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

Optimal. Leaf size=126 \[ c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-b c^3 d^2 \sqrt{c^2 x^2+1}-\frac{b c d^2 \sqrt{c^2 x^2+1}}{6 x^2}-\frac{11}{6} b c^3 d^2 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.158875, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {270, 5730, 12, 1251, 897, 1157, 388, 208} \[ c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-b c^3 d^2 \sqrt{c^2 x^2+1}-\frac{b c d^2 \sqrt{c^2 x^2+1}}{6 x^2}-\frac{11}{6} b c^3 d^2 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 )^2 \left (a+b \sinh ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d^2 \left (-1-6 c^2 x^2+3 c^4 x^4\right )}{3 x^3 \sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{3} \left (b c d^2\right ) \int \frac{-1-6 c^2 x^2+3 c^4 x^4}{x^3 \sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{6} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{-1-6 c^2 x+3 c^4 x^2}{x^2 \sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{8-12 x^2+3 x^4}{\left (-\frac{1}{c^2}+\frac{x^2}{c^2}\right )^2} \, dx,x,\sqrt{1+c^2 x^2}\right )}{3 c}\\ &=-\frac{b c d^2 \sqrt{1+c^2 x^2}}{6 x^2}-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{6} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{-17+6 x^2}{-\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{1+c^2 x^2}\right )\\ &=-b c^3 d^2 \sqrt{1+c^2 x^2}-\frac{b c d^2 \sqrt{1+c^2 x^2}}{6 x^2}-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} \left (11 b c d^2\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 )\\ &=-b c^3 d^2 \sqrt{1+c^2 x^2}-\frac{b c d^2 \sqrt{1+c^2 x^2}}{6 x^2}-\frac{d^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}-\frac{2 c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{11}{6} b c^3 d^2 \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.125521, size = 133, normalized size = 1.06 \[ \frac{d^2 \left (6 a c^4 x^4-12 a c^2 x^2-2 a-6 b c^3 x^3 \sqrt{c^2 x^2+1}-b c x \sqrt{c^2 x^2+1}+11 b c^3 x^3 \log (x)-11 b c^3 x^3 \log \left (\sqrt{c^2 x^2+1}+1\right )+2 b \left (3 c^4 x^4-6 c^2 x^2-1\right ) \sinh ^{-1}(c x)\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-2*a - 12*a*c^2*x^2 + 6*a*c^4*x^4 - b*c*x*Sqrt[1 + c^2*x^2] - 6*b*c^3*x^3*Sqrt[1 + c^2*x^2] + 2*b*(-1 -
6*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x] + 11*b*c^3*x^3*Log[x] - 11*b*c^3*x^3*Log[1 + Sqrt[1 + c^2*x^2]]))/(6*x^3)

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 114, normalized size = 0.9 \begin{align*}{c}^{3} \left ({d}^{2}a \left ( cx-2\,{\frac{1}{cx}}-{\frac{1}{3\,{c}^{3}{x}^{3}}} \right ) +{d}^{2}b \left ({\it Arcsinh} \left ( cx \right ) cx-2\,{\frac{{\it Arcsinh} \left ( cx \right ) }{cx}}-{\frac{{\it Arcsinh} \left ( cx \right ) }{3\,{c}^{3}{x}^{3}}}-\sqrt{{c}^{2}{x}^{2}+1}-{\frac{11}{6}{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) }-{\frac{1}{6\,{c}^{2}{x}^{2}}\sqrt{{c}^{2}{x}^{2}+1}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(d^2*a*(c*x-2/c/x-1/3/c^3/x^3)+d^2*b*(arcsinh(c*x)*c*x-2*arcsinh(c*x)/c/x-1/3*arcsinh(c*x)/c^3/x^3-(c^2*x^
2+1)^(1/2)-11/6*arctanh(1/(c^2*x^2+1)^(1/2))-1/6/c^2/x^2*(c^2*x^2+1)^(1/2)))

________________________________________________________________________________________

Maxima [A]  time = 1.12379, size = 190, normalized size = 1.51 \begin{align*} a c^{4} d^{2} x +{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} b c^{3} d^{2} - 2 \,{\left (c \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arsinh}\left (c x\right )}{x}\right )} b c^{2} d^{2} + \frac{1}{6} \,{\left ({\left (c^{2} \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) - \frac{\sqrt{c^{2} x^{2} + 1}}{x^{2}}\right )} c - \frac{2 \, \operatorname{arsinh}\left (c x\right )}{x^{3}}\right )} b d^{2} - \frac{2 \, a c^{2} d^{2}}{x} - \frac{a d^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^2*(a+b*arcsinh(c*x))/x^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 3.05978, size = 525, normalized size = 4.17 \begin{align*} \frac{6 \, a c^{4} d^{2} x^{4} - 11 \, b c^{3} d^{2} x^{3} \log \left (-c x + \sqrt{c^{2} x^{2} + 1} + 1\right ) + 11 \, b c^{3} d^{2} x^{3} \log \left (-c x + \sqrt{c^{2} x^{2} + 1} - 1\right ) - 12 \, a c^{2} d^{2} x^{2} - 2 \,{\left (3 \, b c^{4} - 6 \, b c^{2} - b\right )} d^{2} x^{3} \log \left (-c x + \sqrt{c^{2} x^{2} + 1}\right ) - 2 \, a d^{2} + 2 \,{\left (3 \, b c^{4} d^{2} x^{4} - 6 \, b c^{2} d^{2} x^{2} -{\left (3 \, b c^{4} - 6 \, b c^{2} - b\right )} d^{2} x^{3} - b d^{2}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (6 \, b c^{3} d^{2} x^{3} + b c d^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^2*(a+b*arcsinh(c*x))/x^4,x, algorithm="fricas")

[Out]

1/6*(6*a*c^4*d^2*x^4 - 11*b*c^3*d^2*x^3*log(-c*x + sqrt(c^2*x^2 + 1) + 1) + 11*b*c^3*d^2*x^3*log(-c*x + sqrt(c
^2*x^2 + 1) - 1) - 12*a*c^2*d^2*x^2 - 2*(3*b*c^4 - 6*b*c^2 - b)*d^2*x^3*log(-c*x + sqrt(c^2*x^2 + 1)) - 2*a*d^
2 + 2*(3*b*c^4*d^2*x^4 - 6*b*c^2*d^2*x^2 - (3*b*c^4 - 6*b*c^2 - b)*d^2*x^3 - b*d^2)*log(c*x + sqrt(c^2*x^2 + 1
)) - (6*b*c^3*d^2*x^3 + b*c*d^2*x)*sqrt(c^2*x^2 + 1))/x^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**2*(a+b*asinh(c*x))/x**4,x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^2*(a+b*arcsinh(c*x))/x^4,x, algorithm="giac")

[Out]

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