Integrand size = 20, antiderivative size = 87 \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=-\frac {3 b d x \sqrt {1+c^2 x^2}}{32 c}-\frac {b d x \left (1+c^2 x^2\right )^{3/2}}{16 c}-\frac {3 b d \text {arcsinh}(c x)}{32 c^2}+\frac {d \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{4 c^2} \] Output:
-3/32*b*d*x*(c^2*x^2+1)^(1/2)/c-1/16*b*d*x*(c^2*x^2+1)^(3/2)/c-3/32*b*d*ar csinh(c*x)/c^2+1/4*d*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))/c^2
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89 \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {d \left (c x \left (8 a c x \left (2+c^2 x^2\right )-b \sqrt {1+c^2 x^2} \left (5+2 c^2 x^2\right )\right )+b \left (5+16 c^2 x^2+8 c^4 x^4\right ) \text {arcsinh}(c x)\right )}{32 c^2} \] Input:
Integrate[x*(d + c^2*d*x^2)*(a + b*ArcSinh[c*x]),x]
Output:
(d*(c*x*(8*a*c*x*(2 + c^2*x^2) - b*Sqrt[1 + c^2*x^2]*(5 + 2*c^2*x^2)) + b* (5 + 16*c^2*x^2 + 8*c^4*x^4)*ArcSinh[c*x]))/(32*c^2)
Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6213, 211, 211, 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.
\(\displaystyle \int x \left (c^2 d x^2+d\right ) (a+b \text {arcsinh}(c x)) \, dx\) |
\(\Big \downarrow \) 6213 |
\(\displaystyle \frac {d \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))}{4 c^2}-\frac {b d \int \left (c^2 x^2+1\right )^{3/2}dx}{4 c}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {d \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))}{4 c^2}-\frac {b d \left (\frac {3}{4} \int \sqrt {c^2 x^2+1}dx+\frac {1}{4} x \left (c^2 x^2+1\right )^{3/2}\right )}{4 c}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {d \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))}{4 c^2}-\frac {b d \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {c^2 x^2+1}}dx+\frac {1}{2} x \sqrt {c^2 x^2+1}\right )+\frac {1}{4} x \left (c^2 x^2+1\right )^{3/2}\right )}{4 c}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {d \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))}{4 c^2}-\frac {b d \left (\frac {3}{4} \left (\frac {\text {arcsinh}(c x)}{2 c}+\frac {1}{2} x \sqrt {c^2 x^2+1}\right )+\frac {1}{4} x \left (c^2 x^2+1\right )^{3/2}\right )}{4 c}\) |
Input:
Int[x*(d + c^2*d*x^2)*(a + b*ArcSinh[c*x]),x]
Output:
(d*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(4*c^2) - (b*d*((x*(1 + c^2*x^2)^ (3/2))/4 + (3*((x*Sqrt[1 + c^2*x^2])/2 + ArcSinh[c*x]/(2*c)))/4))/(4*c)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((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]
Time = 0.68 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(\frac {\frac {a d \left (c^{2} x^{2}+1\right )^{2}}{4}+b d \left (\frac {\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}}{2}+\frac {5 \,\operatorname {arcsinh}\left (x c \right )}{32}-\frac {x c \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}{16}-\frac {3 \sqrt {c^{2} x^{2}+1}\, x c}{32}\right )}{c^{2}}\) | \(85\) |
default | \(\frac {\frac {a d \left (c^{2} x^{2}+1\right )^{2}}{4}+b d \left (\frac {\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}}{2}+\frac {5 \,\operatorname {arcsinh}\left (x c \right )}{32}-\frac {x c \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}{16}-\frac {3 \sqrt {c^{2} x^{2}+1}\, x c}{32}\right )}{c^{2}}\) | \(85\) |
parts | \(\frac {a d \left (c^{2} x^{2}+1\right )^{2}}{4 c^{2}}+\frac {b d \left (\frac {\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}}{2}+\frac {5 \,\operatorname {arcsinh}\left (x c \right )}{32}-\frac {x c \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}{16}-\frac {3 \sqrt {c^{2} x^{2}+1}\, x c}{32}\right )}{c^{2}}\) | \(87\) |
orering | \(\frac {\left (14 c^{4} x^{4}+33 c^{2} x^{2}+10\right ) \left (c^{2} d \,x^{2}+d \right ) \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{32 c^{2} \left (c^{2} x^{2}+1\right )}-\frac {\left (2 c^{2} x^{2}+5\right ) \left (\left (c^{2} d \,x^{2}+d \right ) \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )+2 x^{2} c^{2} d \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )+\frac {x \left (c^{2} d \,x^{2}+d \right ) b c}{\sqrt {c^{2} x^{2}+1}}\right )}{32 c^{2}}\) | \(131\) |
Input:
int(x*(c^2*d*x^2+d)*(a+b*arcsinh(x*c)),x,method=_RETURNVERBOSE)
Output:
1/c^2*(1/4*a*d*(c^2*x^2+1)^2+b*d*(1/4*arcsinh(x*c)*c^4*x^4+1/2*arcsinh(x*c )*c^2*x^2+5/32*arcsinh(x*c)-1/16*x*c*(c^2*x^2+1)^(3/2)-3/32*(c^2*x^2+1)^(1 /2)*x*c))
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13 \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {8 \, a c^{4} d x^{4} + 16 \, a c^{2} d x^{2} + {\left (8 \, b c^{4} d x^{4} + 16 \, b c^{2} d x^{2} + 5 \, b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (2 \, b c^{3} d x^{3} + 5 \, b c d x\right )} \sqrt {c^{2} x^{2} + 1}}{32 \, c^{2}} \] Input:
integrate(x*(c^2*d*x^2+d)*(a+b*arcsinh(c*x)),x, algorithm="fricas")
Output:
1/32*(8*a*c^4*d*x^4 + 16*a*c^2*d*x^2 + (8*b*c^4*d*x^4 + 16*b*c^2*d*x^2 + 5 *b*d)*log(c*x + sqrt(c^2*x^2 + 1)) - (2*b*c^3*d*x^3 + 5*b*c*d*x)*sqrt(c^2* x^2 + 1))/c^2
Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.34 \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\begin {cases} \frac {a c^{2} d x^{4}}{4} + \frac {a d x^{2}}{2} + \frac {b c^{2} d x^{4} \operatorname {asinh}{\left (c x \right )}}{4} - \frac {b c d x^{3} \sqrt {c^{2} x^{2} + 1}}{16} + \frac {b d x^{2} \operatorname {asinh}{\left (c x \right )}}{2} - \frac {5 b d x \sqrt {c^{2} x^{2} + 1}}{32 c} + \frac {5 b d \operatorname {asinh}{\left (c x \right )}}{32 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d x^{2}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*(c**2*d*x**2+d)*(a+b*asinh(c*x)),x)
Output:
Piecewise((a*c**2*d*x**4/4 + a*d*x**2/2 + b*c**2*d*x**4*asinh(c*x)/4 - b*c *d*x**3*sqrt(c**2*x**2 + 1)/16 + b*d*x**2*asinh(c*x)/2 - 5*b*d*x*sqrt(c**2 *x**2 + 1)/(32*c) + 5*b*d*asinh(c*x)/(32*c**2), Ne(c, 0)), (a*d*x**2/2, Tr ue))
Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.46 \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {1}{4} \, a c^{2} d x^{4} + \frac {1}{32} \, {\left (8 \, x^{4} \operatorname {arsinh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac {3 \, \sqrt {c^{2} x^{2} + 1} x}{c^{4}} + \frac {3 \, \operatorname {arsinh}\left (c x\right )}{c^{5}}\right )} c\right )} b c^{2} d + \frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {arsinh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{c^{3}}\right )}\right )} b d \] Input:
integrate(x*(c^2*d*x^2+d)*(a+b*arcsinh(c*x)),x, algorithm="maxima")
Output:
1/4*a*c^2*d*x^4 + 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*c^2*d + 1/2*a*d*x^2 + 1/4*(2*x^2*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x/c^2 - arcsinh(c*x)/c^3 ))*b*d
Exception generated. \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x*(c^2*d*x^2+d)*(a+b*arcsinh(c*x)),x, algorithm="giac")
Output:
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. \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\int x\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\left (d\,c^2\,x^2+d\right ) \,d x \] Input:
int(x*(a + b*asinh(c*x))*(d + c^2*d*x^2),x)
Output:
int(x*(a + b*asinh(c*x))*(d + c^2*d*x^2), x)
Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.18 \[ \int x \left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x)) \, dx=\frac {d \left (8 \mathit {asinh} \left (c x \right ) b \,c^{4} x^{4}+16 \mathit {asinh} \left (c x \right ) b \,c^{2} x^{2}-2 \sqrt {c^{2} x^{2}+1}\, b \,c^{3} x^{3}-5 \sqrt {c^{2} x^{2}+1}\, b c x +5 \,\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x \right ) b +8 a \,c^{4} x^{4}+16 a \,c^{2} x^{2}\right )}{32 c^{2}} \] Input:
int(x*(c^2*d*x^2+d)*(a+b*asinh(c*x)),x)
Output:
(d*(8*asinh(c*x)*b*c**4*x**4 + 16*asinh(c*x)*b*c**2*x**2 - 2*sqrt(c**2*x** 2 + 1)*b*c**3*x**3 - 5*sqrt(c**2*x**2 + 1)*b*c*x + 5*log(sqrt(c**2*x**2 + 1) + c*x)*b + 8*a*c**4*x**4 + 16*a*c**2*x**2))/(32*c**2)