\(\int \frac {x (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [159]

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

Optimal result

Integrand size = 24, antiderivative size = 70 \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {a+b \text {arcsinh}(c x)}{c^2 d \sqrt {d+c^2 d x^2}}+\frac {b \sqrt {1+c^2 x^2} \arctan (c x)}{c^2 d \sqrt {d+c^2 d x^2}} \]

[Out]

(-a-b*arcsinh(c*x))/c^2/d/(c^2*d*x^2+d)^(1/2)+b*arctan(c*x)*(c^2*x^2+1)^(1/2)/c^2/d/(c^2*d*x^2+d)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5798, 209} \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {b \sqrt {c^2 x^2+1} \arctan (c x)}{c^2 d \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{c^2 d \sqrt {c^2 d x^2+d}} \]

[In]

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

[Out]

-((a + b*ArcSinh[c*x])/(c^2*d*Sqrt[d + c^2*d*x^2])) + (b*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(c^2*d*Sqrt[d + c^2*d*
x^2])

Rule 209

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

Rule 5798

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \text {arcsinh}(c x)}{c^2 d \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{c d \sqrt {d+c^2 d x^2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{c^2 d \sqrt {d+c^2 d x^2}}+\frac {b \sqrt {1+c^2 x^2} \arctan (c x)}{c^2 d \sqrt {d+c^2 d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.23 \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (-a \sqrt {1+c^2 x^2}-b \sqrt {1+c^2 x^2} \text {arcsinh}(c x)+\left (b+b c^2 x^2\right ) \arctan (c x)\right )}{c^2 d^2 \left (1+c^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(-(a*Sqrt[1 + c^2*x^2]) - b*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + (b + b*c^2*x^2)*ArcTan[c*x])
)/(c^2*d^2*(1 + c^2*x^2)^(3/2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.79

method result size
default \(-\frac {a}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (-i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )+i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )+\operatorname {arcsinh}\left (c x \right )\right )}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}\) \(125\)
parts \(-\frac {a}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (-i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )+i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )+\operatorname {arcsinh}\left (c x \right )\right )}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}\) \(125\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.83 \[ \int \frac {x (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {{\left (b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 2 \, \sqrt {c^{2} d x^{2} + d} b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {c^{2} d x^{2} + d} a}{2 \, {\left (c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}} \]

[In]

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

[Out]

-1/2*((b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(c^2*d*x^2 + d)*sqrt(c^2*x^2 + 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) + 2*
sqrt(c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 + 1)) + 2*sqrt(c^2*d*x^2 + d)*a)/(c^4*d^2*x^2 + c^2*d^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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