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

Optimal. Leaf size=63 \[ \frac{a \cosh (c+d x)}{d}+\frac{3 b \sinh (c+d x)}{2 d}-\frac{3 b \tan ^{-1}(\sinh (c+d x))}{2 d}-\frac{b \sinh (c+d x) \tanh ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0761299, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3666, 2638, 2592, 288, 321, 203} \[ \frac{a \cosh (c+d x)}{d}+\frac{3 b \sinh (c+d x)}{2 d}-\frac{3 b \tan ^{-1}(\sinh (c+d x))}{2 d}-\frac{b \sinh (c+d x) \tanh ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.120791, size = 72, normalized size = 1.14 \[ \frac{a \sinh (c) \sinh (d x)}{d}+\frac{a \cosh (c) \cosh (d x)}{d}+\frac{b \sinh (c+d x) \tanh ^2(c+d x)}{d}-\frac{3 b \left (\tan ^{-1}(\sinh (c+d x))-\tanh (c+d x) \text{sech}(c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 85, normalized size = 1.4 \begin{align*}{\frac{a\cosh \left ( dx+c \right ) }{d}}+{\frac{b \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{b\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,b{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}-3\,{\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)*(a+b*tanh(d*x+c)^3),x)

[Out]

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

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Maxima [A]  time = 1.61063, size = 142, normalized size = 2.25 \begin{align*} \frac{1}{2} \, 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{a \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),x, algorithm="maxima")

[Out]

1/2*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)))) + a*cosh(d*x + c)/d

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Fricas [B]  time = 2.36054, size = 1473, normalized size = 23.38 \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),x, algorithm="fricas")

[Out]

1/2*((a + b)*cosh(d*x + c)^6 + 6*(a + b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a + b)*sinh(d*x + c)^6 + 3*(a + b)*c
osh(d*x + c)^4 + 3*(5*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c)^4 + 4*(5*(a + b)*cosh(d*x + c)^3 + 3*(a +
b)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a - b)*cosh(d*x + c)^2 + 3*(5*(a + b)*cosh(d*x + c)^4 + 6*(a + b)*cosh
(d*x + c)^2 + a - b)*sinh(d*x + c)^2 - 6*(b*cosh(d*x + c)^5 + 5*b*cosh(d*x + c)*sinh(d*x + c)^4 + b*sinh(d*x +
c)^5 + 2*b*cosh(d*x + c)^3 + 2*(5*b*cosh(d*x + c)^2 + b)*sinh(d*x + c)^3 + 2*(5*b*cosh(d*x + c)^3 + 3*b*cosh(
d*x + c))*sinh(d*x + c)^2 + b*cosh(d*x + c) + (5*b*cosh(d*x + c)^4 + 6*b*cosh(d*x + c)^2 + b)*sinh(d*x + c))*a
rctan(cosh(d*x + c) + sinh(d*x + c)) + 6*((a + b)*cosh(d*x + c)^5 + 2*(a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d
*x + c))*sinh(d*x + c) + a - b)/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5 + 2
*d*cosh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))
*sinh(d*x + c)^2 + d*cosh(d*x + c) + (5*d*cosh(d*x + c)^4 + 6*d*cosh(d*x + c)^2 + d)*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right ) \sinh{\left (c + d x \right )}\, dx \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),x)

[Out]

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

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Giac [A]  time = 1.21657, size = 128, normalized size = 2.03 \begin{align*} -\frac{6 \, b \arctan \left (e^{\left (d x + c\right )}\right ) -{\left (a - b\right )} e^{\left (-d x - c\right )} -{\left (a e^{\left (d x + 8 \, c\right )} + b e^{\left (d x + 8 \, c\right )}\right )} e^{\left (-7 \, c\right )} - \frac{2 \,{\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}}}{2 \, 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),x, algorithm="giac")

[Out]

-1/2*(6*b*arctan(e^(d*x + c)) - (a - b)*e^(-d*x - c) - (a*e^(d*x + 8*c) + b*e^(d*x + 8*c))*e^(-7*c) - 2*(b*e^(
3*d*x + 3*c) - b*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^2)/d