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

Optimal. Leaf size=49 \[ -\frac {a+b \sinh ^{-1}(c+d x)}{d e^2 (c+d x)}-\frac {b \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{d e^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5865, 12, 5661, 266, 63, 207} \[ -\frac {a+b \sinh ^{-1}(c+d x)}{d e^2 (c+d x)}-\frac {b \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 207

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

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 5661

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

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.88 \[ -\frac {\frac {a+b \sinh ^{-1}(c+d x)}{c+d x}+b \tanh ^{-1}\left (\sqrt {(c+d x)^2+1}\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.58, size = 175, normalized size = 3.57 \[ \frac {b d x \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - a c - {\left (b c d x + b c^{2}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} + 1\right ) + {\left (b d x + b c\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + {\left (b c d x + b c^{2}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} - 1\right )}{c d^{2} e^{2} x + c^{2} d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.49, size = 131, normalized size = 2.67 \[ -b {\left (\frac {e^{\left (-1\right )} \log \left (d x + c + \sqrt {{\left (d x + c\right )}^{2} + 1}\right )}{{\left (d x e + c e\right )} d} + \frac {d e^{\left (-2\right )} \log \left (\sqrt {\frac {e^{2}}{{\left (d x e + c e\right )}^{2}} + 1} + \frac {\sqrt {d^{2}} e}{{\left (d x e + c e\right )} d}\right )}{{\left | d \right |}^{2} \mathrm {sgn}\left (\frac {1}{d x e + c e}\right ) \mathrm {sgn}\relax (d)}\right )} - \frac {a e^{\left (-1\right )}}{{\left (d x e + c e\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 54, normalized size = 1.10 \[ \frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arcsinh \left (d x +c \right )}{d x +c}-\arctanh \left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )\right )}{e^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 80, normalized size = 1.63 \[ -b {\left (\frac {\operatorname {arsinh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}} + \frac {\operatorname {arsinh}\left (\frac {d e^{2}}{{\left | d^{2} e^{2} x + c d e^{2} \right |}}\right )}{d e^{2}}\right )} - \frac {a}{d^{2} e^{2} x + c d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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