3.1.23 \(\int (a+b \sinh ^2(c+d x))^3 \, dx\) [23]

Optimal. Leaf size=128 \[ \frac {1}{16} (2 a-b) \left (8 a^2-8 a b+5 b^2\right ) x+\frac {b \left (64 a^2-54 a b+15 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{48 d}+\frac {5 (2 a-b) b^2 \cosh (c+d x) \sinh ^3(c+d x)}{24 d}+\frac {b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^2}{6 d} \]

[Out]

1/16*(2*a-b)*(8*a^2-8*a*b+5*b^2)*x+1/48*b*(64*a^2-54*a*b+15*b^2)*cosh(d*x+c)*sinh(d*x+c)/d+5/24*(2*a-b)*b^2*co
sh(d*x+c)*sinh(d*x+c)^3/d+1/6*b*cosh(d*x+c)*sinh(d*x+c)*(a+b*sinh(d*x+c)^2)^2/d

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Rubi [A]
time = 0.07, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3259, 3248} \begin {gather*} \frac {b \left (64 a^2-54 a b+15 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{48 d}+\frac {1}{16} x (2 a-b) \left (8 a^2-8 a b+5 b^2\right )+\frac {5 b^2 (2 a-b) \sinh ^3(c+d x) \cosh (c+d x)}{24 d}+\frac {b \sinh (c+d x) \cosh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^2}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x]^2)^3,x]

[Out]

((2*a - b)*(8*a^2 - 8*a*b + 5*b^2)*x)/16 + (b*(64*a^2 - 54*a*b + 15*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(48*d) +
(5*(2*a - b)*b^2*Cosh[c + d*x]*Sinh[c + d*x]^3)/(24*d) + (b*Cosh[c + d*x]*Sinh[c + d*x]*(a + b*Sinh[c + d*x]^
2)^2)/(6*d)

Rule 3248

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(4*A
*(2*a + b) + B*(4*a + 3*b))*(x/8), x] + (-Simp[b*B*Cos[e + f*x]*(Sin[e + f*x]^3/(4*f)), x] - Simp[(4*A*b + B*(
4*a + 3*b))*Cos[e + f*x]*(Sin[e + f*x]/(8*f)), x]) /; FreeQ[{a, b, e, f, A, B}, x]

Rule 3259

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p - 1)/(2*f*p)), x] + Dist[1/(2*p), Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(
2*a + b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && GtQ[p, 1]

Rubi steps

\begin {align*} \int \left (a+b \sinh ^2(c+d x)\right )^3 \, dx &=\frac {b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^2}{6 d}+\frac {1}{6} \int \left (a+b \sinh ^2(c+d x)\right ) \left (a (6 a-b)+5 (2 a-b) b \sinh ^2(c+d x)\right ) \, dx\\ &=\frac {1}{16} (2 a-b) \left (8 a^2-8 a b+5 b^2\right ) x+\frac {b \left (64 a^2-54 a b+15 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{48 d}+\frac {5 (2 a-b) b^2 \cosh (c+d x) \sinh ^3(c+d x)}{24 d}+\frac {b \cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^2}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 95, normalized size = 0.74 \begin {gather*} \frac {12 (2 a-b) \left (8 a^2-8 a b+5 b^2\right ) (c+d x)+9 b \left (16 a^2-16 a b+5 b^2\right ) \sinh (2 (c+d x))+9 (2 a-b) b^2 \sinh (4 (c+d x))+b^3 \sinh (6 (c+d x))}{192 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(2*a - b)*(8*a^2 - 8*a*b + 5*b^2)*(c + d*x) + 9*b*(16*a^2 - 16*a*b + 5*b^2)*Sinh[2*(c + d*x)] + 9*(2*a - b
)*b^2*Sinh[4*(c + d*x)] + b^3*Sinh[6*(c + d*x)])/(192*d)

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Maple [A]
time = 0.91, size = 102, normalized size = 0.80

method	result	size
default	\(a^{3} x +\frac {\left (-\frac {3}{16} b^{3}+\frac {3}{8} a \,b^{2}\right ) \sinh \left (4 d x +4 c \right )}{4 d}+\frac {\left (\frac {15}{32} b^{3}-\frac {3}{2} a \,b^{2}+\frac {3}{2} a^{2} b \right ) \sinh \left (2 d x +2 c \right )}{2 d}-\frac {5 b^{3} x}{16}+\frac {9 a \,b^{2} x}{8}-\frac {3 a^{2} b x}{2}+\frac {b^{3} \sinh \left (6 d x +6 c \right )}{192 d}\)	\(102\)
risch	\(a^{3} x -\frac {5 b^{3} x}{16}+\frac {9 a \,b^{2} x}{8}-\frac {3 a^{2} b x}{2}+\frac {b^{3} {\mathrm e}^{6 d x +6 c}}{384 d}-\frac {3 \,{\mathrm e}^{4 d x +4 c} b^{3}}{128 d}+\frac {3 \,{\mathrm e}^{4 d x +4 c} a \,b^{2}}{64 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} a^{2} b}{8 d}-\frac {3 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}}{8 d}+\frac {15 \,{\mathrm e}^{2 d x +2 c} b^{3}}{128 d}-\frac {3 \,{\mathrm e}^{-2 d x -2 c} a^{2} b}{8 d}+\frac {3 \,{\mathrm e}^{-2 d x -2 c} a \,b^{2}}{8 d}-\frac {15 \,{\mathrm e}^{-2 d x -2 c} b^{3}}{128 d}+\frac {3 \,{\mathrm e}^{-4 d x -4 c} b^{3}}{128 d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c} a \,b^{2}}{64 d}-\frac {b^{3} {\mathrm e}^{-6 d x -6 c}}{384 d}\)	\(237\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sinh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

a^3*x+1/4*(-3/16*b^3+3/8*a*b^2)/d*sinh(4*d*x+4*c)+1/2*(15/32*b^3-3/2*a*b^2+3/2*a^2*b)*sinh(2*d*x+2*c)/d-5/16*b
^3*x+9/8*a*b^2*x-3/2*a^2*b*x+1/192*b^3/d*sinh(6*d*x+6*c)

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Maxima [A]
time = 0.28, size = 197, normalized size = 1.54 \begin {gather*} \frac {3}{64} \, a b^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {3}{8} \, a^{2} b {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + a^{3} x - \frac {1}{384} \, b^{3} {\left (\frac {{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {120 \, {\left (d x + c\right )}}{d} + \frac {45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*a*b^2*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) - 3/8*
a^2*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + a^3*x - 1/384*b^3*((9*e^(-2*d*x - 2*c) - 45*e^(-4*d*x -
4*c) - 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*x - 2*c) - 9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))
/d)

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Fricas [A]
time = 0.41, size = 165, normalized size = 1.29 \begin {gather*} \frac {3 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{3} + 9 \, {\left (2 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 6 \, {\left (16 \, a^{3} - 24 \, a^{2} b + 18 \, a b^{2} - 5 \, b^{3}\right )} d x + 3 \, {\left (b^{3} \cosh \left (d x + c\right )^{5} + 6 \, {\left (2 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (16 \, a^{2} b - 16 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*b^3*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(5*b^3*cosh(d*x + c)^3 + 9*(2*a*b^2 - b^3)*cosh(d*x + c))*sinh(d
*x + c)^3 + 6*(16*a^3 - 24*a^2*b + 18*a*b^2 - 5*b^3)*d*x + 3*(b^3*cosh(d*x + c)^5 + 6*(2*a*b^2 - b^3)*cosh(d*x
+ c)^3 + 3*(16*a^2*b - 16*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (119) = 238\).
time = 0.49, size = 350, normalized size = 2.73 \begin {gather*} \begin {cases} a^{3} x + \frac {3 a^{2} b x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac {3 a^{2} b x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{2} b \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} + \frac {9 a b^{2} x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac {9 a b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {9 a b^{2} x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {15 a b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} - \frac {9 a b^{2} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac {5 b^{3} x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac {15 b^{3} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac {15 b^{3} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac {5 b^{3} x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac {11 b^{3} \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{16 d} - \frac {5 b^{3} \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} + \frac {5 b^{3} \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 )^{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*x*sinh(c + d*x)**2/2 - 3*a**2*b*x*cosh(c + d*x)**2/2 + 3*a**2*b*sinh(c + d*x)*cos
h(c + d*x)/(2*d) + 9*a*b**2*x*sinh(c + d*x)**4/8 - 9*a*b**2*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 + 9*a*b**2*x
*cosh(c + d*x)**4/8 + 15*a*b**2*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) - 9*a*b**2*sinh(c + d*x)*cosh(c + d*x)**3
/(8*d) + 5*b**3*x*sinh(c + d*x)**6/16 - 15*b**3*x*sinh(c + d*x)**4*cosh(c + d*x)**2/16 + 15*b**3*x*sinh(c + d*
x)**2*cosh(c + d*x)**4/16 - 5*b**3*x*cosh(c + d*x)**6/16 + 11*b**3*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) - 5*b
**3*sinh(c + d*x)**3*cosh(c + d*x)**3/(6*d) + 5*b**3*sinh(c + d*x)*cosh(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a +
b*sinh(c)**2)**3, True))

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Giac [A]
time = 0.42, size = 177, normalized size = 1.38 \begin {gather*} \frac {b^{3} e^{\left (6 \, d x + 6 \, c\right )}}{384 \, d} - \frac {b^{3} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} + \frac {1}{16} \, {\left (16 \, a^{3} - 24 \, a^{2} b + 18 \, a b^{2} - 5 \, b^{3}\right )} x + \frac {3 \, {\left (2 \, a b^{2} - b^{3}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{128 \, d} + \frac {3 \, {\left (16 \, a^{2} b - 16 \, a b^{2} + 5 \, b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} - \frac {3 \, {\left (16 \, a^{2} b - 16 \, a b^{2} + 5 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} - \frac {3 \, {\left (2 \, a b^{2} - b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{128 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*b^3*e^(6*d*x + 6*c)/d - 1/384*b^3*e^(-6*d*x - 6*c)/d + 1/16*(16*a^3 - 24*a^2*b + 18*a*b^2 - 5*b^3)*x + 3
/128*(2*a*b^2 - b^3)*e^(4*d*x + 4*c)/d + 3/128*(16*a^2*b - 16*a*b^2 + 5*b^3)*e^(2*d*x + 2*c)/d - 3/128*(16*a^2
*b - 16*a*b^2 + 5*b^3)*e^(-2*d*x - 2*c)/d - 3/128*(2*a*b^2 - b^3)*e^(-4*d*x - 4*c)/d

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Mupad [B]
time = 0.75, size = 123, normalized size = 0.96 \begin {gather*} \frac {\frac {45\,b^3\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4}-\frac {9\,b^3\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+\frac {b^3\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )}{4}-36\,a\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+36\,a^2\,b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+\frac {9\,a\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{2}+48\,a^3\,d\,x-15\,b^3\,d\,x+54\,a\,b^2\,d\,x-72\,a^2\,b\,d\,x}{48\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^2)^3,x)

[Out]

((45*b^3*sinh(2*c + 2*d*x))/4 - (9*b^3*sinh(4*c + 4*d*x))/4 + (b^3*sinh(6*c + 6*d*x))/4 - 36*a*b^2*sinh(2*c +
2*d*x) + 36*a^2*b*sinh(2*c + 2*d*x) + (9*a*b^2*sinh(4*c + 4*d*x))/2 + 48*a^3*d*x - 15*b^3*d*x + 54*a*b^2*d*x -
72*a^2*b*d*x)/(48*d)

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