3.1.6 \(\int \frac {(d+c^2 d x^2) (a+b \sinh ^{-1}(c x))}{x} \, dx\) [6]

Optimal. Leaf size=111 \[ -\frac {1}{4} b c d x \sqrt {1+c^2 x^2}-\frac {1}{4} b d \sinh ^{-1}(c x)+\frac {1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )-\frac {1}{2} b d \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right ) \]

[Out]

-1/4*b*d*arcsinh(c*x)+1/2*d*(c^2*x^2+1)*(a+b*arcsinh(c*x))+1/2*d*(a+b*arcsinh(c*x))^2/b+d*(a+b*arcsinh(c*x))*l
n(1-1/(c*x+(c^2*x^2+1)^(1/2))^2)-1/2*b*d*polylog(2,1/(c*x+(c^2*x^2+1)^(1/2))^2)-1/4*b*c*d*x*(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5801, 5775, 3797, 2221, 2317, 2438, 201, 221} \begin {gather*} \frac {1}{2} d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{4} b c d x \sqrt {c^2 x^2+1}-\frac {1}{2} b d \text {Li}_2\left (e^{-2 \sinh ^{-1}(c x)}\right )-\frac {1}{4} b d \sinh ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*c*d*x*Sqrt[1 + c^2*x^2]) - (b*d*ArcSinh[c*x])/4 + (d*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/2 + (d*(a + b
*ArcSinh[c*x])^2)/(2*b) + d*(a + b*ArcSinh[c*x])*Log[1 - E^(-2*ArcSinh[c*x])] - (b*d*PolyLog[2, E^(-2*ArcSinh[
c*x])])/2

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 5801

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.))/(x_), x_Symbol] :> Simp[(d + e*x^2)^p*((
a + b*ArcSinh[c*x])/(2*p)), x] + (Dist[d, Int[(d + e*x^2)^(p - 1)*((a + b*ArcSinh[c*x])/x), x], x] - Dist[b*c*
(d^p/(2*p)), Int[(1 + c^2*x^2)^(p - 1/2), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx &=\frac {1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+d \int \frac {a+b \sinh ^{-1}(c x)}{x} \, dx-\frac {1}{2} (b c d) \int \sqrt {1+c^2 x^2} \, dx\\ &=-\frac {1}{4} b c d x \sqrt {1+c^2 x^2}+\frac {1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+d \text {Subst}\left (\int (a+b x) \coth (x) \, dx,x,\sinh ^{-1}(c x)\right )-\frac {1}{4} (b c d) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {1}{4} b c d x \sqrt {1+c^2 x^2}-\frac {1}{4} b d \sinh ^{-1}(c x)+\frac {1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}-(2 d) \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {1}{4} b c d x \sqrt {1+c^2 x^2}-\frac {1}{4} b d \sinh ^{-1}(c x)+\frac {1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-(b d) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {1}{4} b c d x \sqrt {1+c^2 x^2}-\frac {1}{4} b d \sinh ^{-1}(c x)+\frac {1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\frac {1}{2} (b d) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )\\ &=-\frac {1}{4} b c d x \sqrt {1+c^2 x^2}-\frac {1}{4} b d \sinh ^{-1}(c x)+\frac {1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+\frac {1}{2} b d \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 113, normalized size = 1.02 \begin {gather*} \frac {1}{2} a c^2 d x^2-\frac {1}{4} b c d x \sqrt {1+c^2 x^2}+\frac {1}{4} b d \sinh ^{-1}(c x)+\frac {1}{2} b c^2 d x^2 \sinh ^{-1}(c x)-\frac {1}{2} b d \sinh ^{-1}(c x)^2+b d \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+a d \log (x)+\frac {1}{2} b d \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^2*d*x^2)/2 - (b*c*d*x*Sqrt[1 + c^2*x^2])/4 + (b*d*ArcSinh[c*x])/4 + (b*c^2*d*x^2*ArcSinh[c*x])/2 - (b*d*A
rcSinh[c*x]^2)/2 + b*d*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] + a*d*Log[x] + (b*d*PolyLog[2, E^(2*ArcSinh[c*
x])])/2

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Maple [A]
time = 2.98, size = 162, normalized size = 1.46

method	result	size
derivativedivides	\(\frac {a \,c^{2} d \,x^{2}}{2}+a d \ln \left (c x \right )-\frac {b d \arcsinh \left (c x \right )^{2}}{2}+\frac {b d \arcsinh \left (c x \right ) c^{2} x^{2}}{2}-\frac {b c d x \sqrt {c^{2} x^{2}+1}}{4}+\frac {b d \arcsinh \left (c x \right )}{4}+b d \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+b d \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\)	\(162\)
default	\(\frac {a \,c^{2} d \,x^{2}}{2}+a d \ln \left (c x \right )-\frac {b d \arcsinh \left (c x \right )^{2}}{2}+\frac {b d \arcsinh \left (c x \right ) c^{2} x^{2}}{2}-\frac {b c d x \sqrt {c^{2} x^{2}+1}}{4}+\frac {b d \arcsinh \left (c x \right )}{4}+b d \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+b d \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*c^2*d*x^2+a*d*ln(c*x)-1/2*b*d*arcsinh(c*x)^2+1/2*b*d*arcsinh(c*x)*c^2*x^2-1/4*b*c*d*x*(c^2*x^2+1)^(1/2)+
1/4*b*d*arcsinh(c*x)+b*d*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))+b*d*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+b*d*ar
csinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+b*d*polylog(2,c*x+(c^2*x^2+1)^(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((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x,x, algorithm="maxima")

[Out]

1/2*a*c^2*d*x^2 + a*d*log(x) + integrate(b*c^2*d*x*log(c*x + sqrt(c^2*x^2 + 1)) + b*d*log(c*x + sqrt(c^2*x^2 +
1))/x, 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((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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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\,x\right )\right )\,\left (d\,c^2\,x^2+d\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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