### 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*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)

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Rubi [A]  time = 0.116297, antiderivative size = 99, normalized size of antiderivative = 1., 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 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 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 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 (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*}

Mathematica [A]  time = 0.363403, size = 89, normalized size = 0.9 $\frac{-3 b \left (8 a^2+12 a b+5 b^2\right ) \tan ^{-1}(\sinh (c+d x))+3 b^2 (4 a+3 b) \tanh (c+d x) \text{sech}(c+d x)+8 (a+b)^3 \sinh (c+d x)-2 b^3 \tanh (c+d x) \text{sech}^3(c+d x)}{8 d}$

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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Maple [B]  time = 0.048, size = 257, normalized size = 2.6 \begin{align*}{\frac{{a}^{3}\sinh \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}b\sinh \left ( dx+c \right ) }{d}}-6\,{\frac{{a}^{2}b\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+3\,{\frac{a{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+9\,{\frac{a{b}^{2}\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{9\,a{b}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}-9\,{\frac{a{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}}+5\,{\frac{{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}}+5\,{\frac{{b}^{3}\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}}-{\frac{5\,{b}^{3} \left ({\rm sech} \left (dx+c\right ) \right ) ^{3}\tanh \left ( dx+c \right ) }{4\,d}}-{\frac{15\,{b}^{3}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{8\,d}}-{\frac{15\,{b}^{3}\arctan \left ({{\rm e}^{dx+c}} \right ) }{4\,d}} \end{align*}

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]

1/d*a^3*sinh(d*x+c)+3*a^2*b*sinh(d*x+c)/d-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*si
nh(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*b^3*sech(d
*x+c)^3*tanh(d*x+c)/d-15/8*b^3*sech(d*x+c)*tanh(d*x+c)/d-15/4/d*b^3*arctan(exp(d*x+c))

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Maxima [B]  time = 1.7006, size = 398, normalized size = 4.02 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 2.18889, size = 5933, normalized size = 59.93 \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)*(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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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)*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.83198, size = 347, normalized size = 3.51 \begin{align*} -\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} \end{align*}

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