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

Optimal. Leaf size=78 \[ \frac {b^2 (3 a+2 b) \tanh (c+d x)}{d}+\frac {(a+b)^3 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {1}{2} x (a-5 b) (a+b)^2+\frac {b^3 \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3675, 390, 385, 206} \[ \frac {b^2 (3 a+2 b) \tanh (c+d x)}{d}+\frac {(a+b)^3 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {1}{2} x (a-5 b) (a+b)^2+\frac {b^3 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - 5*b)*(a + b)^2*x)/2 + ((a + b)^3*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) + (b^2*(3*a + 2*b)*Tanh[c + d*x])/d
+ (b^3*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 385

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

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

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Mathematica [A]  time = 0.97, size = 69, normalized size = 0.88 \[ \frac {4 b^2 \tanh (c+d x) \left (9 a-b \text {sech}^2(c+d x)+7 b\right )+6 (a-5 b) (a+b)^2 (c+d x)+3 (a+b)^3 \sinh (2 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(a - 5*b)*(a + b)^2*(c + d*x) + 3*(a + b)^3*Sinh[2*(c + d*x)] + 4*b^2*(9*a + 7*b - b*Sech[c + d*x]^2)*Tanh[
c + d*x])/(12*d)

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fricas [B]  time = 0.41, size = 369, normalized size = 4.73 \[ \frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{5} - 4 \, {\left (18 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (18 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (9 \, a^{3} + 27 \, a^{2} b + 99 \, a b^{2} + 65 \, b^{3} + 30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (18 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 2 \, a^{3} + 6 \, a^{2} b + 30 \, a b^{2} + 10 \, b^{3} + {\left (9 \, a^{3} + 27 \, a^{2} b + 99 \, a b^{2} + 65 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/24*(3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^5 - 4*(18*a*b^2 + 14*b^3 - 3*(a^3 - 3*a^2*b - 9*a*b^2 -
5*b^3)*d*x)*cosh(d*x + c)^3 - 12*(18*a*b^2 + 14*b^3 - 3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c)*s
inh(d*x + c)^2 + (9*a^3 + 27*a^2*b + 99*a*b^2 + 65*b^3 + 30*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*s
inh(d*x + c)^3 - 12*(18*a*b^2 + 14*b^3 - 3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c) + 3*(5*(a^3 +
3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 2*a^3 + 6*a^2*b + 30*a*b^2 + 10*b^3 + (9*a^3 + 27*a^2*b + 99*a*b^2
+ 65*b^3)*cosh(d*x + c)^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 = 0.60, size = 267, normalized size = 3.42 \[ \frac {12 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x - 3 \, {\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, {\left (a^{3} e^{\left (2 \, d x + 12 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, d x + 12 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 12 \, c\right )} + b^{3} e^{\left (2 \, d x + 12 \, c\right )}\right )} e^{\left (-10 \, c\right )} - \frac {16 \, {\left (9 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a b^{2} + 7 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/24*(12*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*d*x - 3*(2*a^3*e^(2*d*x + 2*c) - 6*a^2*b*e^(2*d*x + 2*c) - 18*a*b^2
*e^(2*d*x + 2*c) - 10*b^3*e^(2*d*x + 2*c) + a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-2*d*x - 2*c) + 3*(a^3*e^(2*d*x
+ 12*c) + 3*a^2*b*e^(2*d*x + 12*c) + 3*a*b^2*e^(2*d*x + 12*c) + b^3*e^(2*d*x + 12*c))*e^(-10*c) - 16*(9*a*b^2*
e^(4*d*x + 4*c) + 9*b^3*e^(4*d*x + 4*c) + 18*a*b^2*e^(2*d*x + 2*c) + 12*b^3*e^(2*d*x + 2*c) + 9*a*b^2 + 7*b^3)
/(e^(2*d*x + 2*c) + 1)^3)/d

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maple [B]  time = 0.24, size = 148, normalized size = 1.90 \[ \frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a \,b^{2} \left (\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+b^{3} \left (\frac {\sinh ^{5}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \left (\tanh ^{3}\left (d x +c \right )\right )}{6}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/2*cosh(d*x+c)*sinh(d*x+c)+1/2*d*x+1/2*c)+3*a^2*b*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+3*a*b
^2*(1/2*sinh(d*x+c)^3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*tanh(d*x+c))+b^3*(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 = 0.41, size = 256, normalized size = 3.28 \[ \frac {1}{8} \, a^{3} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {3}{8} \, a^{2} b {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{24} \, b^{3} {\left (\frac {60 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} - \frac {3}{8} \, a b^{2} {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/8*a^3*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) - 3/8*a^2*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)
/d) - 1/24*b^3*(60*(d*x + c)/d + 3*e^(-2*d*x - 2*c)/d - (121*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)))) - 3/8*
a*b^2*(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 = 0.29, size = 243, normalized size = 3.12 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^3}{8\,d}-\frac {\frac {2\,\left (b^3+3\,a\,b^2\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^3+a\,b^2\right )}{d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {2\,\left (b^3+a\,b^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^3}{8\,d}-\frac {\frac {2\,\left (b^3+a\,b^2\right )}{d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^3+3\,a\,b^2\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^3+a\,b^2\right )}{d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {x\,{\left (a+b\right )}^2\,\left (a-5\,b\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**2*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

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

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