3.4.94 \(\int \frac {\cosh (c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) [394]

Optimal. Leaf size=76 \[ -\frac {a^3 \log (a+b \sinh (c+d x))}{b^4 d}+\frac {a^2 \sinh (c+d x)}{b^3 d}-\frac {a \sinh ^2(c+d x)}{2 b^2 d}+\frac {\sinh ^3(c+d x)}{3 b d} \]

[Out]

-a^3*ln(a+b*sinh(d*x+c))/b^4/d+a^2*sinh(d*x+c)/b^3/d-1/2*a*sinh(d*x+c)^2/b^2/d+1/3*sinh(d*x+c)^3/b/d

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Rubi [A]
time = 0.07, antiderivative size = 76, 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 = {2912, 12, 45} \begin {gather*} -\frac {a^3 \log (a+b \sinh (c+d x))}{b^4 d}+\frac {a^2 \sinh (c+d x)}{b^3 d}-\frac {a \sinh ^2(c+d x)}{2 b^2 d}+\frac {\sinh ^3(c+d x)}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*Log[a + b*Sinh[c + d*x]])/(b^4*d)) + (a^2*Sinh[c + d*x])/(b^3*d) - (a*Sinh[c + d*x]^2)/(2*b^2*d) + Sinh
[c + d*x]^3/(3*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 45

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 2912

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

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Mathematica [A]
time = 0.15, size = 71, normalized size = 0.93 \begin {gather*} \frac {-3 a b^2 \cosh (2 (c+d x))-12 a^3 \log (a+b \sinh (c+d x))-3 b \left (-4 a^2+b^2\right ) \sinh (c+d x)+b^3 \sinh (3 (c+d x))}{12 b^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.58, size = 65, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\sinh ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sinh ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sinh \left (d x +c \right )}{b^{3}}-\frac {a^{3} \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(65\)
default	\(\frac {\frac {\frac {\left (\sinh ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sinh ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sinh \left (d x +c \right )}{b^{3}}-\frac {a^{3} \ln \left (a +b \sinh \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(65\)
risch	\(\frac {a^{3} x}{b^{4}}+\frac {{\mathrm e}^{3 d x +3 c}}{24 b d}-\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 b^{2} d}+\frac {{\mathrm e}^{d x +c} a^{2}}{2 b^{3} d}-\frac {{\mathrm e}^{d x +c}}{8 b d}-\frac {{\mathrm e}^{-d x -c} a^{2}}{2 b^{3} d}+\frac {{\mathrm e}^{-d x -c}}{8 b d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 b^{2} d}-\frac {{\mathrm e}^{-3 d x -3 c}}{24 b d}+\frac {2 a^{3} c}{b^{4} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{4} d}\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/b^3*(1/3*sinh(d*x+c)^3*b^2-1/2*b*a*sinh(d*x+c)^2+a^2*sinh(d*x+c))-a^3/b^4*ln(a+b*sinh(d*x+c)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (72) = 144\).
time = 0.27, size = 171, normalized size = 2.25 \begin {gather*} -\frac {{\left (d x + c\right )} a^{3}}{b^{4} d} - \frac {a^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{4} d} - \frac {{\left (3 \, a b e^{\left (-d x - c\right )} - b^{2} - 3 \, {\left (4 \, a^{2} - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{24 \, b^{3} d} - \frac {3 \, a b e^{\left (-2 \, d x - 2 \, c\right )} + b^{2} e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, {\left (4 \, a^{2} - b^{2}\right )} e^{\left (-d x - c\right )}}{24 \, b^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 602 vs. \(2 (72) = 144\).
time = 0.37, size = 602, normalized size = 7.92 \begin {gather*} \frac {b^{3} \cosh \left (d x + c\right )^{6} + b^{3} \sinh \left (d x + c\right )^{6} + 24 \, a^{3} d x \cosh \left (d x + c\right )^{3} - 3 \, a b^{2} \cosh \left (d x + c\right )^{5} + 3 \, {\left (2 \, b^{3} \cosh \left (d x + c\right ) - a b^{2}\right )} \sinh \left (d x + c\right )^{5} + 3 \, {\left (4 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{2} - 5 \, a b^{2} \cosh \left (d x + c\right ) + 4 \, a^{2} b - b^{3}\right )} \sinh \left (d x + c\right )^{4} - 3 \, a b^{2} \cosh \left (d x + c\right ) + 2 \, {\left (10 \, b^{3} \cosh \left (d x + c\right )^{3} + 12 \, a^{3} d x - 15 \, a b^{2} \cosh \left (d x + c\right )^{2} + 6 \, {\left (4 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - b^{3} - 3 \, {\left (4 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 24 \, a^{3} d x \cosh \left (d x + c\right ) - 10 \, a b^{2} \cosh \left (d x + c\right )^{3} - 4 \, a^{2} b + b^{3} + 6 \, {\left (4 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} - 24 \, {\left (a^{3} \cosh \left (d x + c\right )^{3} + 3 \, a^{3} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, a^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + a^{3} \sinh \left (d x + c\right )^{3}\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 ) + 3 \, {\left (2 \, b^{3} \cosh \left (d x + c\right )^{5} + 24 \, a^{3} d x \cosh \left (d x + c\right )^{2} - 5 \, a b^{2} \cosh \left (d x + c\right )^{4} + 4 \, {\left (4 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{3} - a b^{2} - 2 \, {\left (4 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{24 \, {\left (b^{4} d \cosh \left (d x + c\right )^{3} + 3 \, b^{4} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, b^{4} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + b^{4} d \sinh \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b^3*cosh(d*x + c)^6 + b^3*sinh(d*x + c)^6 + 24*a^3*d*x*cosh(d*x + c)^3 - 3*a*b^2*cosh(d*x + c)^5 + 3*(2*
b^3*cosh(d*x + c) - a*b^2)*sinh(d*x + c)^5 + 3*(4*a^2*b - b^3)*cosh(d*x + c)^4 + 3*(5*b^3*cosh(d*x + c)^2 - 5*
a*b^2*cosh(d*x + c) + 4*a^2*b - b^3)*sinh(d*x + c)^4 - 3*a*b^2*cosh(d*x + c) + 2*(10*b^3*cosh(d*x + c)^3 + 12*
a^3*d*x - 15*a*b^2*cosh(d*x + c)^2 + 6*(4*a^2*b - b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - b^3 - 3*(4*a^2*b - b^3
)*cosh(d*x + c)^2 + 3*(5*b^3*cosh(d*x + c)^4 + 24*a^3*d*x*cosh(d*x + c) - 10*a*b^2*cosh(d*x + c)^3 - 4*a^2*b +
b^3 + 6*(4*a^2*b - b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 24*(a^3*cosh(d*x + c)^3 + 3*a^3*cosh(d*x + c)^2*si
nh(d*x + c) + 3*a^3*cosh(d*x + c)*sinh(d*x + c)^2 + a^3*sinh(d*x + c)^3)*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x
+ c) - sinh(d*x + c))) + 3*(2*b^3*cosh(d*x + c)^5 + 24*a^3*d*x*cosh(d*x + c)^2 - 5*a*b^2*cosh(d*x + c)^4 + 4*
(4*a^2*b - b^3)*cosh(d*x + c)^3 - a*b^2 - 2*(4*a^2*b - b^3)*cosh(d*x + c))*sinh(d*x + c))/(b^4*d*cosh(d*x + c)
^3 + 3*b^4*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*b^4*d*cosh(d*x + c)*sinh(d*x + c)^2 + b^4*d*sinh(d*x + c)^3)

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Sympy [A]
time = 0.60, size = 105, normalized size = 1.38 \begin {gather*} \begin {cases} \frac {x \sinh ^{3}{\left (c \right )} \cosh {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sinh ^{4}{\left (c + d x \right )}}{4 a d} & \text {for}\: b = 0 \\\frac {x \sinh ^{3}{\left (c \right )} \cosh {\left (c \right )}}{a + b \sinh {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {a^{3} \log {\left (\frac {a}{b} + \sinh {\left (c + d x \right )} \right )}}{b^{4} d} + \frac {a^{2} \sinh {\left (c + d x \right )}}{b^{3} d} - \frac {a \cosh ^{2}{\left (c + d x \right )}}{2 b^{2} d} + \frac {\sinh ^{3}{\left (c + d x \right )}}{3 b d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*sinh(c)**3*cosh(c)/a, Eq(b, 0) & Eq(d, 0)), (sinh(c + d*x)**4/(4*a*d), Eq(b, 0)), (x*sinh(c)**3*c
osh(c)/(a + b*sinh(c)), Eq(d, 0)), (-a**3*log(a/b + sinh(c + d*x))/(b**4*d) + a**2*sinh(c + d*x)/(b**3*d) - a*
cosh(c + d*x)**2/(2*b**2*d) + sinh(c + d*x)**3/(3*b*d), True))

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Giac [A]
time = 0.45, size = 117, normalized size = 1.54 \begin {gather*} -\frac {\frac {24 \, a^{3} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{4}} - \frac {b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 3 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 12 \, a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{b^{3}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.14, size = 63, normalized size = 0.83 \begin {gather*} -\frac {a^3\,\ln \left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )-\frac {b^3\,{\mathrm {sinh}\left (c+d\,x\right )}^3}{3}+\frac {a\,b^2\,{\mathrm {sinh}\left (c+d\,x\right )}^2}{2}-a^2\,b\,\mathrm {sinh}\left (c+d\,x\right )}{b^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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