Integrand size = 22, antiderivative size = 120 \[ \int x^3 \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {b d x \sqrt {1+c^2 x^2}}{24 c^3}-\frac {b d x^3 \sqrt {1+c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1+c^2 x^2}-\frac {b d \text {arcsinh}(c x)}{24 c^4}+\frac {1}{4} d x^4 (a+b \text {arcsinh}(c x))+\frac {1}{6} c^2 d x^6 (a+b \text {arcsinh}(c x)) \]
-1/24*b*d*arcsinh(c*x)/c^4+1/4*d*x^4*(a+b*arcsinh(c*x))+1/6*c^2*d*x^6*(a+b *arcsinh(c*x))+1/24*b*d*x*(c^2*x^2+1)^(1/2)/c^3-1/36*b*d*x^3*(c^2*x^2+1)^( 1/2)/c-1/36*b*c*d*x^5*(c^2*x^2+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.73 \[ \int x^3 \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {d \left (6 a c^4 x^4 \left (3+2 c^2 x^2\right )+b c x \sqrt {1+c^2 x^2} \left (3-2 c^2 x^2-2 c^4 x^4\right )+3 b \left (-1+6 c^4 x^4+4 c^6 x^6\right ) \text {arcsinh}(c x)\right )}{72 c^4} \]
(d*(6*a*c^4*x^4*(3 + 2*c^2*x^2) + b*c*x*Sqrt[1 + c^2*x^2]*(3 - 2*c^2*x^2 - 2*c^4*x^4) + 3*b*(-1 + 6*c^4*x^4 + 4*c^6*x^6)*ArcSinh[c*x]))/(72*c^4)
Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6218, 27, 363, 262, 262, 222}
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.
\(\displaystyle \int x^3 \left (c^2 d x^2+d\right ) (a+b \text {arcsinh}(c x)) \, dx\) |
\(\Big \downarrow \) 6218 |
\(\displaystyle -b c \int \frac {d x^4 \left (2 c^2 x^2+3\right )}{12 \sqrt {c^2 x^2+1}}dx+\frac {1}{6} c^2 d x^6 (a+b \text {arcsinh}(c x))+\frac {1}{4} d x^4 (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{12} b c d \int \frac {x^4 \left (2 c^2 x^2+3\right )}{\sqrt {c^2 x^2+1}}dx+\frac {1}{6} c^2 d x^6 (a+b \text {arcsinh}(c x))+\frac {1}{4} d x^4 (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 363 |
\(\displaystyle -\frac {1}{12} b c d \left (\frac {4}{3} \int \frac {x^4}{\sqrt {c^2 x^2+1}}dx+\frac {1}{3} x^5 \sqrt {c^2 x^2+1}\right )+\frac {1}{6} c^2 d x^6 (a+b \text {arcsinh}(c x))+\frac {1}{4} d x^4 (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {1}{12} b c d \left (\frac {4}{3} \left (\frac {x^3 \sqrt {c^2 x^2+1}}{4 c^2}-\frac {3 \int \frac {x^2}{\sqrt {c^2 x^2+1}}dx}{4 c^2}\right )+\frac {1}{3} x^5 \sqrt {c^2 x^2+1}\right )+\frac {1}{6} c^2 d x^6 (a+b \text {arcsinh}(c x))+\frac {1}{4} d x^4 (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {1}{12} b c d \left (\frac {4}{3} \left (\frac {x^3 \sqrt {c^2 x^2+1}}{4 c^2}-\frac {3 \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\int \frac {1}{\sqrt {c^2 x^2+1}}dx}{2 c^2}\right )}{4 c^2}\right )+\frac {1}{3} x^5 \sqrt {c^2 x^2+1}\right )+\frac {1}{6} c^2 d x^6 (a+b \text {arcsinh}(c x))+\frac {1}{4} d x^4 (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{6} c^2 d x^6 (a+b \text {arcsinh}(c x))+\frac {1}{4} d x^4 (a+b \text {arcsinh}(c x))-\frac {1}{12} b c d \left (\frac {4}{3} \left (\frac {x^3 \sqrt {c^2 x^2+1}}{4 c^2}-\frac {3 \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{4 c^2}\right )+\frac {1}{3} x^5 \sqrt {c^2 x^2+1}\right )\) |
(d*x^4*(a + b*ArcSinh[c*x]))/4 + (c^2*d*x^6*(a + b*ArcSinh[c*x]))/6 - (b*c *d*((x^5*Sqrt[1 + c^2*x^2])/3 + (4*((x^3*Sqrt[1 + c^2*x^2])/(4*c^2) - (3*( (x*Sqrt[1 + c^2*x^2])/(2*c^2) - ArcSinh[c*x]/(2*c^3)))/(4*c^2)))/3))/12
3.1.2.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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 = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.91
method | result | size |
parts | \(d a \left (\frac {1}{6} c^{2} x^{6}+\frac {1}{4} x^{4}\right )+\frac {d b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {\operatorname {arcsinh}\left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{5} x^{5} \sqrt {c^{2} x^{2}+1}}{36}-\frac {c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{36}+\frac {c x \sqrt {c^{2} x^{2}+1}}{24}-\frac {\operatorname {arcsinh}\left (c x \right )}{24}\right )}{c^{4}}\) | \(109\) |
derivativedivides | \(\frac {d a \left (\frac {1}{6} c^{6} x^{6}+\frac {1}{4} c^{4} x^{4}\right )+d b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {\operatorname {arcsinh}\left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{5} x^{5} \sqrt {c^{2} x^{2}+1}}{36}-\frac {c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{36}+\frac {c x \sqrt {c^{2} x^{2}+1}}{24}-\frac {\operatorname {arcsinh}\left (c x \right )}{24}\right )}{c^{4}}\) | \(113\) |
default | \(\frac {d a \left (\frac {1}{6} c^{6} x^{6}+\frac {1}{4} c^{4} x^{4}\right )+d b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{6} x^{6}}{6}+\frac {\operatorname {arcsinh}\left (c x \right ) c^{4} x^{4}}{4}-\frac {c^{5} x^{5} \sqrt {c^{2} x^{2}+1}}{36}-\frac {c^{3} x^{3} \sqrt {c^{2} x^{2}+1}}{36}+\frac {c x \sqrt {c^{2} x^{2}+1}}{24}-\frac {\operatorname {arcsinh}\left (c x \right )}{24}\right )}{c^{4}}\) | \(113\) |
d*a*(1/6*c^2*x^6+1/4*x^4)+d*b/c^4*(1/6*arcsinh(c*x)*c^6*x^6+1/4*arcsinh(c* x)*c^4*x^4-1/36*c^5*x^5*(c^2*x^2+1)^(1/2)-1/36*c^3*x^3*(c^2*x^2+1)^(1/2)+1 /24*c*x*(c^2*x^2+1)^(1/2)-1/24*arcsinh(c*x))
Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.91 \[ \int x^3 \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {12 \, a c^{6} d x^{6} + 18 \, a c^{4} d x^{4} + 3 \, {\left (4 \, b c^{6} d x^{6} + 6 \, b c^{4} d x^{4} - b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (2 \, b c^{5} d x^{5} + 2 \, b c^{3} d x^{3} - 3 \, b c d x\right )} \sqrt {c^{2} x^{2} + 1}}{72 \, c^{4}} \]
1/72*(12*a*c^6*d*x^6 + 18*a*c^4*d*x^4 + 3*(4*b*c^6*d*x^6 + 6*b*c^4*d*x^4 - b*d)*log(c*x + sqrt(c^2*x^2 + 1)) - (2*b*c^5*d*x^5 + 2*b*c^3*d*x^3 - 3*b* c*d*x)*sqrt(c^2*x^2 + 1))/c^4
Time = 0.60 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.15 \[ \int x^3 \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\begin {cases} \frac {a c^{2} d x^{6}}{6} + \frac {a d x^{4}}{4} + \frac {b c^{2} d x^{6} \operatorname {asinh}{\left (c x \right )}}{6} - \frac {b c d x^{5} \sqrt {c^{2} x^{2} + 1}}{36} + \frac {b d x^{4} \operatorname {asinh}{\left (c x \right )}}{4} - \frac {b d x^{3} \sqrt {c^{2} x^{2} + 1}}{36 c} + \frac {b d x \sqrt {c^{2} x^{2} + 1}}{24 c^{3}} - \frac {b d \operatorname {asinh}{\left (c x \right )}}{24 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((a*c**2*d*x**6/6 + a*d*x**4/4 + b*c**2*d*x**6*asinh(c*x)/6 - b*c *d*x**5*sqrt(c**2*x**2 + 1)/36 + b*d*x**4*asinh(c*x)/4 - b*d*x**3*sqrt(c** 2*x**2 + 1)/(36*c) + b*d*x*sqrt(c**2*x**2 + 1)/(24*c**3) - b*d*asinh(c*x)/ (24*c**4), Ne(c, 0)), (a*d*x**4/4, True))
Time = 0.20 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.38 \[ \int x^3 \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {1}{6} \, a c^{2} d x^{6} + \frac {1}{4} \, a d x^{4} + \frac {1}{288} \, {\left (48 \, x^{6} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {8 \, \sqrt {c^{2} x^{2} + 1} x^{5}}{c^{2}} - \frac {10 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {c^{2} x^{2} + 1} x}{c^{6}} - \frac {15 \, \operatorname {arsinh}\left (c x\right )}{c^{7}}\right )} c\right )} b c^{2} d + \frac {1}{32} \, {\left (8 \, x^{4} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac {3 \, \sqrt {c^{2} x^{2} + 1} x}{c^{4}} + \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5}}\right )} c\right )} b d \]
1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 + 1/288*(48*x^6*arcsinh(c*x) - (8*sqrt(c^2*x ^2 + 1)*x^5/c^2 - 10*sqrt(c^2*x^2 + 1)*x^3/c^4 + 15*sqrt(c^2*x^2 + 1)*x/c^ 6 - 15*arcsinh(c*x)/c^7)*c)*b*c^2*d + 1/32*(8*x^4*arcsinh(c*x) - (2*sqrt(c ^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c*x)/c^5)*c)*b *d
Exception generated. \[ \int x^3 \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \]
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. \[ \int x^3 \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\int x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\left (d\,c^2\,x^2+d\right ) \,d x \]