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

Optimal. Leaf size=77 \[ \frac{(a+b)^2 \cosh ^3(c+d x)}{3 d}-\frac{(a+b) (a+3 b) \cosh (c+d x)}{d}-\frac{b (2 a+3 b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0978984, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3664, 448} \[ \frac{(a+b)^2 \cosh ^3(c+d x)}{3 d}-\frac{(a+b) (a+3 b) \cosh (c+d x)}{d}-\frac{b (2 a+3 b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.516529, size = 71, normalized size = 0.92 \[ \frac{-3 \left (3 a^2+14 a b+11 b^2\right ) \cosh (c+d x)+(a+b)^2 \cosh (3 (c+d x))+4 b \text{sech}(c+d x) \left (-6 a+b \text{sech}^2(c+d x)-9 b\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(3*a^2 + 14*a*b + 11*b^2)*Cosh[c + d*x] + (a + b)^2*Cosh[3*(c + d*x)] + 4*b*Sech[c + d*x]*(-6*a - 9*b + b*
Sech[c + d*x]^2))/(12*d)

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Maple [B]  time = 0.052, size = 162, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( dx+c \right ) +2\,ab \left ( 1/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{\cosh \left ( dx+c \right ) }}+4/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{\cosh \left ( dx+c \right ) }}-8/3\,\cosh \left ( dx+c \right ) \right ) +{b}^{2} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-{\frac{8\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{16\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\,\cosh \left ( dx+c \right ) }}-{\frac{16\,\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)^2,x)

[Out]

1/d*(a^2*(-2/3+1/3*sinh(d*x+c)^2)*cosh(d*x+c)+2*a*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))+b^2*(1/3*sinh(d*x+c)^6/cosh(d*x+c)^3-2*sinh(d*x+c)^4/cosh(d*x+c)^3-8/3*sinh(d*x+c)^2/cosh(
d*x+c)^3+16/3*sinh(d*x+c)^2/cosh(d*x+c)-16/3*cosh(d*x+c)))

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Maxima [B]  time = 1.19186, size = 358, normalized size = 4.65 \begin{align*} -\frac{1}{24} \, b^{2}{\left (\frac{33 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{30 \, e^{\left (-2 \, d x - 2 \, c\right )} + 240 \, e^{\left (-4 \, d x - 4 \, c\right )} + 322 \, e^{\left (-6 \, d x - 6 \, c\right )} + 177 \, e^{\left (-8 \, d x - 8 \, c\right )} - 1}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} + 3 \, e^{\left (-7 \, d x - 7 \, c\right )} + e^{\left (-9 \, d x - 9 \, c\right )}\right )}}\right )} - \frac{1}{12} \, a 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^{2}{\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)^2,x, algorithm="maxima")

[Out]

-1/24*b^2*((33*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + (30*e^(-2*d*x - 2*c) + 240*e^(-4*d*x - 4*c) + 322*e^(-6*d*
x - 6*c) + 177*e^(-8*d*x - 8*c) - 1)/(d*(e^(-3*d*x - 3*c) + 3*e^(-5*d*x - 5*c) + 3*e^(-7*d*x - 7*c) + e^(-9*d*
x - 9*c)))) - 1/12*a*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^2*(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 = 2.05485, size = 664, normalized size = 8.62 \begin{align*} \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{6} +{\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{6} - 6 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \,{\left (5 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} - 2 \, a^{2} - 12 \, a b - 10 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} - 3 \,{\left (11 \, a^{2} + 86 \, a b + 91 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 3 \,{\left (5 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} - 12 \,{\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 11 \, a^{2} - 86 \, a b - 91 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} - 26 \, a^{2} - 220 \, a b - 210 \, b^{2}}{24 \,{\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\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)^2,x, algorithm="fricas")

[Out]

1/24*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^6 - 6*(a^2 + 6*a*b + 5*b^2)*cosh
(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - 2*a^2 - 12*a*b - 10*b^2)*sinh(d*x + c)^4 - 3*(11*a^2
+ 86*a*b + 91*b^2)*cosh(d*x + c)^2 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 - 12*(a^2 + 6*a*b + 5*b^2)*cosh(
d*x + c)^2 - 11*a^2 - 86*a*b - 91*b^2)*sinh(d*x + c)^2 - 26*a^2 - 220*a*b - 210*b^2)/(d*cosh(d*x + c)^3 + 3*d*
cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(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)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.64637, size = 392, normalized size = 5.09 \begin{align*} \frac{{\left (a^{2} e^{\left (3 \, d x + 36 \, c\right )} + 2 \, a b e^{\left (3 \, d x + 36 \, c\right )} + b^{2} e^{\left (3 \, d x + 36 \, c\right )} - 9 \, a^{2} e^{\left (d x + 34 \, c\right )} - 42 \, a b e^{\left (d x + 34 \, c\right )} - 33 \, b^{2} e^{\left (d x + 34 \, c\right )}\right )} e^{\left (-33 \, c\right )} - \frac{{\left (9 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 138 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 177 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 26 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 316 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 322 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 24 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 216 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 240 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 36 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 30 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2} - 2 \, a b - b^{2}\right )} e^{\left (-3 \, c\right )}}{{\left (e^{\left (3 \, d x + 2 \, c\right )} + e^{\left (d x\right )}\right )}^{3}}}{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)^2,x, algorithm="giac")

[Out]

1/24*((a^2*e^(3*d*x + 36*c) + 2*a*b*e^(3*d*x + 36*c) + b^2*e^(3*d*x + 36*c) - 9*a^2*e^(d*x + 34*c) - 42*a*b*e^
(d*x + 34*c) - 33*b^2*e^(d*x + 34*c))*e^(-33*c) - (9*a^2*e^(8*d*x + 8*c) + 138*a*b*e^(8*d*x + 8*c) + 177*b^2*e
^(8*d*x + 8*c) + 26*a^2*e^(6*d*x + 6*c) + 316*a*b*e^(6*d*x + 6*c) + 322*b^2*e^(6*d*x + 6*c) + 24*a^2*e^(4*d*x
+ 4*c) + 216*a*b*e^(4*d*x + 4*c) + 240*b^2*e^(4*d*x + 4*c) + 6*a^2*e^(2*d*x + 2*c) + 36*a*b*e^(2*d*x + 2*c) +
30*b^2*e^(2*d*x + 2*c) - a^2 - 2*a*b - b^2)*e^(-3*c)/(e^(3*d*x + 2*c) + e^(d*x))^3)/d