3.7.27 \(\int \frac {(a+b x^4)^2}{x} \, dx\) [627]

Optimal. Leaf size=26 \[ \frac {1}{2} a b x^4+\frac {b^2 x^8}{8}+a^2 \log (x) \]

[Out]

1/2*a*b*x^4+1/8*b^2*x^8+a^2*ln(x)

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Rubi [A]
time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \begin {gather*} a^2 \log (x)+\frac {1}{2} a b x^4+\frac {b^2 x^8}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^2/x,x]

[Out]

(a*b*x^4)/2 + (b^2*x^8)/8 + a^2*Log[x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^4\right )^2}{x} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {(a+b x)^2}{x} \, dx,x,x^4\right )\\ &=\frac {1}{4} \text {Subst}\left (\int \left (2 a b+\frac {a^2}{x}+b^2 x\right ) \, dx,x,x^4\right )\\ &=\frac {1}{2} a b x^4+\frac {b^2 x^8}{8}+a^2 \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{2} a b x^4+\frac {b^2 x^8}{8}+a^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^2/x,x]

[Out]

(a*b*x^4)/2 + (b^2*x^8)/8 + a^2*Log[x]

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Maple [A]
time = 0.14, size = 23, normalized size = 0.88

method	result	size
default	\(\frac {a b \,x^{4}}{2}+\frac {b^{2} x^{8}}{8}+a^{2} \ln \left (x \right )\)	\(23\)
norman	\(\frac {a b \,x^{4}}{2}+\frac {b^{2} x^{8}}{8}+a^{2} \ln \left (x \right )\)	\(23\)
risch	\(\frac {b^{2} x^{8}}{8}+\frac {a b \,x^{4}}{2}+\frac {a^{2}}{2}+a^{2} \ln \left (x \right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*b*x^4+1/8*b^2*x^8+a^2*ln(x)

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Maxima [A]
time = 0.29, size = 25, normalized size = 0.96 \begin {gather*} \frac {1}{8} \, b^{2} x^{8} + \frac {1}{2} \, a b x^{4} + \frac {1}{4} \, a^{2} \log \left (x^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*x^8 + 1/2*a*b*x^4 + 1/4*a^2*log(x^4)

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Fricas [A]
time = 0.35, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{8} \, b^{2} x^{8} + \frac {1}{2} \, a b x^{4} + a^{2} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*x^8 + 1/2*a*b*x^4 + a^2*log(x)

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Sympy [A]
time = 0.02, size = 22, normalized size = 0.85 \begin {gather*} a^{2} \log {\left (x \right )} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{8}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(x) + a*b*x**4/2 + b**2*x**8/8

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Giac [A]
time = 0.57, size = 25, normalized size = 0.96 \begin {gather*} \frac {1}{8} \, b^{2} x^{8} + \frac {1}{2} \, a b x^{4} + \frac {1}{4} \, a^{2} \log \left (x^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*x^8 + 1/2*a*b*x^4 + 1/4*a^2*log(x^4)

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Mupad [B]
time = 0.96, size = 22, normalized size = 0.85 \begin {gather*} a^2\,\ln \left (x\right )+\frac {b^2\,x^8}{8}+\frac {a\,b\,x^4}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^2/x,x)

[Out]

a^2*log(x) + (b^2*x^8)/8 + (a*b*x^4)/2

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