3.295 \(\int \cosh ^2(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=119 \[ \frac{\left (8 a^2-4 a b+b^2\right ) \sinh (c+d x) \cosh (c+d x)}{16 d}+\frac{1}{16} x \left (8 a^2-4 a b+b^2\right )+\frac{b (8 a-3 b) \sinh (c+d x) \cosh ^3(c+d x)}{24 d}+\frac{b \sinh (c+d x) \cosh ^5(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d} \]

[Out]

((8*a^2 - 4*a*b + b^2)*x)/16 + ((8*a^2 - 4*a*b + b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) + ((8*a - 3*b)*b*Cos
h[c + d*x]^3*Sinh[c + d*x])/(24*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2))/(6*d)

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Rubi [A]  time = 0.14271, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3191, 413, 385, 199, 206} \[ \frac{\left (8 a^2-4 a b+b^2\right ) \sinh (c+d x) \cosh (c+d x)}{16 d}+\frac{1}{16} x \left (8 a^2-4 a b+b^2\right )+\frac{b (8 a-3 b) \sinh (c+d x) \cosh ^3(c+d x)}{24 d}+\frac{b \sinh (c+d x) \cosh ^5(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 - 4*a*b + b^2)*x)/16 + ((8*a^2 - 4*a*b + b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) + ((8*a - 3*b)*b*Cos
h[c + d*x]^3*Sinh[c + d*x])/(24*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2))/(6*d)

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-(a-b) x^2\right )^2}{\left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-a (6 a-b)+3 (a-b) (2 a-b) x^2}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{6 d}\\ &=\frac{(8 a-3 b) b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}+\frac{\left (8 a^2-4 a b+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{\left (8 a^2-4 a b+b^2\right ) \cosh (c+d x) \sinh (c+d x)}{16 d}+\frac{(8 a-3 b) b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}+\frac{\left (8 a^2-4 a b+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=\frac{1}{16} \left (8 a^2-4 a b+b^2\right ) x+\frac{\left (8 a^2-4 a b+b^2\right ) \cosh (c+d x) \sinh (c+d x)}{16 d}+\frac{(8 a-3 b) b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.294553, size = 79, normalized size = 0.66 \[ \frac{12 \left (8 a^2-4 a b+b^2\right ) (c+d x)+3 \left (16 a^2-b^2\right ) \sinh (2 (c+d x))+3 b (4 a-b) \sinh (4 (c+d x))+b^2 \sinh (6 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a^2 - 4*a*b + b^2)*(c + d*x) + 3*(16*a^2 - b^2)*Sinh[2*(c + d*x)] + 3*(4*a - b)*b*Sinh[4*(c + d*x)] + b
^2*Sinh[6*(c + d*x)])/(192*d)

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Maple [A]  time = 0.032, size = 134, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3} \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{6}}-{\frac{\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{8}}+{\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{16}}+{\frac{dx}{16}}+{\frac{c}{16}} \right ) +2\,ab \left ( 1/4\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}-1/8\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/8\,dx-c/8 \right ) +{a}^{2} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^2*(1/6*sinh(d*x+c)^3*cosh(d*x+c)^3-1/8*sinh(d*x+c)*cosh(d*x+c)^3+1/16*cosh(d*x+c)*sinh(d*x+c)+1/16*d*x+
1/16*c)+2*a*b*(1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8*d*x-1/8*c)+a^2*(1/2*cosh(d*x+c)*s
inh(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 1.15212, size = 231, normalized size = 1.94 \begin{align*} \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 )} - \frac{1}{384} \, b^{2}{\left (\frac{{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} - \frac{24 \,{\left (d x + c\right )}}{d} - \frac{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 )}}{d}\right )} - \frac{1}{32} \, a b{\left (\frac{8 \,{\left (d x + c\right )}}{d} - \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*sinh(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) - 1/384*b^2*((3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) -
1)*e^(6*d*x + 6*c)/d - 24*(d*x + c)/d - (3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) - e^(-6*d*x - 6*c))/d) - 1/3
2*a*b*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d)

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Fricas [A]  time = 1.4468, size = 347, normalized size = 2.92 \begin{align*} \frac{3 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2 \,{\left (5 \, b^{2} \cosh \left (d x + c\right )^{3} + 3 \,{\left (4 \, a b - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 6 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} d x + 3 \,{\left (b^{2} \cosh \left (d x + c\right )^{5} + 2 \,{\left (4 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{3} +{\left (16 \, a^{2} - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(5*b^2*cosh(d*x + c)^3 + 3*(4*a*b - b^2)*cosh(d*x + c))*sinh(d*x
+ c)^3 + 6*(8*a^2 - 4*a*b + b^2)*d*x + 3*(b^2*cosh(d*x + c)^5 + 2*(4*a*b - b^2)*cosh(d*x + c)^3 + (16*a^2 - b
^2)*cosh(d*x + c))*sinh(d*x + c))/d

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Sympy [A]  time = 4.5846, size = 314, normalized size = 2.64 \begin{align*} \begin{cases} - \frac{a^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{2 d} - \frac{a b x \sinh ^{4}{\left (c + d x \right )}}{4} + \frac{a b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{2} - \frac{a b x \cosh ^{4}{\left (c + d x \right )}}{4} + \frac{a b \sinh ^{3}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{4 d} + \frac{a b \sinh{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{4 d} - \frac{b^{2} x \sinh ^{6}{\left (c + d x \right )}}{16} + \frac{3 b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} - \frac{3 b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} + \frac{b^{2} x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac{b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{16 d} + \frac{b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} - \frac{b^{2} \sinh{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right )^{2} \cosh ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x*sinh(c + d*x)**2/2 + a**2*x*cosh(c + d*x)**2/2 + a**2*sinh(c + d*x)*cosh(c + d*x)/(2*d) - a
*b*x*sinh(c + d*x)**4/4 + a*b*x*sinh(c + d*x)**2*cosh(c + d*x)**2/2 - a*b*x*cosh(c + d*x)**4/4 + a*b*sinh(c +
d*x)**3*cosh(c + d*x)/(4*d) + a*b*sinh(c + d*x)*cosh(c + d*x)**3/(4*d) - b**2*x*sinh(c + d*x)**6/16 + 3*b**2*x
*sinh(c + d*x)**4*cosh(c + d*x)**2/16 - 3*b**2*x*sinh(c + d*x)**2*cosh(c + d*x)**4/16 + b**2*x*cosh(c + d*x)**
6/16 + b**2*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) + b**2*sinh(c + d*x)**3*cosh(c + d*x)**3/(6*d) - b**2*sinh(c
+ d*x)*cosh(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sinh(c)**2)**2*cosh(c)**2, True))

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Giac [A]  time = 1.18555, size = 277, normalized size = 2.33 \begin{align*} \frac{b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 12 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 48 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 24 \,{\left (8 \, a^{2} - 4 \, a b + b^{2}\right )}{\left (d x + c\right )} -{\left (176 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 88 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 22 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 48 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(b^2*e^(6*d*x + 6*c) + 12*a*b*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*x + 4*c) + 48*a^2*e^(2*d*x + 2*c) - 3*b^2*e
^(2*d*x + 2*c) + 24*(8*a^2 - 4*a*b + b^2)*(d*x + c) - (176*a^2*e^(6*d*x + 6*c) - 88*a*b*e^(6*d*x + 6*c) + 22*b
^2*e^(6*d*x + 6*c) + 48*a^2*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*x + 4*c) + 12*a*b*e^(2*d*x + 2*c) - 3*b^2*e^(2*d*x
+ 2*c) + b^2)*e^(-6*d*x - 6*c))/d