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

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

[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)

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Rubi [A]  time = 0.0923167, antiderivative size = 72, normalized size of antiderivative = 1., 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 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]

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 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])

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*}

Mathematica [A]  time = 0.479235, size = 64, normalized size = 0.89 \[ \frac{6 a^2 (a+6 b) (c+d x)+3 a^3 \sinh (2 (c+d x))+4 b^2 \tanh (c+d x) \left (9 a+b \text{sech}^2(c+d x)+2 b\right )}{12 d} \]

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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Maple [A]  time = 0.044, size = 77, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +3\,{a}^{2}b \left ( dx+c \right ) +3\,a{b}^{2}\tanh \left ( dx+c \right ) +{b}^{3} \left ({\frac{2}{3}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{3}} \right ) \tanh \left ( dx+c \right ) \right ) } \end{align*}

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 = 1.13754, size = 216, normalized size = 3. \begin{align*} \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 )}} \end{align*}

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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Fricas [B]  time = 2.30101, size = 670, normalized size = 9.31 \begin{align*} \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 )}} \end{align*}

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

[Out]

Timed out

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Giac [B]  time = 1.18982, size = 215, normalized size = 2.99 \begin{align*} \frac{a^{3} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac{{\left (a^{3} + 6 \, a^{2} b\right )}{\left (d x + c\right )}}{2 \, d} - \frac{{\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 )}}{8 \, d} - \frac{2 \,{\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 )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}} \end{align*}

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]

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