\(\int \frac {(d+c^2 d x^2)^3 (a+b \text {arcsinh}(c x))}{x^2} \, dx\) [25]

Optimal result

Integrand size = 24, antiderivative size = 160 \[ \int \frac {\left (d+c^2 d x^2\right )^3 (a+b \text {arcsinh}(c x))}{x^2} \, dx=-\frac {11}{5} b c d^3 \sqrt {1+c^2 x^2}-\frac {1}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2}-\frac {1}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2}-\frac {d^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))-b c d^3 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right ) \]

[Out]

-1/5*b*c*d^3*(c^2*x^2+1)^(3/2)-1/25*b*c*d^3*(c^2*x^2+1)^(5/2)-d^3*(a+b*arcsinh(c*x))/x+3*c^2*d^3*x*(a+b*arcsin
h(c*x))+c^4*d^3*x^3*(a+b*arcsinh(c*x))+1/5*c^6*d^3*x^5*(a+b*arcsinh(c*x))-b*c*d^3*arctanh((c^2*x^2+1)^(1/2))-1
1/5*b*c*d^3*(c^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 160, 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, 1813, 1634, 65, 214} \[ \int \frac {\left (d+c^2 d x^2\right )^3 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+3 c^2 d^3 x (a+b \text {arcsinh}(c x))-\frac {d^3 (a+b \text {arcsinh}(c x))}{x}-b c d^3 \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )-\frac {1}{25} b c d^3 \left (c^2 x^2+1\right )^{5/2}-\frac {1}{5} b c d^3 \left (c^2 x^2+1\right )^{3/2}-\frac {11}{5} b c d^3 \sqrt {c^2 x^2+1} \]

[In]

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

[Out]

(-11*b*c*d^3*Sqrt[1 + c^2*x^2])/5 - (b*c*d^3*(1 + c^2*x^2)^(3/2))/5 - (b*c*d^3*(1 + c^2*x^2)^(5/2))/25 - (d^3*
(a + b*ArcSinh[c*x]))/x + 3*c^2*d^3*x*(a + b*ArcSinh[c*x]) + c^4*d^3*x^3*(a + b*ArcSinh[c*x]) + (c^6*d^3*x^5*(
a + b*ArcSinh[c*x]))/5 - b*c*d^3*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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 1813

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && 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^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))-(b c) \int \frac {d^3 \left (-5+15 c^2 x^2+5 c^4 x^4+c^6 x^6\right )}{5 x \sqrt {1+c^2 x^2}} \, dx \\ & = -\frac {d^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))-\frac {1}{5} \left (b c d^3\right ) \int \frac {-5+15 c^2 x^2+5 c^4 x^4+c^6 x^6}{x \sqrt {1+c^2 x^2}} \, dx \\ & = -\frac {d^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))-\frac {1}{10} \left (b c d^3\right ) \text {Subst}\left (\int \frac {-5+15 c^2 x+5 c^4 x^2+c^6 x^3}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {d^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))-\frac {1}{10} \left (b c d^3\right ) \text {Subst}\left (\int \left (\frac {11 c^2}{\sqrt {1+c^2 x}}-\frac {5}{x \sqrt {1+c^2 x}}+3 c^2 \sqrt {1+c^2 x}+c^2 \left (1+c^2 x\right )^{3/2}\right ) \, dx,x,x^2\right ) \\ & = -\frac {11}{5} b c d^3 \sqrt {1+c^2 x^2}-\frac {1}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2}-\frac {1}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2}-\frac {d^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))+\frac {1}{2} \left (b c d^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right ) \\ & = -\frac {11}{5} b c d^3 \sqrt {1+c^2 x^2}-\frac {1}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2}-\frac {1}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2}-\frac {d^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))+\frac {\left (b d^3\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 {11}{5} b c d^3 \sqrt {1+c^2 x^2}-\frac {1}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2}-\frac {1}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2}-\frac {d^3 (a+b \text {arcsinh}(c x))}{x}+3 c^2 d^3 x (a+b \text {arcsinh}(c x))+c^4 d^3 x^3 (a+b \text {arcsinh}(c x))+\frac {1}{5} c^6 d^3 x^5 (a+b \text {arcsinh}(c x))-b c d^3 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.02 \[ \int \frac {\left (d+c^2 d x^2\right )^3 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {d^3 \left (-25 a+75 a c^2 x^2+25 a c^4 x^4+5 a c^6 x^6-61 b c x \sqrt {1+c^2 x^2}-7 b c^3 x^3 \sqrt {1+c^2 x^2}-b c^5 x^5 \sqrt {1+c^2 x^2}+5 b \left (-5+15 c^2 x^2+5 c^4 x^4+c^6 x^6\right ) \text {arcsinh}(c x)+25 b c x \log (x)-25 b c x \log \left (1+\sqrt {1+c^2 x^2}\right )\right )}{25 x} \]

[In]

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

[Out]

(d^3*(-25*a + 75*a*c^2*x^2 + 25*a*c^4*x^4 + 5*a*c^6*x^6 - 61*b*c*x*Sqrt[1 + c^2*x^2] - 7*b*c^3*x^3*Sqrt[1 + c^
2*x^2] - b*c^5*x^5*Sqrt[1 + c^2*x^2] + 5*b*(-5 + 15*c^2*x^2 + 5*c^4*x^4 + c^6*x^6)*ArcSinh[c*x] + 25*b*c*x*Log
[x] - 25*b*c*x*Log[1 + Sqrt[1 + c^2*x^2]]))/(25*x)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.93

method	result	size
parts	\(d^{3} a \left (\frac {c^{6} x^{5}}{5}+c^{4} x^{3}+3 c^{2} x -\frac {1}{x}\right )+d^{3} b c \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}+3 \,\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{c x}-\frac {c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{25}-\frac {61 \sqrt {c^{2} x^{2}+1}}{25}-\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\)	\(149\)
derivativedivides	\(c \left (d^{3} a \left (\frac {c^{5} x^{5}}{5}+c^{3} x^{3}+3 c x -\frac {1}{c x}\right )+d^{3} b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}+3 \,\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{c x}-\frac {c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{25}-\frac {61 \sqrt {c^{2} x^{2}+1}}{25}-\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\)	\(151\)
default	\(c \left (d^{3} a \left (\frac {c^{5} x^{5}}{5}+c^{3} x^{3}+3 c x -\frac {1}{c x}\right )+d^{3} b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{5} x^{5}}{5}+\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}+3 \,\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{c x}-\frac {c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{25}-\frac {61 \sqrt {c^{2} x^{2}+1}}{25}-\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\)	\(151\)

[In]

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

[Out]

d^3*a*(1/5*c^6*x^5+c^4*x^3+3*c^2*x-1/x)+d^3*b*c*(1/5*arcsinh(c*x)*c^5*x^5+arcsinh(c*x)*c^3*x^3+3*arcsinh(c*x)*
c*x-arcsinh(c*x)/c/x-1/25*c^4*x^4*(c^2*x^2+1)^(1/2)-7/25*c^2*x^2*(c^2*x^2+1)^(1/2)-61/25*(c^2*x^2+1)^(1/2)-arc
tanh(1/(c^2*x^2+1)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.72 \[ \int \frac {\left (d+c^2 d x^2\right )^3 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {5 \, a c^{6} d^{3} x^{6} + 25 \, a c^{4} d^{3} x^{4} + 75 \, a c^{2} d^{3} x^{2} - 25 \, b c d^{3} x \log \left (-c x + \sqrt {c^{2} x^{2} + 1} + 1\right ) + 25 \, b c d^{3} x \log \left (-c x + \sqrt {c^{2} x^{2} + 1} - 1\right ) - 5 \, {\left (b c^{6} + 5 \, b c^{4} + 15 \, b c^{2} - 5 \, b\right )} d^{3} x \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) - 25 \, a d^{3} + 5 \, {\left (b c^{6} d^{3} x^{6} + 5 \, b c^{4} d^{3} x^{4} + 15 \, b c^{2} d^{3} x^{2} - {\left (b c^{6} + 5 \, b c^{4} + 15 \, b c^{2} - 5 \, b\right )} d^{3} x - 5 \, b d^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (b c^{5} d^{3} x^{5} + 7 \, b c^{3} d^{3} x^{3} + 61 \, b c d^{3} x\right )} \sqrt {c^{2} x^{2} + 1}}{25 \, x} \]

[In]

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

[Out]

1/25*(5*a*c^6*d^3*x^6 + 25*a*c^4*d^3*x^4 + 75*a*c^2*d^3*x^2 - 25*b*c*d^3*x*log(-c*x + sqrt(c^2*x^2 + 1) + 1) +
25*b*c*d^3*x*log(-c*x + sqrt(c^2*x^2 + 1) - 1) - 5*(b*c^6 + 5*b*c^4 + 15*b*c^2 - 5*b)*d^3*x*log(-c*x + sqrt(c
^2*x^2 + 1)) - 25*a*d^3 + 5*(b*c^6*d^3*x^6 + 5*b*c^4*d^3*x^4 + 15*b*c^2*d^3*x^2 - (b*c^6 + 5*b*c^4 + 15*b*c^2
- 5*b)*d^3*x - 5*b*d^3)*log(c*x + sqrt(c^2*x^2 + 1)) - (b*c^5*d^3*x^5 + 7*b*c^3*d^3*x^3 + 61*b*c*d^3*x)*sqrt(c
^2*x^2 + 1))/x

Sympy [F]

\[ \int \frac {\left (d+c^2 d x^2\right )^3 (a+b \text {arcsinh}(c x))}{x^2} \, dx=d^{3} \left (\int 3 a c^{2}\, dx + \int \frac {a}{x^{2}}\, dx + \int 3 a c^{4} x^{2}\, dx + \int a c^{6} x^{4}\, dx + \int 3 b c^{2} \operatorname {asinh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x^{2}}\, dx + \int 3 b c^{4} x^{2} \operatorname {asinh}{\left (c x \right )}\, dx + \int b c^{6} x^{4} \operatorname {asinh}{\left (c x \right )}\, dx\right ) \]

[In]

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

[Out]

d**3*(Integral(3*a*c**2, x) + Integral(a/x**2, x) + Integral(3*a*c**4*x**2, x) + Integral(a*c**6*x**4, x) + In
tegral(3*b*c**2*asinh(c*x), x) + Integral(b*asinh(c*x)/x**2, x) + Integral(3*b*c**4*x**2*asinh(c*x), x) + Inte
gral(b*c**6*x**4*asinh(c*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.44 \[ \int \frac {\left (d+c^2 d x^2\right )^3 (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {1}{5} \, a c^{6} d^{3} x^{5} + \frac {1}{75} \, {\left (15 \, x^{5} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {3 \, \sqrt {c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac {4 \, \sqrt {c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b c^{6} d^{3} + a c^{4} d^{3} x^{3} + \frac {1}{3} \, {\left (3 \, x^{3} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac {2 \, \sqrt {c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{4} d^{3} + 3 \, a c^{2} d^{3} x + 3 \, {\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b c d^{3} - {\left (c \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arsinh}\left (c x\right )}{x}\right )} b d^{3} - \frac {a d^{3}}{x} \]

[In]

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

[Out]

1/5*a*c^6*d^3*x^5 + 1/75*(15*x^5*arcsinh(c*x) - (3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 + 8
*sqrt(c^2*x^2 + 1)/c^6)*c)*b*c^6*d^3 + a*c^4*d^3*x^3 + 1/3*(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^3 + 3*a*c^2*d^3*x + 3*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*c*d^3 - (c*
arcsinh(1/(c*abs(x))) + arcsinh(c*x)/x)*b*d^3 - a*d^3/x

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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