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

Optimal. Leaf size=79 \[ \frac {(a+b) (a+5 b) \tanh (c+d x)}{2 d}+\frac {(a+b)^2 \sinh ^2(c+d x) \tanh (c+d x)}{2 d}-\frac {1}{2} x (a+b) (a+5 b)+\frac {b^2 \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3663, 463, 459, 321, 206} \[ \frac {(a+b) (a+5 b) \tanh (c+d x)}{2 d}+\frac {(a+b)^2 \sinh ^2(c+d x) \tanh (c+d x)}{2 d}-\frac {1}{2} x (a+b) (a+5 b)+\frac {b^2 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 3663

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

Rubi steps

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

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Mathematica [A]  time = 0.90, size = 70, normalized size = 0.89 \[ \frac {-6 \left (a^2+6 a b+5 b^2\right ) (c+d x)+3 (a+b)^2 \sinh (2 (c+d x))+4 b \tanh (c+d x) \left (6 a-b \text {sech}^2(c+d x)+7 b\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(a^2 + 6*a*b + 5*b^2)*(c + d*x) + 3*(a + b)^2*Sinh[2*(c + d*x)] + 4*b*(6*a + 7*b - b*Sech[c + d*x]^2)*Tanh
[c + d*x])/(12*d)

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fricas [B]  time = 0.48, size = 291, normalized size = 3.68 \[ \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{5} - 4 \, {\left (3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} d x + 12 \, a b + 14 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} d x + 12 \, a b + 14 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} + 66 \, a b + 65 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} d x + 12 \, a b + 14 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + {\left (9 \, a^{2} + 66 \, a b + 65 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, a^{2} + 20 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/24*(3*(a^2 + 2*a*b + b^2)*sinh(d*x + c)^5 - 4*(3*(a^2 + 6*a*b + 5*b^2)*d*x + 12*a*b + 14*b^2)*cosh(d*x + c)^
3 - 12*(3*(a^2 + 6*a*b + 5*b^2)*d*x + 12*a*b + 14*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + (30*(a^2 + 2*a*b + b^2)
*cosh(d*x + c)^2 + 9*a^2 + 66*a*b + 65*b^2)*sinh(d*x + c)^3 - 12*(3*(a^2 + 6*a*b + 5*b^2)*d*x + 12*a*b + 14*b^
2)*cosh(d*x + c) + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + (9*a^2 + 66*a*b + 65*b^2)*cosh(d*x + c)^2 + 2*a^
2 + 20*a*b + 10*b^2)*sinh(d*x + c))/(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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giac [B]  time = 0.33, size = 215, normalized size = 2.72 \[ -\frac {12 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} d x - 3 \, {\left (2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2} - 2 \, a b - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, {\left (a^{2} e^{\left (2 \, d x + 12 \, c\right )} + 2 \, a b e^{\left (2 \, d x + 12 \, c\right )} + b^{2} e^{\left (2 \, d x + 12 \, c\right )}\right )} e^{\left (-10 \, c\right )} + \frac {16 \, {\left (6 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b + 7 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-1/24*(12*(a^2 + 6*a*b + 5*b^2)*d*x - 3*(2*a^2*e^(2*d*x + 2*c) + 12*a*b*e^(2*d*x + 2*c) + 10*b^2*e^(2*d*x + 2*
c) - a^2 - 2*a*b - b^2)*e^(-2*d*x - 2*c) - 3*(a^2*e^(2*d*x + 12*c) + 2*a*b*e^(2*d*x + 12*c) + b^2*e^(2*d*x + 1
2*c))*e^(-10*c) + 16*(6*a*b*e^(4*d*x + 4*c) + 9*b^2*e^(4*d*x + 4*c) + 12*a*b*e^(2*d*x + 2*c) + 12*b^2*e^(2*d*x
+ 2*c) + 6*a*b + 7*b^2)/(e^(2*d*x + 2*c) + 1)^3)/d

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maple [A]  time = 0.21, size = 118, normalized size = 1.49 \[ \frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+2 a b \left (\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (\frac {\sinh ^{5}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \left (\tanh ^{3}\left (d x +c \right )\right )}{6}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.35, size = 217, normalized size = 2.75 \[ -\frac {1}{8} \, a^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{24} \, b^{2} {\left (\frac {60 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} - \frac {1}{4} \, a b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

-1/8*a^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/24*b^2*(60*(d*x + c)/d + 3*e^(-2*d*x - 2*c)/d - (1
21*e^(-2*d*x - 2*c) + 201*e^(-4*d*x - 4*c) + 147*e^(-6*d*x - 6*c) + 3)/(d*(e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*
c) + 3*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c)))) - 1/4*a*b*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (17*e^(-2*d*x -
2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c))))

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mupad [B]  time = 1.18, size = 248, normalized size = 3.14 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{8\,d}-x\,\left (\frac {a^2}{2}+3\,a\,b+\frac {5\,b^2}{2}\right )-\frac {\frac {2\,\left (3\,b^2+2\,a\,b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,b^2+2\,a\,b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,\left (3\,b^2+2\,a\,b\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^2}{8\,d}-\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,b^2+2\,a\,b\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2*c + 2*d*x)*(a + b)^2)/(8*d) - x*(3*a*b + a^2/2 + (5*b^2)/2) - ((2*(2*a*b + 3*b^2))/(3*d) + (4*exp(2*c +
2*d*x)*(2*a*b + b^2))/(3*d) + (2*exp(4*c + 4*d*x)*(2*a*b + 3*b^2))/(3*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4
*d*x) + exp(6*c + 6*d*x) + 1) - (2*(2*a*b + 3*b^2))/(3*d*(exp(2*c + 2*d*x) + 1)) - (exp(- 2*c - 2*d*x)*(a + b)
^2)/(8*d) - ((2*(2*a*b + b^2))/(3*d) + (2*exp(2*c + 2*d*x)*(2*a*b + 3*b^2))/(3*d))/(2*exp(2*c + 2*d*x) + exp(4
*c + 4*d*x) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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