3.134 \(\int \frac {(a+b \sinh ^{-1}(c+d x))^2}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=85 \[ -\frac {b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \log (c+d x)}{d e^3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5865, 12, 5661, 5723, 29} \[ -\frac {b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \log (c+d x)}{d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^2/(c*e + d*e*x)^3,x]

[Out]

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

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 5723

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e
*x^2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 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 {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{(c e+d e x)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{x^2 \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \log (c+d x)}{d e^3}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 120, normalized size = 1.41 \[ -\frac {a \left (a+2 b (c+d x) \sqrt {c^2+2 c d x+d^2 x^2+1}\right )+2 b \sinh ^{-1}(c+d x) \left (a+b (c+d x) \sqrt {c^2+2 c d x+d^2 x^2+1}\right )-2 b^2 (c+d x)^2 \log (c+d x)+b^2 \sinh ^{-1}(c+d x)^2}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^2/(c*e + d*e*x)^3,x]

[Out]

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

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fricas [B]  time = 0.66, size = 319, normalized size = 3.75 \[ -\frac {2 \, a b c^{2} d^{2} x^{2} + 4 \, a b c^{3} d x + 2 \, a b c^{4} + b^{2} c^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + a^{2} c^{2} - 2 \, {\left (a b d^{2} x^{2} + 2 \, a b c d x - {\left (b^{2} c^{2} d x + b^{2} c^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 2 \, {\left (b^{2} c^{2} d^{2} x^{2} + 2 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (d x + c\right ) - 2 \, {\left (a b d^{2} x^{2} + 2 \, a b c d x + a b c^{2}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 2 \, {\left (a b c^{2} d x + a b c^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{2 \, {\left (c^{2} d^{3} e^{3} x^{2} + 2 \, c^{3} d^{2} e^{3} x + c^{4} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

-1/2*(2*a*b*c^2*d^2*x^2 + 4*a*b*c^3*d*x + 2*a*b*c^4 + b^2*c^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))
^2 + a^2*c^2 - 2*(a*b*d^2*x^2 + 2*a*b*c*d*x - (b^2*c^2*d*x + b^2*c^3)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d
*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - 2*(b^2*c^2*d^2*x^2 + 2*b^2*c^3*d*x + b^2*c^4)*log(d*x + c) - 2*(
a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2)*log(-d*x - c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + 2*(a*b*c^2*d*x + a*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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^2/(d*e*x + c*e)^3, x)

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maple [B]  time = 0.20, size = 180, normalized size = 2.12 \[ -\frac {a^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \arcsinh \left (d x +c \right )}{d \,e^{3}}-\frac {b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}-\frac {b^{2} \arcsinh \left (d x +c \right )^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}+\frac {b^{2} \ln \left (\left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )^{2}-1\right )}{d \,e^{3}}-\frac {a b \arcsinh \left (d x +c \right )}{d \,e^{3} \left (d x +c \right )^{2}}-\frac {a b \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^3,x)

[Out]

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

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maxima [B]  time = 0.42, size = 230, normalized size = 2.71 \[ -{\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d \operatorname {arsinh}\left (d x + c\right )}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\log \left (d x + c\right )}{d e^{3}}\right )} b^{2} - a b {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac {\operatorname {arsinh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} - \frac {b^{2} \operatorname {arsinh}\left (d x + c\right )^{2}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac {a^{2}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="maxima")

[Out]

-(sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*d*arcsinh(d*x + c)/(d^3*e^3*x + c*d^2*e^3) - log(d*x + c)/(d*e^3))*b^2 - a
*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*b^2*arcsinh(d*x + c)^2/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3) - 1/2*a^2/(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 = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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