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

Optimal. Leaf size=61 \[ \frac {\left (a^2-b^2\right ) \log (a+b \cosh (c+d x))}{b^3 d}-\frac {a \cosh (c+d x)}{b^2 d}+\frac {\cosh ^2(c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2668, 697} \[ \frac {\left (a^2-b^2\right ) \log (a+b \cosh (c+d x))}{b^3 d}-\frac {a \cosh (c+d x)}{b^2 d}+\frac {\cosh ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 697

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 55, normalized size = 0.90 \[ \frac {4 \left (a^2-b^2\right ) \log (a+b \cosh (c+d x))-4 a b \cosh (c+d x)+b^2 \cosh (2 (c+d x))}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.64, size = 340, normalized size = 5.57 \[ \frac {b^{2} \cosh \left (d x + c\right )^{4} + b^{2} \sinh \left (d x + c\right )^{4} - 8 \, {\left (a^{2} - b^{2}\right )} d x \cosh \left (d x + c\right )^{2} - 4 \, a b \cosh \left (d x + c\right )^{3} + 4 \, {\left (b^{2} \cosh \left (d x + c\right ) - a b\right )} \sinh \left (d x + c\right )^{3} - 4 \, a b \cosh \left (d x + c\right ) + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} - 4 \, {\left (a^{2} - b^{2}\right )} d x - 6 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + b^{2} + 8 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} - 4 \, {\left (a^{2} - b^{2}\right )} d x \cosh \left (d x + c\right ) - 3 \, a b \cosh \left (d x + c\right )^{2} - a b\right )} \sinh \left (d x + c\right )}{8 \, {\left (b^{3} d \cosh \left (d x + c\right )^{2} + 2 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} d \sinh \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2*cosh(d*x + c)^4 + b^2*sinh(d*x + c)^4 - 8*(a^2 - b^2)*d*x*cosh(d*x + c)^2 - 4*a*b*cosh(d*x + c)^3 + 4
*(b^2*cosh(d*x + c) - a*b)*sinh(d*x + c)^3 - 4*a*b*cosh(d*x + c) + 2*(3*b^2*cosh(d*x + c)^2 - 4*(a^2 - b^2)*d*
x - 6*a*b*cosh(d*x + c))*sinh(d*x + c)^2 + b^2 + 8*((a^2 - b^2)*cosh(d*x + c)^2 + 2*(a^2 - b^2)*cosh(d*x + c)*
sinh(d*x + c) + (a^2 - b^2)*sinh(d*x + c)^2)*log(2*(b*cosh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + 4*
(b^2*cosh(d*x + c)^3 - 4*(a^2 - b^2)*d*x*cosh(d*x + c) - 3*a*b*cosh(d*x + c)^2 - a*b)*sinh(d*x + c))/(b^3*d*co
sh(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 = 0.15, size = 88, normalized size = 1.44 \[ \frac {\frac {b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} - 4 \, a {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{b^{2}} + \frac {8 \, {\left (a^{2} - b^{2}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{3}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.08, size = 415, normalized size = 6.80 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}{d \,b^{3}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d b}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a}{d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}}{d \,b^{3}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d b}+\frac {\ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right ) a^{3}}{d \,b^{3} \left (a -b \right )}-\frac {\ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right ) a^{2}}{d \,b^{2} \left (a -b \right )}-\frac {\ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right ) a}{d b \left (a -b \right )}+\frac {\ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}{d \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 130, normalized size = 2.13 \[ -\frac {{\left (4 \, a e^{\left (-d x - c\right )} - b\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b^{2} d} + \frac {{\left (a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{3} d} - \frac {4 \, a e^{\left (-d x - c\right )} - b e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{2} d} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} + b\right )}{b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.06, size = 122, normalized size = 2.00 \[ \frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d}-\frac {x\,\left (a^2-b^2\right )}{b^3}+\frac {\ln \left (b+2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^2-b^2\right )}{b^3\,d}-\frac {a\,{\mathrm {e}}^{-c-d\,x}}{2\,b^2\,d}-\frac {a\,{\mathrm {e}}^{c+d\,x}}{2\,b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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