Integrand size = 25, antiderivative size = 100 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=-\frac {b n \sqrt {d+e x^2}}{e^2}+\frac {2 b \sqrt {d} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2} \]
2*b*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))*d^(1/2)/e^2+d*(a+b*ln(c*x^n))/e^2/( e*x^2+d)^(1/2)-b*n*(e*x^2+d)^(1/2)/e^2+(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^2
Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {2 a d-b d n+a e x^2-b e n x^2-2 b \sqrt {d} n \sqrt {d+e x^2} \log (x)+b \left (2 d+e x^2\right ) \log \left (c x^n\right )+2 b \sqrt {d} n \sqrt {d+e x^2} \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{e^2 \sqrt {d+e x^2}} \]
(2*a*d - b*d*n + a*e*x^2 - b*e*n*x^2 - 2*b*Sqrt[d]*n*Sqrt[d + e*x^2]*Log[x ] + b*(2*d + e*x^2)*Log[c*x^n] + 2*b*Sqrt[d]*n*Sqrt[d + e*x^2]*Log[d + Sqr t[d]*Sqrt[d + e*x^2]])/(e^2*Sqrt[d + e*x^2])
Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2792, 27, 354, 90, 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.
\(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2792 |
\(\displaystyle -b n \int \frac {e x^2+2 d}{e^2 x \sqrt {e x^2+d}}dx+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b n \int \frac {e x^2+2 d}{x \sqrt {e x^2+d}}dx}{e^2}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {b n \int \frac {e x^2+2 d}{x^2 \sqrt {e x^2+d}}dx^2}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle -\frac {b n \left (2 d \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2+2 \sqrt {d+e x^2}\right )}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {b n \left (\frac {4 d \int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{e}+2 \sqrt {d+e x^2}\right )}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}-\frac {b n \left (2 \sqrt {d+e x^2}-4 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )}{2 e^2}\) |
-1/2*(b*n*(2*Sqrt[d + e*x^2] - 4*Sqrt[d]*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]) )/e^2 + (d*(a + b*Log[c*x^n]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^2
3.3.88.3.1 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] :> 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])
\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
Time = 0.30 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.45 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\left [\frac {{\left (b e n x^{2} + b d n\right )} \sqrt {d} \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (b d n + {\left (b e n - a e\right )} x^{2} - 2 \, a d - {\left (b e x^{2} + 2 \, b d\right )} \log \left (c\right ) - {\left (b e n x^{2} + 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{e^{3} x^{2} + d e^{2}}, -\frac {2 \, {\left (b e n x^{2} + b d n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (b d n + {\left (b e n - a e\right )} x^{2} - 2 \, a d - {\left (b e x^{2} + 2 \, b d\right )} \log \left (c\right ) - {\left (b e n x^{2} + 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{e^{3} x^{2} + d e^{2}}\right ] \]
[((b*e*n*x^2 + b*d*n)*sqrt(d)*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2* d)/x^2) - (b*d*n + (b*e*n - a*e)*x^2 - 2*a*d - (b*e*x^2 + 2*b*d)*log(c) - (b*e*n*x^2 + 2*b*d*n)*log(x))*sqrt(e*x^2 + d))/(e^3*x^2 + d*e^2), -(2*(b*e *n*x^2 + b*d*n)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (b*d*n + (b*e* n - a*e)*x^2 - 2*a*d - (b*e*x^2 + 2*b*d)*log(c) - (b*e*n*x^2 + 2*b*d*n)*lo g(x))*sqrt(e*x^2 + d))/(e^3*x^2 + d*e^2)]
Time = 24.36 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=a \left (\begin {cases} \frac {d}{e^{2} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {2 \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{e^{2}} + \frac {d}{e^{\frac {5}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {x}{e^{\frac {3}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{4}}{16 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {d}{e^{2} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((d/(e**2*sqrt(d + e*x**2)) + sqrt(d + e*x**2)/e**2, Ne(e, 0)), (x**4/(4*d**(3/2)), True)) - b*n*Piecewise((-2*sqrt(d)*asinh(sqrt(d)/(sqr t(e)*x))/e**2 + d/(e**(5/2)*x*sqrt(d/(e*x**2) + 1)) + x/(e**(3/2)*sqrt(d/( e*x**2) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**4/(16*d**(3/2)), True )) + b*Piecewise((d/(e**2*sqrt(d + e*x**2)) + sqrt(d + e*x**2)/e**2, Ne(e, 0)), (x**4/(4*d**(3/2)), True))*log(c*x**n)
Exception generated. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]