3.1.9 \(\int \frac {\log (c (a+b x^2)^p)}{x^4} \, dx\) [9]

Optimal. Leaf size=60 \[ -\frac {2 b p}{3 a x}-\frac {2 b^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2505, 331, 211} \begin {gather*} -\frac {2 b^{3/2} p \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3}-\frac {2 b p}{3 a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/x^4,x]

[Out]

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

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx &=-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3}+\frac {1}{3} (2 b p) \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx\\ &=-\frac {2 b p}{3 a x}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3}-\frac {\left (2 b^2 p\right ) \int \frac {1}{a+b x^2} \, dx}{3 a}\\ &=-\frac {2 b p}{3 a x}-\frac {2 b^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.00, size = 49, normalized size = 0.82 \begin {gather*} -\frac {2 b p \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {b x^2}{a}\right )}{3 a x}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/x^4,x]

[Out]

(-2*b*p*Hypergeometric2F1[-1/2, 1, 1/2, -((b*x^2)/a)])/(3*a*x) - Log[c*(a + b*x^2)^p]/(3*x^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.32, size = 211, normalized size = 3.52

method	result	size
risch	\(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 x^{3}}-\frac {i \pi a \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}-i \pi a \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )-i \pi a \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}+i \pi a \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )-2 \left (\munderset {\textit {\_R} =\RootOf \left (a^{3} \textit {\_Z}^{2}+b^{3} p^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 a^{3} \textit {\_R}^{2}+2 b^{3} p^{2}\right ) x +a^{2} b p \textit {\_R} \right )\right ) x^{3} a +4 x^{2} p b +2 \ln \left (c \right ) a}{6 x^{3} a}\)	\(211\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3/x^3*ln((b*x^2+a)^p)-1/6*(I*Pi*a*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)*cs
gn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*a*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*a*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-2*sum(_R
*ln((3*_R^2*a^3+2*b^3*p^2)*x+a^2*b*p*_R),_R=RootOf(_Z^2*a^3+b^3*p^2))*x^3*a+4*x^2*p*b+2*ln(c)*a)/x^3/a

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Maxima [A]
time = 0.49, size = 49, normalized size = 0.82 \begin {gather*} -\frac {2}{3} \, b p {\left (\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a} + \frac {1}{a x}\right )} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.41, size = 135, normalized size = 2.25 \begin {gather*} \left [\frac {b p x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 2 \, b p x^{2} - a p \log \left (b x^{2} + a\right ) - a \log \left (c\right )}{3 \, a x^{3}}, -\frac {2 \, b p x^{3} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 2 \, b p x^{2} + a p \log \left (b x^{2} + a\right ) + a \log \left (c\right )}{3 \, a x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^4,x, algorithm="fricas")

[Out]

[1/3*(b*p*x^3*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 2*b*p*x^2 - a*p*log(b*x^2 + a) - a*
log(c))/(a*x^3), -1/3*(2*b*p*x^3*sqrt(b/a)*arctan(x*sqrt(b/a)) + 2*b*p*x^2 + a*p*log(b*x^2 + a) + a*log(c))/(a
*x^3)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (56) = 112\).
time = 40.55, size = 479, normalized size = 7.98 \begin {gather*} \begin {cases} - \frac {\log {\left (0^{p} c \right )}}{3 x^{3}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {\log {\left (a^{p} c \right )}}{3 x^{3}} & \text {for}\: b = 0 \\- \frac {2 p}{9 x^{3}} - \frac {\log {\left (c \left (b x^{2}\right )^{p} \right )}}{3 x^{3}} & \text {for}\: a = 0 \\- \frac {a^{2} \sqrt {- \frac {a}{b}} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{3 a^{2} x^{3} \sqrt {- \frac {a}{b}} + 3 a b x^{5} \sqrt {- \frac {a}{b}}} - \frac {2 a p x^{3} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{\frac {3 a^{2} x^{3} \sqrt {- \frac {a}{b}}}{b} + 3 a x^{5} \sqrt {- \frac {a}{b}}} - \frac {2 a p x^{2} \sqrt {- \frac {a}{b}}}{\frac {3 a^{2} x^{3} \sqrt {- \frac {a}{b}}}{b} + 3 a x^{5} \sqrt {- \frac {a}{b}}} + \frac {a x^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{\frac {3 a^{2} x^{3} \sqrt {- \frac {a}{b}}}{b} + 3 a x^{5} \sqrt {- \frac {a}{b}}} - \frac {a x^{2} \sqrt {- \frac {a}{b}} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{\frac {3 a^{2} x^{3} \sqrt {- \frac {a}{b}}}{b} + 3 a x^{5} \sqrt {- \frac {a}{b}}} - \frac {2 b p x^{5} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{\frac {3 a^{2} x^{3} \sqrt {- \frac {a}{b}}}{b} + 3 a x^{5} \sqrt {- \frac {a}{b}}} - \frac {2 b p x^{4} \sqrt {- \frac {a}{b}}}{\frac {3 a^{2} x^{3} \sqrt {- \frac {a}{b}}}{b} + 3 a x^{5} \sqrt {- \frac {a}{b}}} + \frac {b x^{5} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{\frac {3 a^{2} x^{3} \sqrt {- \frac {a}{b}}}{b} + 3 a x^{5} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)/x**4,x)

[Out]

Piecewise((-log(0**p*c)/(3*x**3), Eq(a, 0) & Eq(b, 0)), (-log(a**p*c)/(3*x**3), Eq(b, 0)), (-2*p/(9*x**3) - lo
g(c*(b*x**2)**p)/(3*x**3), Eq(a, 0)), (-a**2*sqrt(-a/b)*log(c*(a + b*x**2)**p)/(3*a**2*x**3*sqrt(-a/b) + 3*a*b
*x**5*sqrt(-a/b)) - 2*a*p*x**3*log(x - sqrt(-a/b))/(3*a**2*x**3*sqrt(-a/b)/b + 3*a*x**5*sqrt(-a/b)) - 2*a*p*x*
*2*sqrt(-a/b)/(3*a**2*x**3*sqrt(-a/b)/b + 3*a*x**5*sqrt(-a/b)) + a*x**3*log(c*(a + b*x**2)**p)/(3*a**2*x**3*sq
rt(-a/b)/b + 3*a*x**5*sqrt(-a/b)) - a*x**2*sqrt(-a/b)*log(c*(a + b*x**2)**p)/(3*a**2*x**3*sqrt(-a/b)/b + 3*a*x
**5*sqrt(-a/b)) - 2*b*p*x**5*log(x - sqrt(-a/b))/(3*a**2*x**3*sqrt(-a/b)/b + 3*a*x**5*sqrt(-a/b)) - 2*b*p*x**4
*sqrt(-a/b)/(3*a**2*x**3*sqrt(-a/b)/b + 3*a*x**5*sqrt(-a/b)) + b*x**5*log(c*(a + b*x**2)**p)/(3*a**2*x**3*sqrt
(-a/b)/b + 3*a*x**5*sqrt(-a/b)), True))

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Giac [A]
time = 4.61, size = 58, normalized size = 0.97 \begin {gather*} -\frac {2 \, b^{2} p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a} - \frac {p \log \left (b x^{2} + a\right )}{3 \, x^{3}} - \frac {2 \, b p x^{2} + a \log \left (c\right )}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.25, size = 46, normalized size = 0.77 \begin {gather*} -\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{3\,x^3}-\frac {2\,b^{3/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{3\,a^{3/2}}-\frac {2\,b\,p}{3\,a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^2)^p)/x^4,x)

[Out]

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

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