### 3.60 $$\int \sinh (c+d x) (a+b \tanh ^3(c+d x))^2 \, dx$$

Optimal. Leaf size=123 $\frac{a^2 \cosh (c+d x)}{d}+\frac{3 a b \sinh (c+d x)}{d}-\frac{3 a b \tan ^{-1}(\sinh (c+d x))}{d}-\frac{a b \sinh (c+d x) \tanh ^2(c+d x)}{d}+\frac{b^2 \cosh (c+d x)}{d}+\frac{b^2 \text{sech}^5(c+d x)}{5 d}-\frac{b^2 \text{sech}^3(c+d x)}{d}+\frac{3 b^2 \text{sech}(c+d x)}{d}$

[Out]

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

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Rubi [A]  time = 0.146609, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.381, Rules used = {3666, 2638, 2592, 288, 321, 203, 2590, 270} $\frac{a^2 \cosh (c+d x)}{d}+\frac{3 a b \sinh (c+d x)}{d}-\frac{3 a b \tan ^{-1}(\sinh (c+d x))}{d}-\frac{a b \sinh (c+d x) \tanh ^2(c+d x)}{d}+\frac{b^2 \cosh (c+d x)}{d}+\frac{b^2 \text{sech}^5(c+d x)}{5 d}-\frac{b^2 \text{sech}^3(c+d x)}{d}+\frac{3 b^2 \text{sech}(c+d x)}{d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3666

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> Int[ExpandTrig[(d*sin[e + f*x])^m*(a + b*(c*tan[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && IGtQ[p, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

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

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 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sinh (c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx &=-\left (i \int \left (i a^2 \sinh (c+d x)+2 i a b \sinh (c+d x) \tanh ^3(c+d x)+i b^2 \sinh (c+d x) \tanh ^6(c+d x)\right ) \, dx\right )\\ &=a^2 \int \sinh (c+d x) \, dx+(2 a b) \int \sinh (c+d x) \tanh ^3(c+d x) \, dx+b^2 \int \sinh (c+d x) \tanh ^6(c+d x) \, dx\\ &=\frac{a^2 \cosh (c+d x)}{d}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^6} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{a^2 \cosh (c+d x)}{d}-\frac{a b \sinh (c+d x) \tanh ^2(c+d x)}{d}+\frac{(3 a b) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^6}-\frac{3}{x^4}+\frac{3}{x^2}\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{a^2 \cosh (c+d x)}{d}+\frac{b^2 \cosh (c+d x)}{d}+\frac{3 b^2 \text{sech}(c+d x)}{d}-\frac{b^2 \text{sech}^3(c+d x)}{d}+\frac{b^2 \text{sech}^5(c+d x)}{5 d}+\frac{3 a b \sinh (c+d x)}{d}-\frac{a b \sinh (c+d x) \tanh ^2(c+d x)}{d}-\frac{(3 a b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=-\frac{3 a b \tan ^{-1}(\sinh (c+d x))}{d}+\frac{a^2 \cosh (c+d x)}{d}+\frac{b^2 \cosh (c+d x)}{d}+\frac{3 b^2 \text{sech}(c+d x)}{d}-\frac{b^2 \text{sech}^3(c+d x)}{d}+\frac{b^2 \text{sech}^5(c+d x)}{5 d}+\frac{3 a b \sinh (c+d x)}{d}-\frac{a b \sinh (c+d x) \tanh ^2(c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.393281, size = 90, normalized size = 0.73 $\frac{5 \left (a^2+b^2\right ) \cosh (c+d x)+b \left (5 \text{sech}(c+d x) (a \tanh (c+d x)+3 b)+10 a \left (\sinh (c+d x)-3 \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )\right )+b \text{sech}^5(c+d x)-5 b \text{sech}^3(c+d x)\right )}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*(a^2 + b^2)*Cosh[c + d*x] + b*(-5*b*Sech[c + d*x]^3 + b*Sech[c + d*x]^5 + 10*a*(-3*ArcTan[Tanh[(c + d*x)/2]
] + Sinh[c + d*x]) + 5*Sech[c + d*x]*(3*b + a*Tanh[c + d*x])))/(5*d)

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Maple [A]  time = 0.054, size = 225, normalized size = 1.8 \begin{align*}{\frac{{a}^{2}\cosh \left ( dx+c \right ) }{d}}+2\,{\frac{ab \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+6\,{\frac{ab\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-3\,{\frac{ab{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{d}}-6\,{\frac{ab\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+6\,{\frac{{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{24\,{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{5\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{16\,{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{5\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-{\frac{16\,{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{5\,d\cosh \left ( dx+c \right ) }}+{\frac{16\,{b}^{2}\cosh \left ( dx+c \right ) }{5\,d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*cosh(d*x+c)/d+2/d*a*b*sinh(d*x+c)^3/cosh(d*x+c)^2+6/d*a*b*sinh(d*x+c)/cosh(d*x+c)^2-3/d*a*b*sech(d*x+c)*ta
nh(d*x+c)-6/d*a*b*arctan(exp(d*x+c))+1/d*b^2*sinh(d*x+c)^6/cosh(d*x+c)^5+6/d*b^2*sinh(d*x+c)^4/cosh(d*x+c)^5+2
4/5/d*b^2*sinh(d*x+c)^2/cosh(d*x+c)^5-16/5/d*b^2*sinh(d*x+c)^2/cosh(d*x+c)^3-16/5/d*b^2*sinh(d*x+c)^2/cosh(d*x
+c)+16/5*b^2*cosh(d*x+c)/d

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Maxima [B]  time = 1.5844, size = 342, normalized size = 2.78 \begin{align*} a b{\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{1}{10} \, b^{2}{\left (\frac{5 \, e^{\left (-d x - c\right )}}{d} + \frac{85 \, e^{\left (-2 \, d x - 2 \, c\right )} + 210 \, e^{\left (-4 \, d x - 4 \, c\right )} + 314 \, e^{\left (-6 \, d x - 6 \, c\right )} + 185 \, e^{\left (-8 \, d x - 8 \, c\right )} + 65 \, e^{\left (-10 \, d x - 10 \, c\right )} + 5}{d{\left (e^{\left (-d x - c\right )} + 5 \, e^{\left (-3 \, d x - 3 \, c\right )} + 10 \, e^{\left (-5 \, d x - 5 \, c\right )} + 10 \, e^{\left (-7 \, d x - 7 \, c\right )} + 5 \, e^{\left (-9 \, d x - 9 \, c\right )} + e^{\left (-11 \, d x - 11 \, c\right )}\right )}}\right )} + \frac{a^{2} \cosh \left (d x + c\right )}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b*(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)))) + 1/10*b^2*(5*e^(-d*x - c)/d + (85*e^(-2*d*x - 2*c) + 210*e^(-4*d*x
- 4*c) + 314*e^(-6*d*x - 6*c) + 185*e^(-8*d*x - 8*c) + 65*e^(-10*d*x - 10*c) + 5)/(d*(e^(-d*x - c) + 5*e^(-3*
d*x - 3*c) + 10*e^(-5*d*x - 5*c) + 10*e^(-7*d*x - 7*c) + 5*e^(-9*d*x - 9*c) + e^(-11*d*x - 11*c)))) + a^2*cosh
(d*x + c)/d

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Fricas [B]  time = 2.51216, size = 5998, normalized size = 48.76 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^12 + 60*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^11 + 5*(a^2
+ 2*a*b + b^2)*sinh(d*x + c)^12 + 30*(a^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^10 + 30*(11*(a^2 + 2*a*b + b^2)*cosh(
d*x + c)^2 + a^2 + 2*a*b + 3*b^2)*sinh(d*x + c)^10 + 100*(11*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + 3*(a^2 + 2*
a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^9 + 5*(15*a^2 + 18*a*b + 47*b^2)*cosh(d*x + c)^8 + 5*(495*(a^2 + 2*a
*b + b^2)*cosh(d*x + c)^4 + 270*(a^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^2 + 15*a^2 + 18*a*b + 47*b^2)*sinh(d*x + c
)^8 + 40*(99*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 90*(a^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^3 + (15*a^2 + 18*a*b
+ 47*b^2)*cosh(d*x + c))*sinh(d*x + c)^7 + 4*(25*a^2 + 91*b^2)*cosh(d*x + c)^6 + 4*(1155*(a^2 + 2*a*b + b^2)*
cosh(d*x + c)^6 + 1575*(a^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^4 + 35*(15*a^2 + 18*a*b + 47*b^2)*cosh(d*x + c)^2 +
25*a^2 + 91*b^2)*sinh(d*x + c)^6 + 8*(495*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^7 + 945*(a^2 + 2*a*b + 3*b^2)*cos
h(d*x + c)^5 + 35*(15*a^2 + 18*a*b + 47*b^2)*cosh(d*x + c)^3 + 3*(25*a^2 + 91*b^2)*cosh(d*x + c))*sinh(d*x + c
)^5 + 5*(15*a^2 - 18*a*b + 47*b^2)*cosh(d*x + c)^4 + 5*(495*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 1260*(a^2 +
2*a*b + 3*b^2)*cosh(d*x + c)^6 + 70*(15*a^2 + 18*a*b + 47*b^2)*cosh(d*x + c)^4 + 12*(25*a^2 + 91*b^2)*cosh(d*x
+ c)^2 + 15*a^2 - 18*a*b + 47*b^2)*sinh(d*x + c)^4 + 20*(55*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^9 + 180*(a^2 +
2*a*b + 3*b^2)*cosh(d*x + c)^7 + 14*(15*a^2 + 18*a*b + 47*b^2)*cosh(d*x + c)^5 + 4*(25*a^2 + 91*b^2)*cosh(d*x
+ c)^3 + (15*a^2 - 18*a*b + 47*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 30*(a^2 - 2*a*b + 3*b^2)*cosh(d*x + c)^2
+ 10*(33*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^10 + 135*(a^2 + 2*a*b + 3*b^2)*cosh(d*x + c)^8 + 14*(15*a^2 + 18*a*
b + 47*b^2)*cosh(d*x + c)^6 + 6*(25*a^2 + 91*b^2)*cosh(d*x + c)^4 + 3*(15*a^2 - 18*a*b + 47*b^2)*cosh(d*x + c)
^2 + 3*a^2 - 6*a*b + 9*b^2)*sinh(d*x + c)^2 + 5*a^2 - 10*a*b + 5*b^2 - 60*(a*b*cosh(d*x + c)^11 + 11*a*b*cosh(
d*x + c)*sinh(d*x + c)^10 + a*b*sinh(d*x + c)^11 + 5*a*b*cosh(d*x + c)^9 + 5*(11*a*b*cosh(d*x + c)^2 + a*b)*si
nh(d*x + c)^9 + 10*a*b*cosh(d*x + c)^7 + 15*(11*a*b*cosh(d*x + c)^3 + 3*a*b*cosh(d*x + c))*sinh(d*x + c)^8 + 1
0*(33*a*b*cosh(d*x + c)^4 + 18*a*b*cosh(d*x + c)^2 + a*b)*sinh(d*x + c)^7 + 10*a*b*cosh(d*x + c)^5 + 14*(33*a*
b*cosh(d*x + c)^5 + 30*a*b*cosh(d*x + c)^3 + 5*a*b*cosh(d*x + c))*sinh(d*x + c)^6 + 2*(231*a*b*cosh(d*x + c)^6
+ 315*a*b*cosh(d*x + c)^4 + 105*a*b*cosh(d*x + c)^2 + 5*a*b)*sinh(d*x + c)^5 + 5*a*b*cosh(d*x + c)^3 + 10*(33
*a*b*cosh(d*x + c)^7 + 63*a*b*cosh(d*x + c)^5 + 35*a*b*cosh(d*x + c)^3 + 5*a*b*cosh(d*x + c))*sinh(d*x + c)^4
+ 5*(33*a*b*cosh(d*x + c)^8 + 84*a*b*cosh(d*x + c)^6 + 70*a*b*cosh(d*x + c)^4 + 20*a*b*cosh(d*x + c)^2 + a*b)*
sinh(d*x + c)^3 + a*b*cosh(d*x + c) + 5*(11*a*b*cosh(d*x + c)^9 + 36*a*b*cosh(d*x + c)^7 + 42*a*b*cosh(d*x + c
)^5 + 20*a*b*cosh(d*x + c)^3 + 3*a*b*cosh(d*x + c))*sinh(d*x + c)^2 + (11*a*b*cosh(d*x + c)^10 + 45*a*b*cosh(d
*x + c)^8 + 70*a*b*cosh(d*x + c)^6 + 50*a*b*cosh(d*x + c)^4 + 15*a*b*cosh(d*x + c)^2 + a*b)*sinh(d*x + c))*arc
tan(cosh(d*x + c) + sinh(d*x + c)) + 4*(15*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^11 + 75*(a^2 + 2*a*b + 3*b^2)*cos
h(d*x + c)^9 + 10*(15*a^2 + 18*a*b + 47*b^2)*cosh(d*x + c)^7 + 6*(25*a^2 + 91*b^2)*cosh(d*x + c)^5 + 5*(15*a^2
- 18*a*b + 47*b^2)*cosh(d*x + c)^3 + 15*(a^2 - 2*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^
11 + 11*d*cosh(d*x + c)*sinh(d*x + c)^10 + d*sinh(d*x + c)^11 + 5*d*cosh(d*x + c)^9 + 5*(11*d*cosh(d*x + c)^2
+ d)*sinh(d*x + c)^9 + 15*(11*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^8 + 10*d*cosh(d*x + c)^7 +
10*(33*d*cosh(d*x + c)^4 + 18*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^7 + 14*(33*d*cosh(d*x + c)^5 + 30*d*cosh(d*
x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c)^6 + 10*d*cosh(d*x + c)^5 + 2*(231*d*cosh(d*x + c)^6 + 315*d*cosh(d
*x + c)^4 + 105*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 10*(33*d*cosh(d*x + c)^7 + 63*d*cosh(d*x + c)^5 + 3
5*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(33*d*cosh(d*x + c)^8 + 84*
d*cosh(d*x + c)^6 + 70*d*cosh(d*x + c)^4 + 20*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + 5*(11*d*cosh(d*x + c)^9
+ 36*d*cosh(d*x + c)^7 + 42*d*cosh(d*x + c)^5 + 20*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + d
*cosh(d*x + c) + (11*d*cosh(d*x + c)^10 + 45*d*cosh(d*x + c)^8 + 70*d*cosh(d*x + c)^6 + 50*d*cosh(d*x + c)^4 +
15*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(sinh(d*x+c)*(a+b*tanh(d*x+c)**3)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.61689, size = 289, normalized size = 2.35 \begin{align*} -\frac{60 \, a b \arctan \left (e^{\left (d x + c\right )}\right ) - 5 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-d x - c\right )} - 5 \,{\left (a^{2} e^{\left (d x + 14 \, c\right )} + 2 \, a b e^{\left (d x + 14 \, c\right )} + b^{2} e^{\left (d x + 14 \, c\right )}\right )} e^{\left (-13 \, c\right )} - \frac{4 \,{\left (5 \, a b e^{\left (9 \, d x + 9 \, c\right )} + 15 \, b^{2} e^{\left (9 \, d x + 9 \, c\right )} + 10 \, a b e^{\left (7 \, d x + 7 \, c\right )} + 40 \, b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 66 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} - 10 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 40 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 5 \, a b e^{\left (d x + c\right )} + 15 \, b^{2} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{10 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(60*a*b*arctan(e^(d*x + c)) - 5*(a^2 - 2*a*b + b^2)*e^(-d*x - c) - 5*(a^2*e^(d*x + 14*c) + 2*a*b*e^(d*x
+ 14*c) + b^2*e^(d*x + 14*c))*e^(-13*c) - 4*(5*a*b*e^(9*d*x + 9*c) + 15*b^2*e^(9*d*x + 9*c) + 10*a*b*e^(7*d*x
+ 7*c) + 40*b^2*e^(7*d*x + 7*c) + 66*b^2*e^(5*d*x + 5*c) - 10*a*b*e^(3*d*x + 3*c) + 40*b^2*e^(3*d*x + 3*c) - 5
*a*b*e^(d*x + c) + 15*b^2*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^5)/d