Integrand size = 25, antiderivative size = 129 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\frac {2 b d n \sqrt {d+e x^2}}{3 e^2}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^2}-\frac {2 b d^{3/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2} \]
-1/9*b*n*(e*x^2+d)^(3/2)/e^2-2/3*b*d^(3/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/ 2))/e^2+1/3*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e^2+2/3*b*d*n*(e*x^2+d)^(1/2)/ e^2-d*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^2
Time = 0.12 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\frac {-6 a d \sqrt {d+e x^2}+5 b d n \sqrt {d+e x^2}+3 a e x^2 \sqrt {d+e x^2}-b e n x^2 \sqrt {d+e x^2}+6 b d^{3/2} n \log (x)+3 b \left (-2 d+e x^2\right ) \sqrt {d+e x^2} \log \left (c x^n\right )-6 b d^{3/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{9 e^2} \]
(-6*a*d*Sqrt[d + e*x^2] + 5*b*d*n*Sqrt[d + e*x^2] + 3*a*e*x^2*Sqrt[d + e*x ^2] - b*e*n*x^2*Sqrt[d + e*x^2] + 6*b*d^(3/2)*n*Log[x] + 3*b*(-2*d + e*x^2 )*Sqrt[d + e*x^2]*Log[c*x^n] - 6*b*d^(3/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^ 2]])/(9*e^2)
Time = 0.35 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2792, 27, 354, 90, 60, 73, 221}
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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {\left (2 d-e x^2\right ) \sqrt {e x^2+d}}{3 e^2 x}dx+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \int \frac {\left (2 d-e x^2\right ) \sqrt {e x^2+d}}{x}dx}{3 e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {b n \int \frac {\left (2 d-e x^2\right ) \sqrt {e x^2+d}}{x^2}dx^2}{6 e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {b n \left (2 d \int \frac {\sqrt {e x^2+d}}{x^2}dx^2-\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {b n \left (2 d \left (d \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2+2 \sqrt {d+e x^2}\right )-\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {b n \left (2 d \left (\frac {2 d \int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{e}+2 \sqrt {d+e x^2}\right )-\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {b n \left (2 d \left (2 \sqrt {d+e x^2}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )-\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e^2}\) |
(b*n*((-2*(d + e*x^2)^(3/2))/3 + 2*d*(2*Sqrt[d + e*x^2] - 2*Sqrt[d]*ArcTan h[Sqrt[d + e*x^2]/Sqrt[d]])))/(6*e^2) - (d*Sqrt[d + e*x^2]*(a + b*Log[c*x^ n]))/e^2 + ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2)
3.3.77.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x] }, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2] ) || InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x ] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])
\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e \,x^{2}+d}}d x\]
Time = 0.32 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.60 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\left [\frac {3 \, b d^{\frac {3}{2}} n \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + {\left (5 \, b d n - {\left (b e n - 3 \, a e\right )} x^{2} - 6 \, a d + 3 \, {\left (b e x^{2} - 2 \, b d\right )} \log \left (c\right ) + 3 \, {\left (b e n x^{2} - 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, e^{2}}, \frac {6 \, b \sqrt {-d} d n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (5 \, b d n - {\left (b e n - 3 \, a e\right )} x^{2} - 6 \, a d + 3 \, {\left (b e x^{2} - 2 \, b d\right )} \log \left (c\right ) + 3 \, {\left (b e n x^{2} - 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, e^{2}}\right ] \]
[1/9*(3*b*d^(3/2)*n*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + (5*b*d*n - (b*e*n - 3*a*e)*x^2 - 6*a*d + 3*(b*e*x^2 - 2*b*d)*log(c) + 3*(b *e*n*x^2 - 2*b*d*n)*log(x))*sqrt(e*x^2 + d))/e^2, 1/9*(6*b*sqrt(-d)*d*n*ar ctan(sqrt(-d)/sqrt(e*x^2 + d)) + (5*b*d*n - (b*e*n - 3*a*e)*x^2 - 6*a*d + 3*(b*e*x^2 - 2*b*d)*log(c) + 3*(b*e*n*x^2 - 2*b*d*n)*log(x))*sqrt(e*x^2 + d))/e^2]
Time = 12.28 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.83 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=a \left (\begin {cases} - \frac {2 d \sqrt {d + e x^{2}}}{3 e^{2}} + \frac {x^{2} \sqrt {d + e x^{2}}}{3 e} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {2 d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{3 e^{2}} - \frac {2 d^{2}}{3 e^{\frac {5}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {2 d x}{3 e^{\frac {3}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}}{3 e} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{4}}{16 \sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d \sqrt {d + e x^{2}}}{3 e^{2}} + \frac {x^{2} \sqrt {d + e x^{2}}}{3 e} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((-2*d*sqrt(d + e*x**2)/(3*e**2) + x**2*sqrt(d + e*x**2)/(3*e), Ne(e, 0)), (x**4/(4*sqrt(d)), True)) - b*n*Piecewise((2*d**(3/2)*asinh(sq rt(d)/(sqrt(e)*x))/(3*e**2) - 2*d**2/(3*e**(5/2)*x*sqrt(d/(e*x**2) + 1)) - 2*d*x/(3*e**(3/2)*sqrt(d/(e*x**2) + 1)) + Piecewise((d*sqrt(d + e*x**2)/( 3*e) + x**2*sqrt(d + e*x**2)/3, Ne(e, 0)), (sqrt(d)*x**2/2, True))/(3*e), (e > -oo) & (e < oo) & Ne(e, 0)), (x**4/(16*sqrt(d)), True)) + b*Piecewise ((-2*d*sqrt(d + e*x**2)/(3*e**2) + x**2*sqrt(d + e*x**2)/(3*e), Ne(e, 0)), (x**4/(4*sqrt(d)), True))*log(c*x**n)
Exception generated. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{\sqrt {e x^{2} + d}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {e\,x^2+d}} \,d x \]