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

Optimal. Leaf size=93 \[ \frac {a^3 \sinh (c+d x)}{d}+\frac {3 b \left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {3 b^2 (4 a+b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {b^3 \tanh (c+d x) \text {sech}^3(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4147, 390, 1157, 385, 203} \[ \frac {3 b \left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {a^3 \sinh (c+d x)}{d}+\frac {3 b^2 (4 a+b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {b^3 \tanh (c+d x) \text {sech}^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 4147

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + 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 \cosh (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b+a x^2\right )^3}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^3+\frac {b \left (3 a^2+3 a b+b^2\right )+3 a b (2 a+b) x^2+3 a^2 b x^4}{\left (1+x^2\right )^3}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {a^3 \sinh (c+d x)}{d}+\frac {\operatorname {Subst}\left (\int \frac {b \left (3 a^2+3 a b+b^2\right )+3 a b (2 a+b) x^2+3 a^2 b x^4}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {a^3 \sinh (c+d x)}{d}+\frac {b^3 \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {-3 b (2 a+b)^2-12 a^2 b x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{4 d}\\ &=\frac {a^3 \sinh (c+d x)}{d}+\frac {3 b^2 (4 a+b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {b^3 \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}+\frac {\left (3 b \left (8 a^2+4 a b+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{8 d}\\ &=\frac {3 b \left (8 a^2+4 a b+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {a^3 \sinh (c+d x)}{d}+\frac {3 b^2 (4 a+b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {b^3 \text {sech}^3(c+d x) \tanh (c+d x)}{4 d}\\ \end {align*}

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Mathematica [C]  time = 7.88, size = 575, normalized size = 6.18 \[ -\frac {\cosh (c+d x) \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \left (256 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};-\sinh ^2(c+d x)\right )+384 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^2 \left (a \left (5 \sinh ^2(c+d x)+7\right )+7 b\right ) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};-\sinh ^2(c+d x)\right )-21 \left (a^3 \left (4887 \sinh ^{10}(c+d x)+107725 \sinh ^8(c+d x)+491574 \sinh ^6(c+d x)+922986 \sinh ^4(c+d x)+789235 \sinh ^2(c+d x)+252105\right )+3 a^2 b \left (5640 \sinh ^8(c+d x)+120431 \sinh ^6(c+d x)+437991 \sinh ^4(c+d x)+573145 \sinh ^2(c+d x)+252105\right )+3 a b^2 \left (6393 \sinh ^6(c+d x)+133071 \sinh ^4(c+d x)+357055 \sinh ^2(c+d x)+252105\right )+b^3 \left (8226 \sinh ^4(c+d x)+140965 \sinh ^2(c+d x)+252105\right )\right )+\frac {315 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right ) \left (a^3 \left (7 \sinh ^8(c+d x)+1468 \sinh ^6(c+d x)+11562 \sinh ^4(c+d x)+24604 \sinh ^2(c+d x)+16807\right ) \cosh ^4(c+d x)+3 a^2 b \left (8 \sinh ^{10}(c+d x)+1719 \sinh ^8(c+d x)+14956 \sinh ^6(c+d x)+40442 \sinh ^4(c+d x)+43812 \sinh ^2(c+d x)+16807\right )+3 a b^2 \left (9 \sinh ^8(c+d x)+1858 \sinh ^6(c+d x)+15312 \sinh ^4(c+d x)+29406 \sinh ^2(c+d x)+16807\right )+b^3 \left (-62 \sinh ^6(c+d x)+2187 \sinh ^4(c+d x)+15000 \sinh ^2(c+d x)+16807\right )\right )}{\sqrt {-\sinh ^2(c+d x)}}\right )}{7560 d (a \cosh (2 c+2 d x)+a+2 b)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/7560*(Cosh[c + d*x]*Coth[c + d*x]^5*(a + b*Sech[c + d*x]^2)^3*(256*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2},
{1, 1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^8*(a + b + a*Sinh[c + d*x]^2)^3 + 384*HypergeometricPFQ[{3
/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^8*(a + b + a*Sinh[c + d*x]^2)^2*(7*b + a*(7
+ 5*Sinh[c + d*x]^2)) + (315*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^3*(16807 + 15000*Sinh[c + d*x]^2 + 2187*Sinh[c
+ d*x]^4 - 62*Sinh[c + d*x]^6) + a^3*Cosh[c + d*x]^4*(16807 + 24604*Sinh[c + d*x]^2 + 11562*Sinh[c + d*x]^4 +
1468*Sinh[c + d*x]^6 + 7*Sinh[c + d*x]^8) + 3*a*b^2*(16807 + 29406*Sinh[c + d*x]^2 + 15312*Sinh[c + d*x]^4 +
1858*Sinh[c + d*x]^6 + 9*Sinh[c + d*x]^8) + 3*a^2*b*(16807 + 43812*Sinh[c + d*x]^2 + 40442*Sinh[c + d*x]^4 + 1
4956*Sinh[c + d*x]^6 + 1719*Sinh[c + d*x]^8 + 8*Sinh[c + d*x]^10)))/Sqrt[-Sinh[c + d*x]^2] - 21*(b^3*(252105 +
140965*Sinh[c + d*x]^2 + 8226*Sinh[c + d*x]^4) + 3*a*b^2*(252105 + 357055*Sinh[c + d*x]^2 + 133071*Sinh[c + d
*x]^4 + 6393*Sinh[c + d*x]^6) + 3*a^2*b*(252105 + 573145*Sinh[c + d*x]^2 + 437991*Sinh[c + d*x]^4 + 120431*Sin
h[c + d*x]^6 + 5640*Sinh[c + d*x]^8) + a^3*(252105 + 789235*Sinh[c + d*x]^2 + 922986*Sinh[c + d*x]^4 + 491574*
Sinh[c + d*x]^6 + 107725*Sinh[c + d*x]^8 + 4887*Sinh[c + d*x]^10))))/(d*(a + 2*b + a*Cosh[2*c + 2*d*x])^3)

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fricas [B]  time = 0.47, size = 1992, normalized size = 21.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a^3*cosh(d*x + c)^10 + 20*a^3*cosh(d*x + c)*sinh(d*x + c)^9 + 2*a^3*sinh(d*x + c)^10 + 3*(2*a^3 + 4*a*b
^2 + b^3)*cosh(d*x + c)^8 + 3*(30*a^3*cosh(d*x + c)^2 + 2*a^3 + 4*a*b^2 + b^3)*sinh(d*x + c)^8 + 24*(10*a^3*co
sh(d*x + c)^3 + (2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^7 + (4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x
+ c)^6 + (420*a^3*cosh(d*x + c)^4 + 4*a^3 + 12*a*b^2 + 11*b^3 + 84*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*si
nh(d*x + c)^6 + 6*(84*a^3*cosh(d*x + c)^5 + 28*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + (4*a^3 + 12*a*b^2 + 1
1*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 - (4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^4 + (420*a^3*cosh(d*x + c)^6
+ 210*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^4 - 4*a^3 - 12*a*b^2 - 11*b^3 + 15*(4*a^3 + 12*a*b^2 + 11*b^3)*co
sh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(60*a^3*cosh(d*x + c)^7 + 42*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^5 + 5*(4
*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - (4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - 2*a^3
- 3*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^2 + 3*(30*a^3*cosh(d*x + c)^8 + 28*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x
+ c)^6 + 5*(4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^4 - 2*a^3 - 4*a*b^2 - b^3 - 2*(4*a^3 + 12*a*b^2 + 11*b^3
)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*((8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^9 + 9*(8*a^2*b + 4*a*b^2 + b^3
)*cosh(d*x + c)*sinh(d*x + c)^8 + (8*a^2*b + 4*a*b^2 + b^3)*sinh(d*x + c)^9 + 4*(8*a^2*b + 4*a*b^2 + b^3)*cosh
(d*x + c)^7 + 4*(8*a^2*b + 4*a*b^2 + b^3 + 9*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^7 + 28*(
3*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + (8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^6 + 6*(8*
a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^5 + 6*(21*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^2*b + 4*a*b^2 +
b^3 + 14*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 2*(63*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*
x + c)^5 + 70*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + 15*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x
+ c)^4 + 4*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + 4*(21*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^6 + 35*(
8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^2*b + 4*a*b^2 + b^3 + 15*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c
)^2)*sinh(d*x + c)^3 + 12*(3*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^7 + 7*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x
+ c)^5 + 5*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^3 + (8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^
2 + (8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c) + (9*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^8 + 28*(8*a^2*b + 4*a
*b^2 + b^3)*cosh(d*x + c)^6 + 30*(8*a^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^2*b + 4*a*b^2 + b^3 + 12*(8*a
^2*b + 4*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(10*a^3*cosh(d
*x + c)^9 + 12*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c)^7 + 3*(4*a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^5 - 2*(4*
a^3 + 12*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - 3*(2*a^3 + 4*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x
+ c)^9 + 9*d*cosh(d*x + c)*sinh(d*x + c)^8 + d*sinh(d*x + c)^9 + 4*d*cosh(d*x + c)^7 + 4*(9*d*cosh(d*x + c)^2
+ d)*sinh(d*x + c)^7 + 28*(3*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^6 + 6*d*cosh(d*x + c)^5 + 6*(
21*d*cosh(d*x + c)^4 + 14*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^5 + 2*(63*d*cosh(d*x + c)^5 + 70*d*cosh(d*x + c
)^3 + 15*d*cosh(d*x + c))*sinh(d*x + c)^4 + 4*d*cosh(d*x + c)^3 + 4*(21*d*cosh(d*x + c)^6 + 35*d*cosh(d*x + c)
^4 + 15*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + 12*(3*d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)^5 + 5*d*cosh(d*x
+ c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^2 + d*cosh(d*x + c) + (9*d*cosh(d*x + c)^8 + 28*d*cosh(d*x + c)^6 + 30
*d*cosh(d*x + c)^4 + 12*d*cosh(d*x + c)^2 + d)*sinh(d*x + c))

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giac [B]  time = 0.19, size = 199, normalized size = 2.14 \[ \frac {8 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 3 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (8 \, a^{2} b + 4 \, a b^{2} + b^{3}\right )} + \frac {4 \, {\left (12 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 3 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 48 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 20 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(8*a^3*(e^(d*x + c) - e^(-d*x - c)) + 3*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(8*a^2*b
+ 4*a*b^2 + b^3) + 4*(12*a*b^2*(e^(d*x + c) - e^(-d*x - c))^3 + 3*b^3*(e^(d*x + c) - e^(-d*x - c))^3 + 48*a*b^
2*(e^(d*x + c) - e^(-d*x - c)) + 20*b^3*(e^(d*x + c) - e^(-d*x - c)))/((e^(d*x + c) - e^(-d*x - c))^2 + 4)^2)/
d

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maple [A]  time = 0.51, size = 125, normalized size = 1.34 \[ \frac {a^{3} \sinh \left (d x +c \right )}{d}+\frac {6 a^{2} b \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {3 a \,b^{2} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2 d}+\frac {3 a \,b^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {b^{3} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{4 d}+\frac {3 b^{3} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{8 d}+\frac {3 b^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 221, normalized size = 2.38 \[ -\frac {1}{4} \, b^{3} {\left (\frac {3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {3 \, e^{\left (-d x - c\right )} + 11 \, e^{\left (-3 \, d x - 3 \, c\right )} - 11 \, e^{\left (-5 \, d x - 5 \, c\right )} - 3 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} - 3 \, a b^{2} {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} - \frac {6 \, a^{2} b \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {a^{3} \sinh \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b^3*(3*arctan(e^(-d*x - c))/d - (3*e^(-d*x - c) + 11*e^(-3*d*x - 3*c) - 11*e^(-5*d*x - 5*c) - 3*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))) - 3*a*b^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))) -
6*a^2*b*arctan(e^(-d*x - c))/d + a^3*sinh(d*x + c)/d

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mupad [B]  time = 0.19, size = 344, normalized size = 3.70 \[ \frac {a^3\,{\mathrm {e}}^{c+d\,x}}{2\,d}-\frac {a^3\,{\mathrm {e}}^{-c-d\,x}}{2\,d}+\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (b^3\,\sqrt {d^2}+4\,a\,b^2\,\sqrt {d^2}+8\,a^2\,b\,\sqrt {d^2}\right )}{d\,\sqrt {64\,a^4\,b^2+64\,a^3\,b^3+32\,a^2\,b^4+8\,a\,b^5+b^6}}\right )\,\sqrt {64\,a^4\,b^2+64\,a^3\,b^3+32\,a^2\,b^4+8\,a\,b^5+b^6}}{4\,\sqrt {d^2}}-\frac {6\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (12\,a\,b^2-b^3\right )}{2\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {4\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}+\frac {3\,{\mathrm {e}}^{c+d\,x}\,\left (b^3+4\,a\,b^2\right )}{4\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*exp(c + d*x))/(2*d) - (a^3*exp(- c - d*x))/(2*d) + (3*atan((exp(d*x)*exp(c)*(b^3*(d^2)^(1/2) + 4*a*b^2*(d
^2)^(1/2) + 8*a^2*b*(d^2)^(1/2)))/(d*(8*a*b^5 + b^6 + 32*a^2*b^4 + 64*a^3*b^3 + 64*a^4*b^2)^(1/2)))*(8*a*b^5 +
b^6 + 32*a^2*b^4 + 64*a^3*b^3 + 64*a^4*b^2)^(1/2))/(4*(d^2)^(1/2)) - (6*b^3*exp(c + d*x))/(d*(3*exp(2*c + 2*d
*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - (exp(c + d*x)*(12*a*b^2 - b^3))/(2*d*(2*exp(2*c + 2*d*x) +
exp(4*c + 4*d*x) + 1)) + (4*b^3*exp(c + d*x))/(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)) + (3*exp(c + d*x)*(4*a*b^2 + b^3))/(4*d*(exp(2*c + 2*d*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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