Integrand size = 26, antiderivative size = 335 \[ \int \frac {x^2 \left (d+c^2 d x^2\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \text {arcsinh}(c x)}}+\frac {d e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}-\frac {d e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {d e^{\frac {5 a}{b}} \sqrt {5 \pi } \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {d e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}+\frac {d e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}+\frac {d e^{-\frac {5 a}{b}} \sqrt {5 \pi } \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3} \]
1/8*d*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(3/2)/c^3- 1/8*d*erfi((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(3/2)/c^3/exp(a/b) -1/16*d*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)*P i^(1/2)/b^(3/2)/c^3+1/16*d*erfi(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))* 3^(1/2)*Pi^(1/2)/b^(3/2)/c^3/exp(3*a/b)-1/16*d*exp(5*a/b)*erf(5^(1/2)*(a+b *arcsinh(c*x))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/b^(3/2)/c^3+1/16*d*erfi(5^( 1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/b^(3/2)/c^3/exp(5* a/b)-2*d*x^2*(c^2*x^2+1)^(3/2)/b/c/(a+b*arcsinh(c*x))^(1/2)
Time = 0.68 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 \left (d+c^2 d x^2\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx=\frac {d e^{-5 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )} \left (-e^{\frac {5 a}{b}}-e^{\frac {5 a}{b}+2 \text {arcsinh}(c x)}+2 e^{\frac {5 a}{b}+4 \text {arcsinh}(c x)}+2 e^{\frac {5 a}{b}+6 \text {arcsinh}(c x)}-e^{\frac {5 a}{b}+8 \text {arcsinh}(c x)}-e^{\frac {5 a}{b}+10 \text {arcsinh}(c x)}-2 e^{\frac {6 a}{b}+5 \text {arcsinh}(c x)} \sqrt {\frac {a}{b}+\text {arcsinh}(c x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\text {arcsinh}(c x)\right )+\sqrt {5} e^{5 \text {arcsinh}(c x)} \sqrt {-\frac {a+b \text {arcsinh}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )+\sqrt {3} e^{\frac {2 a}{b}+5 \text {arcsinh}(c x)} \sqrt {-\frac {a+b \text {arcsinh}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )-2 e^{\frac {4 a}{b}+5 \text {arcsinh}(c x)} \sqrt {-\frac {a+b \text {arcsinh}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \text {arcsinh}(c x)}{b}\right )+\sqrt {3} e^{\frac {8 a}{b}+5 \text {arcsinh}(c x)} \sqrt {\frac {a}{b}+\text {arcsinh}(c x)} \Gamma \left (\frac {1}{2},\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )+\sqrt {5} e^{5 \left (\frac {2 a}{b}+\text {arcsinh}(c x)\right )} \sqrt {\frac {a}{b}+\text {arcsinh}(c x)} \Gamma \left (\frac {1}{2},\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )\right )}{16 b c^3 \sqrt {a+b \text {arcsinh}(c x)}} \]
(d*(-E^((5*a)/b) - E^((5*a)/b + 2*ArcSinh[c*x]) + 2*E^((5*a)/b + 4*ArcSinh [c*x]) + 2*E^((5*a)/b + 6*ArcSinh[c*x]) - E^((5*a)/b + 8*ArcSinh[c*x]) - E ^((5*a)/b + 10*ArcSinh[c*x]) - 2*E^((6*a)/b + 5*ArcSinh[c*x])*Sqrt[a/b + A rcSinh[c*x]]*Gamma[1/2, a/b + ArcSinh[c*x]] + Sqrt[5]*E^(5*ArcSinh[c*x])*S qrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-5*(a + b*ArcSinh[c*x]))/b] + S qrt[3]*E^((2*a)/b + 5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[ 1/2, (-3*(a + b*ArcSinh[c*x]))/b] - 2*E^((4*a)/b + 5*ArcSinh[c*x])*Sqrt[-( (a + b*ArcSinh[c*x])/b)]*Gamma[1/2, -((a + b*ArcSinh[c*x])/b)] + Sqrt[3]*E ^((8*a)/b + 5*ArcSinh[c*x])*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (3*(a + b* ArcSinh[c*x]))/b] + Sqrt[5]*E^(5*((2*a)/b + ArcSinh[c*x]))*Sqrt[a/b + ArcS inh[c*x]]*Gamma[1/2, (5*(a + b*ArcSinh[c*x]))/b]))/(16*b*c^3*E^(5*(a/b + A rcSinh[c*x]))*Sqrt[a + b*ArcSinh[c*x]])
Time = 1.52 (sec) , antiderivative size = 520, normalized size of antiderivative = 1.55, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6229, 6234, 25, 5971, 2009}
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.
| \(\displaystyle \int \frac {x^2 \left (c^2 d x^2+d\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 6229 |
| \(\displaystyle \frac {4 d \int \frac {x \sqrt {c^2 x^2+1}}{\sqrt {a+b \text {arcsinh}(c x)}}dx}{b c}+\frac {10 c d \int \frac {x^3 \sqrt {c^2 x^2+1}}{\sqrt {a+b \text {arcsinh}(c x)}}dx}{b}-\frac {2 d x^2 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt {a+b \text {arcsinh}(c x)}}\) |
| \(\Big \downarrow \) 6234 |
| \(\displaystyle \frac {10 d \int -\frac {\cosh ^2\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh ^3\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{\sqrt {a+b \text {arcsinh}(c x)}}d(a+b \text {arcsinh}(c x))}{b^2 c^3}+\frac {4 d \int -\frac {\cosh ^2\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{\sqrt {a+b \text {arcsinh}(c x)}}d(a+b \text {arcsinh}(c x))}{b^2 c^3}-\frac {2 d x^2 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt {a+b \text {arcsinh}(c x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {10 d \int \frac {\cosh ^2\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh ^3\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{\sqrt {a+b \text {arcsinh}(c x)}}d(a+b \text {arcsinh}(c x))}{b^2 c^3}-\frac {4 d \int \frac {\cosh ^2\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{\sqrt {a+b \text {arcsinh}(c x)}}d(a+b \text {arcsinh}(c x))}{b^2 c^3}-\frac {2 d x^2 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt {a+b \text {arcsinh}(c x)}}\) |
| \(\Big \downarrow \) 5971 |
| \(\displaystyle -\frac {10 d \int \left (\frac {\sinh \left (\frac {5 a}{b}-\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )}{16 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {\sinh \left (\frac {3 a}{b}-\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )}{16 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {\sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{8 \sqrt {a+b \text {arcsinh}(c x)}}\right )d(a+b \text {arcsinh}(c x))}{b^2 c^3}-\frac {4 d \int \left (\frac {\sinh \left (\frac {3 a}{b}-\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )}{4 \sqrt {a+b \text {arcsinh}(c x)}}+\frac {\sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{4 \sqrt {a+b \text {arcsinh}(c x)}}\right )d(a+b \text {arcsinh}(c x))}{b^2 c^3}-\frac {2 d x^2 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt {a+b \text {arcsinh}(c x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 d \left (-\frac {1}{8} \sqrt {\pi } \sqrt {b} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )-\frac {1}{8} \sqrt {\frac {\pi }{3}} \sqrt {b} e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )+\frac {1}{8} \sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )+\frac {1}{8} \sqrt {\frac {\pi }{3}} \sqrt {b} e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )\right )}{b^2 c^3}+\frac {10 d \left (\frac {1}{16} \sqrt {\pi } \sqrt {b} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )+\frac {1}{32} \sqrt {\frac {\pi }{3}} \sqrt {b} e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )-\frac {1}{32} \sqrt {\frac {\pi }{5}} \sqrt {b} e^{\frac {5 a}{b}} \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )-\frac {1}{16} \sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )-\frac {1}{32} \sqrt {\frac {\pi }{3}} \sqrt {b} e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )+\frac {1}{32} \sqrt {\frac {\pi }{5}} \sqrt {b} e^{-\frac {5 a}{b}} \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )\right )}{b^2 c^3}-\frac {2 d x^2 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt {a+b \text {arcsinh}(c x)}}\) |
(-2*d*x^2*(1 + c^2*x^2)^(3/2))/(b*c*Sqrt[a + b*ArcSinh[c*x]]) + (4*d*(-1/8 *(Sqrt[b]*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]]) - (Sqrt[ b]*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]]) /8 + (Sqrt[b]*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(8*E^(a/b)) + (Sqrt[b]*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/( 8*E^((3*a)/b))))/(b^2*c^3) + (10*d*((Sqrt[b]*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/16 + (Sqrt[b]*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[ 3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/32 - (Sqrt[b]*E^((5*a)/b)*Sqrt[Pi/5 ]*Erf[(Sqrt[5]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/32 - (Sqrt[b]*Sqrt[Pi]* Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(16*E^(a/b)) - (Sqrt[b]*Sqrt[Pi/3] *Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(32*E^((3*a)/b)) + (Sqr t[b]*Sqrt[Pi/5]*Erfi[(Sqrt[5]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(32*E^(( 5*a)/b))))/(b^2*c^3)
3.5.63.3.1 Defintions of rubi rules used
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & & IGtQ[p, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p *((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Simp[f*(m/(b*c*(n + 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] - Simp[c*((m + 2*p + 1)/(b*f* (n + 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] && LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1 , 0] && IGtQ[m, -3]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_) ^2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2* x^2)^p] Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
\[\int \frac {x^{2} \left (c^{2} d \,x^{2}+d \right )}{\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{\frac {3}{2}}}d x\]
Exception generated. \[ \int \frac {x^2 \left (d+c^2 d x^2\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x^2 \left (d+c^2 d x^2\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx=d \left (\int \frac {x^{2}}{a \sqrt {a + b \operatorname {asinh}{\left (c x \right )}} + b \sqrt {a + b \operatorname {asinh}{\left (c x \right )}} \operatorname {asinh}{\left (c x \right )}}\, dx + \int \frac {c^{2} x^{4}}{a \sqrt {a + b \operatorname {asinh}{\left (c x \right )}} + b \sqrt {a + b \operatorname {asinh}{\left (c x \right )}} \operatorname {asinh}{\left (c x \right )}}\, dx\right ) \]
d*(Integral(x**2/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asin h(c*x)), x) + Integral(c**2*x**4/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b* asinh(c*x))*asinh(c*x)), x))
\[ \int \frac {x^2 \left (d+c^2 d x^2\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx=\int { \frac {{\left (c^{2} d x^{2} + d\right )} x^{2}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^2 \left (d+c^2 d x^2\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx=\int { \frac {{\left (c^{2} d x^{2} + d\right )} x^{2}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (d+c^2 d x^2\right )}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx=\int \frac {x^2\,\left (d\,c^2\,x^2+d\right )}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{3/2}} \,d x \]