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

Optimal. Leaf size=112 \[ \frac {a^3}{4 d (1-\tanh (c+d x))}-\frac {a^3}{4 d (\tanh (c+d x)+1)}-\frac {3 a^2 b \tanh (c+d x)}{d}-\frac {1}{2} a^2 x (a-6 b)+\frac {b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^3 \tanh ^5(c+d x)}{5 d} \]

[Out]

-1/2*a^2*(a-6*b)*x+1/4*a^3/d/(1-tanh(d*x+c))-3*a^2*b*tanh(d*x+c)/d+1/3*b^2*(3*a+b)*tanh(d*x+c)^3/d-1/5*b^3*tan
h(d*x+c)^5/d-1/4*a^3/d/(1+tanh(d*x+c))

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Rubi [A]  time = 0.19, antiderivative size = 143, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4132, 467, 528, 388, 206} \[ -\frac {b \left (81 a^2-28 a b-4 b^2\right ) \tanh (c+d x)}{30 d}-\frac {1}{2} a^2 x (a-6 b)-\frac {7 b \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^2}{10 d}-\frac {b (33 a-2 b) \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )}{30 d}+\frac {\sinh (c+d x) \cosh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(a - 6*b)*x)/2 - (b*(81*a^2 - 28*a*b - 4*b^2)*Tanh[c + d*x])/(30*d) - ((33*a - 2*b)*b*Tanh[c + d*x]*(a +
b - b*Tanh[c + d*x]^2))/(30*d) - (7*b*Tanh[c + d*x]*(a + b - b*Tanh[c + d*x]^2)^2)/(10*d) + (Cosh[c + d*x]*Si
nh[c + d*x]*(a + b - b*Tanh[c + d*x]^2)^3)/(2*d)

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 388

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

Rule 467

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)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 528

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

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 \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh ^2(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b-b x^2\right )^3}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-7 b x^2\right ) \left (a+b-b x^2\right )^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {7 b \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-b x^2\right ) \left (-(5 a-2 b) (a+b)+(33 a-2 b) b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{10 d}\\ &=-\frac {(33 a-2 b) b \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )}{30 d}-\frac {7 b \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+b) \left (15 a^2-24 a b-4 b^2\right )-b \left (81 a^2-28 a b-4 b^2\right ) x^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{30 d}\\ &=-\frac {b \left (81 a^2-28 a b-4 b^2\right ) \tanh (c+d x)}{30 d}-\frac {(33 a-2 b) b \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )}{30 d}-\frac {7 b \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\left (a^2 (a-6 b)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {1}{2} a^2 (a-6 b) x-\frac {b \left (81 a^2-28 a b-4 b^2\right ) \tanh (c+d x)}{30 d}-\frac {(33 a-2 b) b \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )}{30 d}-\frac {7 b \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{2 d}\\ \end {align*}

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Mathematica [B]  time = 1.92, size = 480, normalized size = 4.29 \[ \frac {\text {sech}(c) \text {sech}^5(c+d x) \left (75 a^3 \sinh (2 c+d x)+135 a^3 \sinh (2 c+3 d x)+135 a^3 \sinh (4 c+3 d x)+75 a^3 \sinh (4 c+5 d x)+75 a^3 \sinh (6 c+5 d x)+15 a^3 \sinh (6 c+7 d x)+15 a^3 \sinh (8 c+7 d x)-300 a^3 d x \cosh (2 c+3 d x)-300 a^3 d x \cosh (4 c+3 d x)-60 a^3 d x \cosh (4 c+5 d x)-60 a^3 d x \cosh (6 c+5 d x)+75 a^3 \sinh (d x)+2880 a^2 b \sinh (2 c+d x)-2880 a^2 b \sinh (2 c+3 d x)+720 a^2 b \sinh (4 c+3 d x)-720 a^2 b \sinh (4 c+5 d x)-600 a^2 d x (a-6 b) \cosh (2 c+d x)+1800 a^2 b d x \cosh (2 c+3 d x)+1800 a^2 b d x \cosh (4 c+3 d x)+360 a^2 b d x \cosh (4 c+5 d x)+360 a^2 b d x \cosh (6 c+5 d x)-4320 a^2 b \sinh (d x)-600 a^2 d x (a-6 b) \cosh (d x)-1440 a b^2 \sinh (2 c+d x)+480 a b^2 \sinh (2 c+3 d x)-720 a b^2 \sinh (4 c+3 d x)+240 a b^2 \sinh (4 c+5 d x)+960 a b^2 \sinh (d x)-480 b^3 \sinh (2 c+d x)+160 b^3 \sinh (2 c+3 d x)+32 b^3 \sinh (4 c+5 d x)-160 b^3 \sinh (d x)\right )}{3840 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[c]*Sech[c + d*x]^5*(-600*a^2*(a - 6*b)*d*x*Cosh[d*x] - 600*a^2*(a - 6*b)*d*x*Cosh[2*c + d*x] - 300*a^3*d
*x*Cosh[2*c + 3*d*x] + 1800*a^2*b*d*x*Cosh[2*c + 3*d*x] - 300*a^3*d*x*Cosh[4*c + 3*d*x] + 1800*a^2*b*d*x*Cosh[
4*c + 3*d*x] - 60*a^3*d*x*Cosh[4*c + 5*d*x] + 360*a^2*b*d*x*Cosh[4*c + 5*d*x] - 60*a^3*d*x*Cosh[6*c + 5*d*x] +
360*a^2*b*d*x*Cosh[6*c + 5*d*x] + 75*a^3*Sinh[d*x] - 4320*a^2*b*Sinh[d*x] + 960*a*b^2*Sinh[d*x] - 160*b^3*Sin
h[d*x] + 75*a^3*Sinh[2*c + d*x] + 2880*a^2*b*Sinh[2*c + d*x] - 1440*a*b^2*Sinh[2*c + d*x] - 480*b^3*Sinh[2*c +
d*x] + 135*a^3*Sinh[2*c + 3*d*x] - 2880*a^2*b*Sinh[2*c + 3*d*x] + 480*a*b^2*Sinh[2*c + 3*d*x] + 160*b^3*Sinh[
2*c + 3*d*x] + 135*a^3*Sinh[4*c + 3*d*x] + 720*a^2*b*Sinh[4*c + 3*d*x] - 720*a*b^2*Sinh[4*c + 3*d*x] + 75*a^3*
Sinh[4*c + 5*d*x] - 720*a^2*b*Sinh[4*c + 5*d*x] + 240*a*b^2*Sinh[4*c + 5*d*x] + 32*b^3*Sinh[4*c + 5*d*x] + 75*
a^3*Sinh[6*c + 5*d*x] + 15*a^3*Sinh[6*c + 7*d*x] + 15*a^3*Sinh[8*c + 7*d*x]))/(3840*d)

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fricas [B]  time = 0.43, size = 595, normalized size = 5.31 \[ \frac {15 \, a^{3} \sinh \left (d x + c\right )^{7} + 4 \, {\left (90 \, a^{2} b - 30 \, a b^{2} - 4 \, b^{3} - 15 \, {\left (a^{3} - 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right )^{5} + 20 \, {\left (90 \, a^{2} b - 30 \, a b^{2} - 4 \, b^{3} - 15 \, {\left (a^{3} - 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (315 \, a^{3} \cosh \left (d x + c\right )^{2} + 75 \, a^{3} - 360 \, a^{2} b + 120 \, a b^{2} + 16 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 20 \, {\left (90 \, a^{2} b - 30 \, a b^{2} - 4 \, b^{3} - 15 \, {\left (a^{3} - 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (105 \, a^{3} \cosh \left (d x + c\right )^{4} + 27 \, a^{3} - 216 \, a^{2} b - 24 \, a b^{2} + 16 \, b^{3} + 2 \, {\left (75 \, a^{3} - 360 \, a^{2} b + 120 \, a b^{2} + 16 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 20 \, {\left (2 \, {\left (90 \, a^{2} b - 30 \, a b^{2} - 4 \, b^{3} - 15 \, {\left (a^{3} - 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (90 \, a^{2} b - 30 \, a b^{2} - 4 \, b^{3} - 15 \, {\left (a^{3} - 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 40 \, {\left (90 \, a^{2} b - 30 \, a b^{2} - 4 \, b^{3} - 15 \, {\left (a^{3} - 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right ) + 5 \, {\left (21 \, a^{3} \cosh \left (d x + c\right )^{6} + {\left (75 \, a^{3} - 360 \, a^{2} b + 120 \, a b^{2} + 16 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 15 \, a^{3} - 144 \, a^{2} b - 48 \, a b^{2} - 64 \, b^{3} + 3 \, {\left (27 \, a^{3} - 216 \, a^{2} b - 24 \, a b^{2} + 16 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{120 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*a^3*sinh(d*x + c)^7 + 4*(90*a^2*b - 30*a*b^2 - 4*b^3 - 15*(a^3 - 6*a^2*b)*d*x)*cosh(d*x + c)^5 + 20*
(90*a^2*b - 30*a*b^2 - 4*b^3 - 15*(a^3 - 6*a^2*b)*d*x)*cosh(d*x + c)*sinh(d*x + c)^4 + (315*a^3*cosh(d*x + c)^
2 + 75*a^3 - 360*a^2*b + 120*a*b^2 + 16*b^3)*sinh(d*x + c)^5 + 20*(90*a^2*b - 30*a*b^2 - 4*b^3 - 15*(a^3 - 6*a
^2*b)*d*x)*cosh(d*x + c)^3 + 5*(105*a^3*cosh(d*x + c)^4 + 27*a^3 - 216*a^2*b - 24*a*b^2 + 16*b^3 + 2*(75*a^3 -
360*a^2*b + 120*a*b^2 + 16*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 20*(2*(90*a^2*b - 30*a*b^2 - 4*b^3 - 15*(a
^3 - 6*a^2*b)*d*x)*cosh(d*x + c)^3 + 3*(90*a^2*b - 30*a*b^2 - 4*b^3 - 15*(a^3 - 6*a^2*b)*d*x)*cosh(d*x + c))*s
inh(d*x + c)^2 + 40*(90*a^2*b - 30*a*b^2 - 4*b^3 - 15*(a^3 - 6*a^2*b)*d*x)*cosh(d*x + c) + 5*(21*a^3*cosh(d*x
+ c)^6 + (75*a^3 - 360*a^2*b + 120*a*b^2 + 16*b^3)*cosh(d*x + c)^4 + 15*a^3 - 144*a^2*b - 48*a*b^2 - 64*b^3 +
3*(27*a^3 - 216*a^2*b - 24*a*b^2 + 16*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x +
c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d
*cosh(d*x + c))

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giac [B]  time = 0.20, size = 278, normalized size = 2.48 \[ \frac {15 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 60 \, {\left (a^{3} - 6 \, a^{2} b\right )} {\left (d x + c\right )} + 15 \, {\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 12 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - a^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + \frac {16 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 45 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 30 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 60 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 30 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b - 15 \, a b^{2} - 2 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*a^3*e^(2*d*x + 2*c) - 60*(a^3 - 6*a^2*b)*(d*x + c) + 15*(2*a^3*e^(2*d*x + 2*c) - 12*a^2*b*e^(2*d*x +
2*c) - a^3)*e^(-2*d*x - 2*c) + 16*(45*a^2*b*e^(8*d*x + 8*c) - 45*a*b^2*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x +
6*c) - 90*a*b^2*e^(6*d*x + 6*c) - 30*b^3*e^(6*d*x + 6*c) + 270*a^2*b*e^(4*d*x + 4*c) - 60*a*b^2*e^(4*d*x + 4*
c) + 10*b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) - 30*a*b^2*e^(2*d*x + 2*c) - 10*b^3*e^(2*d*x + 2*c) +
45*a^2*b - 15*a*b^2 - 2*b^3)/(e^(2*d*x + 2*c) + 1)^5)/d

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maple [A]  time = 0.42, size = 145, normalized size = 1.29 \[ \frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a^{2} b \left (d x +c -\tanh \left (d x +c \right )\right )+3 a \,b^{2} \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+b^{3} \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+3*a^2*b*(d*x+c-tanh(d*x+c))+3*a*b^2*(-1/2*sinh(d*x+c)/cos
h(d*x+c)^3+1/2*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c))+b^3*(-1/4*sinh(d*x+c)/cosh(d*x+c)^5+1/4*(8/15+1/5*sech(d*x
+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c)))

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maxima [B]  time = 0.34, size = 443, normalized size = 3.96 \[ -\frac {1}{8} \, a^{3} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + \frac {4}{15} \, b^{3} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac {5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 2 \, a b^{2} {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^3*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + 3*a^2*b*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + 4
/15*b^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x -
8*c) + e^(-10*d*x - 10*c) + 1)) - 5*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*
d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 15*e^(-6*d*x - 6*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e
^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*
c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 2*a*b^2*(3*e
^(-4*d*x - 4*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c
) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)))

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mupad [B]  time = 0.23, size = 592, normalized size = 5.29 \[ \frac {\frac {2\,\left (9\,a^2\,b+3\,a\,b^2+4\,b^3\right )}{15\,d}-\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2\,b-b^3\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (9\,a^2\,b+3\,a\,b^2+4\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2\,b-b^3\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (9\,a^2\,b+3\,a\,b^2+4\,b^3\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}-\frac {6\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {6\,\left (a\,b^2-a^2\,b\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a^2\,x\,\left (a-6\,b\right )}{2}-\frac {a^3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*(3*a*b^2 + 9*a^2*b + 4*b^3))/(15*d) - (6*exp(4*c + 4*d*x)*(a*b^2 - a^2*b))/(5*d) + (4*exp(2*c + 2*d*x)*(3*
a^2*b - b^3))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) + ((2*(3*a^2*b - b^3))/(
5*d) - (6*exp(6*c + 6*d*x)*(a*b^2 - a^2*b))/(5*d) + (2*exp(2*c + 2*d*x)*(3*a*b^2 + 9*a^2*b + 4*b^3))/(5*d) + (
6*exp(4*c + 4*d*x)*(3*a^2*b - b^3))/(5*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) + ((2*(3*a^2*b - b^3))/(5*d) - (6*exp(2*c + 2*d*x)*(a*b^2 - a^2*b))/(5*d))/(2*exp(2*c + 2*d
*x) + exp(4*c + 4*d*x) + 1) + ((4*exp(4*c + 4*d*x)*(3*a*b^2 + 9*a^2*b + 4*b^3))/(5*d) - (6*exp(8*c + 8*d*x)*(a
*b^2 - a^2*b))/(5*d) - (6*(a*b^2 - a^2*b))/(5*d) + (8*exp(2*c + 2*d*x)*(3*a^2*b - b^3))/(5*d) + (8*exp(6*c + 6
*d*x)*(3*a^2*b - b^3))/(5*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) - (6*(a*b^2 - a^2*b))/(5*d*(exp(2*c + 2*d*x) + 1)) - (a^2*x*(a - 6*b))/2 - (a^3
*exp(- 2*c - 2*d*x))/(8*d) + (a^3*exp(2*c + 2*d*x))/(8*d)

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

[Out]

Timed out

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