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

Optimal. Leaf size=47 \[ \frac {a^2 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {1}{2} a x (a+4 b)+\frac {b^2 \tanh (c+d x)}{d} \]

[Out]

1/2*a*(a+4*b)*x+1/2*a^2*cosh(d*x+c)*sinh(d*x+c)/d+b^2*tanh(d*x+c)/d

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Rubi [A]  time = 0.08, antiderivative size = 47, 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} \[ \frac {a^2 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {1}{2} a x (a+4 b)+\frac {b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.15, size = 52, normalized size = 1.11 \[ \frac {a^2 (c+d x)}{2 d}+\frac {a^2 \sinh (2 (c+d x))}{4 d}+2 a b x+\frac {b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*b*x + (a^2*(c + d*x))/(2*d) + (a^2*Sinh[2*(c + d*x)])/(4*d) + (b^2*Tanh[c + d*x])/d

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fricas [A]  time = 0.39, size = 80, normalized size = 1.70 \[ \frac {a^{2} \sinh \left (d x + c\right )^{3} + 4 \, {\left ({\left (a^{2} + 4 \, a b\right )} d x - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) + {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + a^{2} + 8 \, b^{2}\right )} \sinh \left (d x + c\right )}{8 \, d \cosh \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 128, normalized size = 2.72 \[ \frac {a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 4 \, {\left (a^{2} + 4 \, a b\right )} {\left (d x + c\right )} - \frac {a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 16 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2}}{e^{\left (4 \, d x + 4 \, c\right )} + e^{\left (2 \, d x + 2 \, c\right )}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2*e^(2*d*x + 2*c) + 4*(a^2 + 4*a*b)*(d*x + c) - (a^2*e^(4*d*x + 4*c) + 4*a*b*e^(4*d*x + 4*c) + 2*a^2*e^
(2*d*x + 2*c) + 4*a*b*e^(2*d*x + 2*c) + 16*b^2*e^(2*d*x + 2*c) + a^2)/(e^(4*d*x + 4*c) + e^(2*d*x + 2*c)))/d

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maple [A]  time = 0.36, size = 51, normalized size = 1.09 \[ \frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (d x +c \right )+b^{2} \tanh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.32, size = 63, normalized size = 1.34 \[ \frac {1}{8} \, a^{2} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 2 \, a b x + \frac {2 \, b^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.16, size = 65, normalized size = 1.38 \[ \frac {a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {a^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {2\,b^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {a\,x\,\left (a+4\,b\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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