3.121 \(\int x^2 \sqrt {d+c^2 d x^2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=181 \[ \frac {x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {c^2 x^2+1}}-\frac {b x^2 \sqrt {c^2 d x^2+d}}{16 c \sqrt {c^2 x^2+1}}-\frac {b c x^4 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5742, 5758, 5675, 30} \[ \frac {1}{4} x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac {x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {c^2 x^2+1}}-\frac {b c x^4 \sqrt {c^2 d x^2+d}}{16 \sqrt {c^2 x^2+1}}-\frac {b x^2 \sqrt {c^2 d x^2+d}}{16 c \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x^2*Sqrt[d + c^2*d*x^2])/(16*c*Sqrt[1 + c^2*x^2]) - (b*c*x^4*Sqrt[d + c^2*d*x^2])/(16*Sqrt[1 + c^2*x^2]) +
(x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(8*c^2) + (x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/4 - (Sq
rt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(16*b*c^3*Sqrt[1 + c^2*x^2])

Rule 30

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

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 5742

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
(f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1
+ c^2*x^2]), Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*
(m + 2)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f
, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 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 x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\sqrt {d+c^2 d x^2} \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{4 \sqrt {1+c^2 x^2}}-\frac {\left (b c \sqrt {d+c^2 d x^2}\right ) \int x^3 \, dx}{4 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\sqrt {d+c^2 d x^2} \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{8 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (b \sqrt {d+c^2 d x^2}\right ) \int x \, dx}{8 c \sqrt {1+c^2 x^2}}\\ &=-\frac {b x^2 \sqrt {d+c^2 d x^2}}{16 c \sqrt {1+c^2 x^2}}-\frac {b c x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.81, size = 129, normalized size = 0.71 \[ -\frac {-16 a c x \left (2 c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}+16 a \sqrt {d} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+c d x\right )+\frac {b \sqrt {c^2 d x^2+d} \left (8 \sinh ^{-1}(c x)^2-4 \sinh \left (4 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{\sqrt {c^2 x^2+1}}}{128 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/128*(-16*a*c*x*(1 + 2*c^2*x^2)*Sqrt[d + c^2*d*x^2] + 16*a*Sqrt[d]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]]
+ (b*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*ArcSinh[c*x]] - 4*ArcSinh[c*x]*Sinh[4*ArcSinh[c*x]]))/Sqrt
[1 + c^2*x^2])/c^3

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c^{2} d x^{2} + d} {\left (b x^{2} \operatorname {arsinh}\left (c x\right ) + a x^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)*x^2, x)

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maple [B]  time = 0.23, size = 320, normalized size = 1.77 \[ \frac {a x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{4 c^{2} d}-\frac {a x \sqrt {c^{2} d \,x^{2}+d}}{8 c^{2}}-\frac {a d \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{8 c^{2} \sqrt {c^{2} d}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}}{128 c^{3} \sqrt {c^{2} x^{2}+1}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2}}{16 \sqrt {c^{2} x^{2}+1}\, c^{3}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c^{2} \arcsinh \left (c x \right ) x^{5}}{4 c^{2} x^{2}+4}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c \,x^{4}}{16 \sqrt {c^{2} x^{2}+1}}+\frac {3 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{3}}{8 \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x^{2}}{16 c \sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x}{8 c^{2} \left (c^{2} x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x)

[Out]

1/4*a*x*(c^2*d*x^2+d)^(3/2)/c^2/d-1/8*a/c^2*x*(c^2*d*x^2+d)^(1/2)-1/8*a/c^2*d*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*
x^2+d)^(1/2))/(c^2*d)^(1/2)-1/128*b*(d*(c^2*x^2+1))^(1/2)/c^3/(c^2*x^2+1)^(1/2)-1/16*b*(d*(c^2*x^2+1))^(1/2)/(
c^2*x^2+1)^(1/2)/c^3*arcsinh(c*x)^2+1/4*b*(d*(c^2*x^2+1))^(1/2)*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^5-1/16*b*(d*(c^
2*x^2+1))^(1/2)*c/(c^2*x^2+1)^(1/2)*x^4+3/8*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)*arcsinh(c*x)*x^3-1/16*b*(d*(c^
2*x^2+1))^(1/2)/c/(c^2*x^2+1)^(1/2)*x^2+1/8*b*(d*(c^2*x^2+1))^(1/2)/c^2/(c^2*x^2+1)*arcsinh(c*x)*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {d\,c^2\,x^2+d} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asinh(c*x))*(d + c^2*d*x^2)^(1/2),x)

[Out]

int(x^2*(a + b*asinh(c*x))*(d + c^2*d*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asinh(c*x))*(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**2*sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x)), x)

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