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

Optimal. Leaf size=160 \[ \frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{1}{25} b c d^3 \left (c^2 x^2+1\right )^{5/2}-\frac{1}{5} b c d^3 \left (c^2 x^2+1\right )^{3/2}-\frac{11}{5} b c d^3 \sqrt{c^2 x^2+1}-b c d^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right ) \]

[Out]

(-11*b*c*d^3*Sqrt[1 + c^2*x^2])/5 - (b*c*d^3*(1 + c^2*x^2)^(3/2))/5 - (b*c*d^3*(1 + c^2*x^2)^(5/2))/25 - (d^3*
(a + b*ArcSinh[c*x]))/x + 3*c^2*d^3*x*(a + b*ArcSinh[c*x]) + c^4*d^3*x^3*(a + b*ArcSinh[c*x]) + (c^6*d^3*x^5*(
a + b*ArcSinh[c*x]))/5 - b*c*d^3*ArcTanh[Sqrt[1 + c^2*x^2]]

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Rubi [A]  time = 0.221344, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {270, 5730, 12, 1799, 1620, 63, 208} \[ \frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{1}{25} b c d^3 \left (c^2 x^2+1\right )^{5/2}-\frac{1}{5} b c d^3 \left (c^2 x^2+1\right )^{3/2}-\frac{11}{5} b c d^3 \sqrt{c^2 x^2+1}-b c d^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*b*c*d^3*Sqrt[1 + c^2*x^2])/5 - (b*c*d^3*(1 + c^2*x^2)^(3/2))/5 - (b*c*d^3*(1 + c^2*x^2)^(5/2))/25 - (d^3*
(a + b*ArcSinh[c*x]))/x + 3*c^2*d^3*x*(a + b*ArcSinh[c*x]) + c^4*d^3*x^3*(a + b*ArcSinh[c*x]) + (c^6*d^3*x^5*(
a + b*ArcSinh[c*x]))/5 - b*c*d^3*ArcTanh[Sqrt[1 + c^2*x^2]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 5730

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (d+c^2 d x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d^3 \left (-5+15 c^2 x^2+5 c^4 x^4+c^6 x^6\right )}{5 x \sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{5} \left (b c d^3\right ) \int \frac{-5+15 c^2 x^2+5 c^4 x^4+c^6 x^6}{x \sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{10} \left (b c d^3\right ) \operatorname{Subst}\left (\int \frac{-5+15 c^2 x+5 c^4 x^2+c^6 x^3}{x \sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{10} \left (b c d^3\right ) \operatorname{Subst}\left (\int \left (\frac{11 c^2}{\sqrt{1+c^2 x}}-\frac{5}{x \sqrt{1+c^2 x}}+3 c^2 \sqrt{1+c^2 x}+c^2 \left (1+c^2 x\right )^{3/2}\right ) \, dx,x,x^2\right )\\ &=-\frac{11}{5} b c d^3 \sqrt{1+c^2 x^2}-\frac{1}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2}-\frac{1}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2}-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{2} \left (b c d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac{11}{5} b c d^3 \sqrt{1+c^2 x^2}-\frac{1}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2}-\frac{1}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2}-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (b d^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{1+c^2 x^2}\right )}{c}\\ &=-\frac{11}{5} b c d^3 \sqrt{1+c^2 x^2}-\frac{1}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2}-\frac{1}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2}-\frac{d^3 \left (a+b \sinh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )+c^4 d^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^6 d^3 x^5 \left (a+b \sinh ^{-1}(c x)\right )-b c d^3 \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.165508, size = 163, normalized size = 1.02 \[ \frac{d^3 \left (5 a c^6 x^6+25 a c^4 x^4+75 a c^2 x^2-25 a-b c^5 x^5 \sqrt{c^2 x^2+1}-7 b c^3 x^3 \sqrt{c^2 x^2+1}-61 b c x \sqrt{c^2 x^2+1}-25 b c x \log \left (\sqrt{c^2 x^2+1}+1\right )+5 b \left (c^6 x^6+5 c^4 x^4+15 c^2 x^2-5\right ) \sinh ^{-1}(c x)+25 b c x \log (x)\right )}{25 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(-25*a + 75*a*c^2*x^2 + 25*a*c^4*x^4 + 5*a*c^6*x^6 - 61*b*c*x*Sqrt[1 + c^2*x^2] - 7*b*c^3*x^3*Sqrt[1 + c^
2*x^2] - b*c^5*x^5*Sqrt[1 + c^2*x^2] + 5*b*(-5 + 15*c^2*x^2 + 5*c^4*x^4 + c^6*x^6)*ArcSinh[c*x] + 25*b*c*x*Log
[x] - 25*b*c*x*Log[1 + Sqrt[1 + c^2*x^2]]))/(25*x)

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Maple [A]  time = 0.008, size = 151, normalized size = 0.9 \begin{align*} c \left ({d}^{3}a \left ({\frac{{c}^{5}{x}^{5}}{5}}+{c}^{3}{x}^{3}+3\,cx-{\frac{1}{cx}} \right ) +{d}^{3}b \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{5}{x}^{5}}{5}}+{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}+3\,{\it Arcsinh} \left ( cx \right ) cx-{\frac{{\it Arcsinh} \left ( cx \right ) }{cx}}-{\frac{{c}^{4}{x}^{4}}{25}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{7\,{c}^{2}{x}^{2}}{25}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{61}{25}\sqrt{{c}^{2}{x}^{2}+1}}-{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(d^3*a*(1/5*c^5*x^5+c^3*x^3+3*c*x-1/c/x)+d^3*b*(1/5*arcsinh(c*x)*c^5*x^5+arcsinh(c*x)*c^3*x^3+3*arcsinh(c*x)
*c*x-arcsinh(c*x)/c/x-1/25*c^4*x^4*(c^2*x^2+1)^(1/2)-7/25*c^2*x^2*(c^2*x^2+1)^(1/2)-61/25*(c^2*x^2+1)^(1/2)-ar
ctanh(1/(c^2*x^2+1)^(1/2))))

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Maxima [A]  time = 1.24493, size = 315, normalized size = 1.97 \begin{align*} \frac{1}{5} \, a c^{6} d^{3} x^{5} + \frac{1}{75} \,{\left (15 \, x^{5} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{3 \, \sqrt{c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac{4 \, \sqrt{c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b c^{6} d^{3} + a c^{4} d^{3} x^{3} + \frac{1}{3} \,{\left (3 \, x^{3} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{4} d^{3} + 3 \, a c^{2} d^{3} x + 3 \,{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} b c d^{3} -{\left (c \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arsinh}\left (c x\right )}{x}\right )} b d^{3} - \frac{a d^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^6*d^3*x^5 + 1/75*(15*x^5*arcsinh(c*x) - (3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 + 8
*sqrt(c^2*x^2 + 1)/c^6)*c)*b*c^6*d^3 + a*c^4*d^3*x^3 + 1/3*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2
- 2*sqrt(c^2*x^2 + 1)/c^4))*b*c^4*d^3 + 3*a*c^2*d^3*x + 3*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*c*d^3 - (c*
arcsinh(1/(sqrt(c^2)*abs(x))) + arcsinh(c*x)/x)*b*d^3 - a*d^3/x

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Fricas [A]  time = 2.77926, size = 616, normalized size = 3.85 \begin{align*} \frac{5 \, a c^{6} d^{3} x^{6} + 25 \, a c^{4} d^{3} x^{4} + 75 \, a c^{2} d^{3} x^{2} - 25 \, b c d^{3} x \log \left (-c x + \sqrt{c^{2} x^{2} + 1} + 1\right ) + 25 \, b c d^{3} x \log \left (-c x + \sqrt{c^{2} x^{2} + 1} - 1\right ) - 5 \,{\left (b c^{6} + 5 \, b c^{4} + 15 \, b c^{2} - 5 \, b\right )} d^{3} x \log \left (-c x + \sqrt{c^{2} x^{2} + 1}\right ) - 25 \, a d^{3} + 5 \,{\left (b c^{6} d^{3} x^{6} + 5 \, b c^{4} d^{3} x^{4} + 15 \, b c^{2} d^{3} x^{2} -{\left (b c^{6} + 5 \, b c^{4} + 15 \, b c^{2} - 5 \, b\right )} d^{3} x - 5 \, b d^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (b c^{5} d^{3} x^{5} + 7 \, b c^{3} d^{3} x^{3} + 61 \, b c d^{3} x\right )} \sqrt{c^{2} x^{2} + 1}}{25 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*(5*a*c^6*d^3*x^6 + 25*a*c^4*d^3*x^4 + 75*a*c^2*d^3*x^2 - 25*b*c*d^3*x*log(-c*x + sqrt(c^2*x^2 + 1) + 1) +
25*b*c*d^3*x*log(-c*x + sqrt(c^2*x^2 + 1) - 1) - 5*(b*c^6 + 5*b*c^4 + 15*b*c^2 - 5*b)*d^3*x*log(-c*x + sqrt(c
^2*x^2 + 1)) - 25*a*d^3 + 5*(b*c^6*d^3*x^6 + 5*b*c^4*d^3*x^4 + 15*b*c^2*d^3*x^2 - (b*c^6 + 5*b*c^4 + 15*b*c^2
- 5*b)*d^3*x - 5*b*d^3)*log(c*x + sqrt(c^2*x^2 + 1)) - (b*c^5*d^3*x^5 + 7*b*c^3*d^3*x^3 + 61*b*c*d^3*x)*sqrt(c
^2*x^2 + 1))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*(Integral(3*a*c**2, x) + Integral(a/x**2, x) + Integral(3*a*c**4*x**2, x) + Integral(a*c**6*x**4, x) + In
tegral(3*b*c**2*asinh(c*x), x) + Integral(b*asinh(c*x)/x**2, x) + Integral(3*b*c**4*x**2*asinh(c*x), x) + Inte
gral(b*c**6*x**4*asinh(c*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}^{3}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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