\(\int \frac {a+b \log (c x^n)}{x^3 (d+e x^2)^{5/2}} \, dx\) [302]

Optimal result

Integrand size = 25, antiderivative size = 337 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\frac {b e n}{3 d^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{4 d^3 x^2}-\frac {31 b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{12 d^{7/2}}-\frac {5 b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{7/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt {d+e x^2}}+\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}+\frac {5 b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{7/2}}+\frac {5 b e n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{4 d^{7/2}} \] Output:

1/3*b*e*n/d^3/(e*x^2+d)^(1/2)-1/4*b*n*(e*x^2+d)^(1/2)/d^3/x^2-31/12*b*e*n* 
arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(7/2)-5/4*b*e*n*arctanh((e*x^2+d)^(1/2) 
/d^(1/2))^2/d^(7/2)-5/6*e*(a+b*ln(c*x^n))/d^2/(e*x^2+d)^(3/2)-1/2*(a+b*ln( 
c*x^n))/d/x^2/(e*x^2+d)^(3/2)-5/2*e*(a+b*ln(c*x^n))/d^3/(e*x^2+d)^(1/2)+5/ 
2*e*arctanh((e*x^2+d)^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(7/2)+5/2*b*e*n*arc 
tanh((e*x^2+d)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x^2+d)^(1/2)))/d^(7 
/2)+5/4*b*e*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(e*x^2+d)^(1/2)))/d^(7/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.35 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.67 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\frac {b n \sqrt {1+\frac {d}{e x^2}} \left (5 \, _3F_2\left (\frac {7}{2},\frac {7}{2},\frac {7}{2};\frac {9}{2},\frac {9}{2};-\frac {d}{e x^2}\right )-7 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2},\frac {9}{2},-\frac {d}{e x^2}\right ) (1+2 \log (x))\right )}{98 e^2 x^6 \sqrt {d+e x^2}}-\frac {\left (3 d^2+20 d e x^2+15 e^2 x^4\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{6 d^3 x^2 \left (d+e x^2\right )^{3/2}}-\frac {5 e \log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{2 d^{7/2}}+\frac {5 e \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{2 d^{7/2}} \] Input:

Integrate[(a + b*Log[c*x^n])/(x^3*(d + e*x^2)^(5/2)),x]
 

Output:

(b*n*Sqrt[1 + d/(e*x^2)]*(5*HypergeometricPFQ[{7/2, 7/2, 7/2}, {9/2, 9/2}, 
 -(d/(e*x^2))] - 7*Hypergeometric2F1[5/2, 7/2, 9/2, -(d/(e*x^2))]*(1 + 2*L 
og[x])))/(98*e^2*x^6*Sqrt[d + e*x^2]) - ((3*d^2 + 20*d*e*x^2 + 15*e^2*x^4) 
*(a - b*n*Log[x] + b*Log[c*x^n]))/(6*d^3*x^2*(d + e*x^2)^(3/2)) - (5*e*Log 
[x]*(a - b*n*Log[x] + b*Log[c*x^n]))/(2*d^(7/2)) + (5*e*(a - b*n*Log[x] + 
b*Log[c*x^n])*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(2*d^(7/2))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2792, 2009}

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.

\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2792

\(\displaystyle -b n \int \left (\frac {5 e \text {arctanh}\left (\frac {\sqrt {e x^2+d}}{\sqrt {d}}\right )}{2 d^{7/2} x}-\frac {15 e^2 x^4+20 d e x^2+3 d^2}{6 d^3 x^3 \left (e x^2+d\right )^{3/2}}\right )dx+\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt {d+e x^2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt {d+e x^2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}-b n \left (\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{7/2}}+\frac {31 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{12 d^{7/2}}-\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{7/2}}-\frac {5 e \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )}{4 d^{7/2}}+\frac {\sqrt {d+e x^2}}{4 d^3 x^2}-\frac {e}{3 d^3 \sqrt {d+e x^2}}\right )\)

Input:

Int[(a + b*Log[c*x^n])/(x^3*(d + e*x^2)^(5/2)),x]
 

Output:

(-5*e*(a + b*Log[c*x^n]))/(6*d^2*(d + e*x^2)^(3/2)) - (a + b*Log[c*x^n])/( 
2*d*x^2*(d + e*x^2)^(3/2)) - (5*e*(a + b*Log[c*x^n]))/(2*d^3*Sqrt[d + e*x^ 
2]) + (5*e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*(a + b*Log[c*x^n]))/(2*d^(7/2) 
) - b*n*(-1/3*e/(d^3*Sqrt[d + e*x^2]) + Sqrt[d + e*x^2]/(4*d^3*x^2) + (31* 
e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(12*d^(7/2)) + (5*e*ArcTanh[Sqrt[d + e 
*x^2]/Sqrt[d]]^2)/(4*d^(7/2)) - (5*e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[ 
(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(7/2)) - (5*e*PolyLog[2, 1 
- (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(4*d^(7/2)))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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]

\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{3} \left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]

Input:

int((a+b*ln(c*x^n))/x^3/(e*x^2+d)^(5/2),x)
 

Output:

int((a+b*ln(c*x^n))/x^3/(e*x^2+d)^(5/2),x)
 

Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{3}} \,d x } \] Input:

integrate((a+b*log(c*x^n))/x^3/(e*x^2+d)^(5/2),x, algorithm="fricas")
 

Output:

integral((sqrt(e*x^2 + d)*b*log(c*x^n) + sqrt(e*x^2 + d)*a)/(e^3*x^9 + 3*d 
*e^2*x^7 + 3*d^2*e*x^5 + d^3*x^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*x**n))/x**3/(e*x**2+d)**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*log(c*x^n))/x^3/(e*x^2+d)^(5/2),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]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{3}} \,d x } \] Input:

integrate((a+b*log(c*x^n))/x^3/(e*x^2+d)^(5/2),x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)/((e*x^2 + d)^(5/2)*x^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \] Input:

int((a + b*log(c*x^n))/(x^3*(d + e*x^2)^(5/2)),x)
 

Output:

int((a + b*log(c*x^n))/(x^3*(d + e*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\frac {-3 \sqrt {e \,x^{2}+d}\, a \,d^{3}-20 \sqrt {e \,x^{2}+d}\, a \,d^{2} e \,x^{2}-15 \sqrt {e \,x^{2}+d}\, a d \,e^{2} x^{4}-15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a \,d^{2} e \,x^{2}-30 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a d \,e^{2} x^{4}-15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a \,e^{3} x^{6}+15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a \,d^{2} e \,x^{2}+30 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a d \,e^{2} x^{4}+15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a \,e^{3} x^{6}+6 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{\sqrt {e \,x^{2}+d}\, d^{2} x^{3}+2 \sqrt {e \,x^{2}+d}\, d e \,x^{5}+\sqrt {e \,x^{2}+d}\, e^{2} x^{7}}d x \right ) b \,d^{6} x^{2}+12 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{\sqrt {e \,x^{2}+d}\, d^{2} x^{3}+2 \sqrt {e \,x^{2}+d}\, d e \,x^{5}+\sqrt {e \,x^{2}+d}\, e^{2} x^{7}}d x \right ) b \,d^{5} e \,x^{4}+6 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{\sqrt {e \,x^{2}+d}\, d^{2} x^{3}+2 \sqrt {e \,x^{2}+d}\, d e \,x^{5}+\sqrt {e \,x^{2}+d}\, e^{2} x^{7}}d x \right ) b \,d^{4} e^{2} x^{6}}{6 d^{4} x^{2} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:

int((a+b*log(c*x^n))/x^3/(e*x^2+d)^(5/2),x)
 

Output:

( - 3*sqrt(d + e*x**2)*a*d**3 - 20*sqrt(d + e*x**2)*a*d**2*e*x**2 - 15*sqr 
t(d + e*x**2)*a*d*e**2*x**4 - 15*sqrt(d)*log((sqrt(d + e*x**2) - sqrt(d) + 
 sqrt(e)*x)/sqrt(d))*a*d**2*e*x**2 - 30*sqrt(d)*log((sqrt(d + e*x**2) - sq 
rt(d) + sqrt(e)*x)/sqrt(d))*a*d*e**2*x**4 - 15*sqrt(d)*log((sqrt(d + e*x** 
2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*a*e**3*x**6 + 15*sqrt(d)*log((sqrt(d + 
e*x**2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*a*d**2*e*x**2 + 30*sqrt(d)*log((sq 
rt(d + e*x**2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*a*d*e**2*x**4 + 15*sqrt(d)* 
log((sqrt(d + e*x**2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*a*e**3*x**6 + 6*int( 
log(x**n*c)/(sqrt(d + e*x**2)*d**2*x**3 + 2*sqrt(d + e*x**2)*d*e*x**5 + sq 
rt(d + e*x**2)*e**2*x**7),x)*b*d**6*x**2 + 12*int(log(x**n*c)/(sqrt(d + e* 
x**2)*d**2*x**3 + 2*sqrt(d + e*x**2)*d*e*x**5 + sqrt(d + e*x**2)*e**2*x**7 
),x)*b*d**5*e*x**4 + 6*int(log(x**n*c)/(sqrt(d + e*x**2)*d**2*x**3 + 2*sqr 
t(d + e*x**2)*d*e*x**5 + sqrt(d + e*x**2)*e**2*x**7),x)*b*d**4*e**2*x**6)/ 
(6*d**4*x**2*(d**2 + 2*d*e*x**2 + e**2*x**4))