\(\int \frac {x^4 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [156]

Optimal result

Integrand size = 26, antiderivative size = 206 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {b x^2 \sqrt {1+c^2 x^2}}{4 c^3 d \sqrt {d+c^2 d x^2}}-\frac {x^3 (a+b \text {arcsinh}(c x))}{c^2 d \sqrt {d+c^2 d x^2}}+\frac {3 x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{2 c^4 d^2}-\frac {3 \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{4 b c^5 d \sqrt {d+c^2 d x^2}}-\frac {b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{2 c^5 d \sqrt {d+c^2 d x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5810, 5812, 5783, 30, 272, 45} \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {x^3 (a+b \text {arcsinh}(c x))}{c^2 d \sqrt {c^2 d x^2+d}}-\frac {3 \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^2}{4 b c^5 d \sqrt {c^2 d x^2+d}}+\frac {3 x \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{2 c^4 d^2}-\frac {b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 c^5 d \sqrt {c^2 d x^2+d}}-\frac {b x^2 \sqrt {c^2 x^2+1}}{4 c^3 d \sqrt {c^2 d x^2+d}} \]

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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]

Rule 5812

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/(e*(m + 2*p + 1))), x] + (-Dist[f^2*((m - 1)/(c^2*
(m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0
]

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.78 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {4 a c \sqrt {d} x \left (3+c^2 x^2\right )-12 a \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+b \sqrt {d} \left (8 c x \text {arcsinh}(c x)-\sqrt {1+c^2 x^2} \left (6 \text {arcsinh}(c x)^2+\cosh (2 \text {arcsinh}(c x))+4 \log \left (1+c^2 x^2\right )-2 \text {arcsinh}(c x) \sinh (2 \text {arcsinh}(c x))\right )\right )}{8 c^5 d^{3/2} \sqrt {d+c^2 d x^2}} \]

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

(4*a*c*Sqrt[d]*x*(3 + c^2*x^2) - 12*a*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + b*Sqrt[d]
*(8*c*x*ArcSinh[c*x] - Sqrt[1 + c^2*x^2]*(6*ArcSinh[c*x]^2 + Cosh[2*ArcSinh[c*x]] + 4*Log[1 + c^2*x^2] - 2*Arc
Sinh[c*x]*Sinh[2*ArcSinh[c*x]])))/(8*c^5*d^(3/2)*Sqrt[d + c^2*d*x^2])

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.33

method	result	size
default	\(\frac {a \,x^{3}}{2 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {3 a x}{2 c^{4} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {3 a \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 c^{4} d \sqrt {c^{2} d}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (-4 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+2 c^{4} x^{4}+6 \operatorname {arcsinh}\left (c x \right )^{2} x^{2} c^{2}-8 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-12 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+3 c^{2} x^{2}+6 \operatorname {arcsinh}\left (c x \right )^{2}-8 \,\operatorname {arcsinh}\left (c x \right )+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )+1\right )}{8 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{5} d^{2}}\)	\(274\)
parts	\(\frac {a \,x^{3}}{2 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {3 a x}{2 c^{4} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {3 a \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 c^{4} d \sqrt {c^{2} d}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (-4 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+2 c^{4} x^{4}+6 \operatorname {arcsinh}\left (c x \right )^{2} x^{2} c^{2}-8 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-12 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+3 c^{2} x^{2}+6 \operatorname {arcsinh}\left (c x \right )^{2}-8 \,\operatorname {arcsinh}\left (c x \right )+8 \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )+1\right )}{8 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{5} d^{2}}\)	\(274\)

[In]

int(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\int \frac {x^{4} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**4*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(3/2),x)

[Out]

Integral(x**4*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(3/2), x)

Maxima [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^4*(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. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{3/2}} \,d x \]

[In]

int((x^4*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2),x)

[Out]

int((x^4*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2), x)