3.1.47 \(\int \frac {x^3 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^3} \, dx\) [47]

Optimal. Leaf size=97 \[ \frac {b x^3}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b x}{4 c^3 d^3 \sqrt {1+c^2 x^2}}-\frac {b \sinh ^{-1}(c x)}{4 c^4 d^3}+\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5800, 294, 221} \begin {gather*} \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {b \sinh ^{-1}(c x)}{4 c^4 d^3}+\frac {b x^3}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac {b x}{4 c^3 d^3 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^3)/(12*c*d^3*(1 + c^2*x^2)^(3/2)) + (b*x)/(4*c^3*d^3*Sqrt[1 + c^2*x^2]) - (b*ArcSinh[c*x])/(4*c^4*d^3) +
(x^4*(a + b*ArcSinh[c*x]))/(4*d^3*(1 + c^2*x^2)^2)

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 5800

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] - Dist[b*c*(n/(f*(m + 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, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^3} \, dx &=\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {(b c) \int \frac {x^4}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 d^3}\\ &=\frac {b x^3}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {b \int \frac {x^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{4 c d^3}\\ &=\frac {b x^3}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b x}{4 c^3 d^3 \sqrt {1+c^2 x^2}}+\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {b \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{4 c^3 d^3}\\ &=\frac {b x^3}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b x}{4 c^3 d^3 \sqrt {1+c^2 x^2}}-\frac {b \sinh ^{-1}(c x)}{4 c^4 d^3}+\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 79, normalized size = 0.81 \begin {gather*} \frac {-3 a \left (1+2 c^2 x^2\right )+b c x \sqrt {1+c^2 x^2} \left (3+4 c^2 x^2\right )-3 \left (b+2 b c^2 x^2\right ) \sinh ^{-1}(c x)}{12 c^4 d^3 \left (1+c^2 x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*(1 + 2*c^2*x^2) + b*c*x*Sqrt[1 + c^2*x^2]*(3 + 4*c^2*x^2) - 3*(b + 2*b*c^2*x^2)*ArcSinh[c*x])/(12*c^4*d^
3*(1 + c^2*x^2)^2)

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Maple [A]
time = 0.53, size = 108, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {\frac {a \left (-\frac {1}{2 \left (c^{2} x^{2}+1\right )}+\frac {1}{4 \left (c^{2} x^{2}+1\right )^{2}}\right )}{d^{3}}+\frac {b \left (-\frac {\arcsinh \left (c x \right )}{2 \left (c^{2} x^{2}+1\right )}+\frac {\arcsinh \left (c x \right )}{4 \left (c^{2} x^{2}+1\right )^{2}}-\frac {c x}{12 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {c x}{3 \sqrt {c^{2} x^{2}+1}}\right )}{d^{3}}}{c^{4}}\)	\(108\)
default	\(\frac {\frac {a \left (-\frac {1}{2 \left (c^{2} x^{2}+1\right )}+\frac {1}{4 \left (c^{2} x^{2}+1\right )^{2}}\right )}{d^{3}}+\frac {b \left (-\frac {\arcsinh \left (c x \right )}{2 \left (c^{2} x^{2}+1\right )}+\frac {\arcsinh \left (c x \right )}{4 \left (c^{2} x^{2}+1\right )^{2}}-\frac {c x}{12 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {c x}{3 \sqrt {c^{2} x^{2}+1}}\right )}{d^{3}}}{c^{4}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*b*((4*c^2*x^2 + 4*(2*c^2*x^2 + 1)*log(c*x + sqrt(c^2*x^2 + 1)) + 3)/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d
^3) - 16*integrate(1/4*(2*c^2*x^2 + 1)/(c^10*d^3*x^7 + 3*c^8*d^3*x^5 + 3*c^6*d^3*x^3 + c^4*d^3*x + (c^9*d^3*x^
6 + 3*c^7*d^3*x^4 + 3*c^5*d^3*x^2 + c^3*d^3)*sqrt(c^2*x^2 + 1)), x)) - 1/4*(2*c^2*x^2 + 1)*a/(c^8*d^3*x^4 + 2*
c^6*d^3*x^2 + c^4*d^3)

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Fricas [A]
time = 0.35, size = 99, normalized size = 1.02 \begin {gather*} \frac {3 \, a c^{4} x^{4} - 3 \, {\left (2 \, b c^{2} x^{2} + b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (4 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt {c^{2} x^{2} + 1}}{12 \, {\left (c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^4*x^4 - 3*(2*b*c^2*x^2 + b)*log(c*x + sqrt(c^2*x^2 + 1)) + (4*b*c^3*x^3 + 3*b*c*x)*sqrt(c^2*x^2 +
1))/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x^{3}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{3} \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x**3/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b*x**3*asinh(c*x)/(c**6*x**6 + 3*c
**4*x**4 + 3*c**2*x**2 + 1), x))/d**3

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^3,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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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