^6 - 15*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^4 - 2*(a^2 + 4*a*b)*d*x - 24*((a^2 + 6*a*b + 8*b^
2)*d*x + a*b + 2*b^2)*cosh(d*x + c)^2 - a^2 - 6*a*b)*sinh(d*x + c)^2 + 2*((3*a^2 + 4*a*b)*cosh(d*x + c)^6 + 6*
(3*a^2 + 4*a*b)*cosh(d*x + c)*sinh(d*x + c)^5 + (3*a^2 + 4*a*b)*sinh(d*x + c)^6 + 2*(3*a^2 + 10*a*b + 8*b^2)*c
osh(d*x + c)^4 + (15*(3*a^2 + 4*a*b)*cosh(d*x + c)^2 + 6*a^2 + 20*a*b + 16*b^2)*sinh(d*x + c)^4 + 4*(5*(3*a^2
+ 4*a*b)*cosh(d*x + c)^3 + 2*(3*a^2 + 10*a*b + 8*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + (3*a^2 + 4*a*b)*cosh(d*
x + c)^2 + (15*(3*a^2 + 4*a*b)*cosh(d*x + c)^4 + 12*(3*a^2 + 10*a*b + 8*b^2)*cosh(d*x + c)^2 + 3*a^2 + 4*a*b)*
sinh(d*x + c)^2 + 2*(3*(3*a^2 + 4*a*b)*cosh(d*x + c)^5 + 4*(3*a^2 + 10*a*b + 8*b^2)*cosh(d*x + c)^3 + (3*a^2 +
4*a*b)*cosh(d*x + c))*sinh(d*x + c))*sqrt(b/(a + b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x
+ c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(
d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) - 4*((a
^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*
a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(
a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b
)*cosh(d*x + c))*sinh(d*x + c) + a)) - a^2 + 4*(2*a^2*cosh(d*x + c)^7 - 3*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a*b)*
cosh(d*x + c)^5 - 8*((a^2 + 6*a*b + 8*b^2)*d*x + a*b + 2*b^2)*cosh(d*x + c)^3 - (2*(a^2 + 4*a*b)*d*x + a^2 + 6
*a*b)*cosh(d*x + c))*sinh(d*x + c))/(a^4*d*cosh(d*x + c)^6 + 6*a^4*d*cosh(d*x + c)*sinh(d*x + c)^5 + a^4*d*sin
h(d*x + c)^6 + a^4*d*cosh(d*x + c)^2 + 2*(a^4 + 2*a^3*b)*d*cosh(d*x + c)^4 + (15*a^4*d*cosh(d*x + c)^2 + 2*(a^
4 + 2*a^3*b)*d)*sinh(d*x + c)^4 + 4*(5*a^4*d*cosh(d*x + c)^3 + 2*(a^4 + 2*a^3*b)*d*cosh(d*x + c))*sinh(d*x + c
)^3 + (15*a^4*d*cosh(d*x + c)^4 + a^4*d + 12*(a^4 + 2*a^3*b)*d*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 2*(3*a^4*d*c
osh(d*x + c)^5 + a^4*d*cosh(d*x + c) + 4*(a^4 + 2*a^3*b)*d*cosh(d*x + c)^3)*sinh(d*x + c)), 1/8*(a^2*cosh(d*x
+ c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 - 2*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a*b)*cos
h(d*x + c)^6 + 2*(14*a^2*cosh(d*x + c)^2 - 2*(a^2 + 4*a*b)*d*x + a^2 + 2*a*b)*sinh(d*x + c)^6 + 4*(14*a^2*cosh
(d*x + c)^3 - 3*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c))*sinh(d*x + c)^5 - 8*((a^2 + 6*a*b + 8*b^2)*
d*x + a*b + 2*b^2)*cosh(d*x + c)^4 + 2*(35*a^2*cosh(d*x + c)^4 - 4*(a^2 + 6*a*b + 8*b^2)*d*x - 15*(2*(a^2 + 4*
a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^2 - 4*a*b - 8*b^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5 - 5*(2*(a^
2 + 4*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^3 - 4*((a^2 + 6*a*b + 8*b^2)*d*x + a*b + 2*b^2)*cosh(d*x + c))*sin
h(d*x + c)^3 - 2*(2*(a^2 + 4*a*b)*d*x + a^2 + 6*a*b)*cosh(d*x + c)^2 + 2*(14*a^2*cosh(d*x + c)^6 - 15*(2*(a^2
+ 4*a*b)*d*x - a^2 - 2*a*b)*cosh(d*x + c)^4 - 2*(a^2 + 4*a*b)*d*x - 24*((a^2 + 6*a*b + 8*b^2)*d*x + a*b + 2*b^
2)*cosh(d*x + c)^2 - a^2 - 6*a*b)*sinh(d*x + c)^2 + 4*((3*a^2 + 4*a*b)*cosh(d*x + c)^6 + 6*(3*a^2 + 4*a*b)*cos
h(d*x + c)*sinh(d*x + c)^5 + (3*a^2 + 4*a*b)*sinh(d*x + c)^6 + 2*(3*a^2 + 10*a*b + 8*b^2)*cosh(d*x + c)^4 + (1
5*(3*a^2 + 4*a*b)*cosh(d*x + c)^2 + 6*a^2 + 20*a*b + 16*b^2)*sinh(d*x + c)^4 + 4*(5*(3*a^2 + 4*a*b)*cosh(d*x +
c)^3 + 2*(3*a^2 + 10*a*b + 8*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + (3*a^2 + 4*a*b)*cosh(d*x + c)^2 + (15*(3*a
^2 + 4*a*b)*cosh(d*x + c)^4 + 12*(3*a^2 + 10*a*b + 8*b^2)*cosh(d*x + c)^2 + 3*a^2 + 4*a*b)*sinh(d*x + c)^2 + 2
*(3*(3*a^2 + 4*a*b)*cosh(d*x + c)^5 + 4*(3*a^2 + 10*a*b + 8*b^2)*cosh(d*x + c)^3 + (3*a^2 + 4*a*b)*cosh(d*x +
c))*sinh(d*x + c))*sqrt(-b/(a + b))*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d
*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b) - a^2 + 4*(2*a^2*cosh(d*x + c)^7 - 3*(2*(a^2 + 4*a*b)*d*x - a^2 - 2*a
*b)*cosh(d*x + c)^5 - 8*((a^2 + 6*a*b + 8*b^2)*d*x + a*b + 2*b^2)*cosh(d*x + c)^3 - (2*(a^2 + 4*a*b)*d*x + a^2
+ 6*a*b)*cosh(d*x + c))*sinh(d*x + c))/(a^4*d*cosh(d*x + c)^6 + 6*a^4*d*cosh(d*x + c)*sinh(d*x + c)^5 + a^4*d
*sinh(d*x + c)^6 + a^4*d*cosh(d*x + c)^2 + 2*(a^4 + 2*a^3*b)*d*cosh(d*x + c)^4 + (15*a^4*d*cosh(d*x + c)^2 + 2
*(a^4 + 2*a^3*b)*d)*sinh(d*x + c)^4 + 4*(5*a^4*d*cosh(d*x + c)^3 + 2*(a^4 + 2*a^3*b)*d*cosh(d*x + c))*sinh(d*x
+ c)^3 + (15*a^4*d*cosh(d*x + c)^4 + a^4*d + 12*(a^4 + 2*a^3*b)*d*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 2*(3*a^4
*d*cosh(d*x + c)^5 + a^4*d*cosh(d*x + c) + 4*(a^4 + 2*a^3*b)*d*cosh(d*x + c)^3)*sinh(d*x + c))]

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giac [A]  time = 2.05, size = 234, normalized size = 1.79 \[ -\frac {\frac {12 \, {\left (d x + c\right )} {\left (a + 4 \, b\right )}}{a^{3}} - \frac {3 \, e^{\left (2 \, d x + 2 \, c\right )}}{a^{2}} - \frac {12 \, {\left (3 \, a b + 4 \, b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{3}} - \frac {2 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 8 \, a b e^{\left (6 \, d x + 6 \, c\right )} + a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 16 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 4 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 28 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a^{2}}{{\left (a e^{\left (6 \, d x + 6 \, c\right )} + 2 \, a e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (4 \, d x + 4 \, c\right )} + a e^{\left (2 \, d x + 2 \, c\right )}\right )} a^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(12*(d*x + c)*(a + 4*b)/a^3 - 3*e^(2*d*x + 2*c)/a^2 - 12*(3*a*b + 4*b^2)*arctan(1/2*(a*e^(2*d*x + 2*c) +
a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^3) - (2*a^2*e^(6*d*x + 6*c) + 8*a*b*e^(6*d*x + 6*c) + a^2*e^(4
*d*x + 4*c) - 16*b^2*e^(4*d*x + 4*c) - 4*a^2*e^(2*d*x + 2*c) - 28*a*b*e^(2*d*x + 2*c) - 3*a^2)/((a*e^(6*d*x +
6*c) + 2*a*e^(4*d*x + 4*c) + 4*b*e^(4*d*x + 4*c) + a*e^(2*d*x + 2*c))*a^3))/d

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maple [B]  time = 0.37, size = 543, normalized size = 4.15 \[ \frac {1}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}+\frac {2 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b}{d \,a^{3}}-\frac {1}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 d \,a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}-\frac {2 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b}{d \,a^{3}}+\frac {b \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {3 \sqrt {b}\, \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{4 d \,a^{2} \sqrt {a +b}}+\frac {3 \sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d \,a^{2} \sqrt {a +b}}-\frac {b^{\frac {3}{2}} \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{d \,a^{3} \sqrt {a +b}}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{d \,a^{3} \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.46, size = 696, normalized size = 5.31 \[ \frac {{\left (3 \, a^{2} b + 12 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {{\left (3 \, a^{2} b + 12 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {a^{2} b + 2 \, a b^{2} + {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{4 \, {\left (a^{5} + a^{4} b + {\left (a^{5} + a^{4} b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} d} + \frac {a^{2} b + 2 \, a b^{2} + {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, {\left (a^{5} + a^{4} b + 2 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{5} + a^{4} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {a b + {\left (a b + 2 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{2 \, {\left (a^{4} + a^{3} b + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + a^{3} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {d x + c}{2 \, a^{2} d} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a^{2} d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a^{2} d} - \frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{2 \, a^{3} d} + \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*a^2*b + 12*a*b^2 + 8*b^3)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a
+ 2*b + 2*sqrt((a + b)*b)))/((a^4 + a^3*b)*sqrt((a + b)*b)*d) - 1/16*(3*a^2*b + 12*a*b^2 + 8*b^3)*log((a*e^(-
2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^4 + a^3*b)
*sqrt((a + b)*b)*d) - 1/8*(3*a*b + 2*b^2)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x
- 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^3 + a^2*b)*sqrt((a + b)*b)*d) - 1/4*(a^2*b + 2*a*b^2 + (a^2*b + 8*a
*b^2 + 8*b^3)*e^(2*d*x + 2*c))/((a^5 + a^4*b + (a^5 + a^4*b)*e^(4*d*x + 4*c) + 2*(a^5 + 3*a^4*b + 2*a^3*b^2)*e
^(2*d*x + 2*c))*d) + 1/4*(a^2*b + 2*a*b^2 + (a^2*b + 8*a*b^2 + 8*b^3)*e^(-2*d*x - 2*c))/((a^5 + a^4*b + 2*(a^5
+ 3*a^4*b + 2*a^3*b^2)*e^(-2*d*x - 2*c) + (a^5 + a^4*b)*e^(-4*d*x - 4*c))*d) + 1/2*(a*b + (a*b + 2*b^2)*e^(-2
*d*x - 2*c))/((a^4 + a^3*b + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*e^(-2*d*x - 2*c) + (a^4 + a^3*b)*e^(-4*d*x - 4*c))*
d) - 1/2*(d*x + c)/(a^2*d) + 1/8*e^(2*d*x + 2*c)/(a^2*d) - 1/8*e^(-2*d*x - 2*c)/(a^2*d) - 1/2*b*log(a*e^(4*d*x
+ 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/(a^3*d) + 1/2*b*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*
c) + a)/(a^3*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\mathrm {sinh}\left (c+d\,x\right )}^2}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cosh(c + d*x)^4*sinh(c + d*x)^2)/(b + a*cosh(c + d*x)^2)^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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