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

Optimal. Leaf size=49 \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{2 a b \sinh ^3(c+d x)}{3 d}+\frac{b^2 \sinh ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0370079, antiderivative size = 49, 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 = {3190, 194} \[ \frac{a^2 \sinh (c+d x)}{d}+\frac{2 a b \sinh ^3(c+d x)}{3 d}+\frac{b^2 \sinh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[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 \cosh (c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b x^2\right )^2 \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{a^2 \sinh (c+d x)}{d}+\frac{2 a b \sinh ^3(c+d x)}{3 d}+\frac{b^2 \sinh ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.0560869, size = 44, normalized size = 0.9 \[ \frac{a^2 \sinh (c+d x)+\frac{2}{3} a b \sinh ^3(c+d x)+\frac{1}{5} b^2 \sinh ^5(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 41, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{2\,a \left ( \sinh \left ( dx+c \right ) \right ) ^{3}b}{3}}+{a}^{2}\sinh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16507, size = 61, normalized size = 1.24 \begin{align*} \frac{b^{2} \sinh \left (d x + c\right )^{5}}{5 \, d} + \frac{2 \, a b \sinh \left (d x + c\right )^{3}}{3 \, d} + \frac{a^{2} \sinh \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.47832, size = 261, normalized size = 5.33 \begin{align*} \frac{3 \, b^{2} \sinh \left (d x + c\right )^{5} + 5 \,{\left (6 \, b^{2} \cosh \left (d x + c\right )^{2} + 8 \, a b - 3 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 15 \,{\left (b^{2} \cosh \left (d x + c\right )^{4} +{\left (8 \, a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 16 \, a^{2} - 8 \, a b + 2 \, b^{2}\right )} \sinh \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*b^2*sinh(d*x + c)^5 + 5*(6*b^2*cosh(d*x + c)^2 + 8*a*b - 3*b^2)*sinh(d*x + c)^3 + 15*(b^2*cosh(d*x +
c)^4 + (8*a*b - 3*b^2)*cosh(d*x + c)^2 + 16*a^2 - 8*a*b + 2*b^2)*sinh(d*x + c))/d

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Sympy [A]  time = 2.01711, size = 58, normalized size = 1.18 \begin{align*} \begin{cases} \frac{a^{2} \sinh{\left (c + d x \right )}}{d} + \frac{2 a b \sinh ^{3}{\left (c + d x \right )}}{3 d} + \frac{b^{2} \sinh ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right )^{2} \cosh{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17326, size = 221, normalized size = 4.51 \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 )} - 15 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 240 \, a^{2} e^{\left (d x + c\right )} - 120 \, a b e^{\left (d x + c\right )} + 30 \, b^{2} e^{\left (d x + c\right )} -{\left (240 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 120 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 30 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 15 \, 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(cosh(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) - 15*b^2*e^(3*d*x + 3*c) + 240*a^2*e^(d*x + c) - 120*a*b
*e^(d*x + c) + 30*b^2*e^(d*x + c) - (240*a^2*e^(4*d*x + 4*c) - 120*a*b*e^(4*d*x + 4*c) + 30*b^2*e^(4*d*x + 4*c
) + 40*a*b*e^(2*d*x + 2*c) - 15*b^2*e^(2*d*x + 2*c) + 3*b^2)*e^(-5*d*x - 5*c))/d