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

Optimal. Leaf size=187 \[ \frac{d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}\right )}{f^2 \sqrt{a^2+b^2}}-\frac{d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )}{f^2 \sqrt{a^2+b^2}}+\frac{(c+d x) \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}+1\right )}{f \sqrt{a^2+b^2}}-\frac{(c+d x) \log \left (\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}+1\right )}{f \sqrt{a^2+b^2}} \]

[Out]

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

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Rubi [A]  time = 0.369, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {3322, 2264, 2190, 2279, 2391} \[ \frac{d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}\right )}{f^2 \sqrt{a^2+b^2}}-\frac{d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )}{f^2 \sqrt{a^2+b^2}}+\frac{(c+d x) \log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}+1\right )}{f \sqrt{a^2+b^2}}-\frac{(c+d x) \log \left (\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}+1\right )}{f \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0365069, size = 142, normalized size = 0.76 \[ \frac{d \text{PolyLog}\left (2,\frac{b e^{e+f x}}{\sqrt{a^2+b^2}-a}\right )-d \text{PolyLog}\left (2,-\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}\right )+f (c+d x) \left (\log \left (\frac{b e^{e+f x}}{a-\sqrt{a^2+b^2}}+1\right )-\log \left (\frac{b e^{e+f x}}{\sqrt{a^2+b^2}+a}+1\right )\right )}{f^2 \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.054, size = 393, normalized size = 2.1 \begin{align*} -2\,{\frac{c}{f\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b{{\rm e}^{fx+e}}+2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+{\frac{dx}{f}\ln \left ({ \left ( -b{{\rm e}^{fx+e}}+\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{de}{{f}^{2}}\ln \left ({ \left ( -b{{\rm e}^{fx+e}}+\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{dx}{f}\ln \left ({ \left ( b{{\rm e}^{fx+e}}+\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{de}{{f}^{2}}\ln \left ({ \left ( b{{\rm e}^{fx+e}}+\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{d}{{f}^{2}}{\it dilog} \left ({ \left ( -b{{\rm e}^{fx+e}}+\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{d}{{f}^{2}}{\it dilog} \left ({ \left ( b{{\rm e}^{fx+e}}+\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{de}{{f}^{2}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b{{\rm e}^{fx+e}}+2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*c/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(f*x+e)+2*a)/(a^2+b^2)^(1/2))+1/f*d/(a^2+b^2)^(1/2)*ln((-b*exp(f*x+
e)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*x+1/f^2*d/(a^2+b^2)^(1/2)*ln((-b*exp(f*x+e)+(a^2+b^2)^(1/2)-a)/(-a
+(a^2+b^2)^(1/2)))*e-1/f*d/(a^2+b^2)^(1/2)*ln((b*exp(f*x+e)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*x-1/f^2*d/
(a^2+b^2)^(1/2)*ln((b*exp(f*x+e)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*e+1/f^2*d/(a^2+b^2)^(1/2)*dilog((-b*e
xp(f*x+e)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))-1/f^2*d/(a^2+b^2)^(1/2)*dilog((b*exp(f*x+e)+(a^2+b^2)^(1/2)
+a)/(a+(a^2+b^2)^(1/2)))+2/f^2*d*e/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(f*x+e)+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((d*x+c)/(a+b*sinh(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.54519, size = 1127, normalized size = 6.03 \begin{align*} \frac{b d \sqrt{\frac{a^{2} + b^{2}}{b^{2}}}{\rm Li}_2\left (\frac{a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) +{\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) - b d \sqrt{\frac{a^{2} + b^{2}}{b^{2}}}{\rm Li}_2\left (\frac{a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) -{\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) +{\left (b d e - b c f\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (f x + e\right ) + 2 \, b \sinh \left (f x + e\right ) + 2 \, b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) -{\left (b d e - b c f\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (f x + e\right ) + 2 \, b \sinh \left (f x + e\right ) - 2 \, b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) +{\left (b d f x + b d e\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \log \left (-\frac{a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) +{\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) -{\left (b d f x + b d e\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \log \left (-\frac{a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) -{\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} - b}{b}\right )}{{\left (a^{2} + b^{2}\right )} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt
((a^2 + b^2)/b^2) - b)/b + 1) - b*d*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f
*x + e) + b*sinh(f*x + e))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + (b*d*e - b*c*f)*sqrt((a^2 + b^2)/b^2)*log(2*b*c
osh(f*x + e) + 2*b*sinh(f*x + e) + 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) - (b*d*e - b*c*f)*sqrt((a^2 + b^2)/b^2)*lo
g(2*b*cosh(f*x + e) + 2*b*sinh(f*x + e) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + (b*d*f*x + b*d*e)*sqrt((a^2 + b^2
)/b^2)*log(-(a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 + b^2)/b^2) - b
)/b) - (b*d*f*x + b*d*e)*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f*x + e) + b*
sinh(f*x + e))*sqrt((a^2 + b^2)/b^2) - b)/b))/((a^2 + b^2)*f^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((d*x+c)/(a+b*sinh(f*x+e)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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