\(\int \frac {\log (c (a+b x^2)^p)}{d+e x} \, dx\) [188]

Optimal result

Integrand size = 20, antiderivative size = 201 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx=-\frac {p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{e}+\frac {\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e} \] Output:

-p*ln(e*((-a)^(1/2)-b^(1/2)*x)/(b^(1/2)*d+(-a)^(1/2)*e))*ln(e*x+d)/e-p*ln( 
-e*((-a)^(1/2)+b^(1/2)*x)/(b^(1/2)*d-(-a)^(1/2)*e))*ln(e*x+d)/e+ln(e*x+d)* 
ln(c*(b*x^2+a)^p)/e-p*polylog(2,b^(1/2)*(e*x+d)/(b^(1/2)*d-(-a)^(1/2)*e))/ 
e-p*polylog(2,b^(1/2)*(e*x+d)/(b^(1/2)*d+(-a)^(1/2)*e))/e
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx=-\frac {p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{e}+\frac {\log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{e} \] Input:

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

Output:

-((p*Log[(e*(Sqrt[-a] - Sqrt[b]*x))/(Sqrt[b]*d + Sqrt[-a]*e)]*Log[d + e*x] 
)/e) - (p*Log[-((e*(Sqrt[-a] + Sqrt[b]*x))/(Sqrt[b]*d - Sqrt[-a]*e))]*Log[ 
d + e*x])/e + (Log[d + e*x]*Log[c*(a + b*x^2)^p])/e - (p*PolyLog[2, (Sqrt[ 
b]*(d + e*x))/(Sqrt[b]*d - Sqrt[-a]*e)])/e - (p*PolyLog[2, (Sqrt[b]*(d + e 
*x))/(Sqrt[b]*d + Sqrt[-a]*e)])/e
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2912, 2863, 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.

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

\(\Big \downarrow \) 2912

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(Log[d + e*x]*Log[c*(a + b*x^2)^p])/e - (2*b*p*((Log[(e*(Sqrt[-a] - Sqrt[b 
]*x))/(Sqrt[b]*d + Sqrt[-a]*e)]*Log[d + e*x])/(2*b) + (Log[-((e*(Sqrt[-a] 
+ Sqrt[b]*x))/(Sqrt[b]*d - Sqrt[-a]*e))]*Log[d + e*x])/(2*b) + PolyLog[2, 
(Sqrt[b]*(d + e*x))/(Sqrt[b]*d - Sqrt[-a]*e)]/(2*b) + PolyLog[2, (Sqrt[b]* 
(d + e*x))/(Sqrt[b]*d + Sqrt[-a]*e)]/(2*b)))/e
 

Defintions of rubi rules used

rule 2009

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

rule 2863

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

rule 2912

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_.) + (g_. 
)*(x_)), x_Symbol] :> Simp[Log[f + g*x]*((a + b*Log[c*(d + e*x^n)^p])/g), x 
] - Simp[b*e*n*(p/g)   Int[x^(n - 1)*(Log[f + g*x]/(d + e*x^n)), x], x] /; 
FreeQ[{a, b, c, d, e, f, g, n, p}, x] && RationalQ[n]
 

Maple [A] (verified)

Time = 1.68 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.98

method	result	size
parts	\(\frac {\ln \left (e x +d \right ) \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{e}-\frac {2 p b \left (\frac {\ln \left (e x +d \right ) \left (\ln \left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )+\ln \left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )\right )}{2 b}+\frac {\operatorname {dilog}\left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )+\operatorname {dilog}\left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )}{2 b}\right )}{e}\)	\(196\)
risch	\(\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (e x +d \right )}{e}-\frac {p \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )}{e}-\frac {p \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )}{e}-\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-a b}-\left (e x +d \right ) b +b d}{e \sqrt {-a b}+b d}\right )}{e}-\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-a b}+\left (e x +d \right ) b -b d}{e \sqrt {-a b}-b d}\right )}{e}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \ln \left (e x +d \right )}{e}\)	\(331\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx=\frac {4 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}{b e \,x^{3}+b d \,x^{2}+a e x +a d}d x \right ) a e p -4 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) x}{b e \,x^{3}+b d \,x^{2}+a e x +a d}d x \right ) b d p +{\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2}}{4 e p} \] Input:

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

Output:

(4*int(log((a + b*x**2)**p*c)/(a*d + a*e*x + b*d*x**2 + b*e*x**3),x)*a*e*p 
 - 4*int((log((a + b*x**2)**p*c)*x)/(a*d + a*e*x + b*d*x**2 + b*e*x**3),x) 
*b*d*p + log((a + b*x**2)**p*c)**2)/(4*e*p)