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

Optimal. Leaf size=170 \[ \frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^2}+\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b d}+\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b d}-\frac{(e+f x)^2}{2 b f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.234474, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5561, 2190, 2279, 2391} \[ \frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^2}+\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b d}+\frac{(e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b d}-\frac{(e+f x)^2}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), 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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0348309, size = 157, normalized size = 0.92 \[ \frac{2 f^2 \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )+2 f^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )-d (e+f x) \left (-2 f \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )-2 f \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )+d e+d f x\right )}{2 b d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.077, size = 412, normalized size = 2.4 \begin{align*} -{\frac{f{x}^{2}}{2\,b}}+{\frac{ex}{b}}-{\frac{cf\ln \left ( b{{\rm e}^{2\,dx+2\,c}}+2\,a{{\rm e}^{dx+c}}-b \right ) }{{d}^{2}b}}+2\,{\frac{cf\ln \left ({{\rm e}^{dx+c}} \right ) }{{d}^{2}b}}+{\frac{e\ln \left ( b{{\rm e}^{2\,dx+2\,c}}+2\,a{{\rm e}^{dx+c}}-b \right ) }{bd}}-2\,{\frac{e\ln \left ({{\rm e}^{dx+c}} \right ) }{bd}}+{\frac{fx}{bd}\ln \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{cf}{{d}^{2}b}\ln \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{fx}{bd}\ln \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{cf}{{d}^{2}b}\ln \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{f}{{d}^{2}b}{\it dilog} \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{f}{{d}^{2}b}{\it dilog} \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ) }-2\,{\frac{fcx}{bd}}-{\frac{f{c}^{2}}{{d}^{2}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, f{\left (\frac{x^{2}}{b} - \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}\right )} + \frac{e \log \left (b \sinh \left (d x + c\right ) + a\right )}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f*(x^2/b - 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)) + e*lo
g(b*sinh(d*x + c) + a)/(b*d)

________________________________________________________________________________________

Fricas [B]  time = 2.16192, size = 964, normalized size = 5.67 \begin{align*} -\frac{d^{2} f x^{2} + 2 \, d^{2} e x - 2 \, f{\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 \, f{\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 \,{\left (d e - c f\right )} \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 e - c f\right )} \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 f x + c f\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 f x + c f\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 \, b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d^2*f*x^2 + 2*d^2*e*x - 2*f*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*f*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sin
h(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - 2*(d*e - c*f)*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*e - c*f)*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*f*x + c*f)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sq
rt((a^2 + b^2)/b^2) - b)/b) - 2*(d*f*x + c*f)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*s
inh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b))/(b*d^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right ) \cosh{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \cosh \left (d x + c\right )}{b \sinh \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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