3.2.34 \(\int \frac {(a+b \sinh ^{-1}(c+d x))^2}{(c e+d e x)^3} \, dx\) [134]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.09, 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 = {5859, 12, 5776, 5800, 29} \begin {gather*} -\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} \end {gather*}

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 5776

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 5800

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/(f*(m + 1)))*Simp[(
d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[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 5859

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

________________________________________________________________________________________

Mathematica [A]
time = 0.16, size = 120, normalized size = 1.41 \begin {gather*} -\frac {a \left (a+2 b (c+d x) \sqrt {1+c^2+2 c d x+d^2 x^2}\right )+2 b \left (a+b (c+d x) \sqrt {1+c^2+2 c d x+d^2 x^2}\right ) \sinh ^{-1}(c+d x)+b^2 \sinh ^{-1}(c+d x)^2-2 b^2 (c+d x)^2 \log (c+d x)}{2 d e^3 (c+d x)^2} \end {gather*}

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)

________________________________________________________________________________________

Maple [A]
time = 4.23, size = 161, normalized size = 1.89

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \arcsinh \left (d x +c \right )}{e^{3}}-\frac {b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {b^{2} \arcsinh \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b^{2} \ln \left (\left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )^{2}-1\right )}{e^{3}}+\frac {2 a b \left (-\frac {\arcsinh \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(161\)
default	\(\frac {-\frac {a^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \arcsinh \left (d x +c \right )}{e^{3}}-\frac {b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {b^{2} \arcsinh \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b^{2} \ln \left (\left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )^{2}-1\right )}{e^{3}}+\frac {2 a b \left (-\frac {\arcsinh \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(161\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (78) = 156\).
time = 0.28, size = 216, normalized size = 2.54 \begin {gather*} -{\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d \operatorname {arsinh}\left (d x + c\right )}{d^{3} x e^{3} + c d^{2} e^{3}} - \frac {e^{\left (-3\right )} \log \left (d x + c\right )}{d}\right )} b^{2} - a b {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d}{d^{3} x e^{3} + c d^{2} e^{3}} + \frac {\operatorname {arsinh}\left (d x + c\right )}{d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + c^{2} d e^{3}}\right )} - \frac {b^{2} \operatorname {arsinh}\left (d x + c\right )^{2}}{2 \, {\left (d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + c^{2} d e^{3}\right )}} - \frac {a^{2}}{2 \, {\left (d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + c^{2} d e^{3}\right )}} \end {gather*}

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (78) = 156\).
time = 0.49, size = 412, normalized size = 4.85 \begin {gather*} -\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 ({\left (c^{2} d^{3} x^{2} + 2 \, c^{3} d^{2} x + c^{4} d\right )} \cosh \left (1\right )^{3} + 3 \, {\left (c^{2} d^{3} x^{2} + 2 \, c^{3} d^{2} x + c^{4} d\right )} \cosh \left (1\right )^{2} \sinh \left (1\right ) + 3 \, {\left (c^{2} d^{3} x^{2} + 2 \, c^{3} d^{2} x + c^{4} d\right )} \cosh \left (1\right ) \sinh \left (1\right )^{2} + {\left (c^{2} d^{3} x^{2} + 2 \, c^{3} d^{2} x + c^{4} d\right )} \sinh \left (1\right )^{3}\right )}} \end {gather*}

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*x^2 + 2*c^3*d^2*x + c^4*d)*cosh(1)^3 + 3*(c^2*d^3*x^2 + 2*c^
3*d^2*x + c^4*d)*cosh(1)^2*sinh(1) + 3*(c^2*d^3*x^2 + 2*c^3*d^2*x + c^4*d)*cosh(1)*sinh(1)^2 + (c^2*d^3*x^2 +
2*c^3*d^2*x + c^4*d)*sinh(1)^3)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}} \end {gather*}

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \end {gather*}

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)

________________________________________________________________________________________