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

Optimal. Leaf size=187 \[ \frac {(c+d x) \log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}+1\right )}{f \sqrt {a^2+b^2}}-\frac {(c+d x) \log \left (\frac {b e^{e+f x}}{\sqrt {a^2+b^2}+a}+1\right )}{f \sqrt {a^2+b^2}}+\frac {d \text {Li}_2\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{f^2 \sqrt {a^2+b^2}}-\frac {d \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{f^2 \sqrt {a^2+b^2}} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 187, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.04, size = 142, normalized size = 0.76 \[ \frac {f (c+d x) \left (\log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}+1\right )-\log \left (\frac {b e^{e+f x}}{\sqrt {a^2+b^2}+a}+1\right )\right )+d \text {Li}_2\left (\frac {b e^{e+f x}}{\sqrt {a^2+b^2}-a}\right )-d \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{f^2 \sqrt {a^2+b^2}} \]

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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fricas [B]  time = 0.47, size = 455, normalized size = 2.43 \[ \frac {b d \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) + {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) - b d \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) - {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + {\left (b d e - b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (f x + e\right ) + 2 \, b \sinh \left (f x + e\right ) + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - {\left (b d e - b c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (2 \, b \cosh \left (f x + e\right ) + 2 \, b \sinh \left (f x + e\right ) - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) + {\left (b d f x + b d e\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) + {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) - {\left (b d f x + b d e\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right ) - {\left (b \cosh \left (f x + e\right ) + b \sinh \left (f x + e\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right )}{{\left (a^{2} + b^{2}\right )} f^{2}} \]

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

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)

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maple [B]  time = 0.09, size = 393, normalized size = 2.10 \[ -\frac {2 c \arctanh \left (\frac {2 b \,{\mathrm e}^{f x +e}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{f \sqrt {a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{f \sqrt {a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {a^{2}+b^{2}}}-\frac {d \ln \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{f \sqrt {a^{2}+b^{2}}}-\frac {d \ln \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {a^{2}+b^{2}}}+\frac {d \dilog \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{f^{2} \sqrt {a^{2}+b^{2}}}-\frac {d \dilog \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{f^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 d e \arctanh \left (\frac {2 b \,{\mathrm e}^{f x +e}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{f^{2} \sqrt {a^{2}+b^{2}}} \]

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

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]

d*integrate(2*x/(b*(e^(f*x + e) - e^(-f*x - e)) + 2*a), x) + c*log((b*e^(-f*x - e) - a - sqrt(a^2 + b^2))/(b*e
^(-f*x - e) - a + sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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