3.167 \(\int \frac {x^4 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=203 \[ -\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {\sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt {c^2 d x^2+d}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {c^2 d x^2+d}}+\frac {b}{6 c^5 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {2 b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 c^5 d^2 \sqrt {c^2 d x^2+d}} \]

[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)

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Rubi [A]  time = 0.30, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5751, 5677, 5675, 260, 266, 43} \[ -\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {c^2 d x^2+d}}+\frac {\sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt {c^2 d x^2+d}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {b}{6 c^5 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {2 b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 c^5 d^2 \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[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 43

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 260

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 266

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 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5677

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/S
qrt[d + e*x^2], Int[(a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e,
c^2*d] &&  !GtQ[d, 0]

Rule 5751

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*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[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] && Gt
Q[m, 1]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {\int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\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 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\int \frac {a+b \sinh ^{-1}(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 ) \operatorname {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 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 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 {\sqrt {1+c^2 x^2} \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \operatorname {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 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^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*}

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Mathematica [A]  time = 0.53, size = 191, normalized size = 0.94 \[ \frac {-2 a c \sqrt {d} x \left (4 c^2 x^2+3\right )+6 a \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+c d x\right )+b \sqrt {d} \left (\sqrt {c^2 x^2+1}-8 c x \left (c^2 x^2+1\right ) \sinh ^{-1}(c x)+\left (c^2 x^2+1\right )^{3/2} \left (4 \log \left (c^2 x^2+1\right )+3 \sinh ^{-1}(c x)^2\right )+2 c x \sinh ^{-1}(c x)\right )}{6 c^5 d^{5/2} \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[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])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} \operatorname {arsinh}\left (c x\right ) + a x^{4}\right )} \sqrt {c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.37, size = 1430, normalized size = 7.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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(x*c^2*d/(c^2*d)^(1/2)+(c^2*d
*x^2+d)^(1/2))/(c^2*d)^(1/2)+1/2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^3*arcsinh(c*x)^2-8/3*b*(d*(c^
2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^3*arcsinh(c*x)-32*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c
^4*x^4+71*c^2*x^2+16)*c^2/d^3*arcsinh(c*x)*x^7+32*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+7
1*c^2*x^2+16)*c/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^6-8/3*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*
c^4*x^4+71*c^2*x^2+16)*c^2/d^3*x^7+8/3*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+1
6)/d^3*(c^2*x^2+1)*x^5-76*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/d^3*arcsin
h(c*x)*x^5+84*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c/d^3*arcsinh(c*x)*(c^
2*x^2+1)^(1/2)*x^4-22/3*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/d^3*x^5+4*b*
(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c/d^3*(c^2*x^2+1)^(1/2)*x^4+14/3*b*(d*
(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^2/d^3*(c^2*x^2+1)*x^3-181/3*b*(d*(c^2*x
^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^2/d^3*arcsinh(c*x)*x^3+220/3*b*(d*(c^2*x^2+1)
)^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^3/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^2-20/3*b*(d
*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^2/d^3*x^3+13/2*b*(d*(c^2*x^2+1))^(1/2)
/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^3/d^3*x^2*(c^2*x^2+1)^(1/2)+2*b*(d*(c^2*x^2+1))^(1/2)/(24
*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^4/d^3*(c^2*x^2+1)*x-16*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87
*c^6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^4/d^3*arcsinh(c*x)*x+64/3*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6
+118*c^4*x^4+71*c^2*x^2+16)/c^5/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-2*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^
6*x^6+118*c^4*x^4+71*c^2*x^2+16)/c^4/d^3*x+8/3*b*(d*(c^2*x^2+1))^(1/2)/(24*c^8*x^8+87*c^6*x^6+118*c^4*x^4+71*c
^2*x^2+16)/c^5/d^3*(c^2*x^2+1)^(1/2)+4/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^3*ln(1+(c*x+(c^2*x^2+
1)^(1/2))^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, {\left (x {\left (\frac {3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} + \frac {2}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} + \frac {x}{\sqrt {c^{2} d x^{2} + d} c^{4} d^{2}} - \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5} d^{\frac {5}{2}}}\right )} a + b \int \frac {x^{4} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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