3.79 \(\int \sinh ^3(a+b x) \tanh ^3(a+b x) \, dx\)

Optimal. Leaf size=66 \[ \frac {5 \sinh ^3(a+b x)}{6 b}-\frac {5 \sinh (a+b x)}{2 b}-\frac {\sinh ^3(a+b x) \tanh ^2(a+b x)}{2 b}+\frac {5 \tan ^{-1}(\sinh (a+b x))}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2592, 288, 302, 203} \[ \frac {5 \sinh ^3(a+b x)}{6 b}-\frac {5 \sinh (a+b x)}{2 b}-\frac {\sinh ^3(a+b x) \tanh ^2(a+b x)}{2 b}+\frac {5 \tan ^{-1}(\sinh (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^3*Tanh[a + b*x]^3,x]

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 65, normalized size = 0.98 \[ \frac {2 \sinh ^3(a+b x) \tanh ^2(a+b x)+15 \tan ^{-1}(\sinh (a+b x))-10 \sinh (a+b x) \tanh ^2(a+b x)-15 \tanh (a+b x) \text {sech}(a+b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^3*Tanh[a + b*x]^3,x]

[Out]

(15*ArcTan[Sinh[a + b*x]] - 15*Sech[a + b*x]*Tanh[a + b*x] - 10*Sinh[a + b*x]*Tanh[a + b*x]^2 + 2*Sinh[a + b*x
]^3*Tanh[a + b*x]^2)/(6*b)

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fricas [B]  time = 0.45, size = 851, normalized size = 12.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(cosh(b*x + a)^10 + 10*cosh(b*x + a)*sinh(b*x + a)^9 + sinh(b*x + a)^10 + 5*(9*cosh(b*x + a)^2 - 5)*sinh(
b*x + a)^8 - 25*cosh(b*x + a)^8 + 40*(3*cosh(b*x + a)^3 - 5*cosh(b*x + a))*sinh(b*x + a)^7 + 10*(21*cosh(b*x +
a)^4 - 70*cosh(b*x + a)^2 - 5)*sinh(b*x + a)^6 - 50*cosh(b*x + a)^6 + 4*(63*cosh(b*x + a)^5 - 350*cosh(b*x +
a)^3 - 75*cosh(b*x + a))*sinh(b*x + a)^5 + 10*(21*cosh(b*x + a)^6 - 175*cosh(b*x + a)^4 - 75*cosh(b*x + a)^2 +
5)*sinh(b*x + a)^4 + 50*cosh(b*x + a)^4 + 40*(3*cosh(b*x + a)^7 - 35*cosh(b*x + a)^5 - 25*cosh(b*x + a)^3 + 5
*cosh(b*x + a))*sinh(b*x + a)^3 + 5*(9*cosh(b*x + a)^8 - 140*cosh(b*x + a)^6 - 150*cosh(b*x + a)^4 + 60*cosh(b
*x + a)^2 + 5)*sinh(b*x + a)^2 + 120*(cosh(b*x + a)^7 + 7*cosh(b*x + a)*sinh(b*x + a)^6 + sinh(b*x + a)^7 + (2
1*cosh(b*x + a)^2 + 2)*sinh(b*x + a)^5 + 2*cosh(b*x + a)^5 + 5*(7*cosh(b*x + a)^3 + 2*cosh(b*x + a))*sinh(b*x
+ a)^4 + (35*cosh(b*x + a)^4 + 20*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + cosh(b*x + a)^3 + (21*cosh(b*x + a)^5
+ 20*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 + (7*cosh(b*x + a)^6 + 10*cosh(b*x + a)^4 + 3*cosh(b*
x + a)^2)*sinh(b*x + a))*arctan(cosh(b*x + a) + sinh(b*x + a)) + 25*cosh(b*x + a)^2 + 10*(cosh(b*x + a)^9 - 20
*cosh(b*x + a)^7 - 30*cosh(b*x + a)^5 + 20*cosh(b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a) - 1)/(b*cosh(b*x +
a)^7 + 7*b*cosh(b*x + a)*sinh(b*x + a)^6 + b*sinh(b*x + a)^7 + 2*b*cosh(b*x + a)^5 + (21*b*cosh(b*x + a)^2 +
2*b)*sinh(b*x + a)^5 + 5*(7*b*cosh(b*x + a)^3 + 2*b*cosh(b*x + a))*sinh(b*x + a)^4 + b*cosh(b*x + a)^3 + (35*b
*cosh(b*x + a)^4 + 20*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^3 + (21*b*cosh(b*x + a)^5 + 20*b*cosh(b*x + a)^3 +
3*b*cosh(b*x + a))*sinh(b*x + a)^2 + (7*b*cosh(b*x + a)^6 + 10*b*cosh(b*x + a)^4 + 3*b*cosh(b*x + a)^2)*sinh(b
*x + a))

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giac [A]  time = 0.18, size = 96, normalized size = 1.45 \[ \frac {{\left (27 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} + {\left (e^{\left (3 \, b x + 30 \, a\right )} - 27 \, e^{\left (b x + 28 \, a\right )}\right )} e^{\left (-27 \, a\right )} - \frac {24 \, {\left (e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} + 120 \, \arctan \left (e^{\left (b x + a\right )}\right )}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.34, size = 92, normalized size = 1.39 \[ \frac {\sinh ^{5}\left (b x +a \right )}{3 b \cosh \left (b x +a \right )^{2}}-\frac {5 \left (\sinh ^{3}\left (b x +a \right )\right )}{3 b \cosh \left (b x +a \right )^{2}}-\frac {5 \sinh \left (b x +a \right )}{b \cosh \left (b x +a \right )^{2}}+\frac {5 \,\mathrm {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2 b}+\frac {5 \arctan \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 116, normalized size = 1.76 \[ \frac {27 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} - \frac {5 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac {25 \, e^{\left (-2 \, b x - 2 \, a\right )} + 77 \, e^{\left (-4 \, b x - 4 \, a\right )} + 3 \, e^{\left (-6 \, b x - 6 \, a\right )} - 1}{24 \, b {\left (e^{\left (-3 \, b x - 3 \, a\right )} + 2 \, e^{\left (-5 \, b x - 5 \, a\right )} + e^{\left (-7 \, b x - 7 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(27*e^(-b*x - a) - e^(-3*b*x - 3*a))/b - 5*arctan(e^(-b*x - a))/b - 1/24*(25*e^(-2*b*x - 2*a) + 77*e^(-4*
b*x - 4*a) + 3*e^(-6*b*x - 6*a) - 1)/(b*(e^(-3*b*x - 3*a) + 2*e^(-5*b*x - 5*a) + e^(-7*b*x - 7*a)))

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mupad [B]  time = 1.52, size = 136, normalized size = 2.06 \[ \frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {9\,{\mathrm {e}}^{a+b\,x}}{8\,b}+\frac {9\,{\mathrm {e}}^{-a-b\,x}}{8\,b}-\frac {{\mathrm {e}}^{-3\,a-3\,b\,x}}{24\,b}+\frac {{\mathrm {e}}^{3\,a+3\,b\,x}}{24\,b}+\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh ^{3}{\left (a + b x \right )} \tanh ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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