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

Optimal. Leaf size=119 \[ -\frac{\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac{b d p \log \left (a+b x^2\right )}{e \left (a e^2+b d^2\right )}-\frac{2 b d p \log (d+e x)}{e \left (a e^2+b d^2\right )}+\frac{2 \sqrt{a} \sqrt{b} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a e^2+b d^2} \]

[Out]

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

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Rubi [A]  time = 0.0901617, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2463, 801, 635, 205, 260} \[ -\frac{\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac{b d p \log \left (a+b x^2\right )}{e \left (a e^2+b d^2\right )}-\frac{2 b d p \log (d+e x)}{e \left (a e^2+b d^2\right )}+\frac{2 \sqrt{a} \sqrt{b} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a e^2+b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0721267, size = 137, normalized size = 1.15 \[ \frac{-b d^2 \log \left (c \left (a+b x^2\right )^p\right )-a e^2 \log \left (c \left (a+b x^2\right )^p\right )+b d^2 p \log \left (a+b x^2\right )+b d e p x \log \left (a+b x^2\right )+2 \sqrt{a} \sqrt{b} e p (d+e x) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )-2 b d p (d+e x) \log (d+e x)}{e (d+e x) \left (a e^2+b d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.548, size = 755, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/e/(e*x+d)*ln((b*x^2+a)^p)+1/2*(-I*Pi*a*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*e^2+I*Pi*b*d^2*csgn(I*c*(b*x^2+a)^
p)^3-I*Pi*b*d^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+I*Pi*a*csgn(I*c*(b*x^2+a)^p)^3*e^2+I*Pi*b*d^2*csgn(I*(b*x^2+
a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*b*d^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*a*csgn(I*(b*
x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2*e^2+I*Pi*a*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)*e^2+2*sum(_R*
ln(((3*a*e^4-b*d^2*e^2)*_R^2-b*d*e*p*_R+2*b*p^2)*x+4*a*d*e^3*_R^2-a*e^2*p*_R),_R=RootOf((a*e^4+b*d^2*e^2)*_Z^2
-2*b*d*e*p*_Z+b*p^2))*a*e^4*x+2*sum(_R*ln(((3*a*e^4-b*d^2*e^2)*_R^2-b*d*e*p*_R+2*b*p^2)*x+4*a*d*e^3*_R^2-a*e^2
*p*_R),_R=RootOf((a*e^4+b*d^2*e^2)*_Z^2-2*b*d*e*p*_Z+b*p^2))*b*d^2*e^2*x-4*ln(e*x+d)*b*d*e*p*x+2*sum(_R*ln(((3
*a*e^4-b*d^2*e^2)*_R^2-b*d*e*p*_R+2*b*p^2)*x+4*a*d*e^3*_R^2-a*e^2*p*_R),_R=RootOf((a*e^4+b*d^2*e^2)*_Z^2-2*b*d
*e*p*_Z+b*p^2))*a*d*e^3+2*sum(_R*ln(((3*a*e^4-b*d^2*e^2)*_R^2-b*d*e*p*_R+2*b*p^2)*x+4*a*d*e^3*_R^2-a*e^2*p*_R)
,_R=RootOf((a*e^4+b*d^2*e^2)*_Z^2-2*b*d*e*p*_Z+b*p^2))*b*d^3*e-4*ln(e*x+d)*b*d^2*p-2*ln(c)*a*e^2-2*ln(c)*b*d^2
)/(e*x+d)/e/(a*e^2+b*d^2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.55649, size = 582, normalized size = 4.89 \begin{align*} \left [\frac{{\left (e^{2} p x + d e p\right )} \sqrt{-a b} \log \left (\frac{b x^{2} + 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right ) +{\left (b d e p x - a e^{2} p\right )} \log \left (b x^{2} + a\right ) - 2 \,{\left (b d e p x + b d^{2} p\right )} \log \left (e x + d\right ) -{\left (b d^{2} + a e^{2}\right )} \log \left (c\right )}{b d^{3} e + a d e^{3} +{\left (b d^{2} e^{2} + a e^{4}\right )} x}, \frac{2 \,{\left (e^{2} p x + d e p\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) +{\left (b d e p x - a e^{2} p\right )} \log \left (b x^{2} + a\right ) - 2 \,{\left (b d e p x + b d^{2} p\right )} \log \left (e x + d\right ) -{\left (b d^{2} + a e^{2}\right )} \log \left (c\right )}{b d^{3} e + a d e^{3} +{\left (b d^{2} e^{2} + a e^{4}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.22342, size = 213, normalized size = 1.79 \begin{align*} \frac{b d p \log \left (b x^{2} + a\right )}{b d^{2} e + a e^{3}} + \frac{2 \, a b p \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{{\left (b d^{2} + a e^{2}\right )} \sqrt{a b}} - \frac{2 \, b d p x e \log \left (x e + d\right ) + b d^{2} p \log \left (b x^{2} + a\right ) + 2 \, b d^{2} p \log \left (x e + d\right ) + a p e^{2} \log \left (b x^{2} + a\right ) + b d^{2} \log \left (c\right ) + a e^{2} \log \left (c\right )}{b d^{2} x e^{2} + b d^{3} e + a x e^{4} + a d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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