\(\int \frac {x^5 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [155]

Optimal result

Integrand size = 26, antiderivative size = 212 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {5 b x \sqrt {d+c^2 d x^2}}{3 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b x^3 \sqrt {d+c^2 d x^2}}{9 c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^2}+\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}+\frac {b \sqrt {d+c^2 d x^2} \arctan (c x)}{c^6 d^2 \sqrt {1+c^2 x^2}} \]

[Out]

1/3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/c^6/d^3+(-a-b*arcsinh(c*x))/c^6/d/(c^2*d*x^2+d)^(1/2)-2*(a+b*arcsin
h(c*x))*(c^2*d*x^2+d)^(1/2)/c^6/d^2+5/3*b*x*(c^2*d*x^2+d)^(1/2)/c^5/d^2/(c^2*x^2+1)^(1/2)-1/9*b*x^3*(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^6/d^2/(c^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 45, 5804, 12, 1167, 209} \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}-\frac {2 \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^2}-\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {c^2 d x^2+d}}+\frac {b \arctan (c x) \sqrt {c^2 d x^2+d}}{c^6 d^2 \sqrt {c^2 x^2+1}}+\frac {5 b x \sqrt {c^2 d x^2+d}}{3 c^5 d^2 \sqrt {c^2 x^2+1}}-\frac {b x^3 \sqrt {c^2 d x^2+d}}{9 c^3 d^2 \sqrt {c^2 x^2+1}} \]

[In]

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

[Out]

(5*b*x*Sqrt[d + c^2*d*x^2])/(3*c^5*d^2*Sqrt[1 + c^2*x^2]) - (b*x^3*Sqrt[d + c^2*d*x^2])/(9*c^3*d^2*Sqrt[1 + c^
2*x^2]) - (a + b*ArcSinh[c*x])/(c^6*d*Sqrt[d + c^2*d*x^2]) - (2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(c^6
*d^2) + ((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c^6*d^3) + (b*Sqrt[d + c^2*d*x^2]*ArcTan[c*x])/(c^6*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 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 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^6 d \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^2}+\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \int \frac {-8-4 c^2 x^2+c^4 x^4}{3 c^6 d^2 \left (1+c^2 x^2\right )} \, dx}{\sqrt {1+c^2 x^2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^2}+\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}-\frac {\left (b \sqrt {d+c^2 d x^2}\right ) \int \frac {-8-4 c^2 x^2+c^4 x^4}{1+c^2 x^2} \, dx}{3 c^5 d^2 \sqrt {1+c^2 x^2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^2}+\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}-\frac {\left (b \sqrt {d+c^2 d x^2}\right ) \int \left (-5+c^2 x^2-\frac {3}{1+c^2 x^2}\right ) \, dx}{3 c^5 d^2 \sqrt {1+c^2 x^2}} \\ & = \frac {5 b x \sqrt {d+c^2 d x^2}}{3 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b x^3 \sqrt {d+c^2 d x^2}}{9 c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^2}+\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}+\frac {\left (b \sqrt {d+c^2 d x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{c^5 d^2 \sqrt {1+c^2 x^2}} \\ & = \frac {5 b x \sqrt {d+c^2 d x^2}}{3 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b x^3 \sqrt {d+c^2 d x^2}}{9 c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^2}+\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}+\frac {b \sqrt {d+c^2 d x^2} \arctan (c x)}{c^6 d^2 \sqrt {1+c^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (15 b c x+14 b c^3 x^3-b c^5 x^5-24 a \sqrt {1+c^2 x^2}-12 a c^2 x^2 \sqrt {1+c^2 x^2}+3 a c^4 x^4 \sqrt {1+c^2 x^2}+3 b \sqrt {1+c^2 x^2} \left (-8-4 c^2 x^2+c^4 x^4\right ) \text {arcsinh}(c x)+9 \left (b+b c^2 x^2\right ) \arctan (c x)\right )}{9 c^6 d^2 \left (1+c^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(15*b*c*x + 14*b*c^3*x^3 - b*c^5*x^5 - 24*a*Sqrt[1 + c^2*x^2] - 12*a*c^2*x^2*Sqrt[1 + c^2
*x^2] + 3*a*c^4*x^4*Sqrt[1 + c^2*x^2] + 3*b*Sqrt[1 + c^2*x^2]*(-8 - 4*c^2*x^2 + c^4*x^4)*ArcSinh[c*x] + 9*(b +
b*c^2*x^2)*ArcTan[c*x]))/(9*c^6*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 = 238, normalized size of antiderivative = 1.12

method	result	size
default	\(a \left (\frac {x^{4}}{3 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {4 \left (\frac {x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}\right )}{3 c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (3 \,\operatorname {arcsinh}\left (c x \right ) c^{4} x^{4}-c^{3} x^{3} \sqrt {c^{2} x^{2}+1}-12 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}-9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )+9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )+15 c x \sqrt {c^{2} x^{2}+1}-24 \,\operatorname {arcsinh}\left (c x \right )\right )}{9 c^{6} d^{2} \left (c^{2} x^{2}+1\right )}\)	\(238\)
parts	\(a \left (\frac {x^{4}}{3 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {4 \left (\frac {x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}\right )}{3 c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (3 \,\operatorname {arcsinh}\left (c x \right ) c^{4} x^{4}-c^{3} x^{3} \sqrt {c^{2} x^{2}+1}-12 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}-9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )+9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )+15 c x \sqrt {c^{2} x^{2}+1}-24 \,\operatorname {arcsinh}\left (c x \right )\right )}{9 c^{6} d^{2} \left (c^{2} x^{2}+1\right )}\)	\(238\)

[In]

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

[Out]

a*(1/3*x^4/c^2/d/(c^2*d*x^2+d)^(1/2)-4/3/c^2*(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)))+1/9*
b*(d*(c^2*x^2+1))^(1/2)*(3*arcsinh(c*x)*c^4*x^4-c^3*x^3*(c^2*x^2+1)^(1/2)-12*arcsinh(c*x)*c^2*x^2-9*I*(c^2*x^2
+1)^(1/2)*ln(c*x+(c^2*x^2+1)^(1/2)-I)+9*I*(c^2*x^2+1)^(1/2)*ln(c*x+(c^2*x^2+1)^(1/2)+I)+15*c*x*(c^2*x^2+1)^(1/
2)-24*arcsinh(c*x))/c^6/d^2/(c^2*x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.93 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {9 \, {\left (b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) - 6 \, {\left (b c^{4} x^{4} - 4 \, b c^{2} x^{2} - 8 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (3 \, a c^{4} x^{4} - 12 \, a c^{2} x^{2} - {\left (b c^{3} x^{3} - 15 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} - 24 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{18 \, {\left (c^{8} d^{2} x^{2} + c^{6} d^{2}\right )}} \]

[In]

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

[Out]

-1/18*(9*(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)) -
6*(b*c^4*x^4 - 4*b*c^2*x^2 - 8*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) - 2*(3*a*c^4*x^4 - 12*a*c^
2*x^2 - (b*c^3*x^3 - 15*b*c*x)*sqrt(c^2*x^2 + 1) - 24*a)*sqrt(c^2*d*x^2 + d))/(c^8*d^2*x^2 + c^6*d^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

1/3*a*(x^4/(sqrt(c^2*d*x^2 + d)*c^2*d) - 4*x^2/(sqrt(c^2*d*x^2 + d)*c^4*d) - 8/(sqrt(c^2*d*x^2 + d)*c^6*d)) +
1/3*b*((c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d)*x^2 - 8*sqrt(d))*log(c*x + sqrt(c^2*x^2 + 1))/(sqrt(c^2*x^2 + 1)*c^6*d
^2) - integrate((c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d)*x^2 - 8*sqrt(d))/(sqrt(c^2*x^2 + 1)*x), x)/(c^6*d^2) + 3*inte
grate(1/3*(c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d)*x^2 - 8*sqrt(d))/(c^9*d^2*x^4 + c^7*d^2*x^2 + (c^8*d^2*x^3 + c^6*d^
2*x)*sqrt(c^2*x^2 + 1)), x))

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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