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

Optimal. Leaf size=98 \[ \frac{a^2 (a+3 b) \sinh ^3(c+d x)}{3 d}+\frac{a^3 \sinh (c+d x)}{d}+\frac{b^2 (3 a+b) \sinh ^7(c+d x)}{7 d}+\frac{3 a b (a+b) \sinh ^5(c+d x)}{5 d}+\frac{b^3 \sinh ^9(c+d x)}{9 d} \]

[Out]

(a^3*Sinh[c + d*x])/d + (a^2*(a + 3*b)*Sinh[c + d*x]^3)/(3*d) + (3*a*b*(a + b)*Sinh[c + d*x]^5)/(5*d) + (b^2*(
3*a + b)*Sinh[c + d*x]^7)/(7*d) + (b^3*Sinh[c + d*x]^9)/(9*d)

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Rubi [A]  time = 0.0902005, antiderivative size = 98, 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} \[ \frac{a^2 (a+3 b) \sinh ^3(c+d x)}{3 d}+\frac{a^3 \sinh (c+d x)}{d}+\frac{b^2 (3 a+b) \sinh ^7(c+d x)}{7 d}+\frac{3 a b (a+b) \sinh ^5(c+d x)}{5 d}+\frac{b^3 \sinh ^9(c+d x)}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sinh[c + d*x])/d + (a^2*(a + 3*b)*Sinh[c + d*x]^3)/(3*d) + (3*a*b*(a + b)*Sinh[c + d*x]^5)/(5*d) + (b^2*(
3*a + b)*Sinh[c + d*x]^7)/(7*d) + (b^3*Sinh[c + d*x]^9)/(9*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 )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \left (a+b x^2\right )^3 \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^3+a^2 (a+3 b) x^2+3 a b (a+b) x^4+b^2 (3 a+b) x^6+b^3 x^8\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{a^3 \sinh (c+d x)}{d}+\frac{a^2 (a+3 b) \sinh ^3(c+d x)}{3 d}+\frac{3 a b (a+b) \sinh ^5(c+d x)}{5 d}+\frac{b^2 (3 a+b) \sinh ^7(c+d x)}{7 d}+\frac{b^3 \sinh ^9(c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.233608, size = 83, normalized size = 0.85 \[ \frac{105 a^2 (a+3 b) \sinh ^3(c+d x)+315 a^3 \sinh (c+d x)+45 b^2 (3 a+b) \sinh ^7(c+d x)+189 a b (a+b) \sinh ^5(c+d x)+35 b^3 \sinh ^9(c+d x)}{315 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(315*a^3*Sinh[c + d*x] + 105*a^2*(a + 3*b)*Sinh[c + d*x]^3 + 189*a*b*(a + b)*Sinh[c + d*x]^5 + 45*b^2*(3*a + b
)*Sinh[c + d*x]^7 + 35*b^3*Sinh[c + d*x]^9)/(315*d)

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Maple [B]  time = 0.035, size = 209, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5} \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}{9}}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3} \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}{63}}+{\frac{\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}{21}}-{\frac{\sinh \left ( dx+c \right ) }{21} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) +3\,a{b}^{2} \left ( 1/7\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3} \left ( \cosh \left ( dx+c \right ) \right ) ^{4}-{\frac{3\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}{35}}+{\frac{ \left ( 2+ \left ( \cosh \left ( dx+c \right ) \right ) ^{2} \right ) \sinh \left ( dx+c \right ) }{35}} \right ) +3\,{a}^{2}b \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}^{3} \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)^3,x)

[Out]

1/d*(b^3*(1/9*sinh(d*x+c)^5*cosh(d*x+c)^4-5/63*sinh(d*x+c)^3*cosh(d*x+c)^4+1/21*sinh(d*x+c)*cosh(d*x+c)^4-1/21
*(2/3+1/3*cosh(d*x+c)^2)*sinh(d*x+c))+3*a*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))+3*a^2*b*(1/5*sinh(d*x+c)*cosh(d*x+c)^4-1/5*(2/3+1/3*cosh(d*x+c)^2)*s
inh(d*x+c))+a^3*(2/3+1/3*cosh(d*x+c)^2)*sinh(d*x+c))

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Maxima [B]  time = 1.09882, size = 471, normalized size = 4.81 \begin{align*} -\frac{1}{32256} \, b^{3}{\left (\frac{{\left (27 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-6 \, d x - 6 \, c\right )} + 378 \, e^{\left (-8 \, d x - 8 \, c\right )} - 7\right )} e^{\left (9 \, d x + 9 \, c\right )}}{d} - \frac{378 \, e^{\left (-d x - c\right )} - 168 \, e^{\left (-3 \, d x - 3 \, c\right )} + 27 \, e^{\left (-7 \, d x - 7 \, c\right )} - 7 \, e^{\left (-9 \, d x - 9 \, c\right )}}{d}\right )} - \frac{3}{4480} \, a 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}{160} \, a^{2} 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^{3}{\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)^3,x, algorithm="maxima")

[Out]

-1/32256*b^3*((27*e^(-2*d*x - 2*c) - 168*e^(-6*d*x - 6*c) + 378*e^(-8*d*x - 8*c) - 7)*e^(9*d*x + 9*c)/d - (378
*e^(-d*x - c) - 168*e^(-3*d*x - 3*c) + 27*e^(-7*d*x - 7*c) - 7*e^(-9*d*x - 9*c))/d) - 3/4480*a*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) + 1/160*a^2*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^3*(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 = 1.52027, size = 787, normalized size = 8.03 \begin{align*} \frac{35 \, b^{3} \sinh \left (d x + c\right )^{9} + 45 \,{\left (28 \, b^{3} \cosh \left (d x + c\right )^{2} + 12 \, a b^{2} - 3 \, b^{3}\right )} \sinh \left (d x + c\right )^{7} + 63 \,{\left (70 \, b^{3} \cosh \left (d x + c\right )^{4} + 48 \, a^{2} b - 12 \, a b^{2} + 45 \,{\left (4 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{5} + 105 \,{\left (28 \, b^{3} \cosh \left (d x + c\right )^{6} + 45 \,{\left (4 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{4} + 64 \, a^{3} + 48 \, a^{2} b - 36 \, a b^{2} + 8 \, b^{3} + 72 \,{\left (4 \, a^{2} b - a b^{2}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 315 \,{\left (b^{3} \cosh \left (d x + c\right )^{8} + 3 \,{\left (4 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{6} + 12 \,{\left (4 \, a^{2} b - a b^{2}\right )} \cosh \left (d x + c\right )^{4} + 192 \, a^{3} - 96 \, a^{2} b + 36 \, a b^{2} - 6 \, b^{3} + 4 \,{\left (16 \, a^{3} + 12 \, a^{2} b - 9 \, a b^{2} + 2 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{80640 \, 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)^3,x, algorithm="fricas")

[Out]

1/80640*(35*b^3*sinh(d*x + c)^9 + 45*(28*b^3*cosh(d*x + c)^2 + 12*a*b^2 - 3*b^3)*sinh(d*x + c)^7 + 63*(70*b^3*
cosh(d*x + c)^4 + 48*a^2*b - 12*a*b^2 + 45*(4*a*b^2 - b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 105*(28*b^3*cosh
(d*x + c)^6 + 45*(4*a*b^2 - b^3)*cosh(d*x + c)^4 + 64*a^3 + 48*a^2*b - 36*a*b^2 + 8*b^3 + 72*(4*a^2*b - a*b^2)
*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 315*(b^3*cosh(d*x + c)^8 + 3*(4*a*b^2 - b^3)*cosh(d*x + c)^6 + 12*(4*a^2*b
- a*b^2)*cosh(d*x + c)^4 + 192*a^3 - 96*a^2*b + 36*a*b^2 - 6*b^3 + 4*(16*a^3 + 12*a^2*b - 9*a*b^2 + 2*b^3)*co
sh(d*x + c)^2)*sinh(d*x + c))/d

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Sympy [A]  time = 19.4941, size = 182, normalized size = 1.86 \begin{align*} \begin{cases} - \frac{2 a^{3} \sinh ^{3}{\left (c + d x \right )}}{3 d} + \frac{a^{3} \sinh{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{d} - \frac{2 a^{2} b \sinh ^{5}{\left (c + d x \right )}}{5 d} + \frac{a^{2} b \sinh ^{3}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{d} - \frac{6 a b^{2} \sinh ^{7}{\left (c + d x \right )}}{35 d} + \frac{3 a b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{5 d} - \frac{2 b^{3} \sinh ^{9}{\left (c + d x \right )}}{63 d} + \frac{b^{3} \sinh ^{7}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{7 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right )^{3} \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)**3,x)

[Out]

Piecewise((-2*a**3*sinh(c + d*x)**3/(3*d) + a**3*sinh(c + d*x)*cosh(c + d*x)**2/d - 2*a**2*b*sinh(c + d*x)**5/
(5*d) + a**2*b*sinh(c + d*x)**3*cosh(c + d*x)**2/d - 6*a*b**2*sinh(c + d*x)**7/(35*d) + 3*a*b**2*sinh(c + d*x)
**5*cosh(c + d*x)**2/(5*d) - 2*b**3*sinh(c + d*x)**9/(63*d) + b**3*sinh(c + d*x)**7*cosh(c + d*x)**2/(7*d), Ne
(d, 0)), (x*(a + b*sinh(c)**2)**3*cosh(c)**3, True))

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Giac [B]  time = 1.26016, size = 506, normalized size = 5.16 \begin{align*} \frac{35 \, b^{3} e^{\left (9 \, d x + 9 \, c\right )} + 540 \, a b^{2} e^{\left (7 \, d x + 7 \, c\right )} - 135 \, b^{3} e^{\left (7 \, d x + 7 \, c\right )} + 3024 \, a^{2} b e^{\left (5 \, d x + 5 \, c\right )} - 756 \, a b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 6720 \, a^{3} e^{\left (3 \, d x + 3 \, c\right )} + 5040 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} - 3780 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 840 \, b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 60480 \, a^{3} e^{\left (d x + c\right )} - 30240 \, a^{2} b e^{\left (d x + c\right )} + 11340 \, a b^{2} e^{\left (d x + c\right )} - 1890 \, b^{3} e^{\left (d x + c\right )} -{\left (60480 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 30240 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 11340 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 1890 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 6720 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} + 5040 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 3780 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 840 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 3024 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 756 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 540 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 135 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 35 \, b^{3}\right )} e^{\left (-9 \, d x - 9 \, c\right )}}{161280 \, 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)^3,x, algorithm="giac")

[Out]

1/161280*(35*b^3*e^(9*d*x + 9*c) + 540*a*b^2*e^(7*d*x + 7*c) - 135*b^3*e^(7*d*x + 7*c) + 3024*a^2*b*e^(5*d*x +
5*c) - 756*a*b^2*e^(5*d*x + 5*c) + 6720*a^3*e^(3*d*x + 3*c) + 5040*a^2*b*e^(3*d*x + 3*c) - 3780*a*b^2*e^(3*d*
x + 3*c) + 840*b^3*e^(3*d*x + 3*c) + 60480*a^3*e^(d*x + c) - 30240*a^2*b*e^(d*x + c) + 11340*a*b^2*e^(d*x + c)
- 1890*b^3*e^(d*x + c) - (60480*a^3*e^(8*d*x + 8*c) - 30240*a^2*b*e^(8*d*x + 8*c) + 11340*a*b^2*e^(8*d*x + 8*
c) - 1890*b^3*e^(8*d*x + 8*c) + 6720*a^3*e^(6*d*x + 6*c) + 5040*a^2*b*e^(6*d*x + 6*c) - 3780*a*b^2*e^(6*d*x +
6*c) + 840*b^3*e^(6*d*x + 6*c) + 3024*a^2*b*e^(4*d*x + 4*c) - 756*a*b^2*e^(4*d*x + 4*c) + 540*a*b^2*e^(2*d*x +
2*c) - 135*b^3*e^(2*d*x + 2*c) + 35*b^3)*e^(-9*d*x - 9*c))/d