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

Optimal. Leaf size=55 \[ \frac {a^2 \log (a+b \sinh (c+d x))}{b^3 d}-\frac {a \sinh (c+d x)}{b^2 d}+\frac {\sinh ^2(c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2833, 12, 43} \[ \frac {a^2 \log (a+b \sinh (c+d x))}{b^3 d}-\frac {a \sinh (c+d x)}{b^2 d}+\frac {\sinh ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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])

Rule 2833

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 49, normalized size = 0.89 \[ \frac {2 a^2 \log (a+b \sinh (c+d x))-2 a b \sinh (c+d x)+b^2 \sinh ^2(c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 309, normalized size = 5.62 \[ -\frac {8 \, a^{2} d x \cosh \left (d x + c\right )^{2} - b^{2} \cosh \left (d x + c\right )^{4} - b^{2} \sinh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right )^{3} - 4 \, {\left (b^{2} \cosh \left (d x + c\right ) - a b\right )} \sinh \left (d x + c\right )^{3} - 4 \, a b \cosh \left (d x + c\right ) + 2 \, {\left (4 \, a^{2} d x - 3 \, b^{2} \cosh \left (d x + c\right )^{2} + 6 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - b^{2} - 8 \, {\left (a^{2} \cosh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} \sinh \left (d x + c\right )^{2}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \, {\left (4 \, a^{2} d x \cosh \left (d x + c\right ) - b^{2} \cosh \left (d x + c\right )^{3} + 3 \, a b \cosh \left (d x + c\right )^{2} - a b\right )} \sinh \left (d x + c\right )}{8 \, {\left (b^{3} d \cosh \left (d x + c\right )^{2} + 2 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} d \sinh \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*a^2*d*x*cosh(d*x + c)^2 - b^2*cosh(d*x + c)^4 - b^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^3 - 4*(b^2*c
osh(d*x + c) - a*b)*sinh(d*x + c)^3 - 4*a*b*cosh(d*x + c) + 2*(4*a^2*d*x - 3*b^2*cosh(d*x + c)^2 + 6*a*b*cosh(
d*x + c))*sinh(d*x + c)^2 - b^2 - 8*(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x +
c)^2)*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(4*a^2*d*x*cosh(d*x + c) - b^2*cosh(d*x
+ c)^3 + 3*a*b*cosh(d*x + c)^2 - a*b)*sinh(d*x + c))/(b^3*d*cosh(d*x + c)^2 + 2*b^3*d*cosh(d*x + c)*sinh(d*x
+ c) + b^3*d*sinh(d*x + c)^2)

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giac [A]  time = 0.26, size = 88, normalized size = 1.60 \[ \frac {\frac {8 \, a^{2} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{3}} + \frac {b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 4 \, a {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{b^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 54, normalized size = 0.98 \[ \frac {a^{2} \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{3} d}-\frac {a \sinh \left (d x +c \right )}{b^{2} d}+\frac {\sinh ^{2}\left (d x +c \right )}{2 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 119, normalized size = 2.16 \[ \frac {{\left (d x + c\right )} a^{2}}{b^{3} d} - \frac {{\left (4 \, a e^{\left (-d x - c\right )} - b\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b^{2} d} + \frac {a^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{3} d} + \frac {4 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 46, normalized size = 0.84 \[ \frac {a^2\,\ln \left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )+\frac {b^2\,{\mathrm {sinh}\left (c+d\,x\right )}^2}{2}-a\,b\,\mathrm {sinh}\left (c+d\,x\right )}{b^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*log(a + b*sinh(c + d*x)) + (b^2*sinh(c + d*x)^2)/2 - a*b*sinh(c + d*x))/(b^3*d)

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sympy [A]  time = 1.34, size = 87, normalized size = 1.58 \[ \begin {cases} \frac {x \sinh ^{2}{\relax (c )} \cosh {\relax (c )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sinh ^{3}{\left (c + d x \right )}}{3 a d} & \text {for}\: b = 0 \\\frac {x \sinh ^{2}{\relax (c )} \cosh {\relax (c )}}{a + b \sinh {\relax (c )}} & \text {for}\: d = 0 \\\frac {a^{2} \log {\left (\frac {a}{b} + \sinh {\left (c + d x \right )} \right )}}{b^{3} d} - \frac {a \sinh {\left (c + d x \right )}}{b^{2} d} + \frac {\cosh ^{2}{\left (c + d x \right )}}{2 b d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*sinh(c)**2*cosh(c)/a, Eq(b, 0) & Eq(d, 0)), (sinh(c + d*x)**3/(3*a*d), Eq(b, 0)), (x*sinh(c)**2*c
osh(c)/(a + b*sinh(c)), Eq(d, 0)), (a**2*log(a/b + sinh(c + d*x))/(b**3*d) - a*sinh(c + d*x)/(b**2*d) + cosh(c
+ d*x)**2/(2*b*d), True))

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