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

Optimal. Leaf size=147 \[ -\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{d \sqrt {c^2 d^2-e^2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d e} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6288, 961, 266, 63, 208, 725, 204} \[ -\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{d \sqrt {c^2 d^2-e^2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 204

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

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 961

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

Rule 6288

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a +
b*ArcSech[c*x]))/(e*(m + 1)), x] + Dist[(b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)])/(e*(m + 1)), Int[(d + e*x)^(m + 1)
/(x*Sqrt[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 {sech}^{-1}(c x)}{(d+e x)^2} \, dx &=-\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x (d+e x) \sqrt {1-c^2 x^2}} \, dx}{e}\\ &=-\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \left (\frac {1}{d x \sqrt {1-c^2 x^2}}-\frac {e}{d (d+e x) \sqrt {1-c^2 x^2}}\right ) \, dx}{e}\\ &=-\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{(d+e x) \sqrt {1-c^2 x^2}} \, dx}{d}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x \sqrt {1-c^2 x^2}} \, dx}{d e}\\ &=-\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{-c^2 d^2+e^2-x^2} \, dx,x,\frac {e+c^2 d x}{\sqrt {1-c^2 x^2}}\right )}{d}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{2 d e}\\ &=-\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{d \sqrt {c^2 d^2-e^2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{c^2 d e}\\ &=-\frac {a+b \text {sech}^{-1}(c x)}{e (d+e x)}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{d \sqrt {c^2 d^2-e^2}}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d e}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 222, normalized size = 1.51 \[ -\frac {a}{e (d+e x)}+\frac {b \log (d+e x)}{d \sqrt {e^2-c^2 d^2}}-\frac {b \log \left (c x \sqrt {\frac {1-c x}{c x+1}} \sqrt {e^2-c^2 d^2}+\sqrt {\frac {1-c x}{c x+1}} \sqrt {e^2-c^2 d^2}+c^2 d x+e\right )}{d \sqrt {e^2-c^2 d^2}}+\frac {b \log \left (c x \sqrt {\frac {1-c x}{c x+1}}+\sqrt {\frac {1-c x}{c x+1}}+1\right )}{d e}-\frac {b \text {sech}^{-1}(c x)}{e (d+e x)}-\frac {b \log (x)}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a/(e*(d + e*x))) - (b*ArcSech[c*x])/(e*(d + e*x)) - (b*Log[x])/(d*e) + (b*Log[d + e*x])/(d*Sqrt[-(c^2*d^2) +
e^2]) + (b*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1 + c*x)]])/(d*e) - (b*Log[e + c^2*d*x + S
qrt[-(c^2*d^2) + e^2]*Sqrt[(1 - c*x)/(1 + c*x)] + c*Sqrt[-(c^2*d^2) + e^2]*x*Sqrt[(1 - c*x)/(1 + c*x)]])/(d*Sq
rt[-(c^2*d^2) + e^2])

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fricas [B]  time = 1.08, size = 578, normalized size = 3.93 \[ \left [-\frac {a c^{2} d^{3} - a d e^{2} + \sqrt {-c^{2} d^{2} + e^{2}} {\left (b e^{2} x + b d e\right )} \log \left (\frac {c^{2} d e x - {\left (c^{3} d^{2} - c e^{2}\right )} x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + e^{2} - \sqrt {-c^{2} d^{2} + e^{2}} {\left (c^{2} d x + c e x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + e\right )}}{e x + d}\right ) + {\left (b c^{2} d^{3} - b d e^{2} + {\left (b c^{2} d^{2} e - b e^{3}\right )} x\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + {\left (b c^{2} d^{3} - b d e^{2}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{c^{2} d^{4} e - d^{2} e^{3} + {\left (c^{2} d^{3} e^{2} - d e^{4}\right )} x}, -\frac {a c^{2} d^{3} - a d e^{2} - 2 \, \sqrt {c^{2} d^{2} - e^{2}} {\left (b e^{2} x + b d e\right )} \arctan \left (-\frac {\sqrt {c^{2} d^{2} - e^{2}} c d x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - \sqrt {c^{2} d^{2} - e^{2}} {\left (e x + d\right )}}{{\left (c^{2} d^{2} - e^{2}\right )} x}\right ) + {\left (b c^{2} d^{3} - b d e^{2} + {\left (b c^{2} d^{2} e - b e^{3}\right )} x\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + {\left (b c^{2} d^{3} - b d e^{2}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{c^{2} d^{4} e - d^{2} e^{3} + {\left (c^{2} d^{3} e^{2} - d e^{4}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(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^2*d*e*x - (c^3*d^2 - c*e^2)*x*sqrt(-(c^
2*x^2 - 1)/(c^2*x^2)) + e^2 - sqrt(-c^2*d^2 + e^2)*(c^2*d*x + c*e*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + e))/(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)) - 1)/x) + (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), -(a*c^2*d^3 - a*d*e^2 - 2*sqrt(c^2*d^2 - e^2)*(b*e^2*x + b*d*e)*arctan(-(sqrt(c^2*d^2 - e^2)*c*d*x*s
qrt(-(c^2*x^2 - 1)/(c^2*x^2)) - sqrt(c^2*d^2 - e^2)*(e*x + d))/((c^2*d^2 - e^2)*x)) + (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)) - 1)/x) + (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 \[ \int \frac {b \operatorname {arsech}\left (c x\right ) + a}{{\left (e x + d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 231, normalized size = 1.57 \[ -\frac {c a}{\left (c x e +c d \right ) e}-\frac {c b \,\mathrm {arcsech}\left (c x \right )}{\left (c x e +c d \right ) e}+\frac {c b \sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{e d \sqrt {-c^{2} x^{2}+1}}-\frac {c b \sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \ln \left (\frac {2 \sqrt {-c^{2} x^{2}+1}\, \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, e +2 c^{2} d x +2 e}{c x e +c d}\right )}{e \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, d \sqrt {-c^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*a/(c*e*x+c*d)/e-c*b/(c*e*x+c*d)/e*arcsech(c*x)+c*b/e*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/d/(-c^2*x^2
+1)^(1/2)*arctanh(1/(-c^2*x^2+1)^(1/2))-c*b/e*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/(-(c^2*d^2-e^2)/e^2)^
(1/2)/d/(-c^2*x^2+1)^(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 \[ {\left (c^{2} \int \frac {x^{2}}{c^{2} d^{2} x^{2} + {\left (c^{2} d^{2} x^{2} - d^{2} + {\left (c^{2} d e x^{2} - d e\right )} x\right )} \sqrt {c x + 1} \sqrt {-c x + 1} - d^{2} + {\left (c^{2} d e x^{2} - d e\right )} x}\,{d x} + \frac {x \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right ) - x \log \relax (c) - x \log \relax (x)}{d e x + d^{2}} - \int \frac {1}{c^{2} d^{2} x^{2} - d^{2} + {\left (c^{2} d e x^{2} - d e\right )} x}\,{d x}\right )} b - \frac {a}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*integrate(x^2/(c^2*d^2*x^2 + (c^2*d^2*x^2 - d^2 + (c^2*d*e*x^2 - d*e)*x)*sqrt(c*x + 1)*sqrt(-c*x + 1) - d
^2 + (c^2*d*e*x^2 - d*e)*x), x) + (x*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1) - x*log(c) - x*log(x))/(d*e*x + d^2
) - integrate(1/(c^2*d^2*x^2 - d^2 + (c^2*d*e*x^2 - d*e)*x), x))*b - a/(e^2*x + d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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