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

Optimal. Leaf size=130 \[ \frac {a^2 \cosh (c+d x)}{d}+\frac {a b \sinh ^3(c+d x) \cosh (c+d x)}{2 d}-\frac {3 a b \sinh (c+d x) \cosh (c+d x)}{4 d}+\frac {3 a b x}{4}+\frac {b^2 \cosh ^7(c+d x)}{7 d}-\frac {3 b^2 \cosh ^5(c+d x)}{5 d}+\frac {b^2 \cosh ^3(c+d x)}{d}-\frac {b^2 \cosh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3220, 2638, 2635, 8, 2633} \[ \frac {a^2 \cosh (c+d x)}{d}+\frac {a b \sinh ^3(c+d x) \cosh (c+d x)}{2 d}-\frac {3 a b \sinh (c+d x) \cosh (c+d x)}{4 d}+\frac {3 a b x}{4}+\frac {b^2 \cosh ^7(c+d x)}{7 d}-\frac {3 b^2 \cosh ^5(c+d x)}{5 d}+\frac {b^2 \cosh ^3(c+d x)}{d}-\frac {b^2 \cosh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b*x)/4 + (a^2*Cosh[c + d*x])/d - (b^2*Cosh[c + d*x])/d + (b^2*Cosh[c + d*x]^3)/d - (3*b^2*Cosh[c + d*x]^5
)/(5*d) + (b^2*Cosh[c + d*x]^7)/(7*d) - (3*a*b*Cosh[c + d*x]*Sinh[c + d*x])/(4*d) + (a*b*Cosh[c + d*x]*Sinh[c
+ d*x]^3)/(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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 92, normalized size = 0.71 \[ \frac {35 \left (64 a^2-35 b^2\right ) \cosh (c+d x)+b (140 a (12 (c+d x)-8 \sinh (2 (c+d x))+\sinh (4 (c+d x)))+245 b \cosh (3 (c+d x))-49 b \cosh (5 (c+d x))+5 b \cosh (7 (c+d x)))}{2240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*(64*a^2 - 35*b^2)*Cosh[c + d*x] + b*(245*b*Cosh[3*(c + d*x)] - 49*b*Cosh[5*(c + d*x)] + 5*b*Cosh[7*(c + d*
x)] + 140*a*(12*(c + d*x) - 8*Sinh[2*(c + d*x)] + Sinh[4*(c + d*x)])))/(2240*d)

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fricas [A]  time = 0.47, size = 220, normalized size = 1.69 \[ \frac {5 \, b^{2} \cosh \left (d x + c\right )^{7} + 35 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - 49 \, b^{2} \cosh \left (d x + c\right )^{5} + 560 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 245 \, b^{2} \cosh \left (d x + c\right )^{3} + 35 \, {\left (5 \, b^{2} \cosh \left (d x + c\right )^{3} - 7 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 1680 \, a b d x + 35 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{5} - 14 \, b^{2} \cosh \left (d x + c\right )^{3} + 21 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 35 \, {\left (64 \, a^{2} - 35 \, b^{2}\right )} \cosh \left (d x + c\right ) + 560 \, {\left (a b \cosh \left (d x + c\right )^{3} - 4 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2240*(5*b^2*cosh(d*x + c)^7 + 35*b^2*cosh(d*x + c)*sinh(d*x + c)^6 - 49*b^2*cosh(d*x + c)^5 + 560*a*b*cosh(d
*x + c)*sinh(d*x + c)^3 + 245*b^2*cosh(d*x + c)^3 + 35*(5*b^2*cosh(d*x + c)^3 - 7*b^2*cosh(d*x + c))*sinh(d*x
+ c)^4 + 1680*a*b*d*x + 35*(3*b^2*cosh(d*x + c)^5 - 14*b^2*cosh(d*x + c)^3 + 21*b^2*cosh(d*x + c))*sinh(d*x +
c)^2 + 35*(64*a^2 - 35*b^2)*cosh(d*x + c) + 560*(a*b*cosh(d*x + c)^3 - 4*a*b*cosh(d*x + c))*sinh(d*x + c))/d

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giac [A]  time = 0.20, size = 219, normalized size = 1.68 \[ \frac {3}{4} \, a b x + \frac {b^{2} e^{\left (7 \, d x + 7 \, c\right )}}{896 \, d} - \frac {7 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )}}{640 \, d} + \frac {a b e^{\left (4 \, d x + 4 \, c\right )}}{32 \, d} + \frac {7 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )}}{128 \, d} - \frac {a b e^{\left (2 \, d x + 2 \, c\right )}}{4 \, d} + \frac {a b e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, d} + \frac {7 \, b^{2} e^{\left (-3 \, d x - 3 \, c\right )}}{128 \, d} - \frac {a b e^{\left (-4 \, d x - 4 \, c\right )}}{32 \, d} - \frac {7 \, b^{2} e^{\left (-5 \, d x - 5 \, c\right )}}{640 \, d} + \frac {b^{2} e^{\left (-7 \, d x - 7 \, c\right )}}{896 \, d} + \frac {{\left (64 \, a^{2} - 35 \, b^{2}\right )} e^{\left (d x + c\right )}}{128 \, d} + \frac {{\left (64 \, a^{2} - 35 \, b^{2}\right )} e^{\left (-d x - c\right )}}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*a*b*x + 1/896*b^2*e^(7*d*x + 7*c)/d - 7/640*b^2*e^(5*d*x + 5*c)/d + 1/32*a*b*e^(4*d*x + 4*c)/d + 7/128*b^2
*e^(3*d*x + 3*c)/d - 1/4*a*b*e^(2*d*x + 2*c)/d + 1/4*a*b*e^(-2*d*x - 2*c)/d + 7/128*b^2*e^(-3*d*x - 3*c)/d - 1
/32*a*b*e^(-4*d*x - 4*c)/d - 7/640*b^2*e^(-5*d*x - 5*c)/d + 1/896*b^2*e^(-7*d*x - 7*c)/d + 1/128*(64*a^2 - 35*
b^2)*e^(d*x + c)/d + 1/128*(64*a^2 - 35*b^2)*e^(-d*x - c)/d

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maple [A]  time = 0.04, size = 96, normalized size = 0.74 \[ \frac {b^{2} \left (-\frac {16}{35}+\frac {\left (\sinh ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sinh ^{4}\left (d x +c \right )\right )}{35}+\frac {8 \left (\sinh ^{2}\left (d x +c \right )\right )}{35}\right ) \cosh \left (d x +c \right )+2 a b \left (\left (\frac {\left (\sinh ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} \cosh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^2*(-16/35+1/7*sinh(d*x+c)^6-6/35*sinh(d*x+c)^4+8/35*sinh(d*x+c)^2)*cosh(d*x+c)+2*a*b*((1/4*sinh(d*x+c)^
3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+a^2*cosh(d*x+c))

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maxima [A]  time = 0.33, size = 180, normalized size = 1.38 \[ \frac {1}{32} \, a b {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {1}{4480} \, b^{2} {\left (\frac {{\left (49 \, e^{\left (-2 \, d x - 2 \, c\right )} - 245 \, e^{\left (-4 \, d x - 4 \, c\right )} + 1225 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5\right )} e^{\left (7 \, d x + 7 \, c\right )}}{d} + \frac {1225 \, e^{\left (-d x - c\right )} - 245 \, e^{\left (-3 \, d x - 3 \, c\right )} + 49 \, e^{\left (-5 \, d x - 5 \, c\right )} - 5 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d}\right )} + \frac {a^{2} \cosh \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*a*b*(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) - 1/4480
*b^2*((49*e^(-2*d*x - 2*c) - 245*e^(-4*d*x - 4*c) + 1225*e^(-6*d*x - 6*c) - 5)*e^(7*d*x + 7*c)/d + (1225*e^(-d
*x - c) - 245*e^(-3*d*x - 3*c) + 49*e^(-5*d*x - 5*c) - 5*e^(-7*d*x - 7*c))/d) + a^2*cosh(d*x + c)/d

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mupad [B]  time = 0.25, size = 104, normalized size = 0.80 \[ \frac {a^2\,\mathrm {cosh}\left (c+d\,x\right )+\frac {\mathrm {sinh}\left (c+d\,x\right )\,a\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{2}-\frac {5\,\mathrm {sinh}\left (c+d\,x\right )\,a\,b\,\mathrm {cosh}\left (c+d\,x\right )}{4}+\frac {3\,d\,x\,a\,b}{4}+\frac {b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^7}{7}-\frac {3\,b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^5}{5}+b^2\,{\mathrm {cosh}\left (c+d\,x\right )}^3-b^2\,\mathrm {cosh}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.16, size = 219, normalized size = 1.68 \[ \begin {cases} \frac {a^{2} \cosh {\left (c + d x \right )}}{d} + \frac {3 a b x \sinh ^{4}{\left (c + d x \right )}}{4} - \frac {3 a b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b x \cosh ^{4}{\left (c + d x \right )}}{4} + \frac {5 a b \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{4 d} - \frac {3 a b \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{4 d} + \frac {b^{2} \sinh ^{6}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {2 b^{2} \sinh ^{4}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{d} + \frac {8 b^{2} \sinh ^{2}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{5 d} - \frac {16 b^{2} \cosh ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{3}{\relax (c )}\right )^{2} \sinh {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*cosh(c + d*x)/d + 3*a*b*x*sinh(c + d*x)**4/4 - 3*a*b*x*sinh(c + d*x)**2*cosh(c + d*x)**2/2 + 3
*a*b*x*cosh(c + d*x)**4/4 + 5*a*b*sinh(c + d*x)**3*cosh(c + d*x)/(4*d) - 3*a*b*sinh(c + d*x)*cosh(c + d*x)**3/
(4*d) + b**2*sinh(c + d*x)**6*cosh(c + d*x)/d - 2*b**2*sinh(c + d*x)**4*cosh(c + d*x)**3/d + 8*b**2*sinh(c + d
*x)**2*cosh(c + d*x)**5/(5*d) - 16*b**2*cosh(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a + b*sinh(c)**3)**2*sinh(c),
True))

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