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

Optimal. Leaf size=161 \[ \frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^2}{2 b} \]

[Out]

-1/2*x^2/b+x*ln(1+b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)))/b/d+x*ln(1+b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b/d+polylog(2
,-b*exp(d*x+c)/(a-(a^2-b^2)^(1/2)))/b/d^2+polylog(2,-b*exp(d*x+c)/(a+(a^2-b^2)^(1/2)))/b/d^2

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Rubi [A]  time = 0.24, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5562, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {\text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}\right )}{b d^2}+\frac {x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^2/(2*b) + (x*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 - b^2])])/(b*d) + (x*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a
^2 - b^2])])/(b*d) + PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 - b^2]))]/(b*d^2) + PolyLog[2, -((b*E^(c + d*x
))/(a + Sqrt[a^2 - b^2]))]/(b*d^2)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 5562

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)])/(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 - b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 - b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {x \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx &=-\frac {x^2}{2 b}+\int \frac {e^{c+d x} x}{a-\sqrt {a^2-b^2}+b e^{c+d x}} \, dx+\int \frac {e^{c+d x} x}{a+\sqrt {a^2-b^2}+b e^{c+d x}} \, dx\\ &=-\frac {x^2}{2 b}+\frac {x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {\int \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d}-\frac {\int \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d}\\ &=-\frac {x^2}{2 b}+\frac {x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b d^2}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b d^2}\\ &=-\frac {x^2}{2 b}+\frac {x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d}+\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 160, normalized size = 0.99 \[ \frac {\text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}-a}\right )}{b d^2}+\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2-b^2}}+1\right )}{b d}+\frac {x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2-b^2}+a}+1\right )}{b d}-\frac {x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x^2/b + (x*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 - b^2])])/(b*d) + (x*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a
^2 - b^2])])/(b*d) + PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 - b^2])]/(b*d^2) + PolyLog[2, -((b*E^(c + d*x))
/(a + Sqrt[a^2 - b^2]))]/(b*d^2)

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fricas [B]  time = 1.63, size = 354, normalized size = 2.20 \[ -\frac {d^{2} x^{2} + 2 \, c \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + 2 \, a\right ) + 2 \, c \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) - 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + 2 \, a\right ) - 2 \, {\left (d x + c\right )} \log \left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) - 2 \, {\left (d x + c\right )} \log \left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) - 2 \, {\rm Li}_2\left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) - 2 \, {\rm Li}_2\left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right )}{2 \, b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d^2*x^2 + 2*c*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) + 2*c*log(2*b
*cosh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) - 2*(d*x + c)*log((a*cosh(d*x + c) + a*s
inh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) - 2*(d*x + c)*log((a*cosh(d*x
+ c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) - 2*dilog(-(a*cosh
(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - 2*dilog(
-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1))/(
b*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (d x + c\right )}{b \cosh \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.21, size = 368, normalized size = 2.29 \[ -\frac {x^{2}}{2 b}+\frac {\ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d b}+\frac {\ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} b}+\frac {\ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d b}+\frac {\ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} b}+\frac {\dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} b}+\frac {\dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} b}-\frac {2 c x}{d b}-\frac {c^{2}}{d^{2} b}-\frac {c \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}+b \right )}{d^{2} b}+\frac {2 c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2/b+1/d/b*ln((-b*exp(d*x+c)+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*x+1/d^2/b*ln((-b*exp(d*x+c)+(a^2-b
^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*c+1/d/b*ln((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*x+1/d^2/b*
ln((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*c+1/d^2/b*dilog((b*exp(d*x+c)+(a^2-b^2)^(1/2)+a)/(a+(
a^2-b^2)^(1/2)))+1/d^2/b*dilog((-b*exp(d*x+c)+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))-2/d/b*c*x-1/d^2/b*c^2-1
/d^2/b*c*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)+b)+2/d^2/b*c*ln(exp(d*x+c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x^{2}}{2 \, b} - \frac {1}{2} \, \int \frac {4 \, {\left (a x e^{\left (d x + c\right )} + b x\right )}}{b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} + b^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2/b - 1/2*integrate(4*(a*x*e^(d*x + c) + b*x)/(b^2*e^(2*d*x + 2*c) + 2*a*b*e^(d*x + c) + b^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\mathrm {sinh}\left (c+d\,x\right )}{a+b\,\mathrm {cosh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh {\left (c + d x \right )}}{a + b \cosh {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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