Integrand size = 23, antiderivative size = 180 \[ \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {5 b c d x^2 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}-\frac {b c^3 d x^4 \sqrt {d+c^2 d x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {3}{8} d x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))+\frac {1}{4} x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {3 d \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{16 b c \sqrt {1+c^2 x^2}} \]
1/4*x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))+3/8*d*x*(a+b*arcsinh(c*x))*(c ^2*d*x^2+d)^(1/2)-5/16*b*c*d*x^2*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/1 6*b*c^3*d*x^4*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+3/16*d*(a+b*arcsinh(c* x))^2*(c^2*d*x^2+d)^(1/2)/b/c/(c^2*x^2+1)^(1/2)
Time = 0.70 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.11 \[ \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {1}{8} a d x \left (5+2 c^2 x^2\right ) \sqrt {d+c^2 d x^2}+\frac {3 a d^{3/2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{8 c}+\frac {b d \sqrt {d+c^2 d x^2} (-\cosh (2 \text {arcsinh}(c x))+2 \text {arcsinh}(c x) (\text {arcsinh}(c x)+\sinh (2 \text {arcsinh}(c x))))}{8 c \sqrt {1+c^2 x^2}}-\frac {b d \sqrt {d+c^2 d x^2} \left (8 \text {arcsinh}(c x)^2+\cosh (4 \text {arcsinh}(c x))-4 \text {arcsinh}(c x) \sinh (4 \text {arcsinh}(c x))\right )}{128 c \sqrt {1+c^2 x^2}} \]
(a*d*x*(5 + 2*c^2*x^2)*Sqrt[d + c^2*d*x^2])/8 + (3*a*d^(3/2)*Log[c*d*x + S qrt[d]*Sqrt[d + c^2*d*x^2]])/(8*c) + (b*d*Sqrt[d + c^2*d*x^2]*(-Cosh[2*Arc Sinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]])))/(8*c*S qrt[1 + c^2*x^2]) - (b*d*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*Ar cSinh[c*x]] - 4*ArcSinh[c*x]*Sinh[4*ArcSinh[c*x]]))/(128*c*Sqrt[1 + c^2*x^ 2])
Time = 0.56 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6201, 244, 2009, 6200, 15, 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.
\(\displaystyle \int \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx\) |
\(\Big \downarrow \) 6201 |
\(\displaystyle \frac {3}{4} d \int \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))dx-\frac {b c d \sqrt {c^2 d x^2+d} \int x \left (c^2 x^2+1\right )dx}{4 \sqrt {c^2 x^2+1}}+\frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {3}{4} d \int \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))dx-\frac {b c d \sqrt {c^2 d x^2+d} \int \left (c^2 x^3+x\right )dx}{4 \sqrt {c^2 x^2+1}}+\frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{4} d \int \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))dx+\frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {b c d \left (\frac {c^2 x^4}{4}+\frac {x^2}{2}\right ) \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 6200 |
\(\displaystyle \frac {3}{4} d \left (\frac {\sqrt {c^2 d x^2+d} \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx}{2 \sqrt {c^2 x^2+1}}-\frac {b c \sqrt {c^2 d x^2+d} \int xdx}{2 \sqrt {c^2 x^2+1}}+\frac {1}{2} x \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))\right )+\frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {b c d \left (\frac {c^2 x^4}{4}+\frac {x^2}{2}\right ) \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {3}{4} d \left (\frac {\sqrt {c^2 d x^2+d} \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx}{2 \sqrt {c^2 x^2+1}}+\frac {1}{2} x \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))-\frac {b c x^2 \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}\right )+\frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))-\frac {b c d \left (\frac {c^2 x^4}{4}+\frac {x^2}{2}\right ) \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 6198 |
\(\displaystyle \frac {1}{4} x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {3}{4} d \left (\frac {1}{2} x \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))+\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2}{4 b c \sqrt {c^2 x^2+1}}-\frac {b c x^2 \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}\right )-\frac {b c d \left (\frac {c^2 x^4}{4}+\frac {x^2}{2}\right ) \sqrt {c^2 d x^2+d}}{4 \sqrt {c^2 x^2+1}}\) |
-1/4*(b*c*d*Sqrt[d + c^2*d*x^2]*(x^2/2 + (c^2*x^4)/4))/Sqrt[1 + c^2*x^2] + (x*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/4 + (3*d*(-1/4*(b*c*x^2*Sq rt[d + c^2*d*x^2])/Sqrt[1 + c^2*x^2] + (x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSi nh[c*x]))/2 + (Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(4*b*c*Sqrt[1 + c^2*x^2])))/4
3.2.31.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_ Symbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSinh[c*x])^n/2), x] + (Simp[(1 /2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]] Int[(a + b*ArcSinh[c*x])^n/Sq rt[1 + c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2* x^2]] Int[x*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e }, x] && EqQ[e, c^2*d] && GtQ[n, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[x*(d + e*x^2)^p*((a + b*ArcSinh[c*x])^n/(2*p + 1)), x] + (Simp[2*d*(p/(2*p + 1)) Int[(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x ], x] - Simp[b*c*(n/(2*p + 1))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p] Int[x* (1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(495\) vs. \(2(154)=308\).
Time = 0.18 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.76
method | result | size |
default | \(\frac {x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} a}{4}+\frac {3 a d x \sqrt {c^{2} d \,x^{2}+d}}{8}+\frac {3 a \,d^{2} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{8 \sqrt {c^{2} d}}+b \left (\frac {3 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )^{2} d}{16 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}+8 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+12 c^{3} x^{3}+8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+4 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+2 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{16 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+2 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{16 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}-8 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+12 c^{3} x^{3}-8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+4 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}+1\right )}\right )\) | \(496\) |
parts | \(\frac {x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} a}{4}+\frac {3 a d x \sqrt {c^{2} d \,x^{2}+d}}{8}+\frac {3 a \,d^{2} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{8 \sqrt {c^{2} d}}+b \left (\frac {3 \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )^{2} d}{16 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}+8 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+12 c^{3} x^{3}+8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+4 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+2 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{16 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+2 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{16 c \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (8 c^{5} x^{5}-8 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+12 c^{3} x^{3}-8 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+4 \,\operatorname {arcsinh}\left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}+1\right )}\right )\) | \(496\) |
1/4*x*(c^2*d*x^2+d)^(3/2)*a+3/8*a*d*x*(c^2*d*x^2+d)^(1/2)+3/8*a*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)+b*(3/16*(d*(c^2*x^2+ 1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^2*d+1/256*(d*(c^2*x^2+1))^(1/2) *(8*c^5*x^5+8*c^4*x^4*(c^2*x^2+1)^(1/2)+12*c^3*x^3+8*c^2*x^2*(c^2*x^2+1)^( 1/2)+4*c*x+(c^2*x^2+1)^(1/2))*(-1+4*arcsinh(c*x))*d/c/(c^2*x^2+1)+1/16*(d* (c^2*x^2+1))^(1/2)*(2*c^3*x^3+2*c^2*x^2*(c^2*x^2+1)^(1/2)+2*c*x+(c^2*x^2+1 )^(1/2))*(-1+2*arcsinh(c*x))*d/c/(c^2*x^2+1)+1/16*(d*(c^2*x^2+1))^(1/2)*(2 *c^3*x^3-2*c^2*x^2*(c^2*x^2+1)^(1/2)+2*c*x-(c^2*x^2+1)^(1/2))*(1+2*arcsinh (c*x))*d/c/(c^2*x^2+1)+1/256*(d*(c^2*x^2+1))^(1/2)*(8*c^5*x^5-8*c^4*x^4*(c ^2*x^2+1)^(1/2)+12*c^3*x^3-8*c^2*x^2*(c^2*x^2+1)^(1/2)+4*c*x-(c^2*x^2+1)^( 1/2))*(1+4*arcsinh(c*x))*d/c/(c^2*x^2+1))
\[ \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} \,d x } \]
\[ \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )\, dx \]
Exception generated. \[ \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \]