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

Optimal. Leaf size=174 \[ \frac {2 \sqrt {a} b^{3/2} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\left (a e^2+b d^2\right )^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2}+\frac {b p \left (b d^2-a e^2\right ) \log \left (a+b x^2\right )}{2 e \left (a e^2+b d^2\right )^2}+\frac {b d p}{e (d+e x) \left (a e^2+b d^2\right )}-\frac {b p \left (b d^2-a e^2\right ) \log (d+e x)}{e \left (a e^2+b d^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2463, 801, 635, 205, 260} \[ \frac {2 \sqrt {a} b^{3/2} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\left (a e^2+b d^2\right )^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2}+\frac {b p \left (b d^2-a e^2\right ) \log \left (a+b x^2\right )}{2 e \left (a e^2+b d^2\right )^2}+\frac {b d p}{e (d+e x) \left (a e^2+b d^2\right )}-\frac {b p \left (b d^2-a e^2\right ) \log (d+e x)}{e \left (a e^2+b d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2463

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] - Dist[(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]

Rubi steps

\begin {align*} \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^3} \, dx &=-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2}+\frac {(b p) \int \frac {x}{(d+e x)^2 \left (a+b x^2\right )} \, dx}{e}\\ &=-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2}+\frac {(b p) \int \left (-\frac {d e}{\left (b d^2+a e^2\right ) (d+e x)^2}+\frac {e \left (-b d^2+a e^2\right )}{\left (b d^2+a e^2\right )^2 (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 )^2 \left (a+b x^2\right )}\right ) \, dx}{e}\\ &=\frac {b d p}{e \left (b d^2+a e^2\right ) (d+e x)}-\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 {\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2}+\frac {\left (b^2 p\right ) \int \frac {2 a d e+\left (b d^2-a e^2\right ) x}{a+b x^2} \, dx}{e \left (b d^2+a e^2\right )^2}\\ &=\frac {b d p}{e \left (b d^2+a e^2\right ) (d+e x)}-\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 {\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2}+\frac {\left (2 a b^2 d p\right ) \int \frac {1}{a+b x^2} \, dx}{\left (b d^2+a e^2\right )^2}+\frac {\left (b^2 \left (b d^2-a e^2\right ) p\right ) \int \frac {x}{a+b x^2} \, dx}{e \left (b d^2+a e^2\right )^2}\\ &=\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 \tan ^{-1}\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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.57, size = 217, normalized size = 1.25 \[ \frac {\frac {b p (d+e x) \left ((d+e x) \left (\sqrt {-a} b d^2+2 a \sqrt {b} d e+(-a)^{3/2} e^2\right ) \log \left (\sqrt {-a}-\sqrt {b} x\right )+(d+e x) \left (\sqrt {-a} b d^2-2 a \sqrt {b} d e+(-a)^{3/2} e^2\right ) \log \left (\sqrt {-a}+\sqrt {b} x\right )+2 \sqrt {-a} \left (-(d+e x) \left (b d^2-a e^2\right ) \log (d+e x)+a d e^2+b d^3\right )\right )}{\sqrt {-a} \left (a e^2+b d^2\right )^2}-\log \left (c \left (a+b x^2\right )^p\right )}{2 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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] + (Sq
rt[-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)

________________________________________________________________________________________

fricas [B]  time = 0.55, size = 744, normalized size = 4.28 \[ \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 \relax (c)}{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 \relax (c)}{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 ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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)]

________________________________________________________________________________________

giac [B]  time = 0.22, size = 420, normalized size = 2.41 \[ \frac {2 \, a b^{2} d p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a b}} + \frac {{\left (b^{2} d^{2} p - a b p e^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} d^{4} e + 2 \, a b d^{2} e^{3} + a^{2} e^{5}\right )}} - \frac {2 \, b^{2} d^{2} p x^{2} e^{2} \log \left (x e + d\right ) + 4 \, b^{2} d^{3} p x e \log \left (x e + d\right ) - 2 \, b^{2} d^{3} p x e + b^{2} d^{4} p \log \left (b x^{2} + a\right ) + 2 \, b^{2} d^{4} p \log \left (x e + d\right ) - 2 \, b^{2} d^{4} p + 2 \, a b d^{2} p e^{2} \log \left (b x^{2} + a\right ) - 2 \, a b p x^{2} e^{4} \log \left (x e + d\right ) - 4 \, a b d p x e^{3} \log \left (x e + d\right ) - 2 \, a b d^{2} p e^{2} \log \left (x e + d\right ) + b^{2} d^{4} \log \relax (c) - 2 \, a b d p x e^{3} - 2 \, a b d^{2} p e^{2} + 2 \, a b d^{2} e^{2} \log \relax (c) + a^{2} p e^{4} \log \left (b x^{2} + a\right ) + a^{2} e^{4} \log \relax (c)}{2 \, {\left (b^{2} d^{4} x^{2} e^{3} + 2 \, b^{2} d^{5} x e^{2} + b^{2} d^{6} e + 2 \, a b d^{2} x^{2} e^{5} + 4 \, a b d^{3} x e^{4} + 2 \, a b d^{4} e^{3} + a^{2} x^{2} e^{7} + 2 \, a^{2} d x e^{6} + a^{2} d^{2} e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.92, size = 2684, normalized size = 15.43 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/e/(e*x+d)^2*ln((b*x^2+a)^p)+1/4*(-2*I*Pi*a*b*d^2*e^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-2*ln(c)*
b^2*d^4-2*ln(c)*a^2*e^4+4*b^2*d^4*p+2*sum(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)*_R^2+(3
*a^2*b*e^5*p+2*a*b^2*d^2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d^5*e^3)*
_R^2+2*a*b^2*d*e*p^2),_R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_Z+b^2*p^
2))*a^2*e^7*x^2+2*sum(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)*_R^2+(3*a^2*b*e^5*p+2*a*b^2
*d^2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d^5*e^3)*_R^2+2*a*b^2*d*e*p^2
),_R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_Z+b^2*p^2))*a^2*d^2*e^5+2*su
m(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)*_R^2+(3*a^2*b*e^5*p+2*a*b^2*d^2*e^3*p-b^3*d^4*e
*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d^5*e^3)*_R^2+2*a*b^2*d*e*p^2),_R=RootOf((a^2*e^6
+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_Z+b^2*p^2))*b^2*d^6*e-4*ln(e*x+d)*b^2*d^4*p+4*a*
d^2*b*p*e^2+I*Pi*a^2*e^4*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*b^2*d^4*csgn(I*c*(b*x^2+a)^p)^3-2*I*Pi*a*b*d^2*e^2*csgn(
I*c*(b*x^2+a)^p)^2*csgn(I*c)+4*sum(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)*_R^2+(3*a^2*b*
e^5*p+2*a*b^2*d^2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d^5*e^3)*_R^2+2*
a*b^2*d*e*p^2),_R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_Z+b^2*p^2))*a*b
*d^2*e^5*x^2+8*sum(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)*_R^2+(3*a^2*b*e^5*p+2*a*b^2*d^
2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d^5*e^3)*_R^2+2*a*b^2*d*e*p^2),_
R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_Z+b^2*p^2))*a*b*d^3*e^4*x+4*ln(
e*x+d)*a*b*e^4*p*x^2-4*ln(e*x+d)*b^2*d^2*e^2*p*x^2-8*ln(e*x+d)*b^2*d^3*e*p*x+4*ln(e*x+d)*a*b*d^2*e^2*p+4*a*d*p
*e^3*x*b+2*I*Pi*a*b*d^2*e^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+2*sum(_R*ln(((3*a^3*e^8+5*a^2*
b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)*_R^2+(3*a^2*b*e^5*p+2*a*b^2*d^2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4
*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d^5*e^3)*_R^2+2*a*b^2*d*e*p^2),_R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2
)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_Z+b^2*p^2))*b^2*d^4*e^3*x^2+4*sum(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*
d^4*e^4-b^3*d^6*e^2)*_R^2+(3*a^2*b*e^5*p+2*a*b^2*d^2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2
*b*d^3*e^5+4*a*b^2*d^5*e^3)*_R^2+2*a*b^2*d*e*p^2),_R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^
3*p-2*b^2*d^2*e*p)*_Z+b^2*p^2))*a^2*d*e^6*x+4*sum(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)
*_R^2+(3*a^2*b*e^5*p+2*a*b^2*d^2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d
^5*e^3)*_R^2+2*a*b^2*d*e*p^2),_R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_
Z+b^2*p^2))*b^2*d^5*e^2*x+4*sum(_R*ln(((3*a^3*e^8+5*a^2*b*d^2*e^6+a*b^2*d^4*e^4-b^3*d^6*e^2)*_R^2+(3*a^2*b*e^5
*p+2*a*b^2*d^2*e^3*p-b^3*d^4*e*p)*_R+2*b^3*d^2*p^2)*x+(4*a^3*d*e^7+8*a^2*b*d^3*e^5+4*a*b^2*d^5*e^3)*_R^2+2*a*b
^2*d*e*p^2),_R=RootOf((a^2*e^6+2*a*b*d^2*e^4+b^2*d^4*e^2)*_Z^2+(2*a*b*e^3*p-2*b^2*d^2*e*p)*_Z+b^2*p^2))*a*b*d^
4*e^3-4*ln(c)*a*b*d^2*e^2+4*d^3*p*x*b^2*e+I*Pi*b^2*d^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+8*l
n(e*x+d)*a*b*d*e^3*p*x-I*Pi*b^2*d^4*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-I*Pi*a^2*e^4*csgn(I*(b*x^2+a)^p)*csgn(I*
c*(b*x^2+a)^p)^2-I*Pi*a^2*e^4*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-I*Pi*b^2*d^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x
^2+a)^p)^2+2*I*Pi*a*b*d^2*e^2*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*a^2*e^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*c
sgn(I*c))/(e*x+d)^2/e/(a*e^2+b*d^2)^2

________________________________________________________________________________________

maxima [A]  time = 1.07, size = 206, normalized size = 1.18 \[ \frac {{\left (\frac {4 \, a b d e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a b}} + \frac {{\left (b d^{2} - a e^{2}\right )} \log \left (b x^{2} + a\right )}{b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}} - \frac {2 \, {\left (b d^{2} - a e^{2}\right )} \log \left (e x + d\right )}{b^{2} d^{4} + 2 \, a b d^{2} e^{2} + a^{2} e^{4}} + \frac {2 \, d}{b d^{3} + a d e^{2} + {\left (b d^{2} e + a e^{3}\right )} x}\right )} b p}{2 \, e} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{2 \, {\left (e x + d\right )}^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

________________________________________________________________________________________

mupad [B]  time = 0.98, size = 272, normalized size = 1.56 \[ \frac {\ln \left (b^2\,x+\sqrt {-a\,b^3}\right )\,\left (b^2\,d^2\,p-a\,b\,e^2\,p+2\,d\,e\,p\,\sqrt {-a\,b^3}\right )}{2\,\left (a^2\,e^5+2\,a\,b\,d^2\,e^3+b^2\,d^4\,e\right )}-\frac {\ln \left (d+e\,x\right )\,\left (b^2\,d^2\,p-a\,b\,e^2\,p\right )}{a^2\,e^5+2\,a\,b\,d^2\,e^3+b^2\,d^4\,e}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2\,e\,\left (d^2+2\,d\,e\,x+e^2\,x^2\right )}-\frac {\ln \left (b^2\,x-\sqrt {-a\,b^3}\right )\,\left (a\,b\,e^2\,p-b^2\,d^2\,p+2\,d\,e\,p\,\sqrt {-a\,b^3}\right )}{2\,\left (a^2\,e^5+2\,a\,b\,d^2\,e^3+b^2\,d^4\,e\right )}+\frac {b\,d\,p}{\left (x\,e^2+d\,e\right )\,\left (b\,d^2+a\,e^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________