3.2.10 \(\int \frac {x^5}{(a+b x^3) (c+d x^3)} \, dx\) [110]

Optimal. Leaf size=53 \[ -\frac {a \log \left (a+b x^3\right )}{3 b (b c-a d)}+\frac {c \log \left (c+d x^3\right )}{3 d (b c-a d)} \]

[Out]

-1/3*a*ln(b*x^3+a)/b/(-a*d+b*c)+1/3*c*ln(d*x^3+c)/d/(-a*d+b*c)

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Rubi [A]
time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 78} \begin {gather*} \frac {c \log \left (c+d x^3\right )}{3 d (b c-a d)}-\frac {a \log \left (a+b x^3\right )}{3 b (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/((a + b*x^3)*(c + d*x^3)),x]

[Out]

-1/3*(a*Log[a + b*x^3])/(b*(b*c - a*d)) + (c*Log[c + d*x^3])/(3*d*(b*c - a*d))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^5}{\left (a+b x^3\right ) \left (c+d x^3\right )} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x}{(a+b x) (c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {a}{(b c-a d) (a+b x)}+\frac {c}{(b c-a d) (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {a \log \left (a+b x^3\right )}{3 b (b c-a d)}+\frac {c \log \left (c+d x^3\right )}{3 d (b c-a d)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.81 \begin {gather*} -\frac {a d \log \left (a+b x^3\right )-b c \log \left (c+d x^3\right )}{3 b^2 c d-3 a b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/((a + b*x^3)*(c + d*x^3)),x]

[Out]

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

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Maple [A]
time = 0.34, size = 50, normalized size = 0.94

method	result	size
default	\(\frac {a \ln \left (b \,x^{3}+a \right )}{3 \left (a d -b c \right ) b}-\frac {c \ln \left (d \,x^{3}+c \right )}{3 \left (a d -b c \right ) d}\)	\(50\)
norman	\(\frac {a \ln \left (b \,x^{3}+a \right )}{3 \left (a d -b c \right ) b}-\frac {c \ln \left (d \,x^{3}+c \right )}{3 \left (a d -b c \right ) d}\)	\(50\)
risch	\(-\frac {c \ln \left (-d \,x^{3}-c \right )}{3 d \left (a d -b c \right )}+\frac {a \ln \left (b \,x^{3}+a \right )}{3 \left (a d -b c \right ) b}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^3+a)/(d*x^3+c),x,method=_RETURNVERBOSE)

[Out]

1/3*a/(a*d-b*c)/b*ln(b*x^3+a)-1/3*c/(a*d-b*c)/d*ln(d*x^3+c)

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Maxima [A]
time = 0.28, size = 49, normalized size = 0.92 \begin {gather*} -\frac {a \log \left (b x^{3} + a\right )}{3 \, {\left (b^{2} c - a b d\right )}} + \frac {c \log \left (d x^{3} + c\right )}{3 \, {\left (b c d - a d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)/(d*x^3+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.37, size = 42, normalized size = 0.79 \begin {gather*} -\frac {a d \log \left (b x^{3} + a\right ) - b c \log \left (d x^{3} + c\right )}{3 \, {\left (b^{2} c d - a b d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)/(d*x^3+c),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (39) = 78\).
time = 5.88, size = 144, normalized size = 2.72 \begin {gather*} \frac {a \log {\left (x^{3} + \frac {\frac {a^{3} d^{2}}{b \left (a d - b c\right )} - \frac {2 a^{2} c d}{a d - b c} + \frac {a b c^{2}}{a d - b c} + 2 a c}{a d + b c} \right )}}{3 b \left (a d - b c\right )} - \frac {c \log {\left (x^{3} + \frac {- \frac {a^{2} c d}{a d - b c} + \frac {2 a b c^{2}}{a d - b c} + 2 a c - \frac {b^{2} c^{3}}{d \left (a d - b c\right )}}{a d + b c} \right )}}{3 d \left (a d - b c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**3+a)/(d*x**3+c),x)

[Out]

a*log(x**3 + (a**3*d**2/(b*(a*d - b*c)) - 2*a**2*c*d/(a*d - b*c) + a*b*c**2/(a*d - b*c) + 2*a*c)/(a*d + b*c))/
(3*b*(a*d - b*c)) - c*log(x**3 + (-a**2*c*d/(a*d - b*c) + 2*a*b*c**2/(a*d - b*c) + 2*a*c - b**2*c**3/(d*(a*d -
b*c)))/(a*d + b*c))/(3*d*(a*d - b*c))

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Giac [A]
time = 0.63, size = 51, normalized size = 0.96 \begin {gather*} -\frac {a \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, {\left (b^{2} c - a b d\right )}} + \frac {c \log \left ({\left | d x^{3} + c \right |}\right )}{3 \, {\left (b c d - a d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)/(d*x^3+c),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.31, size = 51, normalized size = 0.96 \begin {gather*} -\frac {a\,\ln \left (b\,x^3+a\right )}{3\,b^2\,c-3\,a\,b\,d}-\frac {c\,\ln \left (d\,x^3+c\right )}{3\,a\,d^2-3\,b\,c\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((a + b*x^3)*(c + d*x^3)),x)

[Out]

- (a*log(a + b*x^3))/(3*b^2*c - 3*a*b*d) - (c*log(c + d*x^3))/(3*a*d^2 - 3*b*c*d)

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