3.2.53 \(\int (a+b \sinh ^3(c+d x))^2 \, dx\) [153]

Optimal. Leaf size=114 \[ a^2 x-\frac {5 b^2 x}{16}-\frac {2 a b \cosh (c+d x)}{d}+\frac {2 a b \cosh ^3(c+d x)}{3 d}+\frac {5 b^2 \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac {5 b^2 \cosh (c+d x) \sinh ^3(c+d x)}{24 d}+\frac {b^2 \cosh (c+d x) \sinh ^5(c+d x)}{6 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3292, 2713, 2715, 8} \begin {gather*} a^2 x+\frac {2 a b \cosh ^3(c+d x)}{3 d}-\frac {2 a b \cosh (c+d x)}{d}+\frac {b^2 \sinh ^5(c+d x) \cosh (c+d x)}{6 d}-\frac {5 b^2 \sinh ^3(c+d x) \cosh (c+d x)}{24 d}+\frac {5 b^2 \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac {5 b^2 x}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x]^3)^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3292

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 94, normalized size = 0.82 \begin {gather*} \frac {192 a^2 c-60 b^2 c+192 a^2 d x-60 b^2 d x-288 a b \cosh (c+d x)+32 a b \cosh (3 (c+d x))+45 b^2 \sinh (2 (c+d x))-9 b^2 \sinh (4 (c+d x))+b^2 \sinh (6 (c+d x))}{192 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(192*a^2*c - 60*b^2*c + 192*a^2*d*x - 60*b^2*d*x - 288*a*b*Cosh[c + d*x] + 32*a*b*Cosh[3*(c + d*x)] + 45*b^2*S
inh[2*(c + d*x)] - 9*b^2*Sinh[4*(c + d*x)] + b^2*Sinh[6*(c + d*x)])/(192*d)

________________________________________________________________________________________

Maple [A]
time = 1.32, size = 93, normalized size = 0.82

method	result	size
default	\(a^{2} x -\frac {5 b^{2} x}{16}+\frac {15 b^{2} \sinh \left (2 d x +2 c \right )}{64 d}-\frac {3 b^{2} \sinh \left (4 d x +4 c \right )}{64 d}+\frac {b^{2} \sinh \left (6 d x +6 c \right )}{192 d}-\frac {3 a b \cosh \left (d x +c \right )}{2 d}+\frac {a b \cosh \left (3 d x +3 c \right )}{6 d}\)	\(93\)
risch	\(a^{2} x -\frac {5 b^{2} x}{16}+\frac {b^{2} {\mathrm e}^{6 d x +6 c}}{384 d}-\frac {3 \,{\mathrm e}^{4 d x +4 c} b^{2}}{128 d}+\frac {a b \,{\mathrm e}^{3 d x +3 c}}{12 d}+\frac {15 \,{\mathrm e}^{2 d x +2 c} b^{2}}{128 d}-\frac {3 a b \,{\mathrm e}^{d x +c}}{4 d}-\frac {3 a b \,{\mathrm e}^{-d x -c}}{4 d}-\frac {15 \,{\mathrm e}^{-2 d x -2 c} b^{2}}{128 d}+\frac {a b \,{\mathrm e}^{-3 d x -3 c}}{12 d}+\frac {3 \,{\mathrm e}^{-4 d x -4 c} b^{2}}{128 d}-\frac {b^{2} {\mathrm e}^{-6 d x -6 c}}{384 d}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x-5/16*b^2*x+15/64*b^2*sinh(2*d*x+2*c)/d-3/64*b^2*sinh(4*d*x+4*c)/d+1/192*b^2*sinh(6*d*x+6*c)/d-3/2*a*b*co
sh(d*x+c)/d+1/6*a*b/d*cosh(3*d*x+3*c)

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 151, normalized size = 1.32 \begin {gather*} a^{2} x - \frac {1}{384} \, b^{2} {\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 )} + \frac {1}{12} \, a b {\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x - 1/384*b^2*((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) + 1/12*a*b*(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)

________________________________________________________________________________________

Fricas [A]
time = 0.42, size = 160, normalized size = 1.40 \begin {gather*} \frac {3 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 16 \, a b \cosh \left (d x + c\right )^{3} + 48 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \, {\left (5 \, b^{2} \cosh \left (d x + c\right )^{3} - 9 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 6 \, {\left (16 \, a^{2} - 5 \, b^{2}\right )} d x - 144 \, a b \cosh \left (d x + c\right ) + 3 \, {\left (b^{2} \cosh \left (d x + c\right )^{5} - 6 \, b^{2} \cosh \left (d x + c\right )^{3} + 15 \, b^{2} \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)^3)^2,x, algorithm="fricas")

[Out]

1/96*(3*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + 16*a*b*cosh(d*x + c)^3 + 48*a*b*cosh(d*x + c)*sinh(d*x + c)^2 + 2*
(5*b^2*cosh(d*x + c)^3 - 9*b^2*cosh(d*x + c))*sinh(d*x + c)^3 + 6*(16*a^2 - 5*b^2)*d*x - 144*a*b*cosh(d*x + c)
+ 3*(b^2*cosh(d*x + c)^5 - 6*b^2*cosh(d*x + c)^3 + 15*b^2*cosh(d*x + c))*sinh(d*x + c))/d

________________________________________________________________________________________

Sympy [A]
time = 0.44, size = 212, normalized size = 1.86 \begin {gather*} \begin {cases} a^{2} x + \frac {2 a b \sinh ^{2}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {4 a b \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 b^{2} x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac {15 b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac {15 b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac {5 b^{2} x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac {11 b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{16 d} - \frac {5 b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} + \frac {5 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 ^{3}{\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.42, size = 178, normalized size = 1.56 \begin {gather*} \frac {1}{16} \, {\left (16 \, a^{2} - 5 \, b^{2}\right )} x + \frac {b^{2} e^{\left (6 \, d x + 6 \, c\right )}}{384 \, d} - \frac {3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )}}{128 \, d} + \frac {a b e^{\left (3 \, d x + 3 \, c\right )}}{12 \, d} + \frac {15 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} - \frac {3 \, a b e^{\left (d x + c\right )}}{4 \, d} - \frac {3 \, a b e^{\left (-d x - c\right )}}{4 \, d} - \frac {15 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} + \frac {a b e^{\left (-3 \, d x - 3 \, c\right )}}{12 \, d} + \frac {3 \, b^{2} e^{\left (-4 \, d x - 4 \, c\right )}}{128 \, d} - \frac {b^{2} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a^2 - 5*b^2)*x + 1/384*b^2*e^(6*d*x + 6*c)/d - 3/128*b^2*e^(4*d*x + 4*c)/d + 1/12*a*b*e^(3*d*x + 3*c)
/d + 15/128*b^2*e^(2*d*x + 2*c)/d - 3/4*a*b*e^(d*x + c)/d - 3/4*a*b*e^(-d*x - c)/d - 15/128*b^2*e^(-2*d*x - 2*
c)/d + 1/12*a*b*e^(-3*d*x - 3*c)/d + 3/128*b^2*e^(-4*d*x - 4*c)/d - 1/384*b^2*e^(-6*d*x - 6*c)/d

________________________________________________________________________________________

Mupad [B]
time = 0.46, size = 85, normalized size = 0.75 \begin {gather*} \frac {\frac {45\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4}-\frac {9\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+\frac {b^2\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )}{4}-72\,a\,b\,\mathrm {cosh}\left (c+d\,x\right )+8\,a\,b\,\mathrm {cosh}\left (3\,c+3\,d\,x\right )+48\,a^2\,d\,x-15\,b^2\,d\,x}{48\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^3)^2,x)

[Out]

((45*b^2*sinh(2*c + 2*d*x))/4 - (9*b^2*sinh(4*c + 4*d*x))/4 + (b^2*sinh(6*c + 6*d*x))/4 - 72*a*b*cosh(c + d*x)
+ 8*a*b*cosh(3*c + 3*d*x) + 48*a^2*d*x - 15*b^2*d*x)/(48*d)

________________________________________________________________________________________