\(\int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx\) [29]

Optimal result

Integrand size = 24, antiderivative size = 135 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=-\frac {b x \sqrt {1+c^2 x^2}}{4 c^3 d}+\frac {b \text {arcsinh}(c x)}{4 c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}+\frac {(a+b \text {arcsinh}(c x))^2}{2 b c^4 d}-\frac {(a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{c^4 d}-\frac {b \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{2 c^4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5812, 5797, 3799, 2221, 2317, 2438, 327, 221} \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {(a+b \text {arcsinh}(c x))^2}{2 b c^4 d}-\frac {\log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{2 c^4 d}+\frac {b \text {arcsinh}(c x)}{4 c^4 d}-\frac {b x \sqrt {c^2 x^2+1}}{4 c^3 d} \]

[In]

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

[Out]

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

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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 5812

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/(e*(m + 2*p + 1))), x] + (-Dist[f^2*((m - 1)/(c^2*
(m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0
]

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {\int \frac {x (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx}{c^2}-\frac {b \int \frac {x^2}{\sqrt {1+c^2 x^2}} \, dx}{2 c d} \\ & = -\frac {b x \sqrt {1+c^2 x^2}}{4 c^3 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {\text {Subst}(\int (a+b x) \tanh (x) \, dx,x,\text {arcsinh}(c x))}{c^4 d}+\frac {b \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{4 c^3 d} \\ & = -\frac {b x \sqrt {1+c^2 x^2}}{4 c^3 d}+\frac {b \text {arcsinh}(c x)}{4 c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}+\frac {(a+b \text {arcsinh}(c x))^2}{2 b c^4 d}-\frac {2 \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\text {arcsinh}(c x)\right )}{c^4 d} \\ & = -\frac {b x \sqrt {1+c^2 x^2}}{4 c^3 d}+\frac {b \text {arcsinh}(c x)}{4 c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}+\frac {(a+b \text {arcsinh}(c x))^2}{2 b c^4 d}-\frac {(a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{c^4 d}+\frac {b \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text {arcsinh}(c x)\right )}{c^4 d} \\ & = -\frac {b x \sqrt {1+c^2 x^2}}{4 c^3 d}+\frac {b \text {arcsinh}(c x)}{4 c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}+\frac {(a+b \text {arcsinh}(c x))^2}{2 b c^4 d}-\frac {(a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{c^4 d}+\frac {b \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \text {arcsinh}(c x)}\right )}{2 c^4 d} \\ & = -\frac {b x \sqrt {1+c^2 x^2}}{4 c^3 d}+\frac {b \text {arcsinh}(c x)}{4 c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}+\frac {(a+b \text {arcsinh}(c x))^2}{2 b c^4 d}-\frac {(a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{c^4 d}-\frac {b \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{2 c^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=-\frac {-2 a c^2 x^2+b c x \sqrt {1+c^2 x^2}-b \text {arcsinh}(c x)-2 b c^2 x^2 \text {arcsinh}(c x)-2 b \text {arcsinh}(c x)^2+4 b \text {arcsinh}(c x) \log \left (1+\frac {c e^{\text {arcsinh}(c x)}}{\sqrt {-c^2}}\right )+4 b \text {arcsinh}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\text {arcsinh}(c x)}}{c}\right )+2 a \log \left (1+c^2 x^2\right )+4 b \operatorname {PolyLog}\left (2,\frac {c e^{\text {arcsinh}(c x)}}{\sqrt {-c^2}}\right )+4 b \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\text {arcsinh}(c x)}}{c}\right )}{4 c^4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {a \left (\frac {c^{2} x^{2}}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}\right )}{d}+\frac {b \operatorname {arcsinh}\left (c x \right )^{2}}{2 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}}{2 d}-\frac {b c x \sqrt {c^{2} x^{2}+1}}{4 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right )}{4 d}-\frac {b \,\operatorname {arcsinh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b \operatorname {polylog}\left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d}}{c^{4}}\)	\(146\)
default	\(\frac {\frac {a \left (\frac {c^{2} x^{2}}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}\right )}{d}+\frac {b \operatorname {arcsinh}\left (c x \right )^{2}}{2 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}}{2 d}-\frac {b c x \sqrt {c^{2} x^{2}+1}}{4 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right )}{4 d}-\frac {b \,\operatorname {arcsinh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b \operatorname {polylog}\left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d}}{c^{4}}\)	\(146\)
parts	\(\frac {a \left (\frac {x^{2}}{2 c^{2}}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c^{4}}\right )}{d}+\frac {b \operatorname {arcsinh}\left (c x \right )^{2}}{2 d \,c^{4}}+\frac {b \,\operatorname {arcsinh}\left (c x \right ) x^{2}}{2 d \,c^{2}}-\frac {b x \sqrt {c^{2} x^{2}+1}}{4 c^{3} d}+\frac {b \,\operatorname {arcsinh}\left (c x \right )}{4 c^{4} d}-\frac {b \,\operatorname {arcsinh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d \,c^{4}}-\frac {b \operatorname {polylog}\left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 c^{4} d}\)	\(159\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} + d} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {\int \frac {a x^{3}}{c^{2} x^{2} + 1}\, dx + \int \frac {b x^{3} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} + d} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{d\,c^2\,x^2+d} \,d x \]

[In]

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

[Out]

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