3.1 \(\int \sinh ^4(c+d x) (a+b \sinh ^2(c+d x)) \, dx\)

Optimal. Leaf size=89 \[ \frac{(6 a-5 b) \sinh ^3(c+d x) \cosh (c+d x)}{24 d}-\frac{(6 a-5 b) \sinh (c+d x) \cosh (c+d x)}{16 d}+\frac{1}{16} x (6 a-5 b)+\frac{b \sinh ^5(c+d x) \cosh (c+d x)}{6 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0548921, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3014, 2635, 8} \[ \frac{(6 a-5 b) \sinh ^3(c+d x) \cosh (c+d x)}{24 d}-\frac{(6 a-5 b) \sinh (c+d x) \cosh (c+d x)}{16 d}+\frac{1}{16} x (6 a-5 b)+\frac{b \sinh ^5(c+d x) \cosh (c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^4*(a + b*Sinh[c + d*x]^2),x]

[Out]

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

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.108146, size = 68, normalized size = 0.76 \[ \frac{(45 b-48 a) \sinh (2 (c+d x))+(6 a-9 b) \sinh (4 (c+d x))+72 a c+72 a d x+b \sinh (6 (c+d x))-60 b c-60 b d x}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^4*(a + b*Sinh[c + d*x]^2),x]

[Out]

(72*a*c - 60*b*c + 72*a*d*x - 60*b*d*x + (-48*a + 45*b)*Sinh[2*(c + d*x)] + (6*a - 9*b)*Sinh[4*(c + d*x)] + b*
Sinh[6*(c + d*x)])/(192*d)

________________________________________________________________________________________

Maple [A]  time = 0.014, size = 88, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( b \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{6}}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{5\,\sinh \left ( dx+c \right ) }{16}} \right ) \cosh \left ( dx+c \right ) -{\frac{5\,dx}{16}}-{\frac{5\,c}{16}} \right ) +a \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4*(a+b*sinh(d*x+c)^2),x)

[Out]

1/d*(b*((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)+a*((1/4*sinh(d*x+
c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c))

________________________________________________________________________________________

Maxima [A]  time = 1.034, size = 203, normalized size = 2.28 \begin{align*} \frac{1}{64} \, a{\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}{384} \, b{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*sinh(d*x+c)^2),x, algorithm="maxima")

[Out]

1/64*a*(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) - 1/384*b*
((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)

________________________________________________________________________________________

Fricas [A]  time = 1.95221, size = 319, normalized size = 3.58 \begin{align*} \frac{3 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2 \,{\left (5 \, b \cosh \left (d x + c\right )^{3} + 3 \,{\left (2 \, a - 3 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 6 \,{\left (6 \, a - 5 \, b\right )} d x + 3 \,{\left (b \cosh \left (d x + c\right )^{5} + 2 \,{\left (2 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{3} -{\left (16 \, a - 15 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*sinh(d*x+c)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 4.7777, size = 258, normalized size = 2.9 \begin{align*} \begin{cases} \frac{3 a x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac{3 a x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac{3 a x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac{5 a \sinh ^{3}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{8 d} - \frac{3 a \sinh{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac{5 b x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac{15 b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac{15 b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac{5 b x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac{11 b \sinh ^{5}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{16 d} - \frac{5 b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} + \frac{5 b \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 ) \sinh ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**4*(a+b*sinh(d*x+c)**2),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.32322, size = 221, normalized size = 2.48 \begin{align*} \frac{24 \,{\left (d x + c\right )}{\left (6 \, a - 5 \, b\right )} + b e^{\left (6 \, d x + 6 \, c\right )} + 6 \, a e^{\left (4 \, d x + 4 \, c\right )} - 9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 48 \, a e^{\left (2 \, d x + 2 \, c\right )} + 45 \, b e^{\left (2 \, d x + 2 \, c\right )} -{\left (132 \, a e^{\left (6 \, d x + 6 \, c\right )} - 110 \, b e^{\left (6 \, d x + 6 \, c\right )} - 48 \, a e^{\left (4 \, d x + 4 \, c\right )} + 45 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 9 \, b e^{\left (2 \, d x + 2 \, c\right )} + b\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*sinh(d*x+c)^2),x, algorithm="giac")

[Out]

1/384*(24*(d*x + c)*(6*a - 5*b) + b*e^(6*d*x + 6*c) + 6*a*e^(4*d*x + 4*c) - 9*b*e^(4*d*x + 4*c) - 48*a*e^(2*d*
x + 2*c) + 45*b*e^(2*d*x + 2*c) - (132*a*e^(6*d*x + 6*c) - 110*b*e^(6*d*x + 6*c) - 48*a*e^(4*d*x + 4*c) + 45*b
*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) - 9*b*e^(2*d*x + 2*c) + b)*e^(-6*d*x - 6*c))/d