3.67 \(\int \cosh ^2(c+d x) (a+b \text {sech}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=72 \[ \frac {a^3 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {1}{2} a^2 x (a+6 b)+\frac {b^2 (3 a+b) \tanh (c+d x)}{d}-\frac {b^3 \tanh ^3(c+d x)}{3 d} \]

[Out]

1/2*a^2*(a+6*b)*x+1/2*a^3*cosh(d*x+c)*sinh(d*x+c)/d+b^2*(3*a+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 = 72, 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 = {4146, 390, 385, 206} \[ \frac {1}{2} a^2 x (a+6 b)+\frac {a^3 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {b^2 (3 a+b) \tanh (c+d x)}{d}-\frac {b^3 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a + 6*b)*x)/2 + (a^3*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) + (b^2*(3*a + 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 4146

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

Rubi steps

\begin {align*} \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-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+b)-b^3 x^2+\frac {a^2 (a+3 b)-3 a^2 b x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^2 (3 a+b) \tanh (c+d x)}{d}-\frac {b^3 \tanh ^3(c+d x)}{3 d}+\frac {\operatorname {Subst}\left (\int \frac {a^2 (a+3 b)-3 a^2 b x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {a^3 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 (3 a+b) \tanh (c+d x)}{d}-\frac {b^3 \tanh ^3(c+d x)}{3 d}+\frac {\left (a^2 (a+6 b)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac {1}{2} a^2 (a+6 b) x+\frac {a^3 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 (3 a+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 = 0.51, size = 64, normalized size = 0.89 \[ \frac {3 a^3 \sinh (2 (c+d x))+6 a^2 (a+6 b) (c+d x)+4 b^2 \tanh (c+d x) \left (9 a+b \text {sech}^2(c+d x)+2 b\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*(a + 6*b)*(c + d*x) + 3*a^3*Sinh[2*(c + d*x)] + 4*b^2*(9*a + 2*b + b*Sech[c + d*x]^2)*Tanh[c + d*x])/(1
2*d)

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fricas [B]  time = 0.43, size = 270, normalized size = 3.75 \[ \frac {3 \, a^{3} \sinh \left (d x + c\right )^{5} - 4 \, {\left (18 \, a b^{2} + 4 \, b^{3} - 3 \, {\left (a^{3} + 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (18 \, a b^{2} + 4 \, b^{3} - 3 \, {\left (a^{3} + 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (30 \, a^{3} \cosh \left (d x + c\right )^{2} + 9 \, a^{3} + 72 \, a b^{2} + 16 \, b^{3}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (18 \, a b^{2} + 4 \, b^{3} - 3 \, {\left (a^{3} + 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, a^{3} \cosh \left (d x + c\right )^{4} + 2 \, a^{3} + 24 \, a b^{2} + 16 \, b^{3} + {\left (9 \, a^{3} + 72 \, a b^{2} + 16 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*a^3*sinh(d*x + c)^5 - 4*(18*a*b^2 + 4*b^3 - 3*(a^3 + 6*a^2*b)*d*x)*cosh(d*x + c)^3 - 12*(18*a*b^2 + 4*
b^3 - 3*(a^3 + 6*a^2*b)*d*x)*cosh(d*x + c)*sinh(d*x + c)^2 + (30*a^3*cosh(d*x + c)^2 + 9*a^3 + 72*a*b^2 + 16*b
^3)*sinh(d*x + c)^3 - 12*(18*a*b^2 + 4*b^3 - 3*(a^3 + 6*a^2*b)*d*x)*cosh(d*x + c) + 3*(5*a^3*cosh(d*x + c)^4 +
2*a^3 + 24*a*b^2 + 16*b^3 + (9*a^3 + 72*a*b^2 + 16*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 = 0.20, size = 152, normalized size = 2.11 \[ \frac {3 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, {\left (a^{3} + 6 \, a^{2} b\right )} {\left (d x + c\right )} - 3 \, {\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - \frac {16 \, {\left (9 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a b^{2} + 2 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.46, size = 77, normalized size = 1.07 \[ \frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \left (d x +c \right )+3 a \,b^{2} \tanh \left (d x +c \right )+b^{3} \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [B]  time = 0.32, size = 160, normalized size = 2.22 \[ \frac {1}{8} \, a^{3} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 3 \, a^{2} b x + \frac {4}{3} \, b^{3} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {6 \, a b^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*sech(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*a^2*b*x + 4/3*b^3*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d
*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-
6*d*x - 6*c) + 1))) + 6*a*b^2/(d*(e^(-2*d*x - 2*c) + 1))

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mupad [B]  time = 0.17, size = 221, normalized size = 3.07 \[ \frac {a^2\,x\,\left (a+6\,b\right )}{2}-\frac {\frac {2\,a\,b^2}{d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,b^3+3\,a\,b^2\right )}{3\,d}+\frac {2\,a\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,\left (2\,b^3+3\,a\,b^2\right )}{3\,d}+\frac {2\,a\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {a^3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {2\,a\,b^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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