3.1.1 \(\int x^4 \log (c (a+b x^2)^p) \, dx\) [1]

Optimal. Leaf size=80 \[ -\frac {2 a^2 p x}{5 b^2}+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25}+\frac {2 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right ) \]

[Out]

-2/5*a^2*p*x/b^2+2/15*a*p*x^3/b-2/25*p*x^5+2/5*a^(5/2)*p*arctan(x*b^(1/2)/a^(1/2))/b^(5/2)+1/5*x^5*ln(c*(b*x^2
+a)^p)

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Rubi [A]
time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2505, 308, 211} \begin {gather*} \frac {2 a^{5/2} p \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}-\frac {2 a^2 p x}{5 b^2}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*Log[c*(a + b*x^2)^p],x]

[Out]

(-2*a^2*p*x)/(5*b^2) + (2*a*p*x^3)/(15*b) - (2*p*x^5)/25 + (2*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(5*b^(5/2
)) + (x^5*Log[c*(a + b*x^2)^p])/5

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{5} (2 b p) \int \frac {x^6}{a+b x^2} \, dx\\ &=\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{5} (2 b p) \int \left (\frac {a^2}{b^3}-\frac {a x^2}{b^2}+\frac {x^4}{b}-\frac {a^3}{b^3 \left (a+b x^2\right )}\right ) \, dx\\ &=-\frac {2 a^2 p x}{5 b^2}+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {\left (2 a^3 p\right ) \int \frac {1}{a+b x^2} \, dx}{5 b^2}\\ &=-\frac {2 a^2 p x}{5 b^2}+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25}+\frac {2 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 74, normalized size = 0.92 \begin {gather*} \frac {1}{75} \left (-\frac {30 a^2 p x}{b^2}+\frac {10 a p x^3}{b}-6 p x^5+\frac {30 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}+15 x^5 \log \left (c \left (a+b x^2\right )^p\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[c*(a + b*x^2)^p],x]

[Out]

((-30*a^2*p*x)/b^2 + (10*a*p*x^3)/b - 6*p*x^5 + (30*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(5/2) + 15*x^5*Lo
g[c*(a + b*x^2)^p])/75

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.37, size = 229, normalized size = 2.86

method	result	size
risch	\(\frac {x^{5} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5}+\frac {i \pi \,x^{5} \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{10}-\frac {i \pi \,x^{5} \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{10}-\frac {i \pi \,x^{5} \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}}{10}+\frac {i \pi \,x^{5} \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{10}+\frac {\ln \left (c \right ) x^{5}}{5}-\frac {2 p \,x^{5}}{25}+\frac {2 a p \,x^{3}}{15 b}+\frac {\sqrt {-b a}\, a^{2} p \ln \left (-\sqrt {-b a}\, x +a \right )}{5 b^{3}}-\frac {\sqrt {-b a}\, a^{2} p \ln \left (\sqrt {-b a}\, x +a \right )}{5 b^{3}}-\frac {2 a^{2} p x}{5 b^{2}}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*ln((b*x^2+a)^p)+1/10*I*Pi*x^5*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/10*I*Pi*x^5*csgn(I*(b*x^2+
a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-1/10*I*Pi*x^5*csgn(I*c*(b*x^2+a)^p)^3+1/10*I*Pi*x^5*csgn(I*c*(b*x^2+a)^p
)^2*csgn(I*c)+1/5*ln(c)*x^5-2/25*p*x^5+2/15*a*p*x^3/b+1/5/b^3*(-b*a)^(1/2)*a^2*p*ln(-(-b*a)^(1/2)*x+a)-1/5/b^3
*(-b*a)^(1/2)*a^2*p*ln((-b*a)^(1/2)*x+a)-2/5*a^2*p*x/b^2

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Maxima [A]
time = 0.49, size = 72, normalized size = 0.90 \begin {gather*} \frac {1}{5} \, x^{5} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) + \frac {2}{75} \, b p {\left (\frac {15 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {3 \, b^{2} x^{5} - 5 \, a b x^{3} + 15 \, a^{2} x}{b^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*log((b*x^2 + a)^p*c) + 2/75*b*p*(15*a^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - (3*b^2*x^5 - 5*a*b*x^3
+ 15*a^2*x)/b^3)

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Fricas [A]
time = 0.43, size = 188, normalized size = 2.35 \begin {gather*} \left [\frac {15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 15 \, a^{2} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}, \frac {15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 30 \, a^{2} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/75*(15*b^2*p*x^5*log(b*x^2 + a) - 6*b^2*p*x^5 + 15*b^2*x^5*log(c) + 10*a*b*p*x^3 + 15*a^2*p*sqrt(-a/b)*log(
(b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) - 30*a^2*p*x)/b^2, 1/75*(15*b^2*p*x^5*log(b*x^2 + a) - 6*b^2*p*x^5
+ 15*b^2*x^5*log(c) + 10*a*b*p*x^3 + 30*a^2*p*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) - 30*a^2*p*x)/b^2]

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Sympy [A]
time = 34.42, size = 156, normalized size = 1.95 \begin {gather*} \begin {cases} \frac {x^{5} \log {\left (0^{p} c \right )}}{5} & \text {for}\: a = 0 \wedge b = 0 \\\frac {x^{5} \log {\left (a^{p} c \right )}}{5} & \text {for}\: b = 0 \\- \frac {2 p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{5} & \text {for}\: a = 0 \\\frac {2 a^{3} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{5 b^{3} \sqrt {- \frac {a}{b}}} - \frac {a^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{5 b^{3} \sqrt {- \frac {a}{b}}} - \frac {2 a^{2} p x}{5 b^{2}} + \frac {2 a p x^{3}}{15 b} - \frac {2 p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**5*log(0**p*c)/5, Eq(a, 0) & Eq(b, 0)), (x**5*log(a**p*c)/5, Eq(b, 0)), (-2*p*x**5/25 + x**5*log(
c*(b*x**2)**p)/5, Eq(a, 0)), (2*a**3*p*log(x - sqrt(-a/b))/(5*b**3*sqrt(-a/b)) - a**3*log(c*(a + b*x**2)**p)/(
5*b**3*sqrt(-a/b)) - 2*a**2*p*x/(5*b**2) + 2*a*p*x**3/(15*b) - 2*p*x**5/25 + x**5*log(c*(a + b*x**2)**p)/5, Tr
ue))

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Giac [A]
time = 3.51, size = 71, normalized size = 0.89 \begin {gather*} \frac {1}{5} \, p x^{5} \log \left (b x^{2} + a\right ) - \frac {1}{25} \, {\left (2 \, p - 5 \, \log \left (c\right )\right )} x^{5} + \frac {2 \, a p x^{3}}{15 \, b} + \frac {2 \, a^{3} p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{5 \, \sqrt {a b} b^{2}} - \frac {2 \, a^{2} p x}{5 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*p*x^5*log(b*x^2 + a) - 1/25*(2*p - 5*log(c))*x^5 + 2/15*a*p*x^3/b + 2/5*a^3*p*arctan(b*x/sqrt(a*b))/(sqrt(
a*b)*b^2) - 2/5*a^2*p*x/b^2

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Mupad [B]
time = 0.22, size = 62, normalized size = 0.78 \begin {gather*} \frac {x^5\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{5}-\frac {2\,p\,x^5}{25}+\frac {2\,a^{5/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{5\,b^{5/2}}+\frac {2\,a\,p\,x^3}{15\,b}-\frac {2\,a^2\,p\,x}{5\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*log(c*(a + b*x^2)^p),x)

[Out]

(x^5*log(c*(a + b*x^2)^p))/5 - (2*p*x^5)/25 + (2*a^(5/2)*p*atan((b^(1/2)*x)/a^(1/2)))/(5*b^(5/2)) + (2*a*p*x^3
)/(15*b) - (2*a^2*p*x)/(5*b^2)

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