\(\int \frac {x^4 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [167]

Optimal result

Integrand size = 26, antiderivative size = 203 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {b}{6 c^5 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{2 b c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {2 b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 c^5 d^2 \sqrt {d+c^2 d x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5810, 5783, 266, 272, 45} \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {\sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^2}{2 b c^5 d^2 \sqrt {c^2 d x^2+d}}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d^2 \sqrt {c^2 d x^2+d}}+\frac {b}{6 c^5 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {2 b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 c^5 d^2 \sqrt {c^2 d x^2+d}} \]

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

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

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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 5783

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

Rule 5810

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

Rubi steps \begin{align*} \text {integral}& = -\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {\int \frac {x^2 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{c^2 d}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {x^3}{\left (1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d+c^2 d x^2}} \\ & = -\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+c^2 d x^2}} \, dx}{c^4 d^2}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {x}{1+c^2 x^2} \, dx}{c^3 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 c d^2 \sqrt {d+c^2 d x^2}} \\ & = -\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{2 b c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{2 c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \left (-\frac {1}{c^2 \left (1+c^2 x\right )^2}+\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 c d^2 \sqrt {d+c^2 d x^2}} \\ & = \frac {b}{6 c^5 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{2 b c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {2 b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 c^5 d^2 \sqrt {d+c^2 d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {-2 a c \sqrt {d} x \left (3+4 c^2 x^2\right )+b \sqrt {d} \left (\sqrt {1+c^2 x^2}+2 c x \text {arcsinh}(c x)-8 c x \left (1+c^2 x^2\right ) \text {arcsinh}(c x)+\left (1+c^2 x^2\right )^{3/2} \left (3 \text {arcsinh}(c x)^2+4 \log \left (1+c^2 x^2\right )\right )\right )+6 a \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{6 c^5 d^{5/2} \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}} \]

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

(-2*a*c*Sqrt[d]*x*(3 + 4*c^2*x^2) + b*Sqrt[d]*(Sqrt[1 + c^2*x^2] + 2*c*x*ArcSinh[c*x] - 8*c*x*(1 + c^2*x^2)*Ar
cSinh[c*x] + (1 + c^2*x^2)^(3/2)*(3*ArcSinh[c*x]^2 + 4*Log[1 + c^2*x^2])) + 6*a*(1 + c^2*x^2)*Sqrt[d + c^2*d*x
^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/(6*c^5*d^(5/2)*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.70

method	result	size
default	\(-\frac {a \,x^{3}}{3 c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {a x}{c^{4} d^{2} \sqrt {c^{2} d \,x^{2}+d}}+\frac {a \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{c^{4} d^{2} \sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \sqrt {c^{2} x^{2}+1}\, \left (3 \operatorname {arcsinh}\left (c x \right )^{2} x^{4} c^{4}-8 \,\operatorname {arcsinh}\left (c x \right ) c^{4} x^{4}+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}-8 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+6 \operatorname {arcsinh}\left (c x \right )^{2} x^{2} c^{2}-16 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+16 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-6 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+c^{2} x^{2}+3 \operatorname {arcsinh}\left (c x \right )^{2}-8 \,\operatorname {arcsinh}\left (c x \right )+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )+1\right )}{6 \left (c^{6} x^{6}+3 c^{4} x^{4}+3 c^{2} x^{2}+1\right ) c^{5} d^{3}}\)	\(345\)
parts	\(-\frac {a \,x^{3}}{3 c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {a x}{c^{4} d^{2} \sqrt {c^{2} d \,x^{2}+d}}+\frac {a \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{c^{4} d^{2} \sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \sqrt {c^{2} x^{2}+1}\, \left (3 \operatorname {arcsinh}\left (c x \right )^{2} x^{4} c^{4}-8 \,\operatorname {arcsinh}\left (c x \right ) c^{4} x^{4}+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}-8 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+6 \operatorname {arcsinh}\left (c x \right )^{2} x^{2} c^{2}-16 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+16 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-6 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+c^{2} x^{2}+3 \operatorname {arcsinh}\left (c x \right )^{2}-8 \,\operatorname {arcsinh}\left (c x \right )+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )+1\right )}{6 \left (c^{6} x^{6}+3 c^{4} x^{4}+3 c^{2} x^{2}+1\right ) c^{5} d^{3}}\)	\(345\)

[In]

int(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*a*x^3/c^2/d/(c^2*d*x^2+d)^(3/2)-a/c^4/d^2*x/(c^2*d*x^2+d)^(1/2)+a/c^4/d^2*ln(c^2*d*x/(c^2*d)^(1/2)+(c^2*d
*x^2+d)^(1/2))/(c^2*d)^(1/2)+1/6*b*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+1)^(1/2)/(c^6*x^6+3*c^4*x^4+3*c^2*x^2+1)/c^5
/d^3*(3*arcsinh(c*x)^2*x^4*c^4-8*arcsinh(c*x)*c^4*x^4+8*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)*x^4*c^4-8*arcsinh(c*x)
*(c^2*x^2+1)^(1/2)*x^3*c^3+6*arcsinh(c*x)^2*x^2*c^2-16*arcsinh(c*x)*c^2*x^2+16*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)
*x^2*c^2-6*arcsinh(c*x)*c*x*(c^2*x^2+1)^(1/2)+c^2*x^2+3*arcsinh(c*x)^2-8*arcsinh(c*x)+8*ln(1+(c*x+(c^2*x^2+1)^
(1/2))^2)+1)

Fricas [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral((b*x^4*arcsinh(c*x) + a*x^4)*sqrt(c^2*d*x^2 + d)/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3),
x)

Sympy [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^{4} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**4*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**4*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(5/2), x)

Maxima [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^4/(c^2*d*x^2 + d)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

[In]

int((x^4*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2),x)

[Out]

int((x^4*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2), x)