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

Optimal. Leaf size=85 \[ \frac{\left (a^2+b^2\right ) \sinh (c+d x)}{b^3 d}-\frac{a \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^4 d}-\frac{a \sinh ^2(c+d x)}{2 b^2 d}+\frac{\sinh ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.12166, antiderivative size = 85, normalized size of antiderivative = 1., 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 = {2837, 12, 772} \[ \frac{\left (a^2+b^2\right ) \sinh (c+d x)}{b^3 d}-\frac{a \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^4 d}-\frac{a \sinh ^2(c+d x)}{2 b^2 d}+\frac{\sinh ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/(b^4*d)) + ((a^2 + b^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 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 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.164234, size = 75, normalized size = 0.88 \[ \frac{6 b \left (a^2+b^2\right ) \sinh (c+d x)-6 a \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))-3 a b^2 \sinh ^2(c+d x)+2 b^3 \sinh ^3(c+d x)}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.039, size = 428, normalized size = 5. \begin{align*} -{\frac{1}{3\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{{a}^{2}}{d{b}^{3}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{a}^{3}}{d{b}^{4}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{3\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{{a}^{2}}{d{b}^{3}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{a}{2\,d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{a}^{3}}{d{b}^{4}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{{a}^{3}}{d{b}^{4}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }-{\frac{a}{d{b}^{2}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.06732, size = 247, normalized size = 2.91 \begin{align*} -\frac{{\left (3 \, a b e^{\left (-d x - c\right )} - b^{2} - 3 \,{\left (4 \, a^{2} + 3 \, 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{{\left (a^{3} + a b^{2}\right )}{\left (d x + c\right )}}{b^{4} 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} + 3 \, b^{2}\right )} e^{\left (-d x - c\right )}}{24 \, b^{3} d} - \frac{{\left (a^{3} + a 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^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.20706, size = 1613, normalized size = 18.98 \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)/(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 - 3*a*b^2*cosh(d*x + c)^5 + 24*(a^3 + a*b^2)*d*x*cosh(d*x + c)
^3 + 3*(2*b^3*cosh(d*x + c) - a*b^2)*sinh(d*x + c)^5 + 3*(4*a^2*b + 3*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 + 3*b^3)*sinh(d*x + c)^4 - 3*a*b^2*cosh(d*x + c) + 2*(10*b^3*cosh(d*
x + c)^3 - 15*a*b^2*cosh(d*x + c)^2 + 12*(a^3 + a*b^2)*d*x + 6*(4*a^2*b + 3*b^3)*cosh(d*x + c))*sinh(d*x + c)^
3 - b^3 - 3*(4*a^2*b + 3*b^3)*cosh(d*x + c)^2 + 3*(5*b^3*cosh(d*x + c)^4 - 10*a*b^2*cosh(d*x + c)^3 + 24*(a^3
+ a*b^2)*d*x*cosh(d*x + c) - 4*a^2*b - 3*b^3 + 6*(4*a^2*b + 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 24*((a^3
+ a*b^2)*cosh(d*x + c)^3 + 3*(a^3 + a*b^2)*cosh(d*x + c)^2*sinh(d*x + c) + 3*(a^3 + a*b^2)*cosh(d*x + c)*sinh
(d*x + c)^2 + (a^3 + a*b^2)*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 - 5*a*b^2*cosh(d*x + c)^4 + 24*(a^3 + a*b^2)*d*x*cosh(d*x + c)^2 + 4*(4*a^2*b + 3*b^3)*
cosh(d*x + c)^3 - a*b^2 - 2*(4*a^2*b + 3*b^3)*cosh(d*x + c))*sinh(d*x + c))/(b^4*d*cosh(d*x + c)^3 + 3*b^4*d*c
osh(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 [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)/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.15208, size = 215, normalized size = 2.53 \begin{align*} -\frac{{\left (a^{3} + a 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^{4} d} + \frac{b^{2} d^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 3 \, a b d^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 12 \, a^{2} d^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 12 \, b^{2} d^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{24 \, b^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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