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

Optimal. Leaf size=57 \[ \frac{2 b (a-b) \cosh ^3(c+d x)}{3 d}+\frac{(a-b)^2 \cosh (c+d x)}{d}+\frac{b^2 \cosh ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0600295, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3186, 194} \[ \frac{2 b (a-b) \cosh ^3(c+d x)}{3 d}+\frac{(a-b)^2 \cosh (c+d x)}{d}+\frac{b^2 \cosh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0362276, size = 111, normalized size = 1.95 \[ \frac{a^2 \sinh (c) \sinh (d x)}{d}+\frac{a^2 \cosh (c) \cosh (d x)}{d}-\frac{3 a b \cosh (c+d x)}{2 d}+\frac{a b \cosh (3 (c+d x))}{6 d}+\frac{5 b^2 \cosh (c+d x)}{8 d}-\frac{5 b^2 \cosh (3 (c+d x))}{48 d}+\frac{b^2 \cosh (5 (c+d x))}{80 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cosh[c]*Cosh[d*x])/d - (3*a*b*Cosh[c + d*x])/(2*d) + (5*b^2*Cosh[c + d*x])/(8*d) + (a*b*Cosh[3*(c + d*x)]
)/(6*d) - (5*b^2*Cosh[3*(c + d*x)])/(48*d) + (b^2*Cosh[5*(c + d*x)])/(80*d) + (a^2*Sinh[c]*Sinh[d*x])/d

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Maple [A]  time = 0.015, size = 70, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{8}{15}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \cosh \left ( dx+c \right ) +2\,ab \left ( -2/3+1/3\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2} \right ) \cosh \left ( dx+c \right ) +{a}^{2}\cosh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.04599, size = 212, normalized size = 3.72 \begin{align*} \frac{1}{480} \, b^{2}{\left (\frac{3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} - \frac{25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac{150 \, e^{\left (d x + c\right )}}{d} + \frac{150 \, e^{\left (-d x - c\right )}}{d} - \frac{25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{3 \, e^{\left (-5 \, d x - 5 \, 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 )} + \frac{a^{2} \cosh \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*b^2*(3*e^(5*d*x + 5*c)/d - 25*e^(3*d*x + 3*c)/d + 150*e^(d*x + c)/d + 150*e^(-d*x - c)/d - 25*e^(-3*d*x
- 3*c)/d + 3*e^(-5*d*x - 5*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) + a^2*cosh(d*x + c)/d

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Fricas [B]  time = 2.01757, size = 309, normalized size = 5.42 \begin{align*} \frac{3 \, b^{2} \cosh \left (d x + c\right )^{5} + 15 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \,{\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 15 \,{\left (2 \, b^{2} \cosh \left (d x + c\right )^{3} +{\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 30 \,{\left (8 \, a^{2} - 12 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*b^2*cosh(d*x + c)^5 + 15*b^2*cosh(d*x + c)*sinh(d*x + c)^4 + 5*(8*a*b - 5*b^2)*cosh(d*x + c)^3 + 15*(
2*b^2*cosh(d*x + c)^3 + (8*a*b - 5*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 30*(8*a^2 - 12*a*b + 5*b^2)*cosh(d*x
+ c))/d

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Sympy [A]  time = 2.35011, size = 128, normalized size = 2.25 \begin{align*} \begin{cases} \frac{a^{2} \cosh{\left (c + d x \right )}}{d} + \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{b^{2} \sinh ^{4}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{d} - \frac{4 b^{2} \sinh ^{2}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac{8 b^{2} \cosh ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right )^{2} \sinh{\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)*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

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

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Giac [B]  time = 1.25033, size = 220, normalized size = 3.86 \begin{align*} \frac{3 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 40 \, a b e^{\left (3 \, d x + 3 \, c\right )} - 25 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 240 \, a^{2} e^{\left (d x + c\right )} - 360 \, a b e^{\left (d x + c\right )} + 150 \, b^{2} e^{\left (d x + c\right )} +{\left (240 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 360 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 150 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 25 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{2}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*b^2*e^(5*d*x + 5*c) + 40*a*b*e^(3*d*x + 3*c) - 25*b^2*e^(3*d*x + 3*c) + 240*a^2*e^(d*x + c) - 360*a*b
*e^(d*x + c) + 150*b^2*e^(d*x + c) + (240*a^2*e^(4*d*x + 4*c) - 360*a*b*e^(4*d*x + 4*c) + 150*b^2*e^(4*d*x + 4
*c) + 40*a*b*e^(2*d*x + 2*c) - 25*b^2*e^(2*d*x + 2*c) + 3*b^2)*e^(-5*d*x - 5*c))/d