3.2.32 \(\int \frac {(a+b \sinh ^{-1}(c+d x))^2}{c e+d e x} \, dx\) [132]

Optimal. Leaf size=116 \[ \frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac {b \left (a+b \sinh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac {b^2 \text {PolyLog}\left (3,e^{-2 \sinh ^{-1}(c+d x)}\right )}{2 d e} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5859, 12, 5775, 3797, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {b \text {Li}_2\left (e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}+\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e}-\frac {b^2 \text {Li}_3\left (e^{-2 \sinh ^{-1}(c+d x)}\right )}{2 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

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
]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {\text {Subst}\left (\int (a+b x)^2 \coth (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 b d e}-\frac {2 \text {Subst}\left (\int \frac {e^{2 x} (a+b x)^2}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac {b \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac {b^2 \text {Subst}\left (\int \text {Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac {b \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac {b^2 \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 b d e}+\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac {b \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac {b^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 100, normalized size = 0.86 \begin {gather*} \frac {-2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \left (a+b \sinh ^{-1}(c+d x)-3 b \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )\right )+6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right )-3 b^3 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right )}{6 b d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSinh[c + d*x])^2*(a + b*ArcSinh[c + d*x] - 3*b*Log[1 - E^(2*ArcSinh[c + d*x])]) + 6*b^2*(a + b*A
rcSinh[c + d*x])*PolyLog[2, E^(2*ArcSinh[c + d*x])] - 3*b^3*PolyLog[3, E^(2*ArcSinh[c + d*x])])/(6*b*d*e)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(368\) vs. \(2(136)=272\).
time = 2.52, size = 369, normalized size = 3.18

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \ln \left (d x +c \right )}{e}-\frac {b^{2} \arcsinh \left (d x +c \right )^{3}}{3 e}+\frac {b^{2} \arcsinh \left (d x +c \right )^{2} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \arcsinh \left (d x +c \right ) \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}-\frac {2 b^{2} \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {b^{2} \arcsinh \left (d x +c \right )^{2} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \arcsinh \left (d x +c \right ) \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}-\frac {2 b^{2} \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}-\frac {a b \arcsinh \left (d x +c \right )^{2}}{e}+\frac {2 a b \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}}{d}\)	\(369\)
default	\(\frac {\frac {a^{2} \ln \left (d x +c \right )}{e}-\frac {b^{2} \arcsinh \left (d x +c \right )^{3}}{3 e}+\frac {b^{2} \arcsinh \left (d x +c \right )^{2} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \arcsinh \left (d x +c \right ) \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}-\frac {2 b^{2} \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {b^{2} \arcsinh \left (d x +c \right )^{2} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \arcsinh \left (d x +c \right ) \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}-\frac {2 b^{2} \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}-\frac {a b \arcsinh \left (d x +c \right )^{2}}{e}+\frac {2 a b \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e}}{d}\)	\(369\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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),x, algorithm="maxima")

[Out]

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

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Fricas [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),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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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),x, algorithm="giac")

[Out]

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

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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}{c\,e+d\,e\,x} \,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),x)

[Out]

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

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