3.1027 \(\int \cosh (x) \sinh ^3(x) (a+b \sinh ^2(x))^3 \, dx\)

Optimal. Leaf size=36 \[ \frac{\left (a+b \sinh ^2(x)\right )^5}{10 b^2}-\frac{a \left (a+b \sinh ^2(x)\right )^4}{8 b^2} \]

[Out]

-(a*(a + b*Sinh[x]^2)^4)/(8*b^2) + (a + b*Sinh[x]^2)^5/(10*b^2)

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Rubi [A]  time = 0.093683, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3198, 266, 43} \[ \frac{\left (a+b \sinh ^2(x)\right )^5}{10 b^2}-\frac{a \left (a+b \sinh ^2(x)\right )^4}{8 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a + b*Sinh[x]^2)^4)/(8*b^2) + (a + b*Sinh[x]^2)^5/(10*b^2)

Rule 3198

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^
2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^n*(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, d, e, f, n, p}, x] && IntegerQ[
(m - 1)/2]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cosh (x) \sinh ^3(x) \left (a+b \sinh ^2(x)\right )^3 \, dx &=\operatorname{Subst}\left (\int x^3 \left (a+b x^2\right )^3 \, dx,x,\sinh (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b x)^3 \, dx,x,\sinh ^2(x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{a (a+b x)^3}{b}+\frac{(a+b x)^4}{b}\right ) \, dx,x,\sinh ^2(x)\right )\\ &=-\frac{a \left (a+b \sinh ^2(x)\right )^4}{8 b^2}+\frac{\left (a+b \sinh ^2(x)\right )^5}{10 b^2}\\ \end{align*}

Mathematica [B]  time = 0.583876, size = 114, normalized size = 3.17 \[ \frac{-20 \left (64 a^3+24 a b^2-7 b^3\right ) \cosh (2 x)+20 \left (16 a^3+18 a b^2-5 b^3\right ) \cosh (4 x)+b \left (320 \sinh ^6(x) \left ((b-4 a)^2-b^2 \cosh (2 x)\right )-10 b (16 a-5 b) \cosh (6 x)+15 b (2 a-b) \cosh (8 x)+2 b^2 \cosh (10 x)\right )}{10240} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-20*(64*a^3 + 24*a*b^2 - 7*b^3)*Cosh[2*x] + 20*(16*a^3 + 18*a*b^2 - 5*b^3)*Cosh[4*x] + b*(-10*(16*a - 5*b)*b*
Cosh[6*x] + 15*(2*a - b)*b*Cosh[8*x] + 2*b^2*Cosh[10*x] + 320*((-4*a + b)^2 - b^2*Cosh[2*x])*Sinh[x]^6))/10240

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Maple [A]  time = 0.01, size = 40, normalized size = 1.1 \begin{align*}{\frac{{b}^{3} \left ( \sinh \left ( x \right ) \right ) ^{10}}{10}}+{\frac{3\,a{b}^{2} \left ( \sinh \left ( x \right ) \right ) ^{8}}{8}}+{\frac{{a}^{2}b \left ( \sinh \left ( x \right ) \right ) ^{6}}{2}}+{\frac{{a}^{3} \left ( \sinh \left ( x \right ) \right ) ^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*sinh(x)^10+3/8*a*b^2*sinh(x)^8+1/2*a^2*b*sinh(x)^6+1/4*a^3*sinh(x)^4

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Maxima [A]  time = 1.02097, size = 53, normalized size = 1.47 \begin{align*} \frac{1}{10} \, b^{3} \sinh \left (x\right )^{10} + \frac{3}{8} \, a b^{2} \sinh \left (x\right )^{8} + \frac{1}{2} \, a^{2} b \sinh \left (x\right )^{6} + \frac{1}{4} \, a^{3} \sinh \left (x\right )^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*sinh(x)^10 + 3/8*a*b^2*sinh(x)^8 + 1/2*a^2*b*sinh(x)^6 + 1/4*a^3*sinh(x)^4

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Fricas [B]  time = 2.10862, size = 1041, normalized size = 28.92 \begin{align*} \frac{1}{5120} \, b^{3} \cosh \left (x\right )^{10} + \frac{1}{5120} \, b^{3} \sinh \left (x\right )^{10} + \frac{1}{1024} \,{\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{8} + \frac{1}{1024} \,{\left (9 \, b^{3} \cosh \left (x\right )^{2} + 3 \, a b^{2} - 2 \, b^{3}\right )} \sinh \left (x\right )^{8} + \frac{1}{1024} \,{\left (16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3}\right )} \cosh \left (x\right )^{6} + \frac{1}{1024} \,{\left (42 \, b^{3} \cosh \left (x\right )^{4} + 16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3} + 28 \,{\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{6} + \frac{1}{256} \,{\left (8 \, a^{3} - 24 \, a^{2} b + 21 \, a b^{2} - 6 \, b^{3}\right )} \cosh \left (x\right )^{4} + \frac{1}{1024} \,{\left (42 \, b^{3} \cosh \left (x\right )^{6} + 70 \,{\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{4} + 32 \, a^{3} - 96 \, a^{2} b + 84 \, a b^{2} - 24 \, b^{3} + 15 \,{\left (16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{4} - \frac{1}{512} \,{\left (64 \, a^{3} - 120 \, a^{2} b + 84 \, a b^{2} - 21 \, b^{3}\right )} \cosh \left (x\right )^{2} + \frac{1}{1024} \,{\left (9 \, b^{3} \cosh \left (x\right )^{8} + 28 \,{\left (3 \, a b^{2} - 2 \, b^{3}\right )} \cosh \left (x\right )^{6} + 15 \,{\left (16 \, a^{2} b - 24 \, a b^{2} + 9 \, b^{3}\right )} \cosh \left (x\right )^{4} - 128 \, a^{3} + 240 \, a^{2} b - 168 \, a b^{2} + 42 \, b^{3} + 24 \,{\left (8 \, a^{3} - 24 \, a^{2} b + 21 \, a b^{2} - 6 \, b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5120*b^3*cosh(x)^10 + 1/5120*b^3*sinh(x)^10 + 1/1024*(3*a*b^2 - 2*b^3)*cosh(x)^8 + 1/1024*(9*b^3*cosh(x)^2 +
3*a*b^2 - 2*b^3)*sinh(x)^8 + 1/1024*(16*a^2*b - 24*a*b^2 + 9*b^3)*cosh(x)^6 + 1/1024*(42*b^3*cosh(x)^4 + 16*a
^2*b - 24*a*b^2 + 9*b^3 + 28*(3*a*b^2 - 2*b^3)*cosh(x)^2)*sinh(x)^6 + 1/256*(8*a^3 - 24*a^2*b + 21*a*b^2 - 6*b
^3)*cosh(x)^4 + 1/1024*(42*b^3*cosh(x)^6 + 70*(3*a*b^2 - 2*b^3)*cosh(x)^4 + 32*a^3 - 96*a^2*b + 84*a*b^2 - 24*
b^3 + 15*(16*a^2*b - 24*a*b^2 + 9*b^3)*cosh(x)^2)*sinh(x)^4 - 1/512*(64*a^3 - 120*a^2*b + 84*a*b^2 - 21*b^3)*c
osh(x)^2 + 1/1024*(9*b^3*cosh(x)^8 + 28*(3*a*b^2 - 2*b^3)*cosh(x)^6 + 15*(16*a^2*b - 24*a*b^2 + 9*b^3)*cosh(x)
^4 - 128*a^3 + 240*a^2*b - 168*a*b^2 + 42*b^3 + 24*(8*a^3 - 24*a^2*b + 21*a*b^2 - 6*b^3)*cosh(x)^2)*sinh(x)^2

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Sympy [B]  time = 20.6006, size = 82, normalized size = 2.28 \begin{align*} \frac{a^{3} \sinh ^{4}{\left (x \right )}}{4} + \frac{3 a^{2} b \sinh ^{4}{\left (x \right )} \cosh ^{2}{\left (x \right )}}{2} - \frac{3 a^{2} b \sinh ^{2}{\left (x \right )} \cosh ^{4}{\left (x \right )}}{2} + \frac{a^{2} b \cosh ^{6}{\left (x \right )}}{2} + \frac{3 a b^{2} \sinh ^{8}{\left (x \right )}}{8} + \frac{b^{3} \sinh ^{10}{\left (x \right )}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*sinh(x)**4/4 + 3*a**2*b*sinh(x)**4*cosh(x)**2/2 - 3*a**2*b*sinh(x)**2*cosh(x)**4/2 + a**2*b*cosh(x)**6/2
+ 3*a*b**2*sinh(x)**8/8 + b**3*sinh(x)**10/10

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Giac [B]  time = 1.23311, size = 302, normalized size = 8.39 \begin{align*} \frac{1}{10240} \, b^{3}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{5} + \frac{3}{2048} \, a b^{2}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{4} - \frac{1}{1024} \, b^{3}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{4} + \frac{1}{128} \, a^{2} b{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{3} - \frac{3}{256} \, a b^{2}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{3} + \frac{1}{256} \, b^{3}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{3} + \frac{1}{64} \, a^{3}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} - \frac{3}{64} \, a^{2} b{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} + \frac{9}{256} \, a b^{2}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} - \frac{1}{128} \, b^{3}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}^{2} - \frac{1}{16} \, a^{3}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} + \frac{3}{32} \, a^{2} b{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} - \frac{3}{64} \, a b^{2}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} + \frac{1}{128} \, b^{3}{\left (e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10240*b^3*(e^(2*x) + e^(-2*x))^5 + 3/2048*a*b^2*(e^(2*x) + e^(-2*x))^4 - 1/1024*b^3*(e^(2*x) + e^(-2*x))^4 +
1/128*a^2*b*(e^(2*x) + e^(-2*x))^3 - 3/256*a*b^2*(e^(2*x) + e^(-2*x))^3 + 1/256*b^3*(e^(2*x) + e^(-2*x))^3 +
1/64*a^3*(e^(2*x) + e^(-2*x))^2 - 3/64*a^2*b*(e^(2*x) + e^(-2*x))^2 + 9/256*a*b^2*(e^(2*x) + e^(-2*x))^2 - 1/1
28*b^3*(e^(2*x) + e^(-2*x))^2 - 1/16*a^3*(e^(2*x) + e^(-2*x)) + 3/32*a^2*b*(e^(2*x) + e^(-2*x)) - 3/64*a*b^2*(
e^(2*x) + e^(-2*x)) + 1/128*b^3*(e^(2*x) + e^(-2*x))