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

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Rubi [A]  time = 0.09, antiderivative size = 119, 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} \[ -\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac {b d p \log \left (a+b x^2\right )}{e \left (a e^2+b d^2\right )}-\frac {2 b d p \log (d+e x)}{e \left (a e^2+b d^2\right )}+\frac {2 \sqrt {a} \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a e^2+b d^2} \]

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 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)^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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 0.47, size = 261, normalized size = 2.19 \[ \left [\frac {{\left (e^{2} p x + d e p\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + {\left (b d e p x - a e^{2} p\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b d e p x + b d^{2} p\right )} \log \left (e x + d\right ) - {\left (b d^{2} + a e^{2}\right )} \log \relax (c)}{b d^{3} e + a d e^{3} + {\left (b d^{2} e^{2} + a e^{4}\right )} x}, \frac {2 \, {\left (e^{2} p x + d e p\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (b d e p x - a e^{2} p\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b d e p x + b d^{2} p\right )} \log \left (e x + d\right ) - {\left (b d^{2} + a e^{2}\right )} \log \relax (c)}{b d^{3} e + a d e^{3} + {\left (b d^{2} e^{2} + a e^{4}\right )} x}\right ] \]

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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giac [A]  time = 0.22, size = 158, normalized size = 1.33 \[ \frac {b d p \log \left (b x^{2} + a\right )}{b d^{2} e + a e^{3}} + \frac {2 \, a b p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b d^{2} + a e^{2}\right )} \sqrt {a b}} - \frac {2 \, b d p x e \log \left (x e + d\right ) + b d^{2} p \log \left (b x^{2} + a\right ) + 2 \, b d^{2} p \log \left (x e + d\right ) + a p e^{2} \log \left (b x^{2} + a\right ) + b d^{2} \log \relax (c) + a e^{2} \log \relax (c)}{b d^{2} x e^{2} + b d^{3} e + a x e^{4} + a d e^{3}} \]

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)

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maple [C]  time = 0.61, size = 1233, normalized size = 10.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/e/(e*x+d)*ln((b*x^2+a)^p)-1/2*b*(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*b*d^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-I*Pi*a*csgn(I*(b*x^2+a)^p)*cs
gn(I*c*(b*x^2+a)^p)*csgn(I*c)*e^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*(b*x^2+a)^p)*
csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*b*d^2*csgn(I*c*(b*x^2+a)^p)^3-I*Pi*a*csgn(I*c*(b*x^2+a)^p)^3*e^2-2*(-a*b)
^(1/2)*ln((-a^2*b*e^3+7*a*b^2*d^2*e+5*(-a*b)^(1/2)*a*b*d*e^2-3*(-a*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^2-(-a*b)^(1
/2)*a^2*e^3+7*(-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d^3)*e^2*p*x-2*ln((-a^2*b*e^3+7*a*b^2*d^2*e+5*(-a*b)^(1/2)*a*b*d*
e^2-3*(-a*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^2-(-a*b)^(1/2)*a^2*e^3+7*(-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d^3)*b*d*e*p
*x+2*(-a*b)^(1/2)*ln((-a^2*b*e^3+7*a*b^2*d^2*e-5*(-a*b)^(1/2)*a*b*d*e^2+3*(-a*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^
2+(-a*b)^(1/2)*a^2*e^3-7*(-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d^3)*e^2*p*x-2*ln((-a^2*b*e^3+7*a*b^2*d^2*e-5*(-a*b)^(
1/2)*a*b*d*e^2+3*(-a*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^2+(-a*b)^(1/2)*a^2*e^3-7*(-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d
^3)*b*d*e*p*x+4*ln(e*x+d)*b*d*e*p*x-2*(-a*b)^(1/2)*ln((-a^2*b*e^3+7*a*b^2*d^2*e+5*(-a*b)^(1/2)*a*b*d*e^2-3*(-a
*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^2-(-a*b)^(1/2)*a^2*e^3+7*(-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d^3)*d*e*p-2*ln((-a^2
*b*e^3+7*a*b^2*d^2*e+5*(-a*b)^(1/2)*a*b*d*e^2-3*(-a*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^2-(-a*b)^(1/2)*a^2*e^3+7*(
-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d^3)*b*d^2*p+2*(-a*b)^(1/2)*ln((-a^2*b*e^3+7*a*b^2*d^2*e-5*(-a*b)^(1/2)*a*b*d*e^
2+3*(-a*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^2+(-a*b)^(1/2)*a^2*e^3-7*(-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d^3)*d*e*p-2*l
n((-a^2*b*e^3+7*a*b^2*d^2*e-5*(-a*b)^(1/2)*a*b*d*e^2+3*(-a*b)^(1/2)*b^2*d^3)*x-5*a^2*b*d*e^2+(-a*b)^(1/2)*a^2*
e^3-7*(-a*b)^(1/2)*a*b*d^2*e+3*a*b^2*d^3)*b*d^2*p+4*ln(e*x+d)*b*d^2*p+2*ln(c)*a*e^2+2*b*d^2*ln(c))/(e*x+d)/(b*
d-(-a*b)^(1/2)*e)/(b*d+(-a*b)^(1/2)*e)/e

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maxima [A]  time = 1.51, size = 108, normalized size = 0.91 \[ \frac {{\left (\frac {2 \, a e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b d^{2} + a e^{2}\right )} \sqrt {a b}} + \frac {d \log \left (b x^{2} + a\right )}{b d^{2} + a e^{2}} - \frac {2 \, d \log \left (e x + d\right )}{b d^{2} + a e^{2}}\right )} b p}{e} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{{\left (e x + d\right )} e} \]

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]

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

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mupad [B]  time = 1.26, size = 337, normalized size = 2.83 \[ \frac {\ln \left (\frac {4\,b^3\,p^2\,x}{e}-\frac {p\,\left (b\,d+e\,\sqrt {-a\,b}\right )\,\left (2\,a\,b^2\,e\,p+2\,b^3\,d\,p\,x-\frac {2\,b^2\,e\,p\,\left (b\,d+e\,\sqrt {-a\,b}\right )\,\left (-b\,x\,d^2+4\,a\,d\,e+3\,a\,x\,e^2\right )}{b\,d^2\,e+a\,e^3}\right )}{b\,d^2\,e+a\,e^3}\right )\,\left (b\,d\,p+e\,p\,\sqrt {-a\,b}\right )}{b\,d^2\,e+a\,e^3}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{e\,\left (d+e\,x\right )}+\frac {\ln \left (\frac {4\,b^3\,p^2\,x}{e}-\frac {p\,\left (b\,d-e\,\sqrt {-a\,b}\right )\,\left (2\,a\,b^2\,e\,p+2\,b^3\,d\,p\,x-\frac {2\,b^2\,e\,p\,\left (b\,d-e\,\sqrt {-a\,b}\right )\,\left (-b\,x\,d^2+4\,a\,d\,e+3\,a\,x\,e^2\right )}{b\,d^2\,e+a\,e^3}\right )}{b\,d^2\,e+a\,e^3}\right )\,\left (b\,d\,p-e\,p\,\sqrt {-a\,b}\right )}{b\,d^2\,e+a\,e^3}-\frac {2\,b\,d\,p\,\ln \left (d+e\,x\right )}{b\,d^2\,e+a\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**2,x)

[Out]

Timed out

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