3.2.89 \(\int \frac {\log (c (a+b x^2)^p)}{(d+e x)^2} \, dx\) [189]

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

[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 = 0.07, 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 = {2513, 815, 649, 211, 266} \begin {gather*} \frac {2 \sqrt {a} \sqrt {b} p \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a e^2+b d^2}-\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 )} \end {gather*}

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 211

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

Rule 266

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 649

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 815

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 2513

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 = 0.06, size = 137, normalized size = 1.15 \begin {gather*} \frac {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)+b d^2 p \log \left (a+b x^2\right )+b d e p x \log \left (a+b x^2\right )-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 )}{e \left (b d^2+a e^2\right ) (d+e x)} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.91, size = 755, normalized size = 6.34

method	result	size
risch	\(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{e \left (e x +d \right )}+\frac {-i \pi b \,d^{2} \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}+i \pi a \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} e^{2}-i \pi b \,d^{2} \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )+i \pi a \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) e^{2}-i \pi a \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) e^{2}-i \pi a \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} e^{2}+i \pi b \,d^{2} \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}+i \pi b \,d^{2} \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )+2 \left (\munderset {\textit {\_R} =\RootOf \left (\left (a \,e^{4}+b \,d^{2} e^{2}\right ) \textit {\_Z}^{2}-2 b d e p \textit {\_Z} +b \,p^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a \,e^{4}-b \,d^{2} e^{2}\right ) \textit {\_R}^{2}-b d e p \textit {\_R} +2 b \,p^{2}\right ) x +4 a d \,e^{3} \textit {\_R}^{2}-a \,e^{2} p \textit {\_R} \right )\right ) a \,e^{4} x +2 \left (\munderset {\textit {\_R} =\RootOf \left (\left (a \,e^{4}+b \,d^{2} e^{2}\right ) \textit {\_Z}^{2}-2 b d e p \textit {\_Z} +b \,p^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a \,e^{4}-b \,d^{2} e^{2}\right ) \textit {\_R}^{2}-b d e p \textit {\_R} +2 b \,p^{2}\right ) x +4 a d \,e^{3} \textit {\_R}^{2}-a \,e^{2} p \textit {\_R} \right )\right ) b \,d^{2} e^{2} x -4 \ln \left (e x +d \right ) b d e p x +2 \left (\munderset {\textit {\_R} =\RootOf \left (\left (a \,e^{4}+b \,d^{2} e^{2}\right ) \textit {\_Z}^{2}-2 b d e p \textit {\_Z} +b \,p^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a \,e^{4}-b \,d^{2} e^{2}\right ) \textit {\_R}^{2}-b d e p \textit {\_R} +2 b \,p^{2}\right ) x +4 a d \,e^{3} \textit {\_R}^{2}-a \,e^{2} p \textit {\_R} \right )\right ) a d \,e^{3}+2 \left (\munderset {\textit {\_R} =\RootOf \left (\left (a \,e^{4}+b \,d^{2} e^{2}\right ) \textit {\_Z}^{2}-2 b d e p \textit {\_Z} +b \,p^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a \,e^{4}-b \,d^{2} e^{2}\right ) \textit {\_R}^{2}-b d e p \textit {\_R} +2 b \,p^{2}\right ) x +4 a d \,e^{3} \textit {\_R}^{2}-a \,e^{2} p \textit {\_R} \right )\right ) b \,d^{3} e -4 \ln \left (e x +d \right ) b \,d^{2} p -2 \ln \left (c \right ) a \,e^{2}-2 d^{2} b \ln \left (c \right )}{2 \left (e x +d \right ) e \left (a \,e^{2}+b \,d^{2}\right )}\)	\(755\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/e/(e*x+d)*ln((b*x^2+a)^p)+1/2*(-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*c*(b*x
^2+a)^p)^3*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)*csgn(I*c*(b*x^2+a)^p)*c
sgn(I*c)*e^2-I*Pi*a*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)*e^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)^3+I*Pi*b*d^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*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*d^2*b*ln(c)
)/(e*x+d)/e/(a*e^2+b*d^2)

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Maxima [A]
time = 0.57, size = 106, normalized size = 0.89 \begin {gather*} {\left (\frac {2 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right ) e}{{\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 (x e + d\right )}{b d^{2} + a e^{2}}\right )} b p e^{\left (-1\right )} - \frac {e^{\left (-1\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{x e + d} \end {gather*}

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

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Fricas [A]
time = 0.42, size = 257, normalized size = 2.16 \begin {gather*} \left [\frac {{\left (p x e^{2} + d p e\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + {\left (b d p x e - a p e^{2}\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b d p x e + b d^{2} p\right )} \log \left (x e + d\right ) - {\left (b d^{2} + a e^{2}\right )} \log \left (c\right )}{b d^{2} x e^{2} + b d^{3} e + a x e^{4} + a d e^{3}}, \frac {2 \, {\left (p x e^{2} + d p e\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (b d p x e - a p e^{2}\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b d p x e + b d^{2} p\right )} \log \left (x e + d\right ) - {\left (b d^{2} + a e^{2}\right )} \log \left (c\right )}{b d^{2} x e^{2} + b d^{3} e + a x e^{4} + a d e^{3}}\right ] \end {gather*}

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]

[((p*x*e^2 + d*p*e)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + (b*d*p*x*e - a*p*e^2)*log(b*x^2
+ a) - 2*(b*d*p*x*e + b*d^2*p)*log(x*e + d) - (b*d^2 + a*e^2)*log(c))/(b*d^2*x*e^2 + b*d^3*e + a*x*e^4 + a*d*
e^3), (2*(p*x*e^2 + d*p*e)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + (b*d*p*x*e - a*p*e^2)*log(b*x^2 + a) - 2*(b*d*p*x
*e + b*d^2*p)*log(x*e + d) - (b*d^2 + 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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: AttributeError} \end {gather*}

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]

Exception raised: AttributeError >> 'NoneType' object has no attribute 'primitive'

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Giac [A]
time = 4.72, size = 158, normalized size = 1.33 \begin {gather*} \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 {gather*}

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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Mupad [B]
time = 1.26, size = 337, normalized size = 2.83 \begin {gather*} \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} \end {gather*}

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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