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

Optimal. Leaf size=82 \[ -\frac {a+b \sinh ^{-1}(c x)}{e (d+e x)}-\frac {b c \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}}\right )}{e \sqrt {c^2 d^2+e^2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5801, 725, 206} \[ -\frac {a+b \sinh ^{-1}(c x)}{e (d+e x)}-\frac {b c \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}}\right )}{e \sqrt {c^2 d^2+e^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSinh[c*x])/(e*(d + e*x))) - (b*c*ArcTanh[(e - c^2*d*x)/(Sqrt[c^2*d^2 + e^2]*Sqrt[1 + c^2*x^2])])/(
e*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 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 5801

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)
*(a + b*ArcSinh[c*x])^n)/(e*(m + 1)), x] - Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcSinh[c*x
])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 79, normalized size = 0.96 \[ -\frac {\frac {a+b \sinh ^{-1}(c x)}{d+e x}+\frac {b c \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}}\right )}{\sqrt {c^2 d^2+e^2}}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 253, normalized size = 3.09 \[ -\frac {a c^{2} d^{3} + a d e^{2} - {\left (b c^{2} d^{2} e + b e^{3}\right )} x \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (b c d e x + b c d^{2}\right )} \sqrt {c^{2} d^{2} + e^{2}} \log \left (-\frac {c^{3} d^{2} x - c d e + \sqrt {c^{2} d^{2} + e^{2}} {\left (c^{2} d x - e\right )} + {\left (c^{2} d^{2} + \sqrt {c^{2} d^{2} + e^{2}} c d + e^{2}\right )} \sqrt {c^{2} x^{2} + 1}}{e x + d}\right ) - {\left (b c^{2} d^{3} + b d e^{2} + {\left (b c^{2} d^{2} e + b e^{3}\right )} x\right )} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{2} d^{4} e + d^{2} e^{3} + {\left (c^{2} d^{3} e^{2} + d e^{4}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.86, size = 214, normalized size = 2.61 \[ {\left (\frac {c e^{\left (-1\right )} \log \left (-c^{2} d + \sqrt {c^{2} d^{2} + e^{2}} {\left | c \right |}\right ) \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{\sqrt {c^{2} d^{2} + e^{2}}} - \frac {c e^{\left (-1\right )} \log \left (-c^{2} d + \sqrt {c^{2} d^{2} + e^{2}} {\left (\sqrt {c^{2} - \frac {2 \, c^{2} d}{x e + d} + \frac {c^{2} d^{2}}{{\left (x e + d\right )}^{2}} + \frac {e^{2}}{{\left (x e + d\right )}^{2}}} + \frac {\sqrt {c^{2} d^{2} e^{2} + e^{4}} e^{\left (-1\right )}}{x e + d}\right )}\right )}{\sqrt {c^{2} d^{2} + e^{2}} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {e^{\left (-1\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{x e + d}\right )} b - \frac {a e^{\left (-1\right )}}{x e + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*e^(-1)*log(-c^2*d + sqrt(c^2*d^2 + e^2)*abs(c))*sgn(1/(x*e + d))/sqrt(c^2*d^2 + e^2) - c*e^(-1)*log(-c^2*d
+ sqrt(c^2*d^2 + e^2)*(sqrt(c^2 - 2*c^2*d/(x*e + d) + c^2*d^2/(x*e + d)^2 + e^2/(x*e + d)^2) + sqrt(c^2*d^2*e^
2 + e^4)*e^(-1)/(x*e + d)))/(sqrt(c^2*d^2 + e^2)*sgn(1/(x*e + d))) - e^(-1)*log(c*x + sqrt(c^2*x^2 + 1))/(x*e
+ d))*b - a*e^(-1)/(x*e + d)

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maple [B]  time = 0.02, size = 178, normalized size = 2.17 \[ -\frac {c a}{\left (c e x +c d \right ) e}-\frac {c b \arcsinh \left (c x \right )}{\left (c e x +c d \right ) e}-\frac {c b \ln \left (\frac {\frac {2 c^{2} d^{2}+2 e^{2}}{e^{2}}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{c x +\frac {c d}{e}}\right )}{e^{2} \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 94, normalized size = 1.15 \[ -b {\left (\frac {\operatorname {arsinh}\left (c x\right )}{e^{2} x + d e} - \frac {c \operatorname {arsinh}\left (\frac {c d e x}{{\left | e^{2} x + d e \right |}} - \frac {e^{2}}{c {\left | e^{2} x + d e \right |}}\right )}{\sqrt {\frac {c^{2} d^{2}}{e^{2}} + 1} e^{2}}\right )} - \frac {a}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*(arcsinh(c*x)/(e^2*x + d*e) - c*arcsinh(c*d*e*x/abs(e^2*x + d*e) - e^2/(c*abs(e^2*x + d*e)))/(sqrt(c^2*d^2/
e^2 + 1)*e^2)) - a/(e^2*x + d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(d + e*x)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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