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

Optimal. Leaf size=110 \[ \frac{\left (8 a^2-20 a b+11 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} x \left (8 a^2-12 a b+5 b^2\right )+\frac{b (4 a-3 b) \sinh (c+d x) \cosh ^3(c+d x)}{8 d}+\frac{b^2 \sinh ^3(c+d x) \cosh ^3(c+d x)}{6 d} \]

[Out]

-((8*a^2 - 12*a*b + 5*b^2)*x)/16 + ((8*a^2 - 20*a*b + 11*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) + ((4*a - 3*
b)*b*Cosh[c + d*x]^3*Sinh[c + d*x])/(8*d) + (b^2*Cosh[c + d*x]^3*Sinh[c + d*x]^3)/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.110067, antiderivative size = 117, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3170, 3169} \[ \frac{\left (16 a^2-36 a b+15 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{48 d}-\frac{1}{16} x \left (8 a^2-12 a b+5 b^2\right )+\frac{b (4 a-5 b) \sinh ^3(c+d x) \cosh (c+d x)}{24 d}+\frac{\sinh (c+d x) \cosh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 - 12*a*b + 5*b^2)*x)/16 + ((16*a^2 - 36*a*b + 15*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(48*d) + ((4*a - 5
*b)*b*Cosh[c + d*x]*Sinh[c + d*x]^3)/(24*d) + (Cosh[c + d*x]*Sinh[c + d*x]*(a + b*Sinh[c + d*x]^2)^2)/(6*d)

Rule 3170

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

Rule 3169

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]

Rubi steps

\begin{align*} \int \sinh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\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-(4 a-5 b) \sinh ^2(c+d x)\right ) \left (a+b \sinh ^2(c+d x)\right ) \, dx\\ &=-\frac{1}{16} \left (8 a^2-12 a b+5 b^2\right ) x+\frac{\left (16 a^2-36 a b+15 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{48 d}+\frac{(4 a-5 b) b \cosh (c+d x) \sinh ^3(c+d x)}{24 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^2}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.191873, size = 99, normalized size = 0.9 \[ \frac{\left (48 a^2-96 a b+45 b^2\right ) \sinh (2 (c+d x))-96 a^2 c-96 a^2 d x+3 b (4 a-3 b) \sinh (4 (c+d x))+144 a b c+144 a b d x+b^2 \sinh (6 (c+d x))-60 b^2 c-60 b^2 d x}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-96*a^2*c + 144*a*b*c - 60*b^2*c - 96*a^2*d*x + 144*a*b*d*x - 60*b^2*d*x + (48*a^2 - 96*a*b + 45*b^2)*Sinh[2*
(c + d*x)] + 3*(4*a - 3*b)*b*Sinh[4*(c + d*x)] + b^2*Sinh[6*(c + d*x)])/(192*d)

________________________________________________________________________________________

Maple [A]  time = 0.017, size = 118, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({b}^{2} \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 ) +2\,ab \left ( \left ( 1/4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}-3/8\,\sinh \left ( dx+c \right ) \right ) \cosh \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{a}^{2} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)+2*a*b*((1/4*sin
h(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+a^2*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c))

________________________________________________________________________________________

Maxima [A]  time = 1.04317, size = 255, normalized size = 2.32 \begin{align*} \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}{8} \, a^{2}{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2*(a+b*sinh(d*x+c)^2)^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) - 1/8*a^
2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 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)

________________________________________________________________________________________

Fricas [A]  time = 2.14473, size = 373, normalized size = 3.39 \begin{align*} \frac{3 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2 \,{\left (5 \, b^{2} \cosh \left (d x + c\right )^{3} + 3 \,{\left (4 \, a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 6 \,{\left (8 \, a^{2} - 12 \, a b + 5 \, b^{2}\right )} d x + 3 \,{\left (b^{2} \cosh \left (d x + c\right )^{5} + 2 \,{\left (4 \, a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} +{\left (16 \, a^{2} - 32 \, a b + 15 \, b^{2}\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)^2*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/96*(3*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(5*b^2*cosh(d*x + c)^3 + 3*(4*a*b - 3*b^2)*cosh(d*x + c))*sinh(d
*x + c)^3 - 6*(8*a^2 - 12*a*b + 5*b^2)*d*x + 3*(b^2*cosh(d*x + c)^5 + 2*(4*a*b - 3*b^2)*cosh(d*x + c)^3 + (16*
a^2 - 32*a*b + 15*b^2)*cosh(d*x + c))*sinh(d*x + c))/d

________________________________________________________________________________________

Sympy [A]  time = 4.87993, size = 332, normalized size = 3.02 \begin{align*} \begin{cases} \frac{a^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac{a^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{2 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{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 ^{2}{\left (c \right )}\right )^{2} \sinh ^{2}{\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)**2*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Piecewise((a**2*x*sinh(c + d*x)**2/2 - a**2*x*cosh(c + d*x)**2/2 + a**2*sinh(c + d*x)*cosh(c + d*x)/(2*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*si
nh(c + d*x)**3*cosh(c + d*x)/(4*d) - 3*a*b*sinh(c + d*x)*cosh(c + d*x)**3/(4*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)**2)**2*sinh(c)**2, True))

________________________________________________________________________________________

Giac [B]  time = 1.35627, size = 316, normalized size = 2.87 \begin{align*} \frac{b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 12 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 9 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 48 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 96 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 45 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 24 \,{\left (8 \, a^{2} - 12 \, a b + 5 \, b^{2}\right )}{\left (d x + c\right )} +{\left (176 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 264 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 110 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 48 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 96 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 45 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 9 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}\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)^2*(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/384*(b^2*e^(6*d*x + 6*c) + 12*a*b*e^(4*d*x + 4*c) - 9*b^2*e^(4*d*x + 4*c) + 48*a^2*e^(2*d*x + 2*c) - 96*a*b*
e^(2*d*x + 2*c) + 45*b^2*e^(2*d*x + 2*c) - 24*(8*a^2 - 12*a*b + 5*b^2)*(d*x + c) + (176*a^2*e^(6*d*x + 6*c) -
264*a*b*e^(6*d*x + 6*c) + 110*b^2*e^(6*d*x + 6*c) - 48*a^2*e^(4*d*x + 4*c) + 96*a*b*e^(4*d*x + 4*c) - 45*b^2*e
^(4*d*x + 4*c) - 12*a*b*e^(2*d*x + 2*c) + 9*b^2*e^(2*d*x + 2*c) - b^2)*e^(-6*d*x - 6*c))/d