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

Optimal. Leaf size=47 \[ \frac{(a+b) \cosh ^3(c+d x)}{3 d}-\frac{(a+2 b) \cosh (c+d x)}{d}-\frac{b \text{sech}(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0546025, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3664, 448} \[ \frac{(a+b) \cosh ^3(c+d x)}{3 d}-\frac{(a+2 b) \cosh (c+d x)}{d}-\frac{b \text{sech}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.0503273, size = 73, normalized size = 1.55 \[ -\frac{3 a \cosh (c+d x)}{4 d}+\frac{a \cosh (3 (c+d x))}{12 d}-\frac{7 b \cosh (c+d x)}{4 d}+\frac{b \cosh (3 (c+d x))}{12 d}-\frac{b \text{sech}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*Cosh[c + d*x])/(4*d) - (7*b*Cosh[c + d*x])/(4*d) + (a*Cosh[3*(c + d*x)])/(12*d) + (b*Cosh[3*(c + d*x)])/
(12*d) - (b*Sech[c + d*x])/d

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Maple [A]  time = 0.034, size = 73, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( dx+c \right ) +b \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{3\,\cosh \left ( dx+c \right ) }}+{\frac{4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\,\cosh \left ( dx+c \right ) }}-{\frac{8\,\cosh \left ( dx+c \right ) }{3}} \right ) \right ) } \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)^2),x)

[Out]

1/d*(a*(-2/3+1/3*sinh(d*x+c)^2)*cosh(d*x+c)+b*(1/3*sinh(d*x+c)^4/cosh(d*x+c)+4/3*sinh(d*x+c)^2/cosh(d*x+c)-8/3
*cosh(d*x+c)))

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Maxima [B]  time = 1.15218, size = 184, normalized size = 3.91 \begin{align*} -\frac{1}{24} \, b{\left (\frac{21 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{20 \, e^{\left (-2 \, d x - 2 \, c\right )} + 69 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + e^{\left (-5 \, d x - 5 \, 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)^2),x, algorithm="maxima")

[Out]

-1/24*b*((21*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + (20*e^(-2*d*x - 2*c) + 69*e^(-4*d*x - 4*c) - 1)/(d*(e^(-3*d*
x - 3*c) + e^(-5*d*x - 5*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 = 1.99928, size = 246, normalized size = 5.23 \begin{align*} \frac{{\left (a + b\right )} \cosh \left (d x + c\right )^{4} +{\left (a + b\right )} \sinh \left (d x + c\right )^{4} - 4 \,{\left (2 \, a + 5 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{2} - 4 \, a - 10 \, b\right )} \sinh \left (d x + c\right )^{2} - 9 \, a - 45 \, b}{24 \, d \cosh \left (d x + c\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)^2),x, algorithm="fricas")

[Out]

1/24*((a + b)*cosh(d*x + c)^4 + (a + b)*sinh(d*x + c)^4 - 4*(2*a + 5*b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*
x + c)^2 - 4*a - 10*b)*sinh(d*x + c)^2 - 9*a - 45*b)/(d*cosh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \sinh ^{3}{\left (c + d x \right )}\, dx \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)**2),x)

[Out]

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

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Giac [B]  time = 1.30233, size = 162, normalized size = 3.45 \begin{align*} -\frac{{\left (9 \, a e^{\left (2 \, d x + 2 \, c\right )} + 21 \, 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 + 24 \, c\right )} + b e^{\left (3 \, d x + 24 \, c\right )} - 9 \, a e^{\left (d x + 22 \, c\right )} - 21 \, b e^{\left (d x + 22 \, c\right )}\right )} e^{\left (-21 \, c\right )} + \frac{48 \, b e^{\left (d x + c\right )}}{e^{\left (2 \, d x + 2 \, c\right )} + 1}}{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)^2),x, algorithm="giac")

[Out]

-1/24*((9*a*e^(2*d*x + 2*c) + 21*b*e^(2*d*x + 2*c) - a - b)*e^(-3*d*x - 3*c) - (a*e^(3*d*x + 24*c) + b*e^(3*d*
x + 24*c) - 9*a*e^(d*x + 22*c) - 21*b*e^(d*x + 22*c))*e^(-21*c) + 48*b*e^(d*x + c)/(e^(2*d*x + 2*c) + 1))/d