\(\int \sinh ^3(c+d x) (a+b \tanh ^3(c+d x)) \, dx\) [50]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {5 b \arctan (\sinh (c+d x))}{2 d}-\frac {a \cosh (c+d x)}{d}+\frac {a \cosh ^3(c+d x)}{3 d}-\frac {5 b \sinh (c+d x)}{2 d}+\frac {5 b \sinh ^3(c+d x)}{6 d}-\frac {b \sinh ^3(c+d x) \tanh ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3747, 2713, 2672, 294, 308, 209} \[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {a \cosh ^3(c+d x)}{3 d}-\frac {a \cosh (c+d x)}{d}+\frac {5 b \arctan (\sinh (c+d x))}{2 d}+\frac {5 b \sinh ^3(c+d x)}{6 d}-\frac {5 b \sinh (c+d x)}{2 d}-\frac {b \sinh ^3(c+d x) \tanh ^2(c+d x)}{2 d} \]

[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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 294

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 308

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 2672

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 2713

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 3747

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]

Rubi steps \begin{align*} \text {integral}& = 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 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (c+d x)\right )}{d}+\frac {b \text {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) \text {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) \text {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) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{2 d} \\ & = \frac {5 b \arctan (\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] (verified)

Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14 \[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {5 b \arctan (\sinh (c+d x))}{2 d}-\frac {3 a \cosh (c+d x)}{4 d}+\frac {a \cosh (3 (c+d x))}{12 d}-\frac {5 b \text {sech}(c+d x) \tanh (c+d x)}{2 d}-\frac {5 b \sinh (c+d x) \tanh ^2(c+d x)}{3 d}+\frac {b \sinh ^3(c+d x) \tanh ^2(c+d x)}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {a \left (-\frac {2}{3}+\frac {\sinh \left (d x +c \right )^{2}}{3}\right ) \cosh \left (d x +c \right )+b \left (\frac {\sinh \left (d x +c \right )^{5}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {5 \,\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+5 \arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}\)	\(104\)
default	\(\frac {a \left (-\frac {2}{3}+\frac {\sinh \left (d x +c \right )^{2}}{3}\right ) \cosh \left (d x +c \right )+b \left (\frac {\sinh \left (d x +c \right )^{5}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{2}}-\frac {5 \sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {5 \,\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+5 \arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}\)	\(104\)
risch	\(\frac {{\mathrm e}^{3 d x +3 c} a}{24 d}+\frac {{\mathrm e}^{3 d x +3 c} b}{24 d}-\frac {3 \,{\mathrm e}^{d x +c} a}{8 d}-\frac {9 \,{\mathrm e}^{d x +c} b}{8 d}-\frac {3 \,{\mathrm e}^{-d x -c} a}{8 d}+\frac {9 \,{\mathrm e}^{-d x -c} b}{8 d}+\frac {{\mathrm e}^{-3 d x -3 c} a}{24 d}-\frac {{\mathrm e}^{-3 d x -3 c} b}{24 d}-\frac {b \,{\mathrm e}^{d x +c} \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}+\frac {5 i b \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 d}-\frac {5 i b \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 d}\)	\(186\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1070 vs. \(2 (88) = 176\).

Time = 0.28 (sec) , antiderivative size = 1070, normalized size of antiderivative = 10.92 \[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\text {Too large to display} \]

[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))

Sympy [F]

\[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right ) \sinh ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.78 \[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {1}{24} \, b {\left (\frac {27 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} - \frac {120 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {25 \, e^{\left (-2 \, d x - 2 \, c\right )} + 77 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d {\left (e^{\left (-3 \, d x - 3 \, c\right )} + 2 \, e^{\left (-5 \, d x - 5 \, c\right )} + e^{\left (-7 \, d x - 7 \, c\right )}\right )}}\right )} + \frac {1}{24} \, a {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} + \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.35 \[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {120 \, b \arctan \left (e^{\left (d x + c\right )}\right ) + a e^{\left (3 \, d x + 3 \, c\right )} + b e^{\left (3 \, d x + 3 \, c\right )} - 9 \, a e^{\left (d x + c\right )} - 27 \, b e^{\left (d x + c\right )} - {\left (9 \, a e^{\left (2 \, d x + 2 \, c\right )} - 27 \, b e^{\left (2 \, d x + 2 \, c\right )} - a + b\right )} e^{\left (-3 \, d x - 3 \, c\right )} - \frac {24 \, {\left (b e^{\left (3 \, d x + 3 \, c\right )} - b e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{24 \, d} \]

[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)) + a*e^(3*d*x + 3*c) + b*e^(3*d*x + 3*c) - 9*a*e^(d*x + c) - 27*b*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) - 24*(b*e^(3*d*x + 3*c) - b*e^(d*x + c)
)/(e^(2*d*x + 2*c) + 1)^2)/d

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.74 \[ \int \sinh ^3(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (a+b\right )}{24\,d}+\frac {5\,\mathrm {atan}\left (\frac {b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {d^2}}{d\,\sqrt {b^2}}\right )\,\sqrt {b^2}}{\sqrt {d^2}}+\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,\left (a-b\right )}{24\,d}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (3\,a+9\,b\right )}{8\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (3\,a-9\,b\right )}{8\,d}-\frac {b\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]

[In]

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

[Out]

(exp(3*c + 3*d*x)*(a + b))/(24*d) + (5*atan((b*exp(d*x)*exp(c)*(d^2)^(1/2))/(d*(b^2)^(1/2)))*(b^2)^(1/2))/(d^2
)^(1/2) + (exp(- 3*c - 3*d*x)*(a - b))/(24*d) - (exp(c + d*x)*(3*a + 9*b))/(8*d) - (exp(- c - d*x)*(3*a - 9*b)
)/(8*d) - (b*exp(c + d*x))/(d*(exp(2*c + 2*d*x) + 1)) + (2*b*exp(c + d*x))/(d*(2*exp(2*c + 2*d*x) + exp(4*c +
4*d*x) + 1))