\(\int \frac {(d+c^2 d x^2)^2 (a+b \text {arcsinh}(c x))}{x^2} \, dx\) [16]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 120 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^2} \, dx=-\frac {5}{3} b c d^2 \sqrt {1+c^2 x^2}-\frac {1}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2}-\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))-b c d^2 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right ) \]

[Out]

-1/9*b*c*d^2*(c^2*x^2+1)^(3/2)-d^2*(a+b*arcsinh(c*x))/x+2*c^2*d^2*x*(a+b*arcsinh(c*x))+1/3*c^4*d^2*x^3*(a+b*ar
csinh(c*x))-b*c*d^2*arctanh((c^2*x^2+1)^(1/2))-5/3*b*c*d^2*(c^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {276, 5803, 12, 1265, 911, 1167, 214} \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))+2 c^2 d^2 x (a+b \text {arcsinh}(c x))-\frac {d^2 (a+b \text {arcsinh}(c x))}{x}-b c d^2 \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )-\frac {1}{9} b c d^2 \left (c^2 x^2+1\right )^{3/2}-\frac {5}{3} b c d^2 \sqrt {c^2 x^2+1} \]

[In]

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

[Out]

(-5*b*c*d^2*Sqrt[1 + c^2*x^2])/3 - (b*c*d^2*(1 + c^2*x^2)^(3/2))/9 - (d^2*(a + b*ArcSinh[c*x]))/x + 2*c^2*d^2*
x*(a + b*ArcSinh[c*x]) + (c^4*d^2*x^3*(a + b*ArcSinh[c*x]))/3 - b*c*d^2*ArcTanh[Sqrt[1 + c^2*x^2]]

Rule 12

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
 x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
 p] && FractionQ[m]

Rule 1167

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
 IntegerQ[(m - 1)/2]

Rule 5803

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))-(b c) \int \frac {d^2 \left (-3+6 c^2 x^2+c^4 x^4\right )}{3 x \sqrt {1+c^2 x^2}} \, dx \\ & = -\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))-\frac {1}{3} \left (b c d^2\right ) \int \frac {-3+6 c^2 x^2+c^4 x^4}{x \sqrt {1+c^2 x^2}} \, dx \\ & = -\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))-\frac {1}{6} \left (b c d^2\right ) \text {Subst}\left (\int \frac {-3+6 c^2 x+c^4 x^2}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))-\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {-8+4 x^2+x^4}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{3 c} \\ & = -\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))-\frac {\left (b d^2\right ) \text {Subst}\left (\int \left (5 c^2+c^2 x^2-\frac {3}{-\frac {1}{c^2}+\frac {x^2}{c^2}}\right ) \, dx,x,\sqrt {1+c^2 x^2}\right )}{3 c} \\ & = -\frac {5}{3} b c d^2 \sqrt {1+c^2 x^2}-\frac {1}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2}-\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))+\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{c} \\ & = -\frac {5}{3} b c d^2 \sqrt {1+c^2 x^2}-\frac {1}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2}-\frac {d^2 (a+b \text {arcsinh}(c x))}{x}+2 c^2 d^2 x (a+b \text {arcsinh}(c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \text {arcsinh}(c x))-b c d^2 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.03 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {d^2 \left (-9 a+18 a c^2 x^2+3 a c^4 x^4-16 b c x \sqrt {1+c^2 x^2}-b c^3 x^3 \sqrt {1+c^2 x^2}+3 b \left (-3+6 c^2 x^2+c^4 x^4\right ) \text {arcsinh}(c x)+9 b c x \log (x)-9 b c x \log \left (1+\sqrt {1+c^2 x^2}\right )\right )}{9 x} \]

[In]

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

[Out]

(d^2*(-9*a + 18*a*c^2*x^2 + 3*a*c^4*x^4 - 16*b*c*x*Sqrt[1 + c^2*x^2] - b*c^3*x^3*Sqrt[1 + c^2*x^2] + 3*b*(-3 +
 6*c^2*x^2 + c^4*x^4)*ArcSinh[c*x] + 9*b*c*x*Log[x] - 9*b*c*x*Log[1 + Sqrt[1 + c^2*x^2]]))/(9*x)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.93

method result size
parts \(d^{2} a \left (\frac {c^{4} x^{3}}{3}+2 c^{2} x -\frac {1}{x}\right )+d^{2} b c \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}}{3}+2 \,\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{c x}-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {c^{2} x^{2}+1}}{9}-\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\) \(112\)
derivativedivides \(c \left (d^{2} a \left (\frac {c^{3} x^{3}}{3}+2 c x -\frac {1}{c x}\right )+d^{2} b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}}{3}+2 \,\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{c x}-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {c^{2} x^{2}+1}}{9}-\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\) \(114\)
default \(c \left (d^{2} a \left (\frac {c^{3} x^{3}}{3}+2 c x -\frac {1}{c x}\right )+d^{2} b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}}{3}+2 \,\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{c x}-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {c^{2} x^{2}+1}}{9}-\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\) \(114\)

[In]

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

[Out]

d^2*a*(1/3*c^4*x^3+2*c^2*x-1/x)+d^2*b*c*(1/3*arcsinh(c*x)*c^3*x^3+2*arcsinh(c*x)*c*x-arcsinh(c*x)/c/x-1/9*c^2*
x^2*(c^2*x^2+1)^(1/2)-16/9*(c^2*x^2+1)^(1/2)-arctanh(1/(c^2*x^2+1)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (108) = 216\).

Time = 0.29 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.90 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {3 \, a c^{4} d^{2} x^{4} + 18 \, a c^{2} d^{2} x^{2} - 9 \, b c d^{2} x \log \left (-c x + \sqrt {c^{2} x^{2} + 1} + 1\right ) + 9 \, b c d^{2} x \log \left (-c x + \sqrt {c^{2} x^{2} + 1} - 1\right ) - 3 \, {\left (b c^{4} + 6 \, b c^{2} - 3 \, b\right )} d^{2} x \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) - 9 \, a d^{2} + 3 \, {\left (b c^{4} d^{2} x^{4} + 6 \, b c^{2} d^{2} x^{2} - {\left (b c^{4} + 6 \, b c^{2} - 3 \, b\right )} d^{2} x - 3 \, b d^{2}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (b c^{3} d^{2} x^{3} + 16 \, b c d^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}{9 \, x} \]

[In]

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

[Out]

1/9*(3*a*c^4*d^2*x^4 + 18*a*c^2*d^2*x^2 - 9*b*c*d^2*x*log(-c*x + sqrt(c^2*x^2 + 1) + 1) + 9*b*c*d^2*x*log(-c*x
 + sqrt(c^2*x^2 + 1) - 1) - 3*(b*c^4 + 6*b*c^2 - 3*b)*d^2*x*log(-c*x + sqrt(c^2*x^2 + 1)) - 9*a*d^2 + 3*(b*c^4
*d^2*x^4 + 6*b*c^2*d^2*x^2 - (b*c^4 + 6*b*c^2 - 3*b)*d^2*x - 3*b*d^2)*log(c*x + sqrt(c^2*x^2 + 1)) - (b*c^3*d^
2*x^3 + 16*b*c*d^2*x)*sqrt(c^2*x^2 + 1))/x

Sympy [F]

\[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^2} \, dx=d^{2} \left (\int 2 a c^{2}\, dx + \int \frac {a}{x^{2}}\, dx + \int a c^{4} x^{2}\, dx + \int 2 b c^{2} \operatorname {asinh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x^{2}}\, dx + \int b c^{4} x^{2} \operatorname {asinh}{\left (c x \right )}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {1}{3} \, a c^{4} d^{2} x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{4} d^{2} + 2 \, a c^{2} d^{2} x + 2 \, {\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b c d^{2} - {\left (c \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arsinh}\left (c x\right )}{x}\right )} b d^{2} - \frac {a d^{2}}{x} \]

[In]

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

[Out]

1/3*a*c^4*d^2*x^3 + 1/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*b*c^4*d
^2 + 2*a*c^2*d^2*x + 2*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*c*d^2 - (c*arcsinh(1/(c*abs(x))) + arcsinh(c*x
)/x)*b*d^2 - a*d^2/x

Giac [F(-2)]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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