Integrand size = 24, antiderivative size = 126 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=-b c^3 d^2 \sqrt {1+c^2 x^2}-\frac {b c d^2 \sqrt {1+c^2 x^2}}{6 x^2}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {11}{6} b c^3 d^2 \text {arctanh}\left (\sqrt {1+c^2 x^2}\right ) \]
-1/3*d^2*(a+b*arcsinh(c*x))/x^3-2*c^2*d^2*(a+b*arcsinh(c*x))/x+c^4*d^2*x*( a+b*arcsinh(c*x))-11/6*b*c^3*d^2*arctanh((c^2*x^2+1)^(1/2))-b*c^3*d^2*(c^2 *x^2+1)^(1/2)-1/6*b*c*d^2*(c^2*x^2+1)^(1/2)/x^2
Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\frac {d^2 \left (-2 a-12 a c^2 x^2+6 a c^4 x^4-b c x \sqrt {1+c^2 x^2}-6 b c^3 x^3 \sqrt {1+c^2 x^2}+2 b \left (-1-6 c^2 x^2+3 c^4 x^4\right ) \text {arcsinh}(c x)+11 b c^3 x^3 \log (x)-11 b c^3 x^3 \log \left (1+\sqrt {1+c^2 x^2}\right )\right )}{6 x^3} \]
(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)
Time = 0.42 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6218, 27, 1578, 1192, 25, 1471, 25, 27, 299, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (c^2 d x^2+d\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx\) |
\(\Big \downarrow \) 6218 |
\(\displaystyle -b c \int -\frac {d^2 \left (-3 c^4 x^4+6 c^2 x^2+1\right )}{3 x^3 \sqrt {c^2 x^2+1}}dx+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} b c d^2 \int \frac {-3 c^4 x^4+6 c^2 x^2+1}{x^3 \sqrt {c^2 x^2+1}}dx+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{6} b c d^2 \int \frac {-3 c^4 x^4+6 c^2 x^2+1}{x^4 \sqrt {c^2 x^2+1}}dx^2+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 1192 |
\(\displaystyle \frac {b d^2 \int -\frac {3 c^4 x^8-12 c^4 x^4+8 c^4}{\left (1-x^4\right )^2}d\sqrt {c^2 x^2+1}}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {b d^2 \int \frac {3 c^4 x^8-12 c^4 x^4+8 c^4}{\left (1-x^4\right )^2}d\sqrt {c^2 x^2+1}}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {b d^2 \left (\frac {1}{2} \int -\frac {c^4 \left (17-6 x^4\right )}{1-x^4}d\sqrt {c^2 x^2+1}+\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} \int \frac {c^4 \left (17-6 x^4\right )}{1-x^4}d\sqrt {c^2 x^2+1}\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \int \frac {17-6 x^4}{1-x^4}d\sqrt {c^2 x^2+1}\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \left (11 \int \frac {1}{1-x^4}d\sqrt {c^2 x^2+1}+6 \sqrt {c^2 x^2+1}\right )\right )}{3 c}+c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle c^4 d^2 x (a+b \text {arcsinh}(c x))-\frac {2 c^2 d^2 (a+b \text {arcsinh}(c x))}{x}-\frac {d^2 (a+b \text {arcsinh}(c x))}{3 x^3}+\frac {b d^2 \left (\frac {c^4 \sqrt {c^2 x^2+1}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \left (11 \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )+6 \sqrt {c^2 x^2+1}\right )\right )}{3 c}\) |
-1/3*(d^2*(a + b*ArcSinh[c*x]))/x^3 - (2*c^2*d^2*(a + b*ArcSinh[c*x]))/x + c^4*d^2*x*(a + b*ArcSinh[c*x]) + (b*d^2*((c^4*Sqrt[1 + c^2*x^2])/(2*(1 - x^4)) - (c^4*(6*Sqrt[1 + c^2*x^2] + 11*ArcTanh[Sqrt[1 + c^2*x^2]]))/2))/(3 *c)
3.1.18.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(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] + Simp[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] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[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] && Int egerQ[(m - 1)/2]
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]}, Simp [(a + b*ArcSinh[c*x]) u, x] - Simp[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]
Time = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.89
method | result | size |
parts | \(d^{2} a \left (c^{4} x -\frac {2 c^{2}}{x}-\frac {1}{3 x^{3}}\right )+d^{2} b \,c^{3} \left (\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {2 \,\operatorname {arcsinh}\left (c x \right )}{c x}-\sqrt {c^{2} x^{2}+1}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}\right )\) | \(112\) |
derivativedivides | \(c^{3} \left (d^{2} a \left (c x -\frac {1}{3 c^{3} x^{3}}-\frac {2}{c x}\right )+d^{2} b \left (\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {2 \,\operatorname {arcsinh}\left (c x \right )}{c x}-\sqrt {c^{2} x^{2}+1}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}\right )\right )\) | \(114\) |
default | \(c^{3} \left (d^{2} a \left (c x -\frac {1}{3 c^{3} x^{3}}-\frac {2}{c x}\right )+d^{2} b \left (\operatorname {arcsinh}\left (c x \right ) c x -\frac {\operatorname {arcsinh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {2 \,\operatorname {arcsinh}\left (c x \right )}{c x}-\sqrt {c^{2} x^{2}+1}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}\right )\right )\) | \(114\) |
d^2*a*(c^4*x-2*c^2/x-1/3/x^3)+d^2*b*c^3*(arcsinh(c*x)*c*x-1/3*arcsinh(c*x) /c^3/x^3-2*arcsinh(c*x)/c/x-(c^2*x^2+1)^(1/2)-1/6/c^2/x^2*(c^2*x^2+1)^(1/2 )-11/6*arctanh(1/(c^2*x^2+1)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (114) = 228\).
Time = 0.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.93 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\frac {6 \, a c^{4} d^{2} x^{4} - 11 \, b c^{3} d^{2} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} + 1\right ) + 11 \, b c^{3} d^{2} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1} - 1\right ) - 12 \, a c^{2} d^{2} x^{2} - 2 \, {\left (3 \, b c^{4} - 6 \, b c^{2} - b\right )} d^{2} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, a d^{2} + 2 \, {\left (3 \, b c^{4} d^{2} x^{4} - 6 \, b c^{2} d^{2} x^{2} - {\left (3 \, b c^{4} - 6 \, b c^{2} - b\right )} d^{2} x^{3} - b d^{2}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (6 \, b c^{3} d^{2} x^{3} + b c d^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}{6 \, x^{3}} \]
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
\[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=d^{2} \left (\int a c^{4}\, dx + \int \frac {a}{x^{4}}\, dx + \int \frac {2 a c^{2}}{x^{2}}\, dx + \int b c^{4} \operatorname {asinh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {2 b c^{2} \operatorname {asinh}{\left (c x \right )}}{x^{2}}\, dx\right ) \]
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) + Int egral(2*b*c**2*asinh(c*x)/x**2, x))
Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.09 \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=a c^{4} d^{2} x + {\left (c x \operatorname {arsinh}\left (c x\right ) - \sqrt {c^{2} x^{2} + 1}\right )} b c^{3} d^{2} - 2 \, {\left (c \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arsinh}\left (c x\right )}{x}\right )} b c^{2} d^{2} + \frac {1}{6} \, {\left ({\left (c^{2} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} + 1}}{x^{2}}\right )} c - \frac {2 \, \operatorname {arsinh}\left (c x\right )}{x^{3}}\right )} b d^{2} - \frac {2 \, a c^{2} d^{2}}{x} - \frac {a d^{2}}{3 \, x^{3}} \]
a*c^4*d^2*x + (c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*c^3*d^2 - 2*(c*arcs inh(1/(c*abs(x))) + arcsinh(c*x)/x)*b*c^2*d^2 + 1/6*((c^2*arcsinh(1/(c*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
Exception generated. \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^2}{x^4} \,d x \]