3.49 \(\int \frac {a+b \text {csch}^{-1}(c x)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=98 \[ -\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}+\frac {b \tanh ^{-1}\left (\frac {c^2 d-\frac {e}{x}}{c \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {c^2 d^2+e^2}}\right )}{d \sqrt {c^2 d^2+e^2}}+\frac {b \text {csch}^{-1}(c x)}{d e} \]

[Out]

b*arccsch(c*x)/d/e+(-a-b*arccsch(c*x))/e/(e*x+d)+b*arctanh((c^2*d-e/x)/c/(c^2*d^2+e^2)^(1/2)/(1+1/c^2/x^2)^(1/
2))/d/(c^2*d^2+e^2)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6290, 1568, 1475, 844, 215, 725, 206} \[ -\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}+\frac {b \tanh ^{-1}\left (\frac {c^2 d-\frac {e}{x}}{c \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {c^2 d^2+e^2}}\right )}{d \sqrt {c^2 d^2+e^2}}+\frac {b \text {csch}^{-1}(c x)}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/(d + e*x)^2,x]

[Out]

(b*ArcCsch[c*x])/(d*e) - (a + b*ArcCsch[c*x])/(e*(d + e*x)) + (b*ArcTanh[(c^2*d - e/x)/(c*Sqrt[c^2*d^2 + e^2]*
Sqrt[1 + 1/(c^2*x^2)])])/(d*Sqrt[c^2*d^2 + e^2])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1475

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x
] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1568

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[x^(m + mn*q
)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (P
osQ[n2] ||  !IntegerQ[p])

Rule 6290

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a +
b*ArcCsch[c*x]))/(e*(m + 1)), x] + Dist[b/(c*e*(m + 1)), Int[(d + e*x)^(m + 1)/(x^2*Sqrt[1 + 1/(c^2*x^2)]), x]
, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \text {csch}^{-1}(c x)}{(d+e x)^2} \, dx &=-\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}-\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^2 (d+e x)} \, dx}{c e}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}-\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} \left (e+\frac {d}{x}\right ) x^3} \, dx}{c e}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}+\frac {b \operatorname {Subst}\left (\int \frac {x}{(e+d x) \sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c e}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}-\frac {b \operatorname {Subst}\left (\int \frac {1}{(e+d x) \sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c d}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c d e}\\ &=\frac {b \text {csch}^{-1}(c x)}{d e}-\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}+\frac {b \operatorname {Subst}\left (\int \frac {1}{d^2+\frac {e^2}{c^2}-x^2} \, dx,x,\frac {d-\frac {e}{c^2 x}}{\sqrt {1+\frac {1}{c^2 x^2}}}\right )}{c d}\\ &=\frac {b \text {csch}^{-1}(c x)}{d e}-\frac {a+b \text {csch}^{-1}(c x)}{e (d+e x)}+\frac {b \tanh ^{-1}\left (\frac {c^2 d-\frac {e}{x}}{c \sqrt {c^2 d^2+e^2} \sqrt {1+\frac {1}{c^2 x^2}}}\right )}{d \sqrt {c^2 d^2+e^2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 134, normalized size = 1.37 \[ -\frac {a}{e (d+e x)}-\frac {b \log \left (c x \left (\sqrt {\frac {1}{c^2 x^2}+1} \sqrt {c^2 d^2+e^2}-c d\right )+e\right )}{d \sqrt {c^2 d^2+e^2}}+\frac {b \log (d+e x)}{d \sqrt {c^2 d^2+e^2}}+\frac {b \sinh ^{-1}\left (\frac {1}{c x}\right )}{d e}-\frac {b \text {csch}^{-1}(c x)}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/(d + e*x)^2,x]

[Out]

-(a/(e*(d + e*x))) - (b*ArcCsch[c*x])/(e*(d + e*x)) + (b*ArcSinh[1/(c*x)])/(d*e) + (b*Log[d + e*x])/(d*Sqrt[c^
2*d^2 + e^2]) - (b*Log[e + c*(-(c*d) + Sqrt[c^2*d^2 + e^2]*Sqrt[1 + 1/(c^2*x^2)])*x])/(d*Sqrt[c^2*d^2 + e^2])

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fricas [B]  time = 0.61, size = 354, normalized size = 3.61 \[ -\frac {a c^{2} d^{3} + a d e^{2} - \sqrt {c^{2} d^{2} + e^{2}} {\left (b e^{2} x + b d e\right )} \log \left (-\frac {c^{3} d^{2} x - c d e + {\left (c^{3} d^{2} + c e^{2}\right )} x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + {\left (c^{2} d x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + c^{2} d x - e\right )} \sqrt {c^{2} d^{2} + e^{2}}}{e x + d}\right ) - {\left (b c^{2} d^{3} + b d e^{2} + {\left (b c^{2} d^{2} e + b e^{3}\right )} x\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) + {\left (b c^{2} d^{3} + b d e^{2} + {\left (b c^{2} d^{2} e + b e^{3}\right )} x\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) + {\left (b c^{2} d^{3} + b d e^{2}\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )}{c^{2} d^{4} e + d^{2} e^{3} + {\left (c^{2} d^{3} e^{2} + d e^{4}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x+d)^2,x, algorithm="fricas")

[Out]

-(a*c^2*d^3 + a*d*e^2 - sqrt(c^2*d^2 + e^2)*(b*e^2*x + b*d*e)*log(-(c^3*d^2*x - c*d*e + (c^3*d^2 + c*e^2)*x*sq
rt((c^2*x^2 + 1)/(c^2*x^2)) + (c^2*d*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + c^2*d*x - e)*sqrt(c^2*d^2 + e^2))/(e*x
+ d)) - (b*c^2*d^3 + b*d*e^2 + (b*c^2*d^2*e + b*e^3)*x)*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x + 1) + (b*
c^2*d^3 + b*d*e^2 + (b*c^2*d^2*e + b*e^3)*x)*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x - 1) + (b*c^2*d^3 + b
*d*e^2)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)))/(c^2*d^4*e + d^2*e^3 + (c^2*d^3*e^2 + d*e^4)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x+d)^2,x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)/(e*x + d)^2, x)

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maple [B]  time = 0.09, size = 208, normalized size = 2.12 \[ -\frac {c a}{\left (c x e +c d \right ) e}-\frac {c b \,\mathrm {arccsch}\left (c x \right )}{\left (c x e +c d \right ) e}+\frac {b \sqrt {c^{2} x^{2}+1}\, \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{c e \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x d}-\frac {b \sqrt {c^{2} x^{2}+1}\, \ln \left (\frac {2 \sqrt {c^{2} x^{2}+1}\, \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, e -2 c^{2} d x +2 e}{c x e +c d}\right )}{c e \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x d \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(c*x))/(e*x+d)^2,x)

[Out]

-c*a/(c*e*x+c*d)/e-c*b/(c*e*x+c*d)/e*arccsch(c*x)+1/c*b/e*(c^2*x^2+1)^(1/2)/((c^2*x^2+1)/c^2/x^2)^(1/2)/x/d*ar
ctanh(1/(c^2*x^2+1)^(1/2))-1/c*b/e*(c^2*x^2+1)^(1/2)/((c^2*x^2+1)/c^2/x^2)^(1/2)/x/d/((c^2*d^2+e^2)/e^2)^(1/2)
*ln(2*((c^2*x^2+1)^(1/2)*((c^2*d^2+e^2)/e^2)^(1/2)*e-c^2*d*x+e)/(c*e*x+c*d))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (2 \, c^{2} \int \frac {x}{c^{2} e^{2} x^{3} + c^{2} d e x^{2} + e^{2} x + d e + {\left (c^{2} e^{2} x^{3} + c^{2} d e x^{2} + e^{2} x + d e\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} + \frac {i \, c {\left (\log \left (i \, c x + 1\right ) - \log \left (-i \, c x + 1\right )\right )}}{c^{2} d^{2} + e^{2}} - \frac {2 \, e \log \left (e x + d\right )}{c^{2} d^{3} + d e^{2}} - \frac {2 \, c^{2} d^{3} \log \relax (c) + 2 \, d e^{2} \log \relax (c) - 2 \, {\left (c^{2} d^{2} e + e^{3}\right )} x \log \relax (x) + {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (c^{2} d^{3} + d e^{2}\right )} \log \left (\sqrt {c^{2} x^{2} + 1} + 1\right )}{c^{2} d^{4} e + d^{2} e^{3} + {\left (c^{2} d^{3} e^{2} + d e^{4}\right )} x}\right )} b - \frac {a}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x+d)^2,x, algorithm="maxima")

[Out]

-1/2*(2*c^2*integrate(x/(c^2*e^2*x^3 + c^2*d*e*x^2 + e^2*x + d*e + (c^2*e^2*x^3 + c^2*d*e*x^2 + e^2*x + d*e)*s
qrt(c^2*x^2 + 1)), x) + I*c*(log(I*c*x + 1) - log(-I*c*x + 1))/(c^2*d^2 + e^2) - 2*e*log(e*x + d)/(c^2*d^3 + d
*e^2) - (2*c^2*d^3*log(c) + 2*d*e^2*log(c) - 2*(c^2*d^2*e + e^3)*x*log(x) + (c^2*d^2*e*x + c^2*d^3)*log(c^2*x^
2 + 1) - 2*(c^2*d^3 + d*e^2)*log(sqrt(c^2*x^2 + 1) + 1))/(c^2*d^4*e + d^2*e^3 + (c^2*d^3*e^2 + d*e^4)*x))*b -
a/(e^2*x + d*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )}{{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/(d + e*x)^2,x)

[Out]

int((a + b*asinh(1/(c*x)))/(d + e*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{\left (d + e x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsch(c*x))/(e*x+d)**2,x)

[Out]

Integral((a + b*acsch(c*x))/(d + e*x)**2, x)

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