\(\int \frac {\log (c (a+b x^2)^p)}{(d+e x)^3} \, dx\) [190]

Optimal result

Integrand size = 20, antiderivative size = 174 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx=\frac {b d p}{e \left (b d^2+a e^2\right ) (d+e x)}+\frac {2 \sqrt {a} b^{3/2} d p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\left (b d^2+a e^2\right )^2}-\frac {b \left (b d^2-a e^2\right ) p \log (d+e x)}{e \left (b d^2+a e^2\right )^2}+\frac {b \left (b d^2-a e^2\right ) p \log \left (a+b x^2\right )}{2 e \left (b d^2+a e^2\right )^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.25 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx=\frac {\frac {b p (d+e x) \left (\left (\sqrt {-a} b d^2+2 a \sqrt {b} d e+(-a)^{3/2} e^2\right ) (d+e x) \log \left (\sqrt {-a}-\sqrt {b} x\right )+\left (\sqrt {-a} b d^2-2 a \sqrt {b} d e+(-a)^{3/2} e^2\right ) (d+e x) \log \left (\sqrt {-a}+\sqrt {b} x\right )+2 \sqrt {-a} \left (b d^3+a d e^2-\left (b d^2-a e^2\right ) (d+e x) \log (d+e x)\right )\right )}{\sqrt {-a} \left (b d^2+a e^2\right )^2}-\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2913, 594, 25, 657, 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.

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

\(\Big \downarrow \) 2913

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

\(\Big \downarrow \) 594

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

rule 2913

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_. 
)*(x_))^(r_.), x_Symbol] :> Simp[(f + g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n 
)^p])/(g*(r + 1))), x] - Simp[b*e*n*(p/(g*(r + 1)))   Int[x^(n - 1)*((f + g 
*x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x 
] && (IGtQ[r, 0] || RationalQ[n]) && NeQ[r, -1]
 

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.84

method	result	size
parts	\(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{2 e \left (e x +d \right )^{2}}+\frac {p b \left (\frac {\left (a \,e^{2}-b \,d^{2}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}+b \,d^{2}\right )^{2}}+\frac {d}{\left (a \,e^{2}+b \,d^{2}\right ) \left (e x +d \right )}+\frac {b \left (\frac {\left (-a \,e^{2}+b \,d^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {2 d e a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{\left (a \,e^{2}+b \,d^{2}\right )^{2}}\right )}{e}\)	\(147\)
risch	\(\text {Expression too large to display}\)	\(3183\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (162) = 324\).

Time = 0.14 (sec) , antiderivative size = 744, normalized size of antiderivative = 4.28 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx=\left [\frac {2 \, {\left (b^{2} d^{3} e + a b d e^{3}\right )} p x + 2 \, {\left (b d e^{3} p x^{2} + 2 \, b d^{2} e^{2} p x + b d^{3} e p\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (b^{2} d^{4} + a b d^{2} e^{2}\right )} p + {\left ({\left (b^{2} d^{2} e^{2} - a b e^{4}\right )} p x^{2} + 2 \, {\left (b^{2} d^{3} e - a b d e^{3}\right )} p x - {\left (3 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} p\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left ({\left (b^{2} d^{2} e^{2} - a b e^{4}\right )} p x^{2} + 2 \, {\left (b^{2} d^{3} e - a b d e^{3}\right )} p x + {\left (b^{2} d^{4} - a b d^{2} e^{2}\right )} p\right )} \log \left (e x + d\right ) - {\left (b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (c\right )}{2 \, {\left (b^{2} d^{6} e + 2 \, a b d^{4} e^{3} + a^{2} d^{2} e^{5} + {\left (b^{2} d^{4} e^{3} + 2 \, a b d^{2} e^{5} + a^{2} e^{7}\right )} x^{2} + 2 \, {\left (b^{2} d^{5} e^{2} + 2 \, a b d^{3} e^{4} + a^{2} d e^{6}\right )} x\right )}}, \frac {2 \, {\left (b^{2} d^{3} e + a b d e^{3}\right )} p x + 4 \, {\left (b d e^{3} p x^{2} + 2 \, b d^{2} e^{2} p x + b d^{3} e p\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 2 \, {\left (b^{2} d^{4} + a b d^{2} e^{2}\right )} p + {\left ({\left (b^{2} d^{2} e^{2} - a b e^{4}\right )} p x^{2} + 2 \, {\left (b^{2} d^{3} e - a b d e^{3}\right )} p x - {\left (3 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} p\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left ({\left (b^{2} d^{2} e^{2} - a b e^{4}\right )} p x^{2} + 2 \, {\left (b^{2} d^{3} e - a b d e^{3}\right )} p x + {\left (b^{2} d^{4} - a b d^{2} e^{2}\right )} p\right )} \log \left (e x + d\right ) - {\left (b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (c\right )}{2 \, {\left (b^{2} d^{6} e + 2 \, a b d^{4} e^{3} + a^{2} d^{2} e^{5} + {\left (b^{2} d^{4} e^{3} + 2 \, a b d^{2} e^{5} + a^{2} e^{7}\right )} x^{2} + 2 \, {\left (b^{2} d^{5} e^{2} + 2 \, a b d^{3} e^{4} + a^{2} d e^{6}\right )} x\right )}}\right ] \] Input:

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

Output:

[1/2*(2*(b^2*d^3*e + a*b*d*e^3)*p*x + 2*(b*d*e^3*p*x^2 + 2*b*d^2*e^2*p*x + 
 b*d^3*e*p)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 2*( 
b^2*d^4 + a*b*d^2*e^2)*p + ((b^2*d^2*e^2 - a*b*e^4)*p*x^2 + 2*(b^2*d^3*e - 
 a*b*d*e^3)*p*x - (3*a*b*d^2*e^2 + a^2*e^4)*p)*log(b*x^2 + a) - 2*((b^2*d^ 
2*e^2 - a*b*e^4)*p*x^2 + 2*(b^2*d^3*e - a*b*d*e^3)*p*x + (b^2*d^4 - a*b*d^ 
2*e^2)*p)*log(e*x + d) - (b^2*d^4 + 2*a*b*d^2*e^2 + a^2*e^4)*log(c))/(b^2* 
d^6*e + 2*a*b*d^4*e^3 + a^2*d^2*e^5 + (b^2*d^4*e^3 + 2*a*b*d^2*e^5 + a^2*e 
^7)*x^2 + 2*(b^2*d^5*e^2 + 2*a*b*d^3*e^4 + a^2*d*e^6)*x), 1/2*(2*(b^2*d^3* 
e + a*b*d*e^3)*p*x + 4*(b*d*e^3*p*x^2 + 2*b*d^2*e^2*p*x + b*d^3*e*p)*sqrt( 
a*b)*arctan(sqrt(a*b)*x/a) + 2*(b^2*d^4 + a*b*d^2*e^2)*p + ((b^2*d^2*e^2 - 
 a*b*e^4)*p*x^2 + 2*(b^2*d^3*e - a*b*d*e^3)*p*x - (3*a*b*d^2*e^2 + a^2*e^4 
)*p)*log(b*x^2 + a) - 2*((b^2*d^2*e^2 - a*b*e^4)*p*x^2 + 2*(b^2*d^3*e - a* 
b*d*e^3)*p*x + (b^2*d^4 - a*b*d^2*e^2)*p)*log(e*x + d) - (b^2*d^4 + 2*a*b* 
d^2*e^2 + a^2*e^4)*log(c))/(b^2*d^6*e + 2*a*b*d^4*e^3 + a^2*d^2*e^5 + (b^2 
*d^4*e^3 + 2*a*b*d^2*e^5 + a^2*e^7)*x^2 + 2*(b^2*d^5*e^2 + 2*a*b*d^3*e^4 + 
 a^2*d*e^6)*x)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx=\frac {{\left (\frac {4 \, a b d e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a b}} + \frac {{\left (b d^{2} - a e^{2}\right )} \log \left (b x^{2} + a\right )}{b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}} - \frac {2 \, {\left (b d^{2} - a e^{2}\right )} \log \left (e x + d\right )}{b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}} + \frac {2 \, d}{b d^{3} + a d e^{2} + {\left (b d^{2} e + a e^{3}\right )} x}\right )} b p}{2 \, e} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{2 \, {\left (e x + d\right )}^{2} e} \] Input:

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

Output:

1/2*(4*a*b*d*e*arctan(b*x/sqrt(a*b))/((b^2*d^4 + 2*a*b*d^2*e^2 + a^2*e^4)* 
sqrt(a*b)) + (b*d^2 - a*e^2)*log(b*x^2 + a)/(b^2*d^4 + 2*a*b*d^2*e^2 + a^2 
*e^4) - 2*(b*d^2 - a*e^2)*log(e*x + d)/(b^2*d^4 + 2*a*b*d^2*e^2 + a^2*e^4) 
 + 2*d/(b*d^3 + a*d*e^2 + (b*d^2*e + a*e^3)*x))*b*p/e - 1/2*log((b*x^2 + a 
)^p*c)/((e*x + d)^2*e)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.60 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx=\frac {2 \, a b^{2} d p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a b}} + \frac {{\left (b^{2} d^{2} p - a b e^{2} p\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} d^{4} e + 2 \, a b d^{2} e^{3} + a^{2} e^{5}\right )}} - \frac {p \log \left (b x^{2} + a\right )}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {{\left (b^{2} d^{2} p - a b e^{2} p\right )} \log \left (e x + d\right )}{b^{2} d^{4} e + 2 \, a b d^{2} e^{3} + a^{2} e^{5}} + \frac {2 \, b d e p x + 2 \, b d^{2} p - b d^{2} \log \left (c\right ) - a e^{2} \log \left (c\right )}{2 \, {\left (b d^{2} e^{3} x^{2} + a e^{5} x^{2} + 2 \, b d^{3} e^{2} x + 2 \, a d e^{4} x + b d^{4} e + a d^{2} e^{3}\right )}} \] Input:

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

Output:

2*a*b^2*d*p*arctan(b*x/sqrt(a*b))/((b^2*d^4 + 2*a*b*d^2*e^2 + a^2*e^4)*sqr 
t(a*b)) + 1/2*(b^2*d^2*p - a*b*e^2*p)*log(b*x^2 + a)/(b^2*d^4*e + 2*a*b*d^ 
2*e^3 + a^2*e^5) - 1/2*p*log(b*x^2 + a)/(e^3*x^2 + 2*d*e^2*x + d^2*e) - (b 
^2*d^2*p - a*b*e^2*p)*log(e*x + d)/(b^2*d^4*e + 2*a*b*d^2*e^3 + a^2*e^5) + 
 1/2*(2*b*d*e*p*x + 2*b*d^2*p - b*d^2*log(c) - a*e^2*log(c))/(b*d^2*e^3*x^ 
2 + a*e^5*x^2 + 2*b*d^3*e^2*x + 2*a*d*e^4*x + b*d^4*e + a*d^2*e^3)
 

Mupad [B] (verification not implemented)

Time = 26.63 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.56 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx=\frac {\ln \left (b^2\,x+\sqrt {-a\,b^3}\right )\,\left (b^2\,d^2\,p-a\,b\,e^2\,p+2\,d\,e\,p\,\sqrt {-a\,b^3}\right )}{2\,\left (a^2\,e^5+2\,a\,b\,d^2\,e^3+b^2\,d^4\,e\right )}-\frac {\ln \left (d+e\,x\right )\,\left (b^2\,d^2\,p-a\,b\,e^2\,p\right )}{a^2\,e^5+2\,a\,b\,d^2\,e^3+b^2\,d^4\,e}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2\,e\,\left (d^2+2\,d\,e\,x+e^2\,x^2\right )}-\frac {\ln \left (b^2\,x-\sqrt {-a\,b^3}\right )\,\left (a\,b\,e^2\,p-b^2\,d^2\,p+2\,d\,e\,p\,\sqrt {-a\,b^3}\right )}{2\,\left (a^2\,e^5+2\,a\,b\,d^2\,e^3+b^2\,d^4\,e\right )}+\frac {b\,d\,p}{\left (x\,e^2+d\,e\right )\,\left (b\,d^2+a\,e^2\right )} \] Input:

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

Output:

(log(b^2*x + (-a*b^3)^(1/2))*(b^2*d^2*p - a*b*e^2*p + 2*d*e*p*(-a*b^3)^(1/ 
2)))/(2*(a^2*e^5 + b^2*d^4*e + 2*a*b*d^2*e^3)) - (log(d + e*x)*(b^2*d^2*p 
- a*b*e^2*p))/(a^2*e^5 + b^2*d^4*e + 2*a*b*d^2*e^3) - log(c*(a + b*x^2)^p) 
/(2*e*(d^2 + e^2*x^2 + 2*d*e*x)) - (log(b^2*x - (-a*b^3)^(1/2))*(a*b*e^2*p 
 - b^2*d^2*p + 2*d*e*p*(-a*b^3)^(1/2)))/(2*(a^2*e^5 + b^2*d^4*e + 2*a*b*d^ 
2*e^3)) + (b*d*p)/((d*e + e^2*x)*(a*e^2 + b*d^2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 600, normalized size of antiderivative = 3.45 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx=\frac {8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,d^{4} e^{2} p x +4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,d^{3} e^{3} p \,x^{2}-\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} d^{2} e^{4} p -\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} e^{6} p \,x^{2}+\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} d^{4} e^{2} x^{2}+a b \,d^{4} e^{2} p -a b \,d^{2} e^{4} p \,x^{2}-b^{2} d^{4} e^{2} p \,x^{2}-2 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} d \,e^{5} p x -3 \,\mathrm {log}\left (b \,x^{2}+a \right ) a b \,d^{4} e^{2} p +2 \,\mathrm {log}\left (e x +d \right ) a b \,d^{4} e^{2} p -4 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{5} e p x -2 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{4} e^{2} p \,x^{2}+4 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a b \,d^{3} e^{3} x +2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a b \,d^{2} e^{4} x^{2}+2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{2} d \,e^{5} x +2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{2} d^{5} e x +4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,d^{5} e p -6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a b \,d^{3} e^{3} p x -3 \,\mathrm {log}\left (b \,x^{2}+a \right ) a b \,d^{2} e^{4} p \,x^{2}+4 \,\mathrm {log}\left (e x +d \right ) a b \,d^{3} e^{3} p x +2 \,\mathrm {log}\left (e x +d \right ) a b \,d^{2} e^{4} p \,x^{2}+\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{2} e^{6} x^{2}+b^{2} d^{6} p -2 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{6} p}{2 d^{2} e \left (a^{2} e^{6} x^{2}+2 a b \,d^{2} e^{4} x^{2}+b^{2} d^{4} e^{2} x^{2}+2 a^{2} d \,e^{5} x +4 a b \,d^{3} e^{3} x +2 b^{2} d^{5} e x +a^{2} d^{2} e^{4}+2 a b \,d^{4} e^{2}+b^{2} d^{6}\right )} \] Input:

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

Output:

(4*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*d**5*e*p + 8*sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*d**4*e**2*p*x + 4*sqrt(b)*sqrt(a)*at 
an((b*x)/(sqrt(b)*sqrt(a)))*b*d**3*e**3*p*x**2 - log(a + b*x**2)*a**2*d**2 
*e**4*p - 2*log(a + b*x**2)*a**2*d*e**5*p*x - log(a + b*x**2)*a**2*e**6*p* 
x**2 - 3*log(a + b*x**2)*a*b*d**4*e**2*p - 6*log(a + b*x**2)*a*b*d**3*e**3 
*p*x - 3*log(a + b*x**2)*a*b*d**2*e**4*p*x**2 + 2*log(d + e*x)*a*b*d**4*e* 
*2*p + 4*log(d + e*x)*a*b*d**3*e**3*p*x + 2*log(d + e*x)*a*b*d**2*e**4*p*x 
**2 - 2*log(d + e*x)*b**2*d**6*p - 4*log(d + e*x)*b**2*d**5*e*p*x - 2*log( 
d + e*x)*b**2*d**4*e**2*p*x**2 + 2*log((a + b*x**2)**p*c)*a**2*d*e**5*x + 
log((a + b*x**2)**p*c)*a**2*e**6*x**2 + 4*log((a + b*x**2)**p*c)*a*b*d**3* 
e**3*x + 2*log((a + b*x**2)**p*c)*a*b*d**2*e**4*x**2 + 2*log((a + b*x**2)* 
*p*c)*b**2*d**5*e*x + log((a + b*x**2)**p*c)*b**2*d**4*e**2*x**2 + a*b*d** 
4*e**2*p - a*b*d**2*e**4*p*x**2 + b**2*d**6*p - b**2*d**4*e**2*p*x**2)/(2* 
d**2*e*(a**2*d**2*e**4 + 2*a**2*d*e**5*x + a**2*e**6*x**2 + 2*a*b*d**4*e** 
2 + 4*a*b*d**3*e**3*x + 2*a*b*d**2*e**4*x**2 + b**2*d**6 + 2*b**2*d**5*e*x 
 + b**2*d**4*e**2*x**2))