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

Optimal. Leaf size=220 \[ -\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2 \sqrt{a^2+b^2}}+\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^2 \sqrt{a^2+b^2}}-\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b d \sqrt{a^2+b^2}}+\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b d \sqrt{a^2+b^2}}+\frac{e x}{b}+\frac{f x^2}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.412666, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5557, 3322, 2264, 2190, 2279, 2391} \[ -\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2 \sqrt{a^2+b^2}}+\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^2 \sqrt{a^2+b^2}}-\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b d \sqrt{a^2+b^2}}+\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b d \sqrt{a^2+b^2}}+\frac{e x}{b}+\frac{f x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5557

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

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 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) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x) \, dx}{b}-\frac{a \int \frac{e+f x}{a+b \sinh (c+d x)} \, dx}{b}\\ &=\frac{e x}{b}+\frac{f x^2}{2 b}-\frac{(2 a) \int \frac{e^{c+d x} (e+f x)}{-b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{b}\\ &=\frac{e x}{b}+\frac{f x^2}{2 b}-\frac{(2 a) \int \frac{e^{c+d x} (e+f x)}{2 a-2 \sqrt{a^2+b^2}+2 b e^{c+d x}} \, dx}{\sqrt{a^2+b^2}}+\frac{(2 a) \int \frac{e^{c+d x} (e+f x)}{2 a+2 \sqrt{a^2+b^2}+2 b e^{c+d x}} \, dx}{\sqrt{a^2+b^2}}\\ &=\frac{e x}{b}+\frac{f x^2}{2 b}-\frac{a (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d}+\frac{a (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d}+\frac{(a f) \int \log \left (1+\frac{2 b e^{c+d x}}{2 a-2 \sqrt{a^2+b^2}}\right ) \, dx}{b \sqrt{a^2+b^2} d}-\frac{(a f) \int \log \left (1+\frac{2 b e^{c+d x}}{2 a+2 \sqrt{a^2+b^2}}\right ) \, dx}{b \sqrt{a^2+b^2} d}\\ &=\frac{e x}{b}+\frac{f x^2}{2 b}-\frac{a (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d}+\frac{a (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d}+\frac{(a f) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 b x}{2 a-2 \sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b \sqrt{a^2+b^2} d^2}-\frac{(a f) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 b x}{2 a+2 \sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b \sqrt{a^2+b^2} d^2}\\ &=\frac{e x}{b}+\frac{f x^2}{2 b}-\frac{a (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d}+\frac{a (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d}-\frac{a f \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d^2}+\frac{a f \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2} d^2}\\ \end{align*}

Mathematica [A]  time = 0.699166, size = 163, normalized size = 0.74 \[ \frac{x (2 e+f x)}{2 b}-\frac{a \left (f \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )-f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+d (e+f x) \left (\log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )-\log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )\right )\right )}{b d^2 \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.069, size = 440, normalized size = 2. \begin{align*}{\frac{f{x}^{2}}{2\,b}}+{\frac{ex}{b}}+2\,{\frac{ae}{bd\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b{{\rm e}^{dx+c}}+2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{afx}{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{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{afc}{{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{1}{\sqrt{{a}^{2}+{b}^{2}}}}}+{\frac{afx}{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{1}{\sqrt{{a}^{2}+{b}^{2}}}}}+{\frac{afc}{{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{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{af}{{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{1}{\sqrt{{a}^{2}+{b}^{2}}}}}+{\frac{af}{{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{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-2\,{\frac{afc}{{d}^{2}b\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b{{\rm e}^{dx+c}}+2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.53014, size = 1243, normalized size = 5.65 \begin{align*} \frac{{\left (a^{2} + b^{2}\right )} d^{2} f x^{2} + 2 \,{\left (a^{2} + b^{2}\right )} d^{2} e x - 2 \, a b f \sqrt{\frac{a^{2} + b^{2}}{b^{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 \, a b f \sqrt{\frac{a^{2} + b^{2}}{b^{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 \,{\left (a b d e - a b c f\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \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 (a b d e - a b c f\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \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 (a b d f x + a b c f\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \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 (a b d f x + a b c f\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \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 (a^{2} b + b^{3}\right )} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2 + b^2)*d^2*f*x^2 + 2*(a^2 + b^2)*d^2*e*x - 2*a*b*f*sqrt((a^2 + b^2)/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*a*b*f*sqrt((a^2 + b^
2)/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*(a*b*d*e - a*b*c*f)*sqrt((a^2 + b^2)/b^2)*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*(a*b*d*e - a*b*c*f)*sqrt((a^2 + b^2)/b^2)*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*(a*b*d*f*x + a*b*c*f)*sqrt((a^2 + b^2)/b^2)*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*(a*b*d*f*x + a*b*c*f)*s
qrt((a^2 + b^2)/b^2)*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))/((a^2*b + b^3)*d^2)

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

[Out]

Timed out

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

[Out]

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