Integrand size = 26, antiderivative size = 184 \[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx=-\frac {b c d \sqrt {d+c^2 d x^2}}{6 x^2 \sqrt {1+c^2 x^2}}-\frac {c^2 d \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{x}-\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 x^3}+\frac {c^3 d \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{2 b \sqrt {1+c^2 x^2}}+\frac {4 b c^3 d \sqrt {d+c^2 d x^2} \log (x)}{3 \sqrt {1+c^2 x^2}} \]
-1/3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/x^3-c^2*d*(a+b*arcsinh(c*x))*( c^2*d*x^2+d)^(1/2)/x-1/6*b*c*d*(c^2*d*x^2+d)^(1/2)/x^2/(c^2*x^2+1)^(1/2)+1 /2*c^3*d*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/b/(c^2*x^2+1)^(1/2)+4/3* b*c^3*d*ln(x)*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)
Time = 0.63 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx=\frac {1}{6} \left (-\frac {2 a d \left (1+4 c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{x^3}-\frac {2 b \left (d+c^2 d x^2\right )^{3/2} \text {arcsinh}(c x)}{x^3}+\frac {3 b c^3 d \sqrt {d+c^2 d x^2} \left (-\frac {2 \sqrt {1+c^2 x^2} \text {arcsinh}(c x)}{c x}+\text {arcsinh}(c x)^2+2 \log (c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {b c d \sqrt {d+c^2 d x^2} \left (-1+2 c^2 x^2 \log (c x)\right )}{x^2 \sqrt {1+c^2 x^2}}+6 a c^3 d^{3/2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )\right ) \]
((-2*a*d*(1 + 4*c^2*x^2)*Sqrt[d + c^2*d*x^2])/x^3 - (2*b*(d + c^2*d*x^2)^( 3/2)*ArcSinh[c*x])/x^3 + (3*b*c^3*d*Sqrt[d + c^2*d*x^2]*((-2*Sqrt[1 + c^2* x^2]*ArcSinh[c*x])/(c*x) + ArcSinh[c*x]^2 + 2*Log[c*x]))/Sqrt[1 + c^2*x^2] + (b*c*d*Sqrt[d + c^2*d*x^2]*(-1 + 2*c^2*x^2*Log[c*x]))/(x^2*Sqrt[1 + c^2 *x^2]) + 6*a*c^3*d^(3/2)*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/6
Time = 0.69 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6222, 244, 2009, 6220, 14, 6198}
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.
| \(\displaystyle \int \frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx\) |
| \(\Big \downarrow \) 6222 |
| \(\displaystyle c^2 d \int \frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{x^2}dx+\frac {b c d \sqrt {c^2 d x^2+d} \int \frac {c^2 x^2+1}{x^3}dx}{3 \sqrt {c^2 x^2+1}}-\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 x^3}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle c^2 d \int \frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{x^2}dx+\frac {b c d \sqrt {c^2 d x^2+d} \int \left (\frac {c^2}{x}+\frac {1}{x^3}\right )dx}{3 \sqrt {c^2 x^2+1}}-\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle c^2 d \int \frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{x^2}dx-\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 x^3}+\frac {b c d \sqrt {c^2 d x^2+d} \left (c^2 \log (x)-\frac {1}{2 x^2}\right )}{3 \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 6220 |
| \(\displaystyle c^2 d \left (\frac {c^2 \sqrt {c^2 d x^2+d} \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx}{\sqrt {c^2 x^2+1}}+\frac {b c \sqrt {c^2 d x^2+d} \int \frac {1}{x}dx}{\sqrt {c^2 x^2+1}}-\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{x}\right )-\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 x^3}+\frac {b c d \sqrt {c^2 d x^2+d} \left (c^2 \log (x)-\frac {1}{2 x^2}\right )}{3 \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 14 |
| \(\displaystyle c^2 d \left (\frac {c^2 \sqrt {c^2 d x^2+d} \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx}{\sqrt {c^2 x^2+1}}-\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{x}+\frac {b c \log (x) \sqrt {c^2 d x^2+d}}{\sqrt {c^2 x^2+1}}\right )-\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 x^3}+\frac {b c d \sqrt {c^2 d x^2+d} \left (c^2 \log (x)-\frac {1}{2 x^2}\right )}{3 \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 6198 |
| \(\displaystyle c^2 d \left (\frac {c \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2}{2 b \sqrt {c^2 x^2+1}}-\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{x}+\frac {b c \log (x) \sqrt {c^2 d x^2+d}}{\sqrt {c^2 x^2+1}}\right )-\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 x^3}+\frac {b c d \sqrt {c^2 d x^2+d} \left (c^2 \log (x)-\frac {1}{2 x^2}\right )}{3 \sqrt {c^2 x^2+1}}\) |
-1/3*((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/x^3 + (b*c*d*Sqrt[d + c^ 2*d*x^2]*(-1/2*1/x^2 + c^2*Log[x]))/(3*Sqrt[1 + c^2*x^2]) + c^2*d*(-((Sqrt [d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/x) + (c*Sqrt[d + c^2*d*x^2]*(a + b*A rcSinh[c*x])^2)/(2*b*Sqrt[1 + c^2*x^2]) + (b*c*Sqrt[d + c^2*d*x^2]*Log[x]) /Sqrt[1 + c^2*x^2])
3.2.35.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_ Symbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*( a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c ^2*d] && NeQ[n, -1]
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*Arc Sinh[c*x])^n/(f*(m + 1))), x] + (-Simp[b*c*(n/(f*(m + 1)))*Simp[Sqrt[d + e* x^2]/Sqrt[1 + c^2*x^2]] Int[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x ], x] - Simp[(c^2/(f^2*(m + 1)))*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]] Int[(f*x)^(m + 2)*((a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2]), x], x]) /; Fr eeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1]
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*((a + b*Arc Sinh[c*x])^n/(f*(m + 1))), x] + (-Simp[2*e*(p/(f^2*(m + 1))) Int[(f*x)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Simp[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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]
Time = 0.25 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.40
| method | result | size |
| default | \(-\frac {a \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{3 d \,x^{3}}-\frac {2 a \,c^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{3 d x}+\frac {2 a \,c^{4} x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{3}+a \,c^{4} d x \sqrt {c^{2} d \,x^{2}+d}+\frac {a \,c^{4} d^{2} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (3 \operatorname {arcsinh}\left (c x \right )^{2} x^{3} c^{3}-8 \,\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}+8 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x^{3} c^{3}-8 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{2} c^{2}-2 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}-c x \right ) d}{6 \sqrt {c^{2} x^{2}+1}\, x^{3}}\) | \(257\) |
| parts | \(-\frac {a \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{3 d \,x^{3}}-\frac {2 a \,c^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{3 d x}+\frac {2 a \,c^{4} x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{3}+a \,c^{4} d x \sqrt {c^{2} d \,x^{2}+d}+\frac {a \,c^{4} d^{2} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (3 \operatorname {arcsinh}\left (c x \right )^{2} x^{3} c^{3}-8 \,\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}+8 \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x^{3} c^{3}-8 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{2} c^{2}-2 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}-c x \right ) d}{6 \sqrt {c^{2} x^{2}+1}\, x^{3}}\) | \(257\) |
-1/3*a/d/x^3*(c^2*d*x^2+d)^(5/2)-2/3*a*c^2/d/x*(c^2*d*x^2+d)^(5/2)+2/3*a*c ^4*x*(c^2*d*x^2+d)^(3/2)+a*c^4*d*x*(c^2*d*x^2+d)^(1/2)+a*c^4*d^2*ln(c^2*d* x/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+1/6*b*(d*(c^2*x^2+1))^( 1/2)/(c^2*x^2+1)^(1/2)/x^3*(3*arcsinh(c*x)^2*x^3*c^3-8*arcsinh(c*x)*c^3*x^ 3+8*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)*x^3*c^3-8*arcsinh(c*x)*(c^2*x^2+1)^(1/ 2)*x^2*c^2-2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-c*x)*d
\[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx=\int { \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
\[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx=\int \frac {\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{x^{4}}\, dx \]
Exception generated. \[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{3/2}}{x^4} \,d x \]