\(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x^2)^{3/2}} \, dx\) [288]

Optimal result

Integrand size = 25, antiderivative size = 100 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=-\frac {b n \sqrt {d+e x^2}}{e^2}+\frac {2 b \sqrt {d} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2} \] Output:

-b*n*(e*x^2+d)^(1/2)/e^2+2*b*d^(1/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^ 
2+d*(a+b*ln(c*x^n))/e^2/(e*x^2+d)^(1/2)+(e*x^2+d)^(1/2)*(a+b*ln(c*x^n))/e^ 
2
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {2 a d-b d n+a e x^2-b e n x^2-2 b \sqrt {d} n \sqrt {d+e x^2} \log (x)+b \left (2 d+e x^2\right ) \log \left (c x^n\right )+2 b \sqrt {d} n \sqrt {d+e x^2} \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{e^2 \sqrt {d+e x^2}} \] Input:

Integrate[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^(3/2),x]
 

Output:

(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])
 

Rubi [A] (verified)

Time = 0.36 (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

\(\displaystyle \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}-\frac {b n \left (2 \sqrt {d+e x^2}-4 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )}{2 e^2}\)

Input:

Int[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^(3/2),x]
 

Output:

-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
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

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]
 

rule 90

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]
 

rule 221

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]
 

rule 354

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]
 

rule 2792

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])
 

Maple [F]

\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]

Input:

int(x^3*(a+b*ln(c*x^n))/(e*x^2+d)^(3/2),x)
 

Output:

int(x^3*(a+b*ln(c*x^n))/(e*x^2+d)^(3/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.48 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\left [\frac {{\left (b e n x^{2} + b d n\right )} \sqrt {d} \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (b d n + {\left (b e n - a e\right )} x^{2} - 2 \, a d - {\left (b e x^{2} + 2 \, b d\right )} \log \left (c\right ) - {\left (b e n x^{2} + 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{e^{3} x^{2} + d e^{2}}, -\frac {2 \, {\left (b e n x^{2} + b d n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x^{2} + d} \sqrt {-d}}{d}\right ) + {\left (b d n + {\left (b e n - a e\right )} x^{2} - 2 \, a d - {\left (b e x^{2} + 2 \, b d\right )} \log \left (c\right ) - {\left (b e n x^{2} + 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{e^{3} x^{2} + d e^{2}}\right ] \] Input:

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(3/2),x, algorithm="fricas")
 

Output:

[((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(e*x^2 + d)*sqrt(-d)/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)* 
log(x))*sqrt(e*x^2 + d))/(e^3*x^2 + d*e^2)]
 

Sympy [A] (verification not implemented)

Time = 24.67 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=a \left (\begin {cases} \frac {d}{e^{2} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {2 \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{e^{2}} + \frac {d}{e^{\frac {5}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {x}{e^{\frac {3}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{4}}{16 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {d}{e^{2} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \] Input:

integrate(x**3*(a+b*ln(c*x**n))/(e*x**2+d)**(3/2),x)
 

Output:

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)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(3/2),x, algorithm="maxima")
 

Output:

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
 

Giac [F]

\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*x^3/(e*x^2 + d)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \] Input:

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2)^(3/2),x)
 

Output:

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 325, normalized size of antiderivative = 3.25 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {e \,x^{2}+d}\, \mathrm {log}\left (\frac {\left (2 \sqrt {e}\, \sqrt {e \,x^{2}+d}\, x +2 e \,x^{2}\right )^{n} c}{e^{\frac {n}{2}} \left (\sqrt {e \,x^{2}+d}+\sqrt {e}\, x \right )^{n} 2^{n}}\right ) b d +\sqrt {e \,x^{2}+d}\, \mathrm {log}\left (\frac {\left (2 \sqrt {e}\, \sqrt {e \,x^{2}+d}\, x +2 e \,x^{2}\right )^{n} c}{e^{\frac {n}{2}} \left (\sqrt {e \,x^{2}+d}+\sqrt {e}\, x \right )^{n} 2^{n}}\right ) b e \,x^{2}+2 \sqrt {e \,x^{2}+d}\, a d +\sqrt {e \,x^{2}+d}\, a e \,x^{2}-\sqrt {e \,x^{2}+d}\, b d n -\sqrt {e \,x^{2}+d}\, b e n \,x^{2}-2 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d n -2 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b e n \,x^{2}+2 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d n +2 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b e n \,x^{2}}{e^{2} \left (e \,x^{2}+d \right )} \] Input:

int(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(3/2),x)
 

Output:

(2*sqrt(d + e*x**2)*log(((2*sqrt(e)*sqrt(d + e*x**2)*x + 2*e*x**2)**n*c)/( 
e**(n/2)*(sqrt(d + e*x**2) + sqrt(e)*x)**n*2**n))*b*d + sqrt(d + e*x**2)*l 
og(((2*sqrt(e)*sqrt(d + e*x**2)*x + 2*e*x**2)**n*c)/(e**(n/2)*(sqrt(d + e* 
x**2) + sqrt(e)*x)**n*2**n))*b*e*x**2 + 2*sqrt(d + e*x**2)*a*d + sqrt(d + 
e*x**2)*a*e*x**2 - sqrt(d + e*x**2)*b*d*n - sqrt(d + e*x**2)*b*e*n*x**2 - 
2*sqrt(d)*log((sqrt(d + e*x**2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d*n - 2* 
sqrt(d)*log((sqrt(d + e*x**2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*b*e*n*x**2 + 
 2*sqrt(d)*log((sqrt(d + e*x**2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d*n + 2 
*sqrt(d)*log((sqrt(d + e*x**2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*b*e*n*x**2) 
/(e**2*(d + e*x**2))