3.294 \(\int \cosh ^3(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=74 \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{b (2 a+b) \sinh ^5(c+d x)}{5 d}+\frac{a (a+2 b) \sinh ^3(c+d x)}{3 d}+\frac{b^2 \sinh ^7(c+d x)}{7 d} \]

[Out]

(a^2*Sinh[c + d*x])/d + (a*(a + 2*b)*Sinh[c + d*x]^3)/(3*d) + (b*(2*a + b)*Sinh[c + d*x]^5)/(5*d) + (b^2*Sinh[
c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.0723943, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3190, 373} \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{b (2 a+b) \sinh ^5(c+d x)}{5 d}+\frac{a (a+2 b) \sinh ^3(c+d x)}{3 d}+\frac{b^2 \sinh ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sinh[c + d*x])/d + (a*(a + 2*b)*Sinh[c + d*x]^3)/(3*d) + (b*(2*a + b)*Sinh[c + d*x]^5)/(5*d) + (b^2*Sinh[
c + d*x]^7)/(7*d)

Rule 3190

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

Rule 373

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

Rubi steps

\begin{align*} \int \cosh ^3(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \left (a+b x^2\right )^2 \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+a (a+2 b) x^2+b (2 a+b) x^4+b^2 x^6\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{a^2 \sinh (c+d x)}{d}+\frac{a (a+2 b) \sinh ^3(c+d x)}{3 d}+\frac{b (2 a+b) \sinh ^5(c+d x)}{5 d}+\frac{b^2 \sinh ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.114626, size = 64, normalized size = 0.86 \[ \frac{105 a^2 \sinh (c+d x)+21 b (2 a+b) \sinh ^5(c+d x)+35 a (a+2 b) \sinh ^3(c+d x)+15 b^2 \sinh ^7(c+d x)}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(105*a^2*Sinh[c + d*x] + 35*a*(a + 2*b)*Sinh[c + d*x]^3 + 21*b*(2*a + b)*Sinh[c + d*x]^5 + 15*b^2*Sinh[c + d*x
]^7)/(105*d)

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Maple [A]  time = 0.031, size = 128, normalized size = 1.7 \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 ) ^{4}}{7}}-{\frac{3\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}{35}}+{\frac{3\,\sinh \left ( dx+c \right ) }{35} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) +2\,ab \left ( 1/5\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{4}-1/5\, \left ( 2/3+1/3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{2} \right ) \sinh \left ( dx+c \right ) \right ) +{a}^{2} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \sinh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^2*(1/7*sinh(d*x+c)^3*cosh(d*x+c)^4-3/35*sinh(d*x+c)*cosh(d*x+c)^4+3/35*(2/3+1/3*cosh(d*x+c)^2)*sinh(d*x
+c))+2*a*b*(1/5*sinh(d*x+c)*cosh(d*x+c)^4-1/5*(2/3+1/3*cosh(d*x+c)^2)*sinh(d*x+c))+a^2*(2/3+1/3*cosh(d*x+c)^2)
*sinh(d*x+c))

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Maxima [B]  time = 1.08167, size = 327, normalized size = 4.42 \begin{align*} -\frac{1}{4480} \, b^{2}{\left (\frac{{\left (7 \, e^{\left (-2 \, d x - 2 \, c\right )} + 35 \, e^{\left (-4 \, d x - 4 \, c\right )} - 105 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5\right )} e^{\left (7 \, d x + 7 \, c\right )}}{d} + \frac{105 \, e^{\left (-d x - c\right )} - 35 \, e^{\left (-3 \, d x - 3 \, c\right )} - 7 \, e^{\left (-5 \, d x - 5 \, c\right )} + 5 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d}\right )} + \frac{1}{240} \, a b{\left (\frac{{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 30 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3\right )} e^{\left (5 \, d x + 5 \, c\right )}}{d} + \frac{30 \, e^{\left (-d x - c\right )} - 5 \, e^{\left (-3 \, d x - 3 \, c\right )} - 3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} + \frac{1}{24} \, a^{2}{\left (\frac{e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac{9 \, e^{\left (d x + c\right )}}{d} - \frac{9 \, e^{\left (-d x - c\right )}}{d} - \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4480*b^2*((7*e^(-2*d*x - 2*c) + 35*e^(-4*d*x - 4*c) - 105*e^(-6*d*x - 6*c) - 5)*e^(7*d*x + 7*c)/d + (105*e^
(-d*x - c) - 35*e^(-3*d*x - 3*c) - 7*e^(-5*d*x - 5*c) + 5*e^(-7*d*x - 7*c))/d) + 1/240*a*b*((5*e^(-2*d*x - 2*c
) - 30*e^(-4*d*x - 4*c) + 3)*e^(5*d*x + 5*c)/d + (30*e^(-d*x - c) - 5*e^(-3*d*x - 3*c) - 3*e^(-5*d*x - 5*c))/d
) + 1/24*a^2*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d - e^(-3*d*x - 3*c)/d)

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Fricas [B]  time = 1.46437, size = 463, normalized size = 6.26 \begin{align*} \frac{15 \, b^{2} \sinh \left (d x + c\right )^{7} + 21 \,{\left (15 \, b^{2} \cosh \left (d x + c\right )^{2} + 8 \, a b - b^{2}\right )} \sinh \left (d x + c\right )^{5} + 35 \,{\left (15 \, b^{2} \cosh \left (d x + c\right )^{4} + 6 \,{\left (8 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 16 \, a^{2} + 8 \, a b - 3 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 105 \,{\left (b^{2} \cosh \left (d x + c\right )^{6} +{\left (8 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{4} +{\left (16 \, a^{2} + 8 \, a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 48 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \sinh \left (d x + c\right )}{6720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6720*(15*b^2*sinh(d*x + c)^7 + 21*(15*b^2*cosh(d*x + c)^2 + 8*a*b - b^2)*sinh(d*x + c)^5 + 35*(15*b^2*cosh(d
*x + c)^4 + 6*(8*a*b - b^2)*cosh(d*x + c)^2 + 16*a^2 + 8*a*b - 3*b^2)*sinh(d*x + c)^3 + 105*(b^2*cosh(d*x + c)
^6 + (8*a*b - b^2)*cosh(d*x + c)^4 + (16*a^2 + 8*a*b - 3*b^2)*cosh(d*x + c)^2 + 48*a^2 - 16*a*b + 3*b^2)*sinh(
d*x + c))/d

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Sympy [A]  time = 6.89745, size = 136, normalized size = 1.84 \begin{align*} \begin{cases} - \frac{2 a^{2} \sinh ^{3}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sinh{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{d} - \frac{4 a b \sinh ^{5}{\left (c + d x \right )}}{15 d} + \frac{2 a b \sinh ^{3}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{3 d} - \frac{2 b^{2} \sinh ^{7}{\left (c + d x \right )}}{35 d} + \frac{b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right )^{2} \cosh ^{3}{\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)**3*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Piecewise((-2*a**2*sinh(c + d*x)**3/(3*d) + a**2*sinh(c + d*x)*cosh(c + d*x)**2/d - 4*a*b*sinh(c + d*x)**5/(15
*d) + 2*a*b*sinh(c + d*x)**3*cosh(c + d*x)**2/(3*d) - 2*b**2*sinh(c + d*x)**7/(35*d) + b**2*sinh(c + d*x)**5*c
osh(c + d*x)**2/(5*d), Ne(d, 0)), (x*(a + b*sinh(c)**2)**2*cosh(c)**3, True))

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Giac [B]  time = 1.22079, size = 332, normalized size = 4.49 \begin{align*} \frac{15 \, b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 168 \, a b e^{\left (5 \, d x + 5 \, c\right )} - 21 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 560 \, a^{2} e^{\left (3 \, d x + 3 \, c\right )} + 280 \, a b e^{\left (3 \, d x + 3 \, c\right )} - 105 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 5040 \, a^{2} e^{\left (d x + c\right )} - 1680 \, a b e^{\left (d x + c\right )} + 315 \, b^{2} e^{\left (d x + c\right )} -{\left (5040 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 1680 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 315 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 560 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 280 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 105 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 168 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 21 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, b^{2}\right )} e^{\left (-7 \, d x - 7 \, c\right )}}{13440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13440*(15*b^2*e^(7*d*x + 7*c) + 168*a*b*e^(5*d*x + 5*c) - 21*b^2*e^(5*d*x + 5*c) + 560*a^2*e^(3*d*x + 3*c) +
280*a*b*e^(3*d*x + 3*c) - 105*b^2*e^(3*d*x + 3*c) + 5040*a^2*e^(d*x + c) - 1680*a*b*e^(d*x + c) + 315*b^2*e^(
d*x + c) - (5040*a^2*e^(6*d*x + 6*c) - 1680*a*b*e^(6*d*x + 6*c) + 315*b^2*e^(6*d*x + 6*c) + 560*a^2*e^(4*d*x +
4*c) + 280*a*b*e^(4*d*x + 4*c) - 105*b^2*e^(4*d*x + 4*c) + 168*a*b*e^(2*d*x + 2*c) - 21*b^2*e^(2*d*x + 2*c) +
15*b^2)*e^(-7*d*x - 7*c))/d