3.100 \(\int \cosh (c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=99 \[ \frac {3 b^2 (4 a+3 b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {(a+b)^3 \sinh (c+d x)}{d}-\frac {3 b \left (4 (a+b)^2+(2 a+b)^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}-\frac {b^3 \tanh (c+d x) \text {sech}^3(c+d x)}{4 d} \]

[Out]

-3/8*b*(4*(a+b)^2+(2*a+b)^2)*arctan(sinh(d*x+c))/d+(a+b)^3*sinh(d*x+c)/d+3/8*b^2*(4*a+3*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 = 0.12, antiderivative size = 99, 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 = {3676, 390, 1157, 385, 203} \[ \frac {3 b^2 (4 a+3 b) \tanh (c+d x) \text {sech}(c+d x)}{8 d}+\frac {(a+b)^3 \sinh (c+d x)}{d}-\frac {3 b \left (4 (a+b)^2+(2 a+b)^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}-\frac {b^3 \tanh (c+d x) \text {sech}^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(4*(a + b)^2 + (2*a + b)^2)*ArcTan[Sinh[c + d*x]])/(8*d) + ((a + b)^3*Sinh[c + d*x])/d + (3*b^2*(4*a + 3
*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 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 \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+(a+b) 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+b)^3-\frac {b \left (3 a^2+3 a b+b^2\right )+3 b (a+b) (2 a+b) x^2+3 b (a+b)^2 x^4}{\left (1+x^2\right )^3}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {(a+b)^3 \sinh (c+d x)}{d}-\frac {\operatorname {Subst}\left (\int \frac {b \left (3 a^2+3 a b+b^2\right )+3 b (a+b) (2 a+b) x^2+3 b (a+b)^2 x^4}{\left (1+x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {(a+b)^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 b (a+b)^2 x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{4 d}\\ &=\frac {(a+b)^3 \sinh (c+d x)}{d}+\frac {3 b^2 (4 a+3 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 (4 (a+b)^2+(2 a+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 (4 (a+b)^2+(2 a+b)^2\right ) \tan ^{-1}(\sinh (c+d x))}{8 d}+\frac {(a+b)^3 \sinh (c+d x)}{d}+\frac {3 b^2 (4 a+3 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 [A]  time = 0.42, size = 89, normalized size = 0.90 \[ \frac {-3 b \left (8 a^2+12 a b+5 b^2\right ) \tan ^{-1}(\sinh (c+d x))+3 b^2 (4 a+3 b) \tanh (c+d x) \text {sech}(c+d x)+8 (a+b)^3 \sinh (c+d x)-2 b^3 \tanh (c+d x) \text {sech}^3(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(8*a^2 + 12*a*b + 5*b^2)*ArcTan[Sinh[c + d*x]] + 8*(a + b)^3*Sinh[c + d*x] + 3*b^2*(4*a + 3*b)*Sech[c +
d*x]*Tanh[c + d*x] - 2*b^3*Sech[c + d*x]^3*Tanh[c + d*x])/(8*d)

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fricas [B]  time = 0.45, size = 2411, normalized size = 24.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^10 + 20*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sin
h(d*x + c)^9 + 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^10 + 3*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)*cos
h(d*x + c)^8 + 3*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3 + 30*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sin
h(d*x + c)^8 + 24*(10*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)*c
osh(d*x + c))*sinh(d*x + c)^7 + (4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + (420*(a^3 + 3*a^2*b +
3*a*b^2 + b^3)*cosh(d*x + c)^4 + 4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3 + 84*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)
*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 6*(84*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^5 + 28*(2*a^3 + 6*a^2*
b + 10*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + (4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 -
(4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + (420*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^6
+ 210*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)*cosh(d*x + c)^4 - 4*a^3 - 12*a^2*b - 24*a*b^2 - 5*b^3 + 15*(4*a^3 +
12*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(60*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*
x + c)^7 + 42*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + 5*(4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*c
osh(d*x + c)^3 - (4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - 2*a^3 - 6*a^2*b - 6*a*
b^2 - 2*b^3 - 3*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)*cosh(d*x + c)^2 + 3*(30*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*c
osh(d*x + c)^8 + 28*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + 5*(4*a^3 + 12*a^2*b + 24*a*b^2 + 5*
b^3)*cosh(d*x + c)^4 - 2*a^3 - 6*a^2*b - 10*a*b^2 - 5*b^3 - 2*(4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^2 - 3*((8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^9 + 9*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh
(d*x + c)*sinh(d*x + c)^8 + (8*a^2*b + 12*a*b^2 + 5*b^3)*sinh(d*x + c)^9 + 4*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh
(d*x + c)^7 + 4*(8*a^2*b + 12*a*b^2 + 5*b^3 + 9*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^7
+ 28*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + (8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x +
c)^6 + 6*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + 6*(21*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 8
*a^2*b + 12*a*b^2 + 5*b^3 + 14*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 2*(63*(8*a^2*b
+ 12*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + 70*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 15*(8*a^2*b + 12*a*b^2
+ 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + 4*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 4*(21*(8*a^2*b + 1
2*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + 35*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 8*a^2*b + 12*a*b^2 + 5*b^
3 + 15*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 12*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh
(d*x + c)^7 + 7*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + 5*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^3
+ (8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + (8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c) + (
9*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^8 + 28*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + 30*(8*a^2*b
+ 12*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 8*a^2*b + 12*a*b^2 + 5*b^3 + 12*(8*a^2*b + 12*a*b^2 + 5*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(10*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x +
c)^9 + 12*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*b^3)*cosh(d*x + c)^7 + 3*(4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*cosh
(d*x + c)^5 - 2*(4*a^3 + 12*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c)^3 - 3*(2*a^3 + 6*a^2*b + 10*a*b^2 + 5*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))*s
inh(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*c
osh(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 = 0.38, size = 257, normalized size = 2.60 \[ -\frac {3 \, {\left (8 \, a^{2} b e^{c} + 12 \, a b^{2} e^{c} + 5 \, b^{3} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )} + 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-d x - c\right )} - 2 \, {\left (a^{3} e^{\left (d x + 12 \, c\right )} + 3 \, a^{2} b e^{\left (d x + 12 \, c\right )} + 3 \, a b^{2} e^{\left (d x + 12 \, c\right )} + b^{3} e^{\left (d x + 12 \, c\right )}\right )} e^{\left (-11 \, c\right )} - \frac {12 \, a b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 9 \, b^{3} e^{\left (7 \, d x + 7 \, c\right )} + 12 \, a b^{2} e^{\left (5 \, d x + 5 \, c\right )} + b^{3} e^{\left (5 \, d x + 5 \, c\right )} - 12 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} - b^{3} e^{\left (3 \, d x + 3 \, c\right )} - 12 \, a b^{2} e^{\left (d x + c\right )} - 9 \, b^{3} e^{\left (d x + c\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*(8*a^2*b*e^c + 12*a*b^2*e^c + 5*b^3*e^c)*arctan(e^(d*x + c))*e^(-c) + 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3
)*e^(-d*x - c) - 2*(a^3*e^(d*x + 12*c) + 3*a^2*b*e^(d*x + 12*c) + 3*a*b^2*e^(d*x + 12*c) + b^3*e^(d*x + 12*c))
*e^(-11*c) - (12*a*b^2*e^(7*d*x + 7*c) + 9*b^3*e^(7*d*x + 7*c) + 12*a*b^2*e^(5*d*x + 5*c) + b^3*e^(5*d*x + 5*c
) - 12*a*b^2*e^(3*d*x + 3*c) - b^3*e^(3*d*x + 3*c) - 12*a*b^2*e^(d*x + c) - 9*b^3*e^(d*x + c))/(e^(2*d*x + 2*c
) + 1)^4)/d

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maple [B]  time = 0.40, size = 257, normalized size = 2.60 \[ \frac {a^{3} \sinh \left (d x +c \right )}{d}+\frac {3 a^{2} b \sinh \left (d x +c \right )}{d}-\frac {6 a^{2} b \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {3 a \,b^{2} \left (\sinh ^{3}\left (d x +c \right )\right )}{d \cosh \left (d x +c \right )^{2}}+\frac {9 a \,b^{2} \sinh \left (d x +c \right )}{d \cosh \left (d x +c \right )^{2}}-\frac {9 a \,b^{2} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2 d}-\frac {9 a \,b^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {b^{3} \left (\sinh ^{5}\left (d x +c \right )\right )}{d \cosh \left (d x +c \right )^{4}}+\frac {5 b^{3} \left (\sinh ^{3}\left (d x +c \right )\right )}{d \cosh \left (d x +c \right )^{4}}+\frac {5 b^{3} \sinh \left (d x +c \right )}{d \cosh \left (d x +c \right )^{4}}-\frac {5 b^{3} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{3}}{4 d}-\frac {15 b^{3} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{8 d}-\frac {15 b^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.70, size = 295, normalized size = 2.98 \[ \frac {1}{4} \, b^{3} {\left (\frac {15 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {2 \, e^{\left (-d x - c\right )}}{d} + \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 13 \, e^{\left (-4 \, d x - 4 \, c\right )} + 7 \, e^{\left (-6 \, d x - 6 \, c\right )} - 7 \, e^{\left (-8 \, d x - 8 \, c\right )} + 2}{d {\left (e^{\left (-d x - c\right )} + 4 \, e^{\left (-3 \, d x - 3 \, c\right )} + 6 \, e^{\left (-5 \, d x - 5 \, c\right )} + 4 \, e^{\left (-7 \, d x - 7 \, c\right )} + e^{\left (-9 \, d x - 9 \, c\right )}\right )}}\right )} + \frac {3}{2} \, a b^{2} {\left (\frac {6 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )}}{d} + \frac {4 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + 1}{d {\left (e^{\left (-d x - c\right )} + 2 \, e^{\left (-3 \, d x - 3 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )}\right )}}\right )} + \frac {3}{2} \, a^{2} b {\left (\frac {4 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {e^{\left (d x + c\right )}}{d} - \frac {e^{\left (-d x - c\right )}}{d}\right )} + \frac {a^{3} \sinh \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^3*(15*arctan(e^(-d*x - c))/d - 2*e^(-d*x - c)/d + (17*e^(-2*d*x - 2*c) + 13*e^(-4*d*x - 4*c) + 7*e^(-6*d
*x - 6*c) - 7*e^(-8*d*x - 8*c) + 2)/(d*(e^(-d*x - c) + 4*e^(-3*d*x - 3*c) + 6*e^(-5*d*x - 5*c) + 4*e^(-7*d*x -
7*c) + e^(-9*d*x - 9*c)))) + 3/2*a*b^2*(6*arctan(e^(-d*x - c))/d - e^(-d*x - c)/d + (4*e^(-2*d*x - 2*c) - e^(
-4*d*x - 4*c) + 1)/(d*(e^(-d*x - c) + 2*e^(-3*d*x - 3*c) + e^(-5*d*x - 5*c)))) + 3/2*a^2*b*(4*arctan(e^(-d*x -
c))/d + e^(d*x + c)/d - e^(-d*x - c)/d) + a^3*sinh(d*x + c)/d

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mupad [B]  time = 0.31, size = 355, normalized size = 3.59 \[ \frac {{\mathrm {e}}^{c+d\,x}\,{\left (a+b\right )}^3}{2\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,{\left (a+b\right )}^3}{2\,d}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (5\,b^3\,\sqrt {d^2}+12\,a\,b^2\,\sqrt {d^2}+8\,a^2\,b\,\sqrt {d^2}\right )}{d\,\sqrt {64\,a^4\,b^2+192\,a^3\,b^3+224\,a^2\,b^4+120\,a\,b^5+25\,b^6}}\right )\,\sqrt {64\,a^4\,b^2+192\,a^3\,b^3+224\,a^2\,b^4+120\,a\,b^5+25\,b^6}}{4\,\sqrt {d^2}}+\frac {3\,{\mathrm {e}}^{c+d\,x}\,\left (3\,b^3+4\,a\,b^2\right )}{4\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {6\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (13\,b^3+12\,a\,b^2\right )}{2\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {4\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(c + d*x)*(a + b)^3)/(2*d) - (exp(- c - d*x)*(a + b)^3)/(2*d) - (3*atan((exp(d*x)*exp(c)*(5*b^3*(d^2)^(1/2
) + 12*a*b^2*(d^2)^(1/2) + 8*a^2*b*(d^2)^(1/2)))/(d*(120*a*b^5 + 25*b^6 + 224*a^2*b^4 + 192*a^3*b^3 + 64*a^4*b
^2)^(1/2)))*(120*a*b^5 + 25*b^6 + 224*a^2*b^4 + 192*a^3*b^3 + 64*a^4*b^2)^(1/2))/(4*(d^2)^(1/2)) + (3*exp(c +
d*x)*(4*a*b^2 + 3*b^3))/(4*d*(exp(2*c + 2*d*x) + 1)) + (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 + 13*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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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