Integrand size = 26, antiderivative size = 267 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {16 b^2 d \sqrt {d+c^2 d x^2}}{75 c^2}+\frac {8 b^2 d \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{225 c^2}+\frac {2 b^2 d \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}}{125 c^2}-\frac {2 b d x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d} \]
1/5*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2/c^2/d+16/75*b^2*d*(c^2*d*x^2+ d)^(1/2)/c^2+8/225*b^2*d*(c^2*x^2+1)*(c^2*d*x^2+d)^(1/2)/c^2+2/125*b^2*d*( c^2*x^2+1)^2*(c^2*d*x^2+d)^(1/2)/c^2-2/5*b*d*x*(a+b*arcsinh(c*x))*(c^2*d*x ^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-4/15*b*c*d*x^3*(a+b*arcsinh(c*x))*(c^2*d*x ^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2/25*b*c^3*d*x^5*(a+b*arcsinh(c*x))*(c^2*d*x ^2+d)^(1/2)/(c^2*x^2+1)^(1/2)
Time = 1.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.74 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {d \sqrt {d+c^2 d x^2} \left (225 a^2 \left (1+c^2 x^2\right )^3-30 a b c x \sqrt {1+c^2 x^2} \left (15+10 c^2 x^2+3 c^4 x^4\right )+2 b^2 \left (149+187 c^2 x^2+47 c^4 x^4+9 c^6 x^6\right )+30 b \left (15 a \left (1+c^2 x^2\right )^3-b c x \sqrt {1+c^2 x^2} \left (15+10 c^2 x^2+3 c^4 x^4\right )\right ) \text {arcsinh}(c x)+225 b^2 \left (1+c^2 x^2\right )^3 \text {arcsinh}(c x)^2\right )}{1125 c^2 \left (1+c^2 x^2\right )} \]
(d*Sqrt[d + c^2*d*x^2]*(225*a^2*(1 + c^2*x^2)^3 - 30*a*b*c*x*Sqrt[1 + c^2* x^2]*(15 + 10*c^2*x^2 + 3*c^4*x^4) + 2*b^2*(149 + 187*c^2*x^2 + 47*c^4*x^4 + 9*c^6*x^6) + 30*b*(15*a*(1 + c^2*x^2)^3 - b*c*x*Sqrt[1 + c^2*x^2]*(15 + 10*c^2*x^2 + 3*c^4*x^4))*ArcSinh[c*x] + 225*b^2*(1 + c^2*x^2)^3*ArcSinh[c *x]^2))/(1125*c^2*(1 + c^2*x^2))
Time = 0.57 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.69, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6213, 6199, 27, 1576, 1140, 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.
| \(\displaystyle \int x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx\) |
| \(\Big \downarrow \) 6213 |
| \(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \int \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))dx}{5 c \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 6199 |
| \(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-b c \int \frac {x \left (3 c^4 x^4+10 c^2 x^2+15\right )}{15 \sqrt {c^2 x^2+1}}dx+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-\frac {1}{15} b c \int \frac {x \left (3 c^4 x^4+10 c^2 x^2+15\right )}{\sqrt {c^2 x^2+1}}dx+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-\frac {1}{30} b c \int \frac {3 c^4 x^4+10 c^2 x^2+15}{\sqrt {c^2 x^2+1}}dx^2+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-\frac {1}{30} b c \int \left (3 \left (c^2 x^2+1\right )^{3/2}+4 \sqrt {c^2 x^2+1}+\frac {8}{\sqrt {c^2 x^2+1}}\right )dx^2+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))-\frac {1}{30} b c \left (\frac {6 \left (c^2 x^2+1\right )^{5/2}}{5 c^2}+\frac {8 \left (c^2 x^2+1\right )^{3/2}}{3 c^2}+\frac {16 \sqrt {c^2 x^2+1}}{c^2}\right )\right )}{5 c \sqrt {c^2 x^2+1}}\) |
((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/(5*c^2*d) - (2*b*d*Sqrt[d + c^2*d*x^2]*(-1/30*(b*c*((16*Sqrt[1 + c^2*x^2])/c^2 + (8*(1 + c^2*x^2)^(3/ 2))/(3*c^2) + (6*(1 + c^2*x^2)^(5/2))/(5*c^2))) + x*(a + b*ArcSinh[c*x]) + (2*c^2*x^3*(a + b*ArcSinh[c*x]))/3 + (c^4*x^5*(a + b*ArcSinh[c*x]))/5))/( 5*c*Sqrt[1 + c^2*x^2])
3.3.68.3.1 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[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(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[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symb ol] :> With[{u = IntHide[(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}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p] Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[ {a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1148\) vs. \(2(233)=466\).
Time = 0.37 (sec) , antiderivative size = 1149, normalized size of antiderivative = 4.30
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1149\) |
| parts | \(\text {Expression too large to display}\) | \(1149\) |
1/5*a^2*(c^2*d*x^2+d)^(5/2)/c^2/d+b^2*(1/4000*(d*(c^2*x^2+1))^(1/2)*(16*c^ 6*x^6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(1/2) +13*c^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+1)*(25*arcsinh(c*x)^2-10*arcsinh(c*x)+ 2)*d/c^2/(c^2*x^2+1)+1/288*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*c^3*x^3*(c^2 *x^2+1)^(1/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1/2)+1)*(9*arcsinh(c*x)^2-6*arc sinh(c*x)+2)*d/c^2/(c^2*x^2+1)+1/16*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^ 2*x^2+1)^(1/2)+1)*(arcsinh(c*x)^2-2*arcsinh(c*x)+2)*d/c^2/(c^2*x^2+1)+1/16 *(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(arcsinh(c*x)^2+2 *arcsinh(c*x)+2)*d/c^2/(c^2*x^2+1)+1/288*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4- 4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2-3*c*x*(c^2*x^2+1)^(1/2)+1)*(9*arcsin h(c*x)^2+6*arcsinh(c*x)+2)*d/c^2/(c^2*x^2+1)+1/4000*(d*(c^2*x^2+1))^(1/2)* (16*c^6*x^6-16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4-20*c^3*x^3*(c^2*x^2+1) ^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^2+1)^(1/2)+1)*(25*arcsinh(c*x)^2+10*arcsinh (c*x)+2)*d/c^2/(c^2*x^2+1))+2*a*b*(1/800*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6 +16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c ^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+5*arcsinh(c*x))*d/c^2/(c^2*x^2+1)+1/ 96*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2+ 3*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+3*arcsinh(c*x))*d/c^2/(c^2*x^2+1)+1/16*(d*( c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(-1+arcsinh(c*x))*d/c^ 2/(c^2*x^2+1)+1/16*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)...
Time = 0.27 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.24 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {225 \, {\left (b^{2} c^{6} d x^{6} + 3 \, b^{2} c^{4} d x^{4} + 3 \, b^{2} c^{2} d x^{2} + b^{2} d\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 30 \, {\left (15 \, a b c^{6} d x^{6} + 45 \, a b c^{4} d x^{4} + 45 \, a b c^{2} d x^{2} + 15 \, a b d - {\left (3 \, b^{2} c^{5} d x^{5} + 10 \, b^{2} c^{3} d x^{3} + 15 \, b^{2} c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (9 \, {\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{6} d x^{6} + {\left (675 \, a^{2} + 94 \, b^{2}\right )} c^{4} d x^{4} + {\left (675 \, a^{2} + 374 \, b^{2}\right )} c^{2} d x^{2} + {\left (225 \, a^{2} + 298 \, b^{2}\right )} d - 30 \, {\left (3 \, a b c^{5} d x^{5} + 10 \, a b c^{3} d x^{3} + 15 \, a b c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d}}{1125 \, {\left (c^{4} x^{2} + c^{2}\right )}} \]
1/1125*(225*(b^2*c^6*d*x^6 + 3*b^2*c^4*d*x^4 + 3*b^2*c^2*d*x^2 + b^2*d)*sq rt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 30*(15*a*b*c^6*d*x^6 + 45*a*b*c^4*d*x^4 + 45*a*b*c^2*d*x^2 + 15*a*b*d - (3*b^2*c^5*d*x^5 + 10*b^2 *c^3*d*x^3 + 15*b^2*c*d*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + (9*(25*a^2 + 2*b^2)*c^6*d*x^6 + (675*a^2 + 94*b^2)* c^4*d*x^4 + (675*a^2 + 374*b^2)*c^2*d*x^2 + (225*a^2 + 298*b^2)*d - 30*(3* a*b*c^5*d*x^5 + 10*a*b*c^3*d*x^3 + 15*a*b*c*d*x)*sqrt(c^2*x^2 + 1))*sqrt(c ^2*d*x^2 + d))/(c^4*x^2 + c^2)
\[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}\, dx \]
Time = 0.23 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.86 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b^{2} \operatorname {arsinh}\left (c x\right )^{2}}{5 \, c^{2} d} + \frac {2}{1125} \, b^{2} {\left (\frac {9 \, \sqrt {c^{2} x^{2} + 1} c^{2} d^{\frac {5}{2}} x^{4} + 38 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {5}{2}} x^{2} + \frac {149 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {5}{2}}}{c^{2}}}{d} - \frac {15 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} + 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} \operatorname {arsinh}\left (c x\right )}{c d}\right )} + \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a b \operatorname {arsinh}\left (c x\right )}{5 \, c^{2} d} + \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a^{2}}{5 \, c^{2} d} - \frac {2 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} + 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} a b}{75 \, c d} \]
1/5*(c^2*d*x^2 + d)^(5/2)*b^2*arcsinh(c*x)^2/(c^2*d) + 2/1125*b^2*((9*sqrt (c^2*x^2 + 1)*c^2*d^(5/2)*x^4 + 38*sqrt(c^2*x^2 + 1)*d^(5/2)*x^2 + 149*sqr t(c^2*x^2 + 1)*d^(5/2)/c^2)/d - 15*(3*c^4*d^(5/2)*x^5 + 10*c^2*d^(5/2)*x^3 + 15*d^(5/2)*x)*arcsinh(c*x)/(c*d)) + 2/5*(c^2*d*x^2 + d)^(5/2)*a*b*arcsi nh(c*x)/(c^2*d) + 1/5*(c^2*d*x^2 + d)^(5/2)*a^2/(c^2*d) - 2/75*(3*c^4*d^(5 /2)*x^5 + 10*c^2*d^(5/2)*x^3 + 15*d^(5/2)*x)*a*b/(c*d)
Exception generated. \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\text {Exception raised: TypeError} \]
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. \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \]