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

Optimal. Leaf size=95 \[ \frac {2 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^3 d}+\frac {x \left (2 a^2+b^2\right )}{2 b^3}-\frac {\cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2865, 2735, 2660, 618, 204} \[ \frac {2 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^3 d}+\frac {x \left (2 a^2+b^2\right )}{2 b^3}-\frac {\cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 + b^2)*x)/(2*b^3) + (2*a*Sqrt[a^2 + b^2]*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b^3*d) -
(Cosh[c + d*x]*(2*a - b*Sinh[c + d*x]))/(2*b^2*d)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2865

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

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 109, normalized size = 1.15 \[ \frac {8 a \sqrt {-a^2-b^2} \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )+4 a^2 c+4 a^2 d x-4 a b \cosh (c+d x)+b^2 \sinh (2 (c+d x))+2 b^2 c+2 b^2 d x}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*c + 2*b^2*c + 4*a^2*d*x + 2*b^2*d*x + 8*a*Sqrt[-a^2 - b^2]*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 -
b^2]] - 4*a*b*Cosh[c + d*x] + b^2*Sinh[2*(c + d*x)])/(4*b^3*d)

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fricas [B]  time = 0.49, size = 446, normalized size = 4.69 \[ \frac {b^{2} \cosh \left (d x + c\right )^{4} + b^{2} \sinh \left (d x + c\right )^{4} + 4 \, {\left (2 \, a^{2} + b^{2}\right )} d x \cosh \left (d x + c\right )^{2} - 4 \, a b \cosh \left (d x + c\right )^{3} + 4 \, {\left (b^{2} \cosh \left (d x + c\right ) - a b\right )} \sinh \left (d x + c\right )^{3} - 4 \, a b \cosh \left (d x + c\right ) + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{2} + b^{2}\right )} d x - 6 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 8 \, {\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) - b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 2 \, {\left (2 \, a^{2} + b^{2}\right )} d x \cosh \left (d x + c\right ) - 3 \, a b \cosh \left (d x + c\right )^{2} - a b\right )} \sinh \left (d x + c\right )}{8 \, {\left (b^{3} d \cosh \left (d x + c\right )^{2} + 2 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} d \sinh \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2*cosh(d*x + c)^4 + b^2*sinh(d*x + c)^4 + 4*(2*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 + 2*(2*a^2 + b^2
)*d*x - 6*a*b*cosh(d*x + c))*sinh(d*x + c)^2 + 8*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh
(d*x + c)^2)*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^
2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*
cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) - b^2 +
4*(b^2*cosh(d*x + c)^3 + 2*(2*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*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 = 0.22, size = 155, normalized size = 1.63 \[ \frac {\frac {4 \, {\left (2 \, a^{2} + b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {b e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a e^{\left (d x + c\right )}}{b^{2}} - \frac {{\left (4 \, a b e^{\left (d x + c\right )} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{b^{3}} - \frac {8 \, {\left (a^{3} + a b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{3}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 260, normalized size = 2.74 \[ \frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a}{d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}{d \,b^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d b}-\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a}{d \,b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}}{d \,b^{3}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d b}-\frac {2 a \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 160, normalized size = 1.68 \[ -\frac {{\left (4 \, a e^{\left (-d x - c\right )} - b\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b^{2} d} - \frac {\sqrt {a^{2} + b^{2}} a \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{b^{3} d} + \frac {{\left (2 \, a^{2} + b^{2}\right )} {\left (d x + c\right )}}{2 \, b^{3} d} - \frac {4 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*sinh(d*x+c)/(a+b*sinh(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) - sqrt(a^2 + b^2)*a*log((b*e^(-d*x - c) - a - sqrt(a^2 + b
^2))/(b*e^(-d*x - c) - a + sqrt(a^2 + b^2)))/(b^3*d) + 1/2*(2*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)

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mupad [B]  time = 0.49, size = 212, normalized size = 2.23 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}+\frac {x\,\left (2\,a^2+b^2\right )}{2\,b^3}-\frac {a\,{\mathrm {e}}^{-c-d\,x}}{2\,b^2\,d}-\frac {a\,{\mathrm {e}}^{c+d\,x}}{2\,b^2\,d}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}\,\left (a^2+b^2\right )}{b^4}-\frac {2\,a\,\sqrt {a^2+b^2}\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^4}\right )\,\sqrt {a^2+b^2}}{b^3\,d}+\frac {a\,\ln \left (\frac {2\,a\,\sqrt {a^2+b^2}\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^4}+\frac {2\,a\,{\mathrm {e}}^{c+d\,x}\,\left (a^2+b^2\right )}{b^4}\right )\,\sqrt {a^2+b^2}}{b^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*c + 2*d*x)/(8*b*d) - exp(- 2*c - 2*d*x)/(8*b*d) + (x*(2*a^2 + b^2))/(2*b^3) - (a*exp(- c - d*x))/(2*b^2*
d) - (a*exp(c + d*x))/(2*b^2*d) - (a*log((2*a*exp(c + d*x)*(a^2 + b^2))/b^4 - (2*a*(a^2 + b^2)^(1/2)*(b - a*ex
p(c + d*x)))/b^4)*(a^2 + b^2)^(1/2))/(b^3*d) + (a*log((2*a*(a^2 + b^2)^(1/2)*(b - a*exp(c + d*x)))/b^4 + (2*a*
exp(c + d*x)*(a^2 + b^2))/b^4)*(a^2 + b^2)^(1/2))/(b^3*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(cosh(d*x+c)**2*sinh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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