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

Optimal. Leaf size=56 \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{b (4 a+b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^2 \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0678789, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4147, 390, 385, 203} \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{b (4 a+b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^2 \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4147

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

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 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 203

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

Rubi steps

\begin{align*} \int \cosh (c+d x) \left (a+b \text{sech}^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+a x^2\right )^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+\frac{b (2 a+b)+2 a b x^2}{\left (1+x^2\right )^2}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{a^2 \sinh (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \frac{b (2 a+b)+2 a b x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{a^2 \sinh (c+d x)}{d}+\frac{b^2 \text{sech}(c+d x) \tanh (c+d x)}{2 d}+\frac{(b (4 a+b)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{2 d}\\ &=\frac{b (4 a+b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{a^2 \sinh (c+d x)}{d}+\frac{b^2 \text{sech}(c+d x) \tanh (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0398912, size = 80, normalized size = 1.43 \[ \frac{a^2 \sinh (c) \cosh (d x)}{d}+\frac{a^2 \cosh (c) \sinh (d x)}{d}+\frac{2 a b \tan ^{-1}(\sinh (c+d x))}{d}+\frac{b^2 \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^2 \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 63, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}\sinh \left ( dx+c \right ) }{d}}+4\,{\frac{ab\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.59969, size = 136, normalized size = 2.43 \begin{align*} -b^{2}{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} - \frac{4 \, a b \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{a^{2} \sinh \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.22962, size = 1678, normalized size = 29.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*sinh(d*x + c)^6 + (a^2 + 2*b^2)*cosh(d*x
+ c)^4 + (15*a^2*cosh(d*x + c)^2 + a^2 + 2*b^2)*sinh(d*x + c)^4 + 4*(5*a^2*cosh(d*x + c)^3 + (a^2 + 2*b^2)*cos
h(d*x + c))*sinh(d*x + c)^3 - (a^2 + 2*b^2)*cosh(d*x + c)^2 + (15*a^2*cosh(d*x + c)^4 + 6*(a^2 + 2*b^2)*cosh(d
*x + c)^2 - a^2 - 2*b^2)*sinh(d*x + c)^2 - a^2 + 2*((4*a*b + b^2)*cosh(d*x + c)^5 + 5*(4*a*b + b^2)*cosh(d*x +
c)*sinh(d*x + c)^4 + (4*a*b + b^2)*sinh(d*x + c)^5 + 2*(4*a*b + b^2)*cosh(d*x + c)^3 + 2*(5*(4*a*b + b^2)*cos
h(d*x + c)^2 + 4*a*b + b^2)*sinh(d*x + c)^3 + 2*(5*(4*a*b + b^2)*cosh(d*x + c)^3 + 3*(4*a*b + b^2)*cosh(d*x +
c))*sinh(d*x + c)^2 + (4*a*b + b^2)*cosh(d*x + c) + (5*(4*a*b + b^2)*cosh(d*x + c)^4 + 6*(4*a*b + b^2)*cosh(d*
x + c)^2 + 4*a*b + b^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(3*a^2*cosh(d*x + c)^5 + 2*(a
^2 + 2*b^2)*cosh(d*x + c)^3 - (a^2 + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x +
c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5 + 2*d*cosh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + 2
*(5*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + d*cosh(d*x + c) + (5*d*cosh(d*x + c)^4 + 6*d*cosh
(d*x + c)^2 + d)*sinh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.17467, size = 157, normalized size = 2.8 \begin{align*} \frac{a^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{2 \, d} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )}{\left (4 \, a b + b^{2}\right )}}{4 \, d} + \frac{b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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