3.156 \(\int \frac {x^4 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=206 \[ -\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {c^2 d x^2+d}}-\frac {3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^5 d \sqrt {c^2 d x^2+d}}+\frac {3 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 c^5 d \sqrt {c^2 d x^2+d}}-\frac {b x^2 \sqrt {c^2 x^2+1}}{4 c^3 d \sqrt {c^2 d x^2+d}} \]

[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

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Rubi [A]  time = 0.28, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5751, 5758, 5677, 5675, 30, 266, 43} \[ \frac {3 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {c^2 d x^2+d}}-\frac {3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^5 d \sqrt {c^2 d x^2+d}}-\frac {b x^2 \sqrt {c^2 x^2+1}}{4 c^3 d \sqrt {c^2 d x^2+d}}-\frac {b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 c^5 d \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x^2*Sqrt[1 + c^2*x^2])/(4*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 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 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]

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*Sqrt[d + e*x^2]
), Int[(f*x)^(m - 1)*(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] && GtQ[m, 1] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.47, size = 161, normalized size = 0.78 \[ \frac {4 a c \sqrt {d} x \left (c^2 x^2+3\right )-12 a \sqrt {c^2 d x^2+d} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+c d x\right )+b \sqrt {d} \left (8 c x \sinh ^{-1}(c x)-\sqrt {c^2 x^2+1} \left (4 \log \left (c^2 x^2+1\right )+6 \sinh ^{-1}(c x)^2-2 \sinh \left (2 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (2 \sinh ^{-1}(c x)\right )\right )\right )}{8 c^5 d^{3/2} \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} \operatorname {arsinh}\left (c x\right ) + a x^{4}\right )} \sqrt {c^{2} d x^{2} + d}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

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)^(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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.33, size = 366, normalized size = 1.78 \[ \frac {a \,x^{3}}{2 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {3 a x}{2 c^{4} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {3 a \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 c^{4} d \sqrt {c^{2} d}}-\frac {3 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2}}{4 \sqrt {c^{2} x^{2}+1}\, c^{5} d^{2}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{3}}{2 c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x^{2}}{4 c^{3} d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {3 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x}{2 c^{4} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{c^{5} d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}}{8 c^{5} d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{\sqrt {c^{2} x^{2}+1}\, c^{5} d^{2}} \]

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)^(3/2),x)

[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(x*c^2*d/(c^2*d)^(1/2)+(c^
2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-3/4*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^2*arcsinh(c*x)^2+1/2*b*(d*
(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*arcsinh(c*x)*x^3-1/4*b*(d*(c^2*x^2+1))^(1/2)/c^3/d^2/(c^2*x^2+1)^(1/2)*
x^2+3/2*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^2/(c^2*x^2+1)*arcsinh(c*x)*x+b*(d*(c^2*x^2+1))^(1/2)/c^5/d^2/(c^2*x^2+1)
^(1/2)*arcsinh(c*x)-1/8*b*(d*(c^2*x^2+1))^(1/2)/c^5/d^2/(c^2*x^2+1)^(1/2)-b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^
(1/2)/c^5/d^2*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 \[ \frac {1}{2} \, a {\left (\frac {x^{3}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} + \frac {3 \, x}{\sqrt {c^{2} d x^{2} + d} c^{4} d} - \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5} d^{\frac {3}{2}}}\right )} + b \int \frac {x^{4} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{3/2}} \,d x \]

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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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)**(3/2),x)

[Out]

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

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