\(\int \frac {x^3 \log (c (a+\frac {b}{x^2})^p)}{d+e x} \, dx\) [247]

Optimal result

Integrand size = 23, antiderivative size = 421 \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {2 b p x}{3 a e}+\frac {2 \sqrt {b} d^2 p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a} e^3}-\frac {2 b^{3/2} p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{3 a^{3/2} e}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{3 e}-\frac {d^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)}{e^4}-\frac {2 d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \log \left (-\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{\sqrt {-a} d-\sqrt {b} e}\right ) \log (d+e x)}{e^4}-\frac {b d p \log \left (b+a x^2\right )}{2 a e^2}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e^4}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e^4}-\frac {2 d^3 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^4} \] Output:

2/3*b*p*x/a/e+2*b^(1/2)*d^2*p*arctan(a^(1/2)*x/b^(1/2))/a^(1/2)/e^3-2/3*b^ 
(3/2)*p*arctan(a^(1/2)*x/b^(1/2))/a^(3/2)/e+d^2*x*ln(c*(a+b/x^2)^p)/e^3-1/ 
2*d*x^2*ln(c*(a+b/x^2)^p)/e^2+1/3*x^3*ln(c*(a+b/x^2)^p)/e-d^3*ln(c*(a+b/x^ 
2)^p)*ln(e*x+d)/e^4-2*d^3*p*ln(-e*x/d)*ln(e*x+d)/e^4+d^3*p*ln(e*(b^(1/2)-( 
-a)^(1/2)*x)/((-a)^(1/2)*d+b^(1/2)*e))*ln(e*x+d)/e^4+d^3*p*ln(-e*(b^(1/2)+ 
(-a)^(1/2)*x)/((-a)^(1/2)*d-b^(1/2)*e))*ln(e*x+d)/e^4-1/2*b*d*p*ln(a*x^2+b 
)/a/e^2+d^3*p*polylog(2,(-a)^(1/2)*(e*x+d)/((-a)^(1/2)*d-b^(1/2)*e))/e^4+d 
^3*p*polylog(2,(-a)^(1/2)*(e*x+d)/((-a)^(1/2)*d+b^(1/2)*e))/e^4-2*d^3*p*po 
lylog(2,1+e*x/d)/e^4
 

Mathematica [C] (verified)

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

Time = 0.28 (sec) , antiderivative size = 404, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{d+e x} \, dx=\frac {-12 \sqrt {a} \sqrt {b} d^2 e p \arctan \left (\frac {\sqrt {b}}{\sqrt {a} x}\right )+4 b e^3 p x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {b}{a x^2}\right )-3 b d e^2 p \log \left (a+\frac {b}{x^2}\right )+6 a d^2 e x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )-3 a d e^2 x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+2 a e^3 x^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )-6 b d e^2 p \log (x)-6 a d^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log (d+e x)-12 a d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)+6 a d^3 p \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)+6 a d^3 p \log \left (\frac {e \left (\sqrt {b}+\sqrt {-a} x\right )}{-\sqrt {-a} d+\sqrt {b} e}\right ) \log (d+e x)+6 a d^3 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )+6 a d^3 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )-12 a d^3 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{6 a e^4} \] Input:

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

Output:

(-12*Sqrt[a]*Sqrt[b]*d^2*e*p*ArcTan[Sqrt[b]/(Sqrt[a]*x)] + 4*b*e^3*p*x*Hyp 
ergeometric2F1[-1/2, 1, 1/2, -(b/(a*x^2))] - 3*b*d*e^2*p*Log[a + b/x^2] + 
6*a*d^2*e*x*Log[c*(a + b/x^2)^p] - 3*a*d*e^2*x^2*Log[c*(a + b/x^2)^p] + 2* 
a*e^3*x^3*Log[c*(a + b/x^2)^p] - 6*b*d*e^2*p*Log[x] - 6*a*d^3*Log[c*(a + b 
/x^2)^p]*Log[d + e*x] - 12*a*d^3*p*Log[-((e*x)/d)]*Log[d + e*x] + 6*a*d^3* 
p*Log[(e*(Sqrt[b] - Sqrt[-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log[d + e*x] + 
6*a*d^3*p*Log[(e*(Sqrt[b] + Sqrt[-a]*x))/(-(Sqrt[-a]*d) + Sqrt[b]*e)]*Log[ 
d + e*x] + 6*a*d^3*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d - Sqrt[b] 
*e)] + 6*a*d^3*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a]*d + Sqrt[b]*e)] 
 - 12*a*d^3*p*PolyLog[2, 1 + (e*x)/d])/(6*a*e^4)
 

Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2916, 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.

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

\(\Big \downarrow \) 2916

\(\displaystyle \int \left (-\frac {d^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e^3 (d+e x)}+\frac {d^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e^3}-\frac {d x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e^2}+\frac {x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b^{3/2} p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{3 a^{3/2} e}+\frac {2 \sqrt {b} d^2 p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a} e^3}-\frac {d^3 \log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e^4}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{3 e}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e^4}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e^4}+\frac {d^3 p \log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{e^4}+\frac {d^3 p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{e^4}-\frac {b d p \log \left (a x^2+b\right )}{2 a e^2}+\frac {2 b p x}{3 a e}-\frac {2 d^3 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e^4}-\frac {2 d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}\)

Input:

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

Output:

(2*b*p*x)/(3*a*e) + (2*Sqrt[b]*d^2*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(Sqrt[a] 
*e^3) - (2*b^(3/2)*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(3*a^(3/2)*e) + (d^2*x*L 
og[c*(a + b/x^2)^p])/e^3 - (d*x^2*Log[c*(a + b/x^2)^p])/(2*e^2) + (x^3*Log 
[c*(a + b/x^2)^p])/(3*e) - (d^3*Log[c*(a + b/x^2)^p]*Log[d + e*x])/e^4 - ( 
2*d^3*p*Log[-((e*x)/d)]*Log[d + e*x])/e^4 + (d^3*p*Log[(e*(Sqrt[b] - Sqrt[ 
-a]*x))/(Sqrt[-a]*d + Sqrt[b]*e)]*Log[d + e*x])/e^4 + (d^3*p*Log[-((e*(Sqr 
t[b] + Sqrt[-a]*x))/(Sqrt[-a]*d - Sqrt[b]*e))]*Log[d + e*x])/e^4 - (b*d*p* 
Log[b + a*x^2])/(2*a*e^2) + (d^3*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[- 
a]*d - Sqrt[b]*e)])/e^4 + (d^3*p*PolyLog[2, (Sqrt[-a]*(d + e*x))/(Sqrt[-a] 
*d + Sqrt[b]*e)])/e^4 - (2*d^3*p*PolyLog[2, 1 + (e*x)/d])/e^4
 

Defintions of rubi rules used

rule 2009

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

rule 2916

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log 
[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e, f, g 
, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]
 

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 411, normalized size of antiderivative = 0.98

method	result	size
parts	\(\frac {x^{3} \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{3 e}-\frac {d \,x^{2} \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{2 e^{2}}+\frac {d^{2} x \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{e^{3}}-\frac {d^{3} \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{4}}+2 p b \,e^{2} \left (\frac {d^{3} \left (-\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b \,e^{2}}-\frac {\left (-\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}+d a -a \left (e x +d \right )}{e \sqrt {-a b}+d a}\right )+\ln \left (\frac {e \sqrt {-a b}-d a +a \left (e x +d \right )}{e \sqrt {-a b}-d a}\right )\right )}{2 a}-\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}+d a -a \left (e x +d \right )}{e \sqrt {-a b}+d a}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}-d a +a \left (e x +d \right )}{e \sqrt {-a b}-d a}\right )}{2 a}\right ) a}{b \,e^{2}}\right )}{e^{4}}+\frac {\frac {2 e x +2 d}{a}+\frac {-\frac {3 d \ln \left (a \,d^{2}-2 a d \left (e x +d \right )+a \left (e x +d \right )^{2}+b \,e^{2}\right )}{2}+\frac {\left (6 a \,d^{2}-2 b \,e^{2}\right ) \arctan \left (\frac {-2 d a +2 a \left (e x +d \right )}{2 e \sqrt {a b}}\right )}{e \sqrt {a b}}}{a}}{6 e^{4}}\right )\)	\(411\)

Input:

int(x^3*ln(c*(a+b/x^2)^p)/(e*x+d),x,method=_RETURNVERBOSE)
 

Output:

1/3*x^3*ln(c*(a+b/x^2)^p)/e-1/2*d*x^2*ln(c*(a+b/x^2)^p)/e^2+d^2*x*ln(c*(a+ 
b/x^2)^p)/e^3-d^3*ln(c*(a+b/x^2)^p)*ln(e*x+d)/e^4+2*p*b*e^2*(1/e^4*d^3*(-( 
dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d))/b/e^2-(-1/2*ln(e*x+d)*(ln((e*(-a*b)^(1 
/2)+d*a-a*(e*x+d))/(e*(-a*b)^(1/2)+d*a))+ln((e*(-a*b)^(1/2)-d*a+a*(e*x+d)) 
/(e*(-a*b)^(1/2)-d*a)))/a-1/2*(dilog((e*(-a*b)^(1/2)+d*a-a*(e*x+d))/(e*(-a 
*b)^(1/2)+d*a))+dilog((e*(-a*b)^(1/2)-d*a+a*(e*x+d))/(e*(-a*b)^(1/2)-d*a)) 
)/a)/b*a/e^2)+1/6/e^4*(2*(e*x+d)/a+1/a*(-3/2*d*ln(a*d^2-2*a*d*(e*x+d)+a*(e 
*x+d)^2+b*e^2)+(6*a*d^2-2*b*e^2)/e/(a*b)^(1/2)*arctan(1/2*(-2*d*a+2*a*(e*x 
+d))/e/(a*b)^(1/2)))))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**3*log(c*(a + b/x**2)**p)/(d + e*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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