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

Optimal. Leaf size=67 \[ \frac {a^3 \sinh (c+d x)}{d}+\frac {a^2 b \sinh ^3(c+d x)}{d}+\frac {3 a b^2 \sinh ^5(c+d x)}{5 d}+\frac {b^3 \sinh ^7(c+d x)}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, 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} \[ \frac {a^2 b \sinh ^3(c+d x)}{d}+\frac {a^3 \sinh (c+d x)}{d}+\frac {3 a b^2 \sinh ^5(c+d x)}{5 d}+\frac {b^3 \sinh ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 59, normalized size = 0.88 \[ \frac {a^3 \sinh (c+d x)+a^2 b \sinh ^3(c+d x)+\frac {3}{5} a b^2 \sinh ^5(c+d x)+\frac {1}{7} b^3 \sinh ^7(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.40, size = 209, normalized size = 3.12 \[ \frac {5 \, b^{3} \sinh \left (d x + c\right )^{7} + 7 \, {\left (15 \, b^{3} \cosh \left (d x + c\right )^{2} + 12 \, a b^{2} - 5 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 35 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 16 \, a^{2} b - 12 \, a b^{2} + 3 \, b^{3} + 2 \, {\left (12 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 35 \, {\left (b^{3} \cosh \left (d x + c\right )^{6} + {\left (12 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 64 \, a^{3} - 48 \, a^{2} b + 24 \, a b^{2} - 5 \, b^{3} + 3 \, {\left (16 \, a^{2} b - 12 \, a b^{2} + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{2240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2240*(5*b^3*sinh(d*x + c)^7 + 7*(15*b^3*cosh(d*x + c)^2 + 12*a*b^2 - 5*b^3)*sinh(d*x + c)^5 + 35*(5*b^3*cosh
(d*x + c)^4 + 16*a^2*b - 12*a*b^2 + 3*b^3 + 2*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 35*(b^3*co
sh(d*x + c)^6 + (12*a*b^2 - 5*b^3)*cosh(d*x + c)^4 + 64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3 + 3*(16*a^2*b - 12*a
*b^2 + 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/d

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giac [B]  time = 0.19, size = 222, normalized size = 3.31 \[ \frac {b^{3} e^{\left (7 \, d x + 7 \, c\right )}}{896 \, d} - \frac {b^{3} e^{\left (-7 \, d x - 7 \, c\right )}}{896 \, d} + \frac {{\left (12 \, a b^{2} - 5 \, b^{3}\right )} e^{\left (5 \, d x + 5 \, c\right )}}{640 \, d} + \frac {{\left (16 \, a^{2} b - 12 \, a b^{2} + 3 \, b^{3}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{128 \, d} + \frac {{\left (64 \, a^{3} - 48 \, a^{2} b + 24 \, a b^{2} - 5 \, b^{3}\right )} e^{\left (d x + c\right )}}{128 \, d} - \frac {{\left (64 \, a^{3} - 48 \, a^{2} b + 24 \, a b^{2} - 5 \, b^{3}\right )} e^{\left (-d x - c\right )}}{128 \, d} - \frac {{\left (16 \, a^{2} b - 12 \, a b^{2} + 3 \, b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{128 \, d} - \frac {{\left (12 \, a b^{2} - 5 \, b^{3}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{640 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/896*b^3*e^(7*d*x + 7*c)/d - 1/896*b^3*e^(-7*d*x - 7*c)/d + 1/640*(12*a*b^2 - 5*b^3)*e^(5*d*x + 5*c)/d + 1/12
8*(16*a^2*b - 12*a*b^2 + 3*b^3)*e^(3*d*x + 3*c)/d + 1/128*(64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*e^(d*x + c)/d
- 1/128*(64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*e^(-d*x - c)/d - 1/128*(16*a^2*b - 12*a*b^2 + 3*b^3)*e^(-3*d*x
- 3*c)/d - 1/640*(12*a*b^2 - 5*b^3)*e^(-5*d*x - 5*c)/d

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maple [A]  time = 0.04, size = 56, normalized size = 0.84 \[ \frac {\frac {b^{3} \left (\sinh ^{7}\left (d x +c \right )\right )}{7}+\frac {3 a \,b^{2} \left (\sinh ^{5}\left (d x +c \right )\right )}{5}+a^{2} b \left (\sinh ^{3}\left (d x +c \right )\right )+a^{3} \sinh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 63, normalized size = 0.94 \[ \frac {b^{3} \sinh \left (d x + c\right )^{7}}{7 \, d} + \frac {3 \, a b^{2} \sinh \left (d x + c\right )^{5}}{5 \, d} + \frac {a^{2} b \sinh \left (d x + c\right )^{3}}{d} + \frac {a^{3} \sinh \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.15, size = 58, normalized size = 0.87 \[ \frac {\mathrm {sinh}\left (c+d\,x\right )\,\left (35\,a^3+35\,a^2\,b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+21\,a\,b^2\,{\mathrm {sinh}\left (c+d\,x\right )}^4+5\,b^3\,{\mathrm {sinh}\left (c+d\,x\right )}^6\right )}{35\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sinh(c + d*x)*(35*a^3 + 5*b^3*sinh(c + d*x)^6 + 35*a^2*b*sinh(c + d*x)^2 + 21*a*b^2*sinh(c + d*x)^4))/(35*d)

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sympy [A]  time = 4.97, size = 75, normalized size = 1.12 \[ \begin {cases} \frac {a^{3} \sinh {\left (c + d x \right )}}{d} + \frac {a^{2} b \sinh ^{3}{\left (c + d x \right )}}{d} + \frac {3 a b^{2} \sinh ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{3} \sinh ^{7}{\left (c + d x \right )}}{7 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\relax (c )}\right )^{3} \cosh {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*sinh(c + d*x)/d + a**2*b*sinh(c + d*x)**3/d + 3*a*b**2*sinh(c + d*x)**5/(5*d) + b**3*sinh(c +
d*x)**7/(7*d), Ne(d, 0)), (x*(a + b*sinh(c)**2)**3*cosh(c), True))

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