3.122 \(\int \frac {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=59 \[ -\frac {a+b \sinh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2}-\frac {b \sqrt {(c+d x)^2+1}}{2 d e^3 (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5865, 12, 5661, 264} \[ -\frac {a+b \sinh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2}-\frac {b \sqrt {(c+d x)^2+1}}{2 d e^3 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^3,x]

[Out]

-(b*Sqrt[1 + (c + d*x)^2])/(2*d*e^3*(c + d*x)) - (a + b*ArcSinh[c + d*x])/(2*d*e^3*(c + d*x)^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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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

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Mathematica [A]  time = 0.06, size = 57, normalized size = 0.97 \[ \frac {\frac {-a-b \sinh ^{-1}(c+d x)}{2 (c+d x)^2}-\frac {b \sqrt {(c+d x)^2+1}}{2 (c+d x)}}{d e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^3,x]

[Out]

(-1/2*(b*Sqrt[1 + (c + d*x)^2])/(c + d*x) + (-a - b*ArcSinh[c + d*x])/(2*(c + d*x)^2))/(d*e^3)

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fricas [B]  time = 0.55, size = 118, normalized size = 2.00 \[ \frac {a d^{2} x^{2} + 2 \, a c d x - b c^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (b c^{2} d x + b c^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{2 \, {\left (c^{2} d^{3} e^{3} x^{2} + 2 \, c^{3} d^{2} e^{3} x + c^{4} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)/(d*e*x + c*e)^3, x)

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maple [A]  time = 0.01, size = 60, normalized size = 1.02 \[ \frac {-\frac {a}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b \left (-\frac {\arcsinh \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^3,x)

[Out]

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

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maxima [B]  time = 0.41, size = 117, normalized size = 1.98 \[ -\frac {1}{2} \, b {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac {\operatorname {arsinh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} - \frac {a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))/(c*e + d*e*x)^3,x)

[Out]

int((a + b*asinh(c + d*x))/(c*e + d*e*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b \operatorname {asinh}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))/(d*e*x+c*e)**3,x)

[Out]

(Integral(a/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(b*asinh(c + d*x)/(c**3 + 3*c**2*d*x
+ 3*c*d**2*x**2 + d**3*x**3), x))/e**3

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