Integrand size = 24, antiderivative size = 135 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=-\frac {b x \sqrt {1+c^2 x^2}}{4 c^3 d}+\frac {b \text {arcsinh}(c x)}{4 c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}+\frac {(a+b \text {arcsinh}(c x))^2}{2 b c^4 d}-\frac {(a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{c^4 d}-\frac {b \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{2 c^4 d} \]
1/4*b*arcsinh(c*x)/c^4/d+1/2*x^2*(a+b*arcsinh(c*x))/c^2/d+1/2*(a+b*arcsinh (c*x))^2/b/c^4/d-(a+b*arcsinh(c*x))*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d- 1/2*b*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d-1/4*b*x*(c^2*x^2+1)^(1/2 )/c^3/d
Time = 0.16 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=-\frac {-2 a c^2 x^2+b c x \sqrt {1+c^2 x^2}-b \text {arcsinh}(c x)-2 b c^2 x^2 \text {arcsinh}(c x)-2 b \text {arcsinh}(c x)^2+4 b \text {arcsinh}(c x) \log \left (1+\frac {c e^{\text {arcsinh}(c x)}}{\sqrt {-c^2}}\right )+4 b \text {arcsinh}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\text {arcsinh}(c x)}}{c}\right )+2 a \log \left (1+c^2 x^2\right )+4 b \operatorname {PolyLog}\left (2,\frac {c e^{\text {arcsinh}(c x)}}{\sqrt {-c^2}}\right )+4 b \operatorname {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\text {arcsinh}(c x)}}{c}\right )}{4 c^4 d} \]
-1/4*(-2*a*c^2*x^2 + b*c*x*Sqrt[1 + c^2*x^2] - b*ArcSinh[c*x] - 2*b*c^2*x^ 2*ArcSinh[c*x] - 2*b*ArcSinh[c*x]^2 + 4*b*ArcSinh[c*x]*Log[1 + (c*E^ArcSin h[c*x])/Sqrt[-c^2]] + 4*b*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x]) /c] + 2*a*Log[1 + c^2*x^2] + 4*b*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 4*b*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c])/(c^4*d)
Result contains complex when optimal does not.
Time = 0.66 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {6227, 27, 262, 222, 6212, 3042, 26, 4201, 2620, 2715, 2838}
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.
\(\displaystyle \int \frac {x^3 (a+b \text {arcsinh}(c x))}{c^2 d x^2+d} \, dx\) |
\(\Big \downarrow \) 6227 |
\(\displaystyle -\frac {\int \frac {x (a+b \text {arcsinh}(c x))}{d \left (c^2 x^2+1\right )}dx}{c^2}-\frac {b \int \frac {x^2}{\sqrt {c^2 x^2+1}}dx}{2 c d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {x (a+b \text {arcsinh}(c x))}{c^2 x^2+1}dx}{c^2 d}-\frac {b \int \frac {x^2}{\sqrt {c^2 x^2+1}}dx}{2 c d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {\int \frac {x (a+b \text {arcsinh}(c x))}{c^2 x^2+1}dx}{c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\int \frac {1}{\sqrt {c^2 x^2+1}}dx}{2 c^2}\right )}{2 c d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle -\frac {\int \frac {x (a+b \text {arcsinh}(c x))}{c^2 x^2+1}dx}{c^2 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
\(\Big \downarrow \) 6212 |
\(\displaystyle -\frac {\int \frac {c x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}d\text {arcsinh}(c x)}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int -i (a+b \text {arcsinh}(c x)) \tan (i \text {arcsinh}(c x))d\text {arcsinh}(c x)}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \int (a+b \text {arcsinh}(c x)) \tan (i \text {arcsinh}(c x))d\text {arcsinh}(c x)}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle \frac {i \left (2 i \int \frac {e^{2 \text {arcsinh}(c x)} (a+b \text {arcsinh}(c x))}{1+e^{2 \text {arcsinh}(c x)}}d\text {arcsinh}(c x)-\frac {i (a+b \text {arcsinh}(c x))^2}{2 b}\right )}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))-\frac {1}{2} b \int \log \left (1+e^{2 \text {arcsinh}(c x)}\right )d\text {arcsinh}(c x)\right )-\frac {i (a+b \text {arcsinh}(c x))^2}{2 b}\right )}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))-\frac {1}{4} b \int e^{-2 \text {arcsinh}(c x)} \log \left (1+e^{2 \text {arcsinh}(c x)}\right )de^{2 \text {arcsinh}(c x)}\right )-\frac {i (a+b \text {arcsinh}(c x))^2}{2 b}\right )}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )\right )-\frac {i (a+b \text {arcsinh}(c x))^2}{2 b}\right )}{c^4 d}+\frac {x^2 (a+b \text {arcsinh}(c x))}{2 c^2 d}-\frac {b \left (\frac {x \sqrt {c^2 x^2+1}}{2 c^2}-\frac {\text {arcsinh}(c x)}{2 c^3}\right )}{2 c d}\) |
(x^2*(a + b*ArcSinh[c*x]))/(2*c^2*d) - (b*((x*Sqrt[1 + c^2*x^2])/(2*c^2) - ArcSinh[c*x]/(2*c^3)))/(2*c*d) + (I*(((-1/2*I)*(a + b*ArcSinh[c*x])^2)/b + (2*I)*(((a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/2 + (b*PolyLog [2, -E^(2*ArcSinh[c*x])])/4)))/(c^4*d)
3.1.29.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x _Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I Int[ (c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[1/e Subst[Int[(a + b*x)^n*Tanh[x], x], x, ArcSinh[c*x] ], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ .)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Simp[f^2*((m - 1)/(c^2*(m + 2*p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Simp[b*f*(n/(c*(m + 2*p + 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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[ m, 1] && NeQ[m + 2*p + 1, 0]
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {\frac {a \left (\frac {c^{2} x^{2}}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}\right )}{d}+\frac {b \operatorname {arcsinh}\left (c x \right )^{2}}{2 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}}{2 d}-\frac {b c x \sqrt {c^{2} x^{2}+1}}{4 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right )}{4 d}-\frac {b \,\operatorname {arcsinh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b \operatorname {polylog}\left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d}}{c^{4}}\) | \(146\) |
default | \(\frac {\frac {a \left (\frac {c^{2} x^{2}}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}\right )}{d}+\frac {b \operatorname {arcsinh}\left (c x \right )^{2}}{2 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}}{2 d}-\frac {b c x \sqrt {c^{2} x^{2}+1}}{4 d}+\frac {b \,\operatorname {arcsinh}\left (c x \right )}{4 d}-\frac {b \,\operatorname {arcsinh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b \operatorname {polylog}\left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d}}{c^{4}}\) | \(146\) |
parts | \(\frac {a \left (\frac {x^{2}}{2 c^{2}}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c^{4}}\right )}{d}+\frac {b \operatorname {arcsinh}\left (c x \right )^{2}}{2 d \,c^{4}}+\frac {b \,\operatorname {arcsinh}\left (c x \right ) x^{2}}{2 d \,c^{2}}-\frac {b x \sqrt {c^{2} x^{2}+1}}{4 c^{3} d}+\frac {b \,\operatorname {arcsinh}\left (c x \right )}{4 c^{4} d}-\frac {b \,\operatorname {arcsinh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d \,c^{4}}-\frac {b \operatorname {polylog}\left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 c^{4} d}\) | \(159\) |
1/c^4*(a/d*(1/2*c^2*x^2-1/2*ln(c^2*x^2+1))+1/2*b/d*arcsinh(c*x)^2+1/2*b/d* arcsinh(c*x)*c^2*x^2-1/4*b/d*c*x*(c^2*x^2+1)^(1/2)+1/4*b/d*arcsinh(c*x)-b/ d*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-1/2*b*polylog(2,-(c*x+(c^2* x^2+1)^(1/2))^2)/d)
\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} + d} \,d x } \]
\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {\int \frac {a x^{3}}{c^{2} x^{2} + 1}\, dx + \int \frac {b x^{3} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \]
\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} + d} \,d x } \]
1/2*a*(x^2/(c^2*d) - log(c^2*x^2 + 1)/(c^4*d)) - 1/8*b*((2*c^2*x^2 - log(c ^2*x^2 + 1)^2 - 4*(c^2*x^2 - log(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1) ) - 2*log(c^2*x^2 + 1))/(c^4*d) - 8*integrate(-1/2*(c^2*x^2 - log(c^2*x^2 + 1))/(c^6*d*x^3 + c^4*d*x + (c^5*d*x^2 + c^3*d)*sqrt(c^2*x^2 + 1)), x))
Exception generated. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{d\,c^2\,x^2+d} \,d x \]