\(\int \frac {x^3 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [168]

Optimal result

Integrand size = 26, antiderivative size = 144 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {b x \sqrt {d+c^2 d x^2}}{6 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 b \sqrt {d+c^2 d x^2} \arctan (c x)}{6 c^4 d^3 \sqrt {1+c^2 x^2}} \]

[Out]

1/3*(a+b*arcsinh(c*x))/c^4/d/(c^2*d*x^2+d)^(3/2)+(-a-b*arcsinh(c*x))/c^4/d^2/(c^2*d*x^2+d)^(1/2)-1/6*b*x*(c^2*
d*x^2+d)^(1/2)/c^3/d^3/(c^2*x^2+1)^(3/2)+5/6*b*arctan(c*x)*(c^2*d*x^2+d)^(1/2)/c^4/d^3/(c^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 144, 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, 393, 209} \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {a+b \text {arcsinh}(c x)}{c^4 d^2 \sqrt {c^2 d x^2+d}}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {5 b \arctan (c x) \sqrt {c^2 d x^2+d}}{6 c^4 d^3 \sqrt {c^2 x^2+1}}-\frac {b x \sqrt {c^2 d x^2+d}}{6 c^3 d^3 \left (c^2 x^2+1\right )^{3/2}} \]

[In]

Int[(x^3*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

-1/6*(b*x*Sqrt[d + c^2*d*x^2])/(c^3*d^3*(1 + c^2*x^2)^(3/2)) + (a + b*ArcSinh[c*x])/(3*c^4*d*(d + c^2*d*x^2)^(
3/2)) - (a + b*ArcSinh[c*x])/(c^4*d^2*Sqrt[d + c^2*d*x^2]) + (5*b*Sqrt[d + c^2*d*x^2]*ArcTan[c*x])/(6*c^4*d^3*
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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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)}{3 c^4 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \int \frac {-2-3 c^2 x^2}{3 c^4 d^3 \left (1+c^2 x^2\right )^2} \, dx}{\sqrt {1+c^2 x^2}} \\ & = \frac {a+b \text {arcsinh}(c x)}{3 c^4 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b \sqrt {d+c^2 d x^2}\right ) \int \frac {-2-3 c^2 x^2}{\left (1+c^2 x^2\right )^2} \, dx}{3 c^3 d^3 \sqrt {1+c^2 x^2}} \\ & = -\frac {b x \sqrt {d+c^2 d x^2}}{6 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (5 b \sqrt {d+c^2 d x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{6 c^3 d^3 \sqrt {1+c^2 x^2}} \\ & = -\frac {b x \sqrt {d+c^2 d x^2}}{6 c^3 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 b \sqrt {d+c^2 d x^2} \arctan (c x)}{6 c^4 d^3 \sqrt {1+c^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d+c^2 d x^2} \left (b c x+b c^3 x^3+4 a \sqrt {1+c^2 x^2}+6 a c^2 x^2 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (2+3 c^2 x^2\right ) \text {arcsinh}(c x)-5 b \left (1+c^2 x^2\right )^2 \arctan (c x)\right )}{6 c^4 d^3 \left (1+c^2 x^2\right )^{5/2}} \]

[In]

Integrate[(x^3*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

-1/6*(Sqrt[d + c^2*d*x^2]*(b*c*x + b*c^3*x^3 + 4*a*Sqrt[1 + c^2*x^2] + 6*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 2*b*Sqr
t[1 + c^2*x^2]*(2 + 3*c^2*x^2)*ArcSinh[c*x] - 5*b*(1 + c^2*x^2)^2*ArcTan[c*x]))/(c^4*d^3*(1 + c^2*x^2)^(5/2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.83

method	result	size
default	\(a \left (-\frac {x^{2}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right ) x^{2}}{\left (c^{2} x^{2}+1\right )^{2} d^{3} c^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c^{3}}-\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{4}}+\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{4} d^{3}}-\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{4} d^{3}}\)	\(263\)
parts	\(a \left (-\frac {x^{2}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right ) x^{2}}{\left (c^{2} x^{2}+1\right )^{2} d^{3} c^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c^{3}}-\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{4}}+\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{4} d^{3}}-\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{4} d^{3}}\)	\(263\)

[In]

int(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

a*(-x^2/c^2/d/(c^2*d*x^2+d)^(3/2)-2/3/d/c^4/(c^2*d*x^2+d)^(3/2))-b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^2/d^3/c^2
*arcsinh(c*x)*x^2-1/6*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(3/2)/d^3/c^3*x-2/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2
+1)^2/d^3/c^4*arcsinh(c*x)+5/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^3*ln(c*x+(c^2*x^2+1)^(1/2)+I)
-5/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^4/d^3*ln(c*x+(c^2*x^2+1)^(1/2)-I)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.31 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {5 \, {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 4 \, {\left (3 \, b c^{2} x^{2} + 2 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (6 \, a c^{2} x^{2} + \sqrt {c^{2} x^{2} + 1} b c x + 4 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{12 \, {\left (c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \]

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(5*(b*c^4*x^4 + 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)) + 4*(3*b*c^2*x^2 + 2*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(6*a*c^2*x^2 + sqrt(
c^2*x^2 + 1)*b*c*x + 4*a)*sqrt(c^2*d*x^2 + d))/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3)

Sympy [F]

\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**3*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**3*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {x}{c^{6} d^{\frac {5}{2}} x^{2} + c^{4} d^{\frac {5}{2}}} - \frac {5 \, \arctan \left (c x\right )}{c^{5} d^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} + \frac {2}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} + \frac {2}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \]

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b*c*(x/(c^6*d^(5/2)*x^2 + c^4*d^(5/2)) - 5*arctan(c*x)/(c^5*d^(5/2))) - 1/3*b*(3*x^2/((c^2*d*x^2 + d)^(3/
2)*c^2*d) + 2/((c^2*d*x^2 + d)^(3/2)*c^4*d))*arcsinh(c*x) - 1/3*a*(3*x^2/((c^2*d*x^2 + d)^(3/2)*c^2*d) + 2/((c
^2*d*x^2 + d)^(3/2)*c^4*d))

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),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. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

[In]

int((x^3*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2),x)

[Out]

int((x^3*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2), x)