\(\int \frac {x^3 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [157]

Optimal result

Integrand size = 26, antiderivative size = 136 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {b x \sqrt {d+c^2 d x^2}}{c^3 d^2 \sqrt {1+c^2 x^2}}+\frac {a+b \text {arcsinh}(c x)}{c^4 d \sqrt {d+c^2 d x^2}}+\frac {\sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^4 d^2}-\frac {b \sqrt {d+c^2 d x^2} \arctan (c x)}{c^4 d^2 \sqrt {1+c^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 45, 5804, 12, 396, 209} \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^4 d^2}+\frac {a+b \text {arcsinh}(c x)}{c^4 d \sqrt {c^2 d x^2+d}}-\frac {b \arctan (c x) \sqrt {c^2 d x^2+d}}{c^4 d^2 \sqrt {c^2 x^2+1}}-\frac {b x \sqrt {c^2 d x^2+d}}{c^3 d^2 \sqrt {c^2 x^2+1}} \]

[In]

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

[Out]

-((b*x*Sqrt[d + c^2*d*x^2])/(c^3*d^2*Sqrt[1 + c^2*x^2])) + (a + b*ArcSinh[c*x])/(c^4*d*Sqrt[d + c^2*d*x^2]) +
(Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(c^4*d^2) - (b*Sqrt[d + c^2*d*x^2]*ArcTan[c*x])/(c^4*d^2*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 396

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

Rule 5804

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p -
1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (-b c x-b c^3 x^3+2 a \sqrt {1+c^2 x^2}+a c^2 x^2 \sqrt {1+c^2 x^2}+b \sqrt {1+c^2 x^2} \left (2+c^2 x^2\right ) \text {arcsinh}(c x)-\left (b+b c^2 x^2\right ) \arctan (c x)\right )}{c^4 d^2 \left (1+c^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.29

method	result	size
default	\(a \left (\frac {x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )-i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )-c x \sqrt {c^{2} x^{2}+1}+2 \,\operatorname {arcsinh}\left (c x \right )\right )}{c^{4} d^{2} \left (c^{2} x^{2}+1\right )}\)	\(176\)
parts	\(a \left (\frac {x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )-i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )-c x \sqrt {c^{2} x^{2}+1}+2 \,\operatorname {arcsinh}\left (c x \right )\right )}{c^{4} d^{2} \left (c^{2} x^{2}+1\right )}\)	\(176\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.22 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {{\left (b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 2 \, {\left (b c^{2} x^{2} + 2 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a c^{2} x^{2} - \sqrt {c^{2} x^{2} + 1} b c x + 2 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{2 \, {\left (c^{6} d^{2} x^{2} + c^{4} d^{2}\right )}} \]

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-b c {\left (\frac {x}{c^{4} d^{\frac {3}{2}}} + \frac {\arctan \left (c x\right )}{c^{5} d^{\frac {3}{2}}}\right )} + b {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} \operatorname {arsinh}\left (c x\right ) + a {\left (\frac {x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {2}{\sqrt {c^{2} d x^{2} + d} c^{4} d}\right )} \]

[In]

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

[Out]

-b*c*(x/(c^4*d^(3/2)) + arctan(c*x)/(c^5*d^(3/2))) + b*(x^2/(sqrt(c^2*d*x^2 + d)*c^2*d) + 2/(sqrt(c^2*d*x^2 +
d)*c^4*d))*arcsinh(c*x) + a*(x^2/(sqrt(c^2*d*x^2 + d)*c^2*d) + 2/(sqrt(c^2*d*x^2 + d)*c^4*d))

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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