3.3.28 \(\int \frac {x^5}{(a+b x^2) (c+d x^2)} \, dx\) [228]

Optimal. Leaf size=70 \[ \frac {x^2}{2 b d}+\frac {a^2 \log \left (a+b x^2\right )}{2 b^2 (b c-a d)}-\frac {c^2 \log \left (c+d x^2\right )}{2 d^2 (b c-a d)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 70, 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, 84} \begin {gather*} \frac {a^2 \log \left (a+b x^2\right )}{2 b^2 (b c-a d)}-\frac {c^2 \log \left (c+d x^2\right )}{2 d^2 (b c-a d)}+\frac {x^2}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/((a + b*x^2)*(c + d*x^2)),x]

[Out]

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

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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^2\right ) \left (c+d x^2\right )} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{(a+b x) (c+d x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{b d}+\frac {a^2}{b (b c-a d) (a+b x)}+\frac {c^2}{d (-b c+a d) (c+d x)}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{2 b d}+\frac {a^2 \log \left (a+b x^2\right )}{2 b^2 (b c-a d)}-\frac {c^2 \log \left (c+d x^2\right )}{2 d^2 (b c-a d)}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 66, normalized size = 0.94 \begin {gather*} \frac {a^2 d^2 \log \left (a+b x^2\right )-b \left (d (-b c+a d) x^2+b c^2 \log \left (c+d x^2\right )\right )}{2 b^2 d^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/((a + b*x^2)*(c + d*x^2)),x]

[Out]

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

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Maple [A]
time = 0.10, size = 65, normalized size = 0.93

method	result	size
default	\(\frac {x^{2}}{2 b d}-\frac {a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{2} \left (a d -b c \right )}+\frac {c^{2} \ln \left (d \,x^{2}+c \right )}{2 d^{2} \left (a d -b c \right )}\)	\(65\)
norman	\(\frac {x^{2}}{2 b d}-\frac {a^{2} \ln \left (b \,x^{2}+a \right )}{2 b^{2} \left (a d -b c \right )}+\frac {c^{2} \ln \left (d \,x^{2}+c \right )}{2 d^{2} \left (a d -b c \right )}\)	\(65\)
risch	\(\frac {x^{2}}{2 b d}+\frac {c^{2} \ln \left (d \,x^{2}+c \right )}{2 d^{2} \left (a d -b c \right )}-\frac {a^{2} \ln \left (-b \,x^{2}-a \right )}{2 b^{2} \left (a d -b c \right )}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2+a)/(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 68, normalized size = 0.97 \begin {gather*} \frac {a^{2} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )}} - \frac {c^{2} \log \left (d x^{2} + c\right )}{2 \, {\left (b c d^{2} - a d^{3}\right )}} + \frac {x^{2}}{2 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)/(d*x^2+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.34, size = 72, normalized size = 1.03 \begin {gather*} \frac {a^{2} d^{2} \log \left (b x^{2} + a\right ) - b^{2} c^{2} \log \left (d x^{2} + c\right ) + {\left (b^{2} c d - a b d^{2}\right )} x^{2}}{2 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)/(d*x^2+c),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**2+a)/(d*x**2+c),x)

[Out]

Timed out

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Giac [A]
time = 0.57, size = 70, normalized size = 1.00 \begin {gather*} \frac {a^{2} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )}} - \frac {c^{2} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b c d^{2} - a d^{3}\right )}} + \frac {x^{2}}{2 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^2+a)/(d*x^2+c),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.19, size = 68, normalized size = 0.97 \begin {gather*} \frac {a^2\,\ln \left (b\,x^2+a\right )}{2\,b^3\,c-2\,a\,b^2\,d}+\frac {c^2\,\ln \left (d\,x^2+c\right )}{2\,a\,d^3-2\,b\,c\,d^2}+\frac {x^2}{2\,b\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((a + b*x^2)*(c + d*x^2)),x)

[Out]

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

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