### 3.50 $$\int \sinh ^3(c+d x) (a+b \tanh ^3(c+d x)) \, dx$$

Optimal. Leaf size=98 $\frac{a \cosh ^3(c+d x)}{3 d}-\frac{a \cosh (c+d x)}{d}+\frac{5 b \sinh ^3(c+d x)}{6 d}-\frac{5 b \sinh (c+d x)}{2 d}-\frac{b \sinh ^3(c+d x) \tanh ^2(c+d x)}{2 d}+\frac{5 b \tan ^{-1}(\sinh (c+d x))}{2 d}$

[Out]

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

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Rubi [A]  time = 0.119697, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {3666, 2633, 2592, 288, 302, 203} $\frac{a \cosh ^3(c+d x)}{3 d}-\frac{a \cosh (c+d x)}{d}+\frac{5 b \sinh ^3(c+d x)}{6 d}-\frac{5 b \sinh (c+d x)}{2 d}-\frac{b \sinh ^3(c+d x) \tanh ^2(c+d x)}{2 d}+\frac{5 b \tan ^{-1}(\sinh (c+d x))}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*b*ArcTan[Sinh[c + d*x]])/(2*d) - (a*Cosh[c + d*x])/d + (a*Cosh[c + d*x]^3)/(3*d) - (5*b*Sinh[c + d*x])/(2*d
) + (5*b*Sinh[c + d*x]^3)/(6*d) - (b*Sinh[c + d*x]^3*Tanh[c + d*x]^2)/(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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [A]  time = 0.257416, size = 104, normalized size = 1.06 $-\frac{3 a \cosh (c+d x)}{4 d}+\frac{a \cosh (3 (c+d x))}{12 d}+\frac{b \sinh ^3(c+d x) \tanh ^2(c+d x)}{3 d}-\frac{5 b \left (2 \sinh (c+d x) \tanh ^2(c+d x)-3 \left (\tan ^{-1}(\sinh (c+d x))-\tanh (c+d x) \text{sech}(c+d x)\right )\right )}{6 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*a*Cosh[c + d*x])/(4*d) + (a*Cosh[3*(c + d*x)])/(12*d) + (b*Sinh[c + d*x]^3*Tanh[c + d*x]^2)/(3*d) - (5*b*(
2*Sinh[c + d*x]*Tanh[c + d*x]^2 - 3*(ArcTan[Sinh[c + d*x]] - Sech[c + d*x]*Tanh[c + d*x])))/(6*d)

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Maple [A]  time = 0.043, size = 129, normalized size = 1.3 \begin{align*} -{\frac{2\,a\cosh \left ( dx+c \right ) }{3\,d}}+{\frac{a\cosh \left ( dx+c \right ) \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{b \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{3\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,b \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-5\,{\frac{b\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,b{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+5\,{\frac{b\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.5411, size = 235, normalized size = 2.4 \begin{align*} \frac{1}{24} \, b{\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{\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(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^3),x, algorithm="maxima")

[Out]

1/24*b*((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*
(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 = 2.33449, size = 2942, normalized size = 30.02 \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)^3*(a+b*tanh(d*x+c)^3),x, algorithm="fricas")

[Out]

1/24*((a + b)*cosh(d*x + c)^10 + 10*(a + b)*cosh(d*x + c)*sinh(d*x + c)^9 + (a + b)*sinh(d*x + c)^10 - (7*a +
25*b)*cosh(d*x + c)^8 + (45*(a + b)*cosh(d*x + c)^2 - 7*a - 25*b)*sinh(d*x + c)^8 + 8*(15*(a + b)*cosh(d*x + c
)^3 - (7*a + 25*b)*cosh(d*x + c))*sinh(d*x + c)^7 - 2*(13*a + 25*b)*cosh(d*x + c)^6 + 2*(105*(a + b)*cosh(d*x
+ c)^4 - 14*(7*a + 25*b)*cosh(d*x + c)^2 - 13*a - 25*b)*sinh(d*x + c)^6 + 4*(63*(a + b)*cosh(d*x + c)^5 - 14*(
7*a + 25*b)*cosh(d*x + c)^3 - 3*(13*a + 25*b)*cosh(d*x + c))*sinh(d*x + c)^5 - 2*(13*a - 25*b)*cosh(d*x + c)^4
+ 2*(105*(a + b)*cosh(d*x + c)^6 - 35*(7*a + 25*b)*cosh(d*x + c)^4 - 15*(13*a + 25*b)*cosh(d*x + c)^2 - 13*a
+ 25*b)*sinh(d*x + c)^4 + 8*(15*(a + b)*cosh(d*x + c)^7 - 7*(7*a + 25*b)*cosh(d*x + c)^5 - 5*(13*a + 25*b)*cos
h(d*x + c)^3 - (13*a - 25*b)*cosh(d*x + c))*sinh(d*x + c)^3 - (7*a - 25*b)*cosh(d*x + c)^2 + (45*(a + b)*cosh(
d*x + c)^8 - 28*(7*a + 25*b)*cosh(d*x + c)^6 - 30*(13*a + 25*b)*cosh(d*x + c)^4 - 12*(13*a - 25*b)*cosh(d*x +
c)^2 - 7*a + 25*b)*sinh(d*x + c)^2 + 120*(b*cosh(d*x + c)^7 + 7*b*cosh(d*x + c)*sinh(d*x + c)^6 + b*sinh(d*x +
c)^7 + 2*b*cosh(d*x + c)^5 + (21*b*cosh(d*x + c)^2 + 2*b)*sinh(d*x + c)^5 + 5*(7*b*cosh(d*x + c)^3 + 2*b*cosh
(d*x + c))*sinh(d*x + c)^4 + b*cosh(d*x + c)^3 + (35*b*cosh(d*x + c)^4 + 20*b*cosh(d*x + c)^2 + b)*sinh(d*x +
c)^3 + (21*b*cosh(d*x + c)^5 + 20*b*cosh(d*x + c)^3 + 3*b*cosh(d*x + c))*sinh(d*x + c)^2 + (7*b*cosh(d*x + c)^
6 + 10*b*cosh(d*x + c)^4 + 3*b*cosh(d*x + c)^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(5*(a
+ b)*cosh(d*x + c)^9 - 4*(7*a + 25*b)*cosh(d*x + c)^7 - 6*(13*a + 25*b)*cosh(d*x + c)^5 - 4*(13*a - 25*b)*cos
h(d*x + c)^3 - (7*a - 25*b)*cosh(d*x + c))*sinh(d*x + c) + a - b)/(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*cosh(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^5 + 5*(7*d*c
osh(d*x + c)^3 + 2*d*cosh(d*x + c))*sinh(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*cosh(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(sinh(d*x+c)**3*(a+b*tanh(d*x+c)**3),x)

[Out]

Timed out

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Giac [A]  time = 1.29955, size = 192, normalized size = 1.96 \begin{align*} \frac{120 \, b \arctan \left (e^{\left (d x + c\right )}\right ) -{\left (9 \, a e^{\left (2 \, d x + 2 \, c\right )} - 27 \, b e^{\left (2 \, d x + 2 \, c\right )} - a + b\right )} e^{\left (-3 \, d x - 3 \, c\right )} +{\left (a e^{\left (3 \, d x + 30 \, c\right )} + b e^{\left (3 \, d x + 30 \, c\right )} - 9 \, a e^{\left (d x + 28 \, c\right )} - 27 \, b e^{\left (d x + 28 \, c\right )}\right )} e^{\left (-27 \, c\right )} - \frac{24 \,{\left (b e^{\left (3 \, d x + 3 \, c\right )} - b 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(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^3),x, algorithm="giac")

[Out]

1/24*(120*b*arctan(e^(d*x + c)) - (9*a*e^(2*d*x + 2*c) - 27*b*e^(2*d*x + 2*c) - a + b)*e^(-3*d*x - 3*c) + (a*e
^(3*d*x + 30*c) + b*e^(3*d*x + 30*c) - 9*a*e^(d*x + 28*c) - 27*b*e^(d*x + 28*c))*e^(-27*c) - 24*(b*e^(3*d*x +
3*c) - b*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^2)/d