3.1.38 \(\int \frac {x^3 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^2} \, dx\) [38]

Optimal. Leaf size=145 \[ -\frac {b x}{2 c^3 d^2 \sqrt {1+c^2 x^2}}+\frac {b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d^2}+\frac {b \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2} \]

[Out]

1/2*b*arcsinh(c*x)/c^4/d^2-1/2*x^2*(a+b*arcsinh(c*x))/c^2/d^2/(c^2*x^2+1)-1/2*(a+b*arcsinh(c*x))^2/b/c^4/d^2+(
a+b*arcsinh(c*x))*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d^2+1/2*b*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d^2-
1/2*b*x/c^3/d^2/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5810, 5797, 3799, 2221, 2317, 2438, 294, 221} \begin {gather*} -\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac {\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2}+\frac {b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac {b x}{2 c^3 d^2 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*x)/(c^3*d^2*Sqrt[1 + c^2*x^2]) + (b*ArcSinh[c*x])/(2*c^4*d^2) - (x^2*(a + b*ArcSinh[c*x]))/(2*c^2*d^2*
(1 + c^2*x^2)) - (a + b*ArcSinh[c*x])^2/(2*b*c^4*d^2) + ((a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(c^
4*d^2) + (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*c^4*d^2)

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 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, 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 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 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))), x]
, x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5797

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

Rule 5810

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] + (-Dist[f^2*((m - 1)/(2*e*(p +
1))), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(2*c*(p + 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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^2} \, dx &=-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {b \int \frac {x^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 c d^2}+\frac {\int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx}{c^2 d}\\ &=-\frac {b x}{2 c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\text {Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d^2}+\frac {b \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{2 c^3 d^2}\\ &=-\frac {b x}{2 c^3 d^2 \sqrt {1+c^2 x^2}}+\frac {b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac {2 \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d^2}\\ &=-\frac {b x}{2 c^3 d^2 \sqrt {1+c^2 x^2}}+\frac {b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d^2}-\frac {b \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d^2}\\ &=-\frac {b x}{2 c^3 d^2 \sqrt {1+c^2 x^2}}+\frac {b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d^2}-\frac {b \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2}\\ &=-\frac {b x}{2 c^3 d^2 \sqrt {1+c^2 x^2}}+\frac {b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d^2}+\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.08, size = 241, normalized size = 1.66 \begin {gather*} \frac {a-b c x \sqrt {1+c^2 x^2}+b \sinh ^{-1}(c x)-b \sinh ^{-1}(c x)^2-b c^2 x^2 \sinh ^{-1}(c x)^2+2 b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+2 b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+2 b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+2 b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+a \log \left (1+c^2 x^2\right )+a c^2 x^2 \log \left (1+c^2 x^2\right )+2 b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )+2 b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c^4 d^2 \left (1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 4.88, size = 187, normalized size = 1.29

method	result	size
derivativedivides	\(\frac {\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}+\frac {a}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )^{2}}{2 d^{2}}-\frac {b c x}{2 d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {b \,c^{2} x^{2}}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right )}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d^{2}}}{c^{4}}\)	\(187\)
default	\(\frac {\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}+\frac {a}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )^{2}}{2 d^{2}}-\frac {b c x}{2 d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {b \,c^{2} x^{2}}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right )}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d^{2}}}{c^{4}}\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(1/2*a/d^2*ln(c^2*x^2+1)+1/2*a/d^2/(c^2*x^2+1)-1/2*b/d^2*arcsinh(c*x)^2-1/2*b*c*x/d^2/(c^2*x^2+1)^(1/2)+
1/2*b/d^2/(c^2*x^2+1)*c^2*x^2+1/2*b/d^2*arcsinh(c*x)/(c^2*x^2+1)+1/2*b/d^2/(c^2*x^2+1)+b/d^2*arcsinh(c*x)*ln(1
+(c*x+(c^2*x^2+1)^(1/2))^2)+1/2*b*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^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(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/8*b*(((c^2*x^2 + 1)*log(c^2*x^2 + 1)^2 - 4*((c^2*x^2 + 1)*log(c^2*x^2 + 1) + 1)*log(c*x + sqrt(c^2*x^2 + 1)
) - 2)/(c^6*d^2*x^2 + c^4*d^2) + 8*integrate(1/2*((c^2*x^2 + 1)*log(c^2*x^2 + 1) + 1)/(c^8*d^2*x^5 + 2*c^6*d^2
*x^3 + c^4*d^2*x + (c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2)*sqrt(c^2*x^2 + 1)), x)) + 1/2*a*(1/(c^6*d^2*x^2 + c
^4*d^2) + log(c^2*x^2 + 1)/(c^4*d^2))

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

[Out]

integral((b*x^3*arcsinh(c*x) + a*x^3)/(c^4*d^2*x^4 + 2*c^2*d^2*x^2 + d^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x^{3}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{3} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x**3/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b*x**3*asinh(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1),
x))/d**2

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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(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,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 {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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