### 3.98 $$\int \cosh ^3(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=87 $\frac{b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac{(a-2 b) (a+b)^2 \sinh (c+d x)}{d}-\frac{b^3 \tanh (c+d x) \text{sech}(c+d x)}{2 d}$

[Out]

(b^2*(6*a + 5*b)*ArcTan[Sinh[c + d*x]])/(2*d) + ((a - 2*b)*(a + b)^2*Sinh[c + d*x])/d + ((a + b)^3*Sinh[c + d*
x]^3)/(3*d) - (b^3*Sech[c + d*x]*Tanh[c + d*x])/(2*d)

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Rubi [A]  time = 0.105108, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.174, Rules used = {3676, 390, 385, 203} $\frac{b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac{(a-2 b) (a+b)^2 \sinh (c+d x)}{d}-\frac{b^3 \tanh (c+d x) \text{sech}(c+d x)}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(b^2*(6*a + 5*b)*ArcTan[Sinh[c + d*x]])/(2*d) + ((a - 2*b)*(a + b)^2*Sinh[c + d*x])/d + ((a + b)^3*Sinh[c + d*
x]^3)/(3*d) - (b^3*Sech[c + d*x]*Tanh[c + d*x])/(2*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 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 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 \cosh ^3(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 )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left ((a-2 b) (a+b)^2+(a+b)^3 x^2+\frac{b^2 (3 a+2 b)+3 b^2 (a+b) x^2}{\left (1+x^2\right )^2}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{(a-2 b) (a+b)^2 \sinh (c+d x)}{d}+\frac{(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac{\operatorname{Subst}\left (\int \frac{b^2 (3 a+2 b)+3 b^2 (a+b) x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{(a-2 b) (a+b)^2 \sinh (c+d x)}{d}+\frac{(a+b)^3 \sinh ^3(c+d x)}{3 d}-\frac{b^3 \text{sech}(c+d x) \tanh (c+d x)}{2 d}+\frac{\left (b^2 (6 a+5 b)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{2 d}\\ &=\frac{b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{(a-2 b) (a+b)^2 \sinh (c+d x)}{d}+\frac{(a+b)^3 \sinh ^3(c+d x)}{3 d}-\frac{b^3 \text{sech}(c+d x) \tanh (c+d x)}{2 d}\\ \end{align*}

Mathematica [C]  time = 6.90058, size = 494, normalized size = 5.68 $\frac{\text{csch}^5(c+d x) \left (-256 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b \sinh ^2(c+d x)\right )^3 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2\right \},\left \{1,1,1,\frac{11}{2}\right \},-\sinh ^2(c+d x)\right )+21 \left (3 a^2 b \left (753 \sinh ^8(c+d x)+18826 \sinh ^6(c+d x)+69728 \sinh ^4(c+d x)+88150 \sinh ^2(c+d x)+36015\right ) \sinh ^2(c+d x)+a^3 \left (753 \sinh ^{10}(c+d x)+19579 \sinh ^8(c+d x)+89514 \sinh ^6(c+d x)+157878 \sinh ^4(c+d x)+124165 \sinh ^2(c+d x)+36015\right )+3 a b^2 \left (753 \sinh ^6(c+d x)+18073 \sinh ^4(c+d x)+50695 \sinh ^2(c+d x)+36015\right ) \sinh ^4(c+d x)+b^3 \left (753 \sinh ^4(c+d x)+17320 \sinh ^2(c+d x)+32415\right ) \sinh ^6(c+d x)\right )-\frac{315 \tanh ^{-1}\left (\sqrt{-\sinh ^2(c+d x)}\right ) \left (3 a^2 b \left (\sinh ^3(c+d x)+\sinh (c+d x)\right )^2 \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right )+a^3 \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right ) \cosh ^6(c+d x)+3 a b^2 \sinh ^4(c+d x) \left (\sinh ^8(c+d x)+244 \sinh ^6(c+d x)+2118 \sinh ^4(c+d x)+4180 \sinh ^2(c+d x)+2401\right )+b^3 \sinh ^6(c+d x) \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2161\right )\right )}{\sqrt{-\sinh ^2(c+d x)}}\right )}{30240 d}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Csch[c + d*x]^5*(-256*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^8
*(a + a*Sinh[c + d*x]^2 + b*Sinh[c + d*x]^2)^3 - (315*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^3*Sinh[c + d*x]^6*(21
61 + 1875*Sinh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + a^3*Cosh[c + d*x]^6*(2401 + 1875*Sinh[c +
d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + 3*a^2*b*(Sinh[c + d*x] + Sinh[c + d*x]^3)^2*(2401 + 1875*Si
nh[c + d*x]^2 + 243*Sinh[c + d*x]^4 + Sinh[c + d*x]^6) + 3*a*b^2*Sinh[c + d*x]^4*(2401 + 4180*Sinh[c + d*x]^2
+ 2118*Sinh[c + d*x]^4 + 244*Sinh[c + d*x]^6 + Sinh[c + d*x]^8)))/Sqrt[-Sinh[c + d*x]^2] + 21*(b^3*Sinh[c + d*
x]^6*(32415 + 17320*Sinh[c + d*x]^2 + 753*Sinh[c + d*x]^4) + 3*a*b^2*Sinh[c + d*x]^4*(36015 + 50695*Sinh[c + d
*x]^2 + 18073*Sinh[c + d*x]^4 + 753*Sinh[c + d*x]^6) + 3*a^2*b*Sinh[c + d*x]^2*(36015 + 88150*Sinh[c + d*x]^2
+ 69728*Sinh[c + d*x]^4 + 18826*Sinh[c + d*x]^6 + 753*Sinh[c + d*x]^8) + a^3*(36015 + 124165*Sinh[c + d*x]^2 +
157878*Sinh[c + d*x]^4 + 89514*Sinh[c + d*x]^6 + 19579*Sinh[c + d*x]^8 + 753*Sinh[c + d*x]^10))))/(30240*d)

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Maple [B]  time = 0.044, size = 227, normalized size = 2.6 \begin{align*}{\frac{2\,{a}^{3}\sinh \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}b \left ( \cosh \left ( dx+c \right ) \right ) ^{2}\sinh \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}b\sinh \left ( dx+c \right ) }{d}}+{\frac{a{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d}}+6\,{\frac{a{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}-3\,{\frac{a{b}^{2}\sinh \left ( dx+c \right ) }{d}}+{\frac{{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{3\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-5\,{\frac{{b}^{3}\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,{b}^{3}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+5\,{\frac{{b}^{3}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.66538, size = 383, normalized size = 4.4 \begin{align*} \frac{a^{2} b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3}}{8 \, d} - \frac{1}{8} \, a b^{2}{\left (\frac{{\left (15 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )} e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac{15 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{48 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} + \frac{1}{24} \, b^{3}{\left (\frac{27 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} - \frac{120 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{25 \, e^{\left (-2 \, d x - 2 \, c\right )} + 77 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + 2 \, e^{\left (-5 \, d x - 5 \, c\right )} + e^{\left (-7 \, d x - 7 \, c\right )}\right )}}\right )} + \frac{1}{24} \, a^{3}{\left (\frac{e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac{9 \, e^{\left (d x + c\right )}}{d} - \frac{9 \, e^{\left (-d x - c\right )}}{d} - \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*b*(e^(d*x + c) - e^(-d*x - c))^3/d - 1/8*a*b^2*((15*e^(-2*d*x - 2*c) - 1)*e^(3*d*x + 3*c)/d - (15*e^(-
d*x - c) - e^(-3*d*x - 3*c))/d + 48*arctan(e^(-d*x - c))/d) + 1/24*b^3*((27*e^(-d*x - c) - e^(-3*d*x - 3*c))/d
- 120*arctan(e^(-d*x - c))/d - (25*e^(-2*d*x - 2*c) + 77*e^(-4*d*x - 4*c) + 3*e^(-6*d*x - 6*c) - 1)/(d*(e^(-3
*d*x - 3*c) + 2*e^(-5*d*x - 5*c) + e^(-7*d*x - 7*c)))) + 1/24*a^3*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)/d - 9*e^(
-d*x - c)/d - e^(-3*d*x - 3*c)/d)

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Fricas [B]  time = 2.20745, size = 4523, normalized size = 51.99 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^10 + 10*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh
(d*x + c)^9 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^10 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d
*x + c)^8 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3 + 45*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d
*x + c)^8 + 8*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cos
h(d*x + c))*sinh(d*x + c)^7 + 2*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^6 + 2*(105*(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3 + 14*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*
b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 4*(63*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^5 + 14*(11*a^3 - 3
*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^3 + 3*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c))*sinh(d*x
+ c)^5 - 2*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^4 + 2*(105*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh
(d*x + c)^6 + 35*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^4 - 5*a^3 + 3*a^2*b + 21*a*b^2 + 25*b^3
+ 15*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(15*(a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*cosh(d*x + c)^7 + 7*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^5 + 5*(5*a^3 - 3*a^2*b - 21*a*b
^2 - 25*b^3)*cosh(d*x + c)^3 - (5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - a^3 - 3*
a^2*b - 3*a*b^2 - b^3 - (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^2 + (45*(a^3 + 3*a^2*b + 3*a*b^2
+ b^3)*cosh(d*x + c)^8 + 28*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^6 + 30*(5*a^3 - 3*a^2*b - 21*
a*b^2 - 25*b^3)*cosh(d*x + c)^4 - 11*a^3 + 3*a^2*b + 39*a*b^2 + 25*b^3 - 12*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b
^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 24*((6*a*b^2 + 5*b^3)*cosh(d*x + c)^7 + 7*(6*a*b^2 + 5*b^3)*cosh(d*x +
c)*sinh(d*x + c)^6 + (6*a*b^2 + 5*b^3)*sinh(d*x + c)^7 + 2*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + (12*a*b^2 + 10*
b^3 + 21*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 5*(7*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 2*(6*a*
b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + (6*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + (35*(6*a*b^2 + 5*b^3)*cosh(d
*x + c)^4 + 6*a*b^2 + 5*b^3 + 20*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + (21*(6*a*b^2 + 5*b^3)*co
sh(d*x + c)^5 + 20*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 3*(6*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + (7
*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + 10*(6*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 3*(6*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*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^9
+ 4*(11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cosh(d*x + c)^7 + 6*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x
+ c)^5 - 4*(5*a^3 - 3*a^2*b - 21*a*b^2 - 25*b^3)*cosh(d*x + c)^3 - (11*a^3 - 3*a^2*b - 39*a*b^2 - 25*b^3)*cos
h(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 2*d*co
sh(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 2*d*cosh(d*x + c))*sin
h(d*x + c)^4 + d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c)^4 + 20*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + (21*d*c
osh(d*x + c)^5 + 20*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + (7*d*cosh(d*x + c)^6 + 10*d*cosh(
d*x + c)^4 + 3*d*cosh(d*x + c)^2)*sinh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 2.4196, size = 377, normalized size = 4.33 \begin{align*} \frac{24 \,{\left (6 \, a b^{2} e^{c} + 5 \, b^{3} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )} -{\left (9 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 9 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 45 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 27 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )} +{\left (a^{3} e^{\left (3 \, d x + 30 \, c\right )} + 3 \, a^{2} b e^{\left (3 \, d x + 30 \, c\right )} + 3 \, a b^{2} e^{\left (3 \, d x + 30 \, c\right )} + b^{3} e^{\left (3 \, d x + 30 \, c\right )} + 9 \, a^{3} e^{\left (d x + 28 \, c\right )} - 9 \, a^{2} b e^{\left (d x + 28 \, c\right )} - 45 \, a b^{2} e^{\left (d x + 28 \, c\right )} - 27 \, b^{3} e^{\left (d x + 28 \, c\right )}\right )} e^{\left (-27 \, c\right )} - \frac{24 \,{\left (b^{3} e^{\left (3 \, d x + 3 \, c\right )} - b^{3} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{24 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(6*a*b^2*e^c + 5*b^3*e^c)*arctan(e^(d*x + c))*e^(-c) - (9*a^3*e^(2*d*x + 2*c) - 9*a^2*b*e^(2*d*x + 2*
c) - 45*a*b^2*e^(2*d*x + 2*c) - 27*b^3*e^(2*d*x + 2*c) + a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-3*d*x - 3*c) + (a^
3*e^(3*d*x + 30*c) + 3*a^2*b*e^(3*d*x + 30*c) + 3*a*b^2*e^(3*d*x + 30*c) + b^3*e^(3*d*x + 30*c) + 9*a^3*e^(d*x
+ 28*c) - 9*a^2*b*e^(d*x + 28*c) - 45*a*b^2*e^(d*x + 28*c) - 27*b^3*e^(d*x + 28*c))*e^(-27*c) - 24*(b^3*e^(3*
d*x + 3*c) - b^3*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^2)/d