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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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]

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]

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 = 0.54, 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)

________________________________________________________________________________________

fricas [B]  time = 0.52, size = 259, normalized size = 3.36 \[ \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{6} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{6} - 6 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} - 2 \, a^{2} - 12 \, a b - 10 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} - 3 \, {\left (11 \, a^{2} + 86 \, a b + 91 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} - 12 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 11 \, a^{2} - 86 \, a b - 91 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} - 26 \, a^{2} - 220 \, a b - 210 \, b^{2}}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]

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))

________________________________________________________________________________________

giac [B]  time = 0.42, size = 290, normalized size = 3.77 \[ \frac {{\left (a^{2} e^{\left (3 \, d x + 36 \, c\right )} + 2 \, a b e^{\left (3 \, d x + 36 \, c\right )} + b^{2} e^{\left (3 \, d x + 36 \, c\right )} - 9 \, a^{2} e^{\left (d x + 34 \, c\right )} - 42 \, a b e^{\left (d x + 34 \, c\right )} - 33 \, b^{2} e^{\left (d x + 34 \, c\right )}\right )} e^{\left (-33 \, c\right )} - \frac {{\left (9 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 138 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 177 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 26 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 316 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 322 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 24 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 216 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 240 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 36 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 30 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2} - 2 \, a b - b^{2}\right )} e^{\left (-3 \, c\right )}}{{\left (e^{\left (3 \, d x + 2 \, c\right )} + e^{\left (d x\right )}\right )}^{3}}}{24 \, d} \]

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

________________________________________________________________________________________

maple [B]  time = 0.29, size = 148, normalized size = 1.92 \[ \frac {a^{2} \left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\left (d x +c \right )\right )}{3}\right ) \cosh \left (d x +c \right )+2 a b \left (\frac {\sinh ^{4}\left (d x +c \right )}{3 \cosh \left (d x +c \right )}-\frac {4 \left (\sinh ^{2}\left (d x +c \right )\right )}{3 \cosh \left (d x +c \right )}-\frac {8}{3 \cosh \left (d x +c \right )}\right )+b^{2} \left (\frac {\sinh ^{6}\left (d x +c \right )}{3 \cosh \left (d x +c \right )^{3}}-\frac {2 \left (\sinh ^{4}\left (d x +c \right )\right )}{\cosh \left (d x +c \right )^{3}}-\frac {8 \left (\sinh ^{2}\left (d x +c \right )\right )}{\cosh \left (d x +c \right )^{3}}-\frac {16}{3 \cosh \left (d x +c \right )^{3}}\right )}{d} \]

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*sinh(d*x+c)^2/cosh(d*
x+c)^3-16/3/cosh(d*x+c)^3))

________________________________________________________________________________________

maxima [B]  time = 0.37, size = 265, normalized size = 3.44 \[ -\frac {1}{24} \, b^{2} {\left (\frac {33 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {30 \, e^{\left (-2 \, d x - 2 \, c\right )} + 240 \, e^{\left (-4 \, d x - 4 \, c\right )} + 322 \, e^{\left (-6 \, d x - 6 \, c\right )} + 177 \, e^{\left (-8 \, d x - 8 \, c\right )} - 1}{d {\left (e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} + 3 \, e^{\left (-7 \, d x - 7 \, c\right )} + e^{\left (-9 \, d x - 9 \, c\right )}\right )}}\right )} - \frac {1}{12} \, a b {\left (\frac {21 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {20 \, e^{\left (-2 \, d x - 2 \, c\right )} + 69 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d {\left (e^{\left (-3 \, d x - 3 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )}\right )}}\right )} + \frac {1}{24} \, a^{2} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} + \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.29, size = 215, normalized size = 2.79 \[ \frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,{\left (a+b\right )}^2}{24\,d}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (3\,a^2+14\,a\,b+11\,b^2\right )}{8\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,{\left (a+b\right )}^2}{24\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (3\,a^2+14\,a\,b+11\,b^2\right )}{8\,d}-\frac {8\,b^2\,{\mathrm {e}}^{c+d\,x}}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (3\,b^2+2\,a\,b\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {8\,b^2\,{\mathrm {e}}^{c+d\,x}}{3\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(- 3*c - 3*d*x)*(a + b)^2)/(24*d) - (exp(c + d*x)*(14*a*b + 3*a^2 + 11*b^2))/(8*d) + (exp(3*c + 3*d*x)*(a
+ b)^2)/(24*d) - (exp(- c - d*x)*(14*a*b + 3*a^2 + 11*b^2))/(8*d) - (8*b^2*exp(c + d*x))/(3*d*(3*exp(2*c + 2*d
*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - (2*exp(c + d*x)*(2*a*b + 3*b^2))/(d*(exp(2*c + 2*d*x) + 1)
) + (8*b^2*exp(c + d*x))/(3*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \sinh ^{3}{\left (c + d x \right )}\, dx \]

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]

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

________________________________________________________________________________________