3.163 \(\int (a+b \sinh ^3(c+d x))^3 \, dx\)

Optimal. Leaf size=204 \[ a^3 x+\frac {a^2 b \cosh ^3(c+d x)}{d}-\frac {3 a^2 b \cosh (c+d x)}{d}+\frac {a b^2 \sinh ^5(c+d x) \cosh (c+d x)}{2 d}-\frac {5 a b^2 \sinh ^3(c+d x) \cosh (c+d x)}{8 d}+\frac {15 a b^2 \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac {15}{16} a b^2 x+\frac {b^3 \cosh ^9(c+d x)}{9 d}-\frac {4 b^3 \cosh ^7(c+d x)}{7 d}+\frac {6 b^3 \cosh ^5(c+d x)}{5 d}-\frac {4 b^3 \cosh ^3(c+d x)}{3 d}+\frac {b^3 \cosh (c+d x)}{d} \]

[Out]

a^3*x-15/16*a*b^2*x-3*a^2*b*cosh(d*x+c)/d+b^3*cosh(d*x+c)/d+a^2*b*cosh(d*x+c)^3/d-4/3*b^3*cosh(d*x+c)^3/d+6/5*
b^3*cosh(d*x+c)^5/d-4/7*b^3*cosh(d*x+c)^7/d+1/9*b^3*cosh(d*x+c)^9/d+15/16*a*b^2*cosh(d*x+c)*sinh(d*x+c)/d-5/8*
a*b^2*cosh(d*x+c)*sinh(d*x+c)^3/d+1/2*a*b^2*cosh(d*x+c)*sinh(d*x+c)^5/d

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Rubi [A]  time = 0.13, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3213, 2633, 2635, 8} \[ \frac {a^2 b \cosh ^3(c+d x)}{d}-\frac {3 a^2 b \cosh (c+d x)}{d}+a^3 x+\frac {a b^2 \sinh ^5(c+d x) \cosh (c+d x)}{2 d}-\frac {5 a b^2 \sinh ^3(c+d x) \cosh (c+d x)}{8 d}+\frac {15 a b^2 \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac {15}{16} a b^2 x+\frac {b^3 \cosh ^9(c+d x)}{9 d}-\frac {4 b^3 \cosh ^7(c+d x)}{7 d}+\frac {6 b^3 \cosh ^5(c+d x)}{5 d}-\frac {4 b^3 \cosh ^3(c+d x)}{3 d}+\frac {b^3 \cosh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x]^3)^3,x]

[Out]

a^3*x - (15*a*b^2*x)/16 - (3*a^2*b*Cosh[c + d*x])/d + (b^3*Cosh[c + d*x])/d + (a^2*b*Cosh[c + d*x]^3)/d - (4*b
^3*Cosh[c + d*x]^3)/(3*d) + (6*b^3*Cosh[c + d*x]^5)/(5*d) - (4*b^3*Cosh[c + d*x]^7)/(7*d) + (b^3*Cosh[c + d*x]
^9)/(9*d) + (15*a*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (5*a*b^2*Cosh[c + d*x]*Sinh[c + d*x]^3)/(8*d) + (a
*b^2*Cosh[c + d*x]*Sinh[c + d*x]^5)/(2*d)

Rule 8

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

Rule 2633

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 2635

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] && Integer
Q[2*n]

Rule 3213

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 )^3 \, dx &=\int \left (a^3+3 a^2 b \sinh ^3(c+d x)+3 a b^2 \sinh ^6(c+d x)+b^3 \sinh ^9(c+d x)\right ) \, dx\\ &=a^3 x+\left (3 a^2 b\right ) \int \sinh ^3(c+d x) \, dx+\left (3 a b^2\right ) \int \sinh ^6(c+d x) \, dx+b^3 \int \sinh ^9(c+d x) \, dx\\ &=a^3 x+\frac {a b^2 \cosh (c+d x) \sinh ^5(c+d x)}{2 d}-\frac {1}{2} \left (5 a b^2\right ) \int \sinh ^4(c+d x) \, dx-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (c+d x)\right )}{d}+\frac {b^3 \operatorname {Subst}\left (\int \left (1-4 x^2+6 x^4-4 x^6+x^8\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=a^3 x-\frac {3 a^2 b \cosh (c+d x)}{d}+\frac {b^3 \cosh (c+d x)}{d}+\frac {a^2 b \cosh ^3(c+d x)}{d}-\frac {4 b^3 \cosh ^3(c+d x)}{3 d}+\frac {6 b^3 \cosh ^5(c+d x)}{5 d}-\frac {4 b^3 \cosh ^7(c+d x)}{7 d}+\frac {b^3 \cosh ^9(c+d x)}{9 d}-\frac {5 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{8 d}+\frac {a b^2 \cosh (c+d x) \sinh ^5(c+d x)}{2 d}+\frac {1}{8} \left (15 a b^2\right ) \int \sinh ^2(c+d x) \, dx\\ &=a^3 x-\frac {3 a^2 b \cosh (c+d x)}{d}+\frac {b^3 \cosh (c+d x)}{d}+\frac {a^2 b \cosh ^3(c+d x)}{d}-\frac {4 b^3 \cosh ^3(c+d x)}{3 d}+\frac {6 b^3 \cosh ^5(c+d x)}{5 d}-\frac {4 b^3 \cosh ^7(c+d x)}{7 d}+\frac {b^3 \cosh ^9(c+d x)}{9 d}+\frac {15 a b^2 \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac {5 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{8 d}+\frac {a b^2 \cosh (c+d x) \sinh ^5(c+d x)}{2 d}-\frac {1}{16} \left (15 a b^2\right ) \int 1 \, dx\\ &=a^3 x-\frac {15}{16} a b^2 x-\frac {3 a^2 b \cosh (c+d x)}{d}+\frac {b^3 \cosh (c+d x)}{d}+\frac {a^2 b \cosh ^3(c+d x)}{d}-\frac {4 b^3 \cosh ^3(c+d x)}{3 d}+\frac {6 b^3 \cosh ^5(c+d x)}{5 d}-\frac {4 b^3 \cosh ^7(c+d x)}{7 d}+\frac {b^3 \cosh ^9(c+d x)}{9 d}+\frac {15 a b^2 \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac {5 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{8 d}+\frac {a b^2 \cosh (c+d x) \sinh ^5(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 159, normalized size = 0.78 \[ \frac {80640 a^3 c+80640 a^3 d x+1260 \left (16 a^2 b-7 b^3\right ) \cosh (3 (c+d x))+5670 b \left (7 b^2-32 a^2\right ) \cosh (c+d x)+56700 a b^2 \sinh (2 (c+d x))-11340 a b^2 \sinh (4 (c+d x))+1260 a b^2 \sinh (6 (c+d x))-75600 a b^2 c-75600 a b^2 d x+2268 b^3 \cosh (5 (c+d x))-405 b^3 \cosh (7 (c+d x))+35 b^3 \cosh (9 (c+d x))}{80640 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(80640*a^3*c - 75600*a*b^2*c + 80640*a^3*d*x - 75600*a*b^2*d*x + 5670*b*(-32*a^2 + 7*b^2)*Cosh[c + d*x] + 1260
*(16*a^2*b - 7*b^3)*Cosh[3*(c + d*x)] + 2268*b^3*Cosh[5*(c + d*x)] - 405*b^3*Cosh[7*(c + d*x)] + 35*b^3*Cosh[9
*(c + d*x)] + 56700*a*b^2*Sinh[2*(c + d*x)] - 11340*a*b^2*Sinh[4*(c + d*x)] + 1260*a*b^2*Sinh[6*(c + d*x)])/(8
0640*d)

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fricas [B]  time = 0.45, size = 380, normalized size = 1.86 \[ \frac {35 \, b^{3} \cosh \left (d x + c\right )^{9} + 315 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{8} - 405 \, b^{3} \cosh \left (d x + c\right )^{7} + 7560 \, a b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2268 \, b^{3} \cosh \left (d x + c\right )^{5} + 105 \, {\left (28 \, b^{3} \cosh \left (d x + c\right )^{3} - 27 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{6} + 315 \, {\left (14 \, b^{3} \cosh \left (d x + c\right )^{5} - 45 \, b^{3} \cosh \left (d x + c\right )^{3} + 36 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 1260 \, {\left (16 \, a^{2} b - 7 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5040 \, {\left (5 \, a b^{2} \cosh \left (d x + c\right )^{3} - 9 \, a b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 5040 \, {\left (16 \, a^{3} - 15 \, a b^{2}\right )} d x + 315 \, {\left (4 \, b^{3} \cosh \left (d x + c\right )^{7} - 27 \, b^{3} \cosh \left (d x + c\right )^{5} + 72 \, b^{3} \cosh \left (d x + c\right )^{3} + 12 \, {\left (16 \, a^{2} b - 7 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5670 \, {\left (32 \, a^{2} b - 7 \, b^{3}\right )} \cosh \left (d x + c\right ) + 7560 \, {\left (a b^{2} \cosh \left (d x + c\right )^{5} - 6 \, a b^{2} \cosh \left (d x + c\right )^{3} + 15 \, a b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{80640 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80640*(35*b^3*cosh(d*x + c)^9 + 315*b^3*cosh(d*x + c)*sinh(d*x + c)^8 - 405*b^3*cosh(d*x + c)^7 + 7560*a*b^2
*cosh(d*x + c)*sinh(d*x + c)^5 + 2268*b^3*cosh(d*x + c)^5 + 105*(28*b^3*cosh(d*x + c)^3 - 27*b^3*cosh(d*x + c)
)*sinh(d*x + c)^6 + 315*(14*b^3*cosh(d*x + c)^5 - 45*b^3*cosh(d*x + c)^3 + 36*b^3*cosh(d*x + c))*sinh(d*x + c)
^4 + 1260*(16*a^2*b - 7*b^3)*cosh(d*x + c)^3 + 5040*(5*a*b^2*cosh(d*x + c)^3 - 9*a*b^2*cosh(d*x + c))*sinh(d*x
+ c)^3 + 5040*(16*a^3 - 15*a*b^2)*d*x + 315*(4*b^3*cosh(d*x + c)^7 - 27*b^3*cosh(d*x + c)^5 + 72*b^3*cosh(d*x
+ c)^3 + 12*(16*a^2*b - 7*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 5670*(32*a^2*b - 7*b^3)*cosh(d*x + c) + 7560*
(a*b^2*cosh(d*x + c)^5 - 6*a*b^2*cosh(d*x + c)^3 + 15*a*b^2*cosh(d*x + c))*sinh(d*x + c))/d

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giac [A]  time = 0.19, size = 327, normalized size = 1.60 \[ \frac {b^{3} e^{\left (9 \, d x + 9 \, c\right )}}{4608 \, d} - \frac {9 \, b^{3} e^{\left (7 \, d x + 7 \, c\right )}}{3584 \, d} + \frac {a b^{2} e^{\left (6 \, d x + 6 \, c\right )}}{128 \, d} + \frac {9 \, b^{3} e^{\left (5 \, d x + 5 \, c\right )}}{640 \, d} - \frac {9 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )}}{128 \, d} + \frac {45 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} - \frac {45 \, a b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} + \frac {9 \, a b^{2} e^{\left (-4 \, d x - 4 \, c\right )}}{128 \, d} + \frac {9 \, b^{3} e^{\left (-5 \, d x - 5 \, c\right )}}{640 \, d} - \frac {a b^{2} e^{\left (-6 \, d x - 6 \, c\right )}}{128 \, d} - \frac {9 \, b^{3} e^{\left (-7 \, d x - 7 \, c\right )}}{3584 \, d} + \frac {b^{3} e^{\left (-9 \, d x - 9 \, c\right )}}{4608 \, d} + \frac {1}{16} \, {\left (16 \, a^{3} - 15 \, a b^{2}\right )} x + \frac {{\left (16 \, a^{2} b - 7 \, b^{3}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{128 \, d} - \frac {9 \, {\left (32 \, a^{2} b - 7 \, b^{3}\right )} e^{\left (d x + c\right )}}{256 \, d} - \frac {9 \, {\left (32 \, a^{2} b - 7 \, b^{3}\right )} e^{\left (-d x - c\right )}}{256 \, d} + \frac {{\left (16 \, a^{2} b - 7 \, b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4608*b^3*e^(9*d*x + 9*c)/d - 9/3584*b^3*e^(7*d*x + 7*c)/d + 1/128*a*b^2*e^(6*d*x + 6*c)/d + 9/640*b^3*e^(5*d
*x + 5*c)/d - 9/128*a*b^2*e^(4*d*x + 4*c)/d + 45/128*a*b^2*e^(2*d*x + 2*c)/d - 45/128*a*b^2*e^(-2*d*x - 2*c)/d
+ 9/128*a*b^2*e^(-4*d*x - 4*c)/d + 9/640*b^3*e^(-5*d*x - 5*c)/d - 1/128*a*b^2*e^(-6*d*x - 6*c)/d - 9/3584*b^3
*e^(-7*d*x - 7*c)/d + 1/4608*b^3*e^(-9*d*x - 9*c)/d + 1/16*(16*a^3 - 15*a*b^2)*x + 1/128*(16*a^2*b - 7*b^3)*e^
(3*d*x + 3*c)/d - 9/256*(32*a^2*b - 7*b^3)*e^(d*x + c)/d - 9/256*(32*a^2*b - 7*b^3)*e^(-d*x - c)/d + 1/128*(16
*a^2*b - 7*b^3)*e^(-3*d*x - 3*c)/d

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maple [A]  time = 0.05, size = 141, normalized size = 0.69 \[ \frac {b^{3} \left (\frac {128}{315}+\frac {\left (\sinh ^{8}\left (d x +c \right )\right )}{9}-\frac {8 \left (\sinh ^{6}\left (d x +c \right )\right )}{63}+\frac {16 \left (\sinh ^{4}\left (d x +c \right )\right )}{105}-\frac {64 \left (\sinh ^{2}\left (d x +c \right )\right )}{315}\right ) \cosh \left (d x +c \right )+3 a \,b^{2} \left (\left (\frac {\left (\sinh ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sinh ^{3}\left (d x +c \right )\right )}{24}+\frac {5 \sinh \left (d x +c \right )}{16}\right ) \cosh \left (d x +c \right )-\frac {5 d x}{16}-\frac {5 c}{16}\right )+3 a^{2} b \left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\left (d x +c \right )\right )}{3}\right ) \cosh \left (d x +c \right )+a^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^3*(128/315+1/9*sinh(d*x+c)^8-8/63*sinh(d*x+c)^6+16/105*sinh(d*x+c)^4-64/315*sinh(d*x+c)^2)*cosh(d*x+c)+
3*a*b^2*((1/6*sinh(d*x+c)^5-5/24*sinh(d*x+c)^3+5/16*sinh(d*x+c))*cosh(d*x+c)-5/16*d*x-5/16*c)+3*a^2*b*(-2/3+1/
3*sinh(d*x+c)^2)*cosh(d*x+c)+a^3*(d*x+c))

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maxima [A]  time = 0.33, size = 280, normalized size = 1.37 \[ a^{3} x - \frac {1}{161280} \, b^{3} {\left (\frac {{\left (405 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2268 \, e^{\left (-4 \, d x - 4 \, c\right )} + 8820 \, e^{\left (-6 \, d x - 6 \, c\right )} - 39690 \, e^{\left (-8 \, d x - 8 \, c\right )} - 35\right )} e^{\left (9 \, d x + 9 \, c\right )}}{d} - \frac {39690 \, e^{\left (-d x - c\right )} - 8820 \, e^{\left (-3 \, d x - 3 \, c\right )} + 2268 \, e^{\left (-5 \, d x - 5 \, c\right )} - 405 \, e^{\left (-7 \, d x - 7 \, c\right )} + 35 \, e^{\left (-9 \, d x - 9 \, c\right )}}{d}\right )} - \frac {1}{128} \, a b^{2} {\left (\frac {{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {120 \, {\left (d x + c\right )}}{d} + \frac {45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} + \frac {1}{8} \, a^{2} b {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} + \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 1/161280*b^3*((405*e^(-2*d*x - 2*c) - 2268*e^(-4*d*x - 4*c) + 8820*e^(-6*d*x - 6*c) - 39690*e^(-8*d*x
- 8*c) - 35)*e^(9*d*x + 9*c)/d - (39690*e^(-d*x - c) - 8820*e^(-3*d*x - 3*c) + 2268*e^(-5*d*x - 5*c) - 405*e^(
-7*d*x - 7*c) + 35*e^(-9*d*x - 9*c))/d) - 1/128*a*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) + 1/8*a^2*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)

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mupad [B]  time = 0.93, size = 164, normalized size = 0.80 \[ \frac {d\,x\,a^3+a^2\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^3-3\,a^2\,b\,\mathrm {cosh}\left (c+d\,x\right )+\frac {\mathrm {sinh}\left (c+d\,x\right )\,a\,b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^5}{2}-\frac {13\,\mathrm {sinh}\left (c+d\,x\right )\,a\,b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{8}+\frac {33\,\mathrm {sinh}\left (c+d\,x\right )\,a\,b^2\,\mathrm {cosh}\left (c+d\,x\right )}{16}-\frac {15\,d\,x\,a\,b^2}{16}+\frac {b^3\,{\mathrm {cosh}\left (c+d\,x\right )}^9}{9}-\frac {4\,b^3\,{\mathrm {cosh}\left (c+d\,x\right )}^7}{7}+\frac {6\,b^3\,{\mathrm {cosh}\left (c+d\,x\right )}^5}{5}-\frac {4\,b^3\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{3}+b^3\,\mathrm {cosh}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^3)^3,x)

[Out]

(b^3*cosh(c + d*x) - (4*b^3*cosh(c + d*x)^3)/3 + (6*b^3*cosh(c + d*x)^5)/5 - (4*b^3*cosh(c + d*x)^7)/7 + (b^3*
cosh(c + d*x)^9)/9 + a^2*b*cosh(c + d*x)^3 - 3*a^2*b*cosh(c + d*x) + a^3*d*x + (33*a*b^2*cosh(c + d*x)*sinh(c
+ d*x))/16 - (15*a*b^2*d*x)/16 - (13*a*b^2*cosh(c + d*x)^3*sinh(c + d*x))/8 + (a*b^2*cosh(c + d*x)^5*sinh(c +
d*x))/2)/d

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sympy [A]  time = 13.92, size = 340, normalized size = 1.67 \[ \begin {cases} a^{3} x + \frac {3 a^{2} b \sinh ^{2}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {2 a^{2} b \cosh ^{3}{\left (c + d x \right )}}{d} + \frac {15 a b^{2} x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac {45 a b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac {45 a b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac {15 a b^{2} x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac {33 a b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{16 d} - \frac {5 a b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{2 d} + \frac {15 a b^{2} \sinh {\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{16 d} + \frac {b^{3} \sinh ^{8}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {8 b^{3} \sinh ^{6}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {16 b^{3} \sinh ^{4}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{5 d} - \frac {64 b^{3} \sinh ^{2}{\left (c + d x \right )} \cosh ^{7}{\left (c + d x \right )}}{35 d} + \frac {128 b^{3} \cosh ^{9}{\left (c + d x \right )}}{315 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{3}{\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*sinh(c + d*x)**2*cosh(c + d*x)/d - 2*a**2*b*cosh(c + d*x)**3/d + 15*a*b**2*x*sinh
(c + d*x)**6/16 - 45*a*b**2*x*sinh(c + d*x)**4*cosh(c + d*x)**2/16 + 45*a*b**2*x*sinh(c + d*x)**2*cosh(c + d*x
)**4/16 - 15*a*b**2*x*cosh(c + d*x)**6/16 + 33*a*b**2*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) - 5*a*b**2*sinh(c
+ d*x)**3*cosh(c + d*x)**3/(2*d) + 15*a*b**2*sinh(c + d*x)*cosh(c + d*x)**5/(16*d) + b**3*sinh(c + d*x)**8*cos
h(c + d*x)/d - 8*b**3*sinh(c + d*x)**6*cosh(c + d*x)**3/(3*d) + 16*b**3*sinh(c + d*x)**4*cosh(c + d*x)**5/(5*d
) - 64*b**3*sinh(c + d*x)**2*cosh(c + d*x)**7/(35*d) + 128*b**3*cosh(c + d*x)**9/(315*d), Ne(d, 0)), (x*(a + b
*sinh(c)**3)**3, True))

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