3.375 \(\int \frac{\cosh ^3(c+d x) \sinh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=113 \[ \frac{\left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^3 d}-\frac{a \left (a^2+b^2\right ) \sinh (c+d x)}{b^4 d}+\frac{a^2 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^5 d}-\frac{a \sinh ^3(c+d x)}{3 b^2 d}+\frac{\sinh ^4(c+d x)}{4 b d} \]

[Out]

(a^2*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/(b^5*d) - (a*(a^2 + b^2)*Sinh[c + d*x])/(b^4*d) + ((a^2 + b^2)*Sinh
[c + d*x]^2)/(2*b^3*d) - (a*Sinh[c + d*x]^3)/(3*b^2*d) + Sinh[c + d*x]^4/(4*b*d)

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Rubi [A]  time = 0.162831, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2837, 12, 894} \[ \frac{\left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^3 d}-\frac{a \left (a^2+b^2\right ) \sinh (c+d x)}{b^4 d}+\frac{a^2 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^5 d}-\frac{a \sinh ^3(c+d x)}{3 b^2 d}+\frac{\sinh ^4(c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/(b^5*d) - (a*(a^2 + b^2)*Sinh[c + d*x])/(b^4*d) + ((a^2 + b^2)*Sinh
[c + d*x]^2)/(2*b^3*d) - (a*Sinh[c + d*x]^3)/(3*b^2*d) + Sinh[c + d*x]^4/(4*b*d)

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.302434, size = 98, normalized size = 0.87 \[ \frac{6 b^2 \left (a^2+b^2\right ) \sinh ^2(c+d x)-12 a b \left (a^2+b^2\right ) \sinh (c+d x)+12 a^2 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))-4 a b^3 \sinh ^3(c+d x)+3 b^4 \sinh ^4(c+d x)}{12 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]] - 12*a*b*(a^2 + b^2)*Sinh[c + d*x] + 6*b^2*(a^2 + b^2)*Sinh[c + d
*x]^2 - 4*a*b^3*Sinh[c + d*x]^3 + 3*b^4*Sinh[c + d*x]^4)/(12*b^5*d)

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Maple [B]  time = 0.046, size = 614, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8/d/b/(tanh(1/2*d*x+1/2*c)+1)^2+5/8/d/b/(tanh(1/2*d*x+1/2*c)-1)^2-1/d*a^4/b^5*ln(tanh(1/2*d*x+1/2*c)-1)+1/d*
a^4/b^5*ln(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)+1/2/d/b^3/(tanh(1/2*d*x+1/2*c)+1)^2*a^2+1/d/b^4/
(tanh(1/2*d*x+1/2*c)+1)*a^3-1/d*a^4/b^5*ln(tanh(1/2*d*x+1/2*c)+1)+1/3/d/b^2/(tanh(1/2*d*x+1/2*c)+1)^3*a+1/3/d/
b^2/(tanh(1/2*d*x+1/2*c)-1)^3*a+1/2/d/b^3/(tanh(1/2*d*x+1/2*c)-1)^2*a^2+1/d/b^4/(tanh(1/2*d*x+1/2*c)-1)*a^3-1/
2/d/b^2/(tanh(1/2*d*x+1/2*c)+1)^2*a-1/2/d/b^3/(tanh(1/2*d*x+1/2*c)+1)*a^2+1/2/d/b^2/(tanh(1/2*d*x+1/2*c)-1)^2*
a+1/2/d/b^3/(tanh(1/2*d*x+1/2*c)-1)*a^2+1/d/b^3*ln(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)*a^2+1/d/
b^2/(tanh(1/2*d*x+1/2*c)+1)*a-1/d/b^3*ln(tanh(1/2*d*x+1/2*c)+1)*a^2+1/d/b^2/(tanh(1/2*d*x+1/2*c)-1)*a-1/d/b^3*
ln(tanh(1/2*d*x+1/2*c)-1)*a^2+1/4/d/b/(tanh(1/2*d*x+1/2*c)-1)^4+1/4/d/b/(tanh(1/2*d*x+1/2*c)+1)^4-1/2/d/b/(tan
h(1/2*d*x+1/2*c)+1)^3+1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)^3-3/8/d/b/(tanh(1/2*d*x+1/2*c)+1)+3/8/d/b/(tanh(1/2*d*x+
1/2*c)-1)

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Maxima [B]  time = 1.04974, size = 316, normalized size = 2.8 \begin{align*} -\frac{{\left (8 \, a b^{2} e^{\left (-d x - c\right )} - 3 \, b^{3} - 12 \,{\left (2 \, a^{2} b + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 24 \,{\left (4 \, a^{3} + 3 \, a b^{2}\right )} e^{\left (-3 \, d x - 3 \, c\right )}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{192 \, b^{4} d} + \frac{{\left (a^{4} + a^{2} b^{2}\right )}{\left (d x + c\right )}}{b^{5} d} + \frac{8 \, a b^{2} e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, b^{3} e^{\left (-4 \, d x - 4 \, c\right )} + 24 \,{\left (4 \, a^{3} + 3 \, a b^{2}\right )} e^{\left (-d x - c\right )} + 12 \,{\left (2 \, a^{2} b + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{192 \, b^{4} d} + \frac{{\left (a^{4} + a^{2} b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(8*a*b^2*e^(-d*x - c) - 3*b^3 - 12*(2*a^2*b + b^3)*e^(-2*d*x - 2*c) + 24*(4*a^3 + 3*a*b^2)*e^(-3*d*x -
3*c))*e^(4*d*x + 4*c)/(b^4*d) + (a^4 + a^2*b^2)*(d*x + c)/(b^5*d) + 1/192*(8*a*b^2*e^(-3*d*x - 3*c) + 3*b^3*e^
(-4*d*x - 4*c) + 24*(4*a^3 + 3*a*b^2)*e^(-d*x - c) + 12*(2*a^2*b + b^3)*e^(-2*d*x - 2*c))/(b^4*d) + (a^4 + a^2
*b^2)*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/(b^5*d)

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Fricas [B]  time = 2.23817, size = 2631, normalized size = 23.28 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*b^4*cosh(d*x + c)^8 + 3*b^4*sinh(d*x + c)^8 - 8*a*b^3*cosh(d*x + c)^7 + 8*(3*b^4*cosh(d*x + c) - a*b^
3)*sinh(d*x + c)^7 - 192*(a^4 + a^2*b^2)*d*x*cosh(d*x + c)^4 + 12*(2*a^2*b^2 + b^4)*cosh(d*x + c)^6 + 4*(21*b^
4*cosh(d*x + c)^2 - 14*a*b^3*cosh(d*x + c) + 6*a^2*b^2 + 3*b^4)*sinh(d*x + c)^6 - 24*(4*a^3*b + 3*a*b^3)*cosh(
d*x + c)^5 + 24*(7*b^4*cosh(d*x + c)^3 - 7*a*b^3*cosh(d*x + c)^2 - 4*a^3*b - 3*a*b^3 + 3*(2*a^2*b^2 + b^4)*cos
h(d*x + c))*sinh(d*x + c)^5 + 8*a*b^3*cosh(d*x + c) + 2*(105*b^4*cosh(d*x + c)^4 - 140*a*b^3*cosh(d*x + c)^3 -
96*(a^4 + a^2*b^2)*d*x + 90*(2*a^2*b^2 + b^4)*cosh(d*x + c)^2 - 60*(4*a^3*b + 3*a*b^3)*cosh(d*x + c))*sinh(d*
x + c)^4 + 3*b^4 + 24*(4*a^3*b + 3*a*b^3)*cosh(d*x + c)^3 + 8*(21*b^4*cosh(d*x + c)^5 - 35*a*b^3*cosh(d*x + c)
^4 + 12*a^3*b + 9*a*b^3 - 96*(a^4 + a^2*b^2)*d*x*cosh(d*x + c) + 30*(2*a^2*b^2 + b^4)*cosh(d*x + c)^3 - 30*(4*
a^3*b + 3*a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 12*(2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 12*(7*b^4*cosh(d*x
+ c)^6 - 14*a*b^3*cosh(d*x + c)^5 - 96*(a^4 + a^2*b^2)*d*x*cosh(d*x + c)^2 + 15*(2*a^2*b^2 + b^4)*cosh(d*x + c
)^4 + 2*a^2*b^2 + b^4 - 20*(4*a^3*b + 3*a*b^3)*cosh(d*x + c)^3 + 6*(4*a^3*b + 3*a*b^3)*cosh(d*x + c))*sinh(d*x
+ c)^2 + 192*((a^4 + a^2*b^2)*cosh(d*x + c)^4 + 4*(a^4 + a^2*b^2)*cosh(d*x + c)^3*sinh(d*x + c) + 6*(a^4 + a^
2*b^2)*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a^4 + a^2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 + a^2*b^2)*sin
h(d*x + c)^4)*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + 8*(3*b^4*cosh(d*x + c)^7 - 7*a*b^
3*cosh(d*x + c)^6 - 96*(a^4 + a^2*b^2)*d*x*cosh(d*x + c)^3 + 9*(2*a^2*b^2 + b^4)*cosh(d*x + c)^5 - 15*(4*a^3*b
+ 3*a*b^3)*cosh(d*x + c)^4 + a*b^3 + 9*(4*a^3*b + 3*a*b^3)*cosh(d*x + c)^2 + 3*(2*a^2*b^2 + b^4)*cosh(d*x + c
))*sinh(d*x + c))/(b^5*d*cosh(d*x + c)^4 + 4*b^5*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*b^5*d*cosh(d*x + c)^2*sin
h(d*x + c)^2 + 4*b^5*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^5*d*sinh(d*x + c)^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.20187, size = 298, normalized size = 2.64 \begin{align*} \frac{{\left (a^{4} + a^{2} b^{2}\right )} \log \left ({\left | b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{5} d} + \frac{3 \, b^{3} d^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{4} - 8 \, a b^{2} d^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 24 \, a^{2} b d^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 24 \, b^{3} d^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 96 \, a^{3} d^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 96 \, a b^{2} d^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{192 \, b^{4} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^4 + a^2*b^2)*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(b^5*d) + 1/192*(3*b^3*d^3*(e^(d*x + c) - e^(-d
*x - c))^4 - 8*a*b^2*d^3*(e^(d*x + c) - e^(-d*x - c))^3 + 24*a^2*b*d^3*(e^(d*x + c) - e^(-d*x - c))^2 + 24*b^3
*d^3*(e^(d*x + c) - e^(-d*x - c))^2 - 96*a^3*d^3*(e^(d*x + c) - e^(-d*x - c)) - 96*a*b^2*d^3*(e^(d*x + c) - e^
(-d*x - c)))/(b^4*d^4)