3.7.26 \(\int (a+b x^4)^2 \, dx\) [626]

Optimal. Leaf size=25 \[ a^2 x+\frac {2}{5} a b x^5+\frac {b^2 x^9}{9} \]

[Out]

a^2*x+2/5*a*b*x^5+1/9*b^2*x^9

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {200} \begin {gather*} a^2 x+\frac {2}{5} a b x^5+\frac {b^2 x^9}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^2,x]

[Out]

a^2*x + (2*a*b*x^5)/5 + (b^2*x^9)/9

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \left (a+b x^4\right )^2 \, dx &=\int \left (a^2+2 a b x^4+b^2 x^8\right ) \, dx\\ &=a^2 x+\frac {2}{5} a b x^5+\frac {b^2 x^9}{9}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 25, normalized size = 1.00 \begin {gather*} a^2 x+\frac {2}{5} a b x^5+\frac {b^2 x^9}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^2,x]

[Out]

a^2*x + (2*a*b*x^5)/5 + (b^2*x^9)/9

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Maple [A]
time = 0.13, size = 22, normalized size = 0.88

method	result	size
gosper	\(a^{2} x +\frac {2}{5} a b \,x^{5}+\frac {1}{9} b^{2} x^{9}\)	\(22\)
default	\(a^{2} x +\frac {2}{5} a b \,x^{5}+\frac {1}{9} b^{2} x^{9}\)	\(22\)
norman	\(a^{2} x +\frac {2}{5} a b \,x^{5}+\frac {1}{9} b^{2} x^{9}\)	\(22\)
risch	\(a^{2} x +\frac {2}{5} a b \,x^{5}+\frac {1}{9} b^{2} x^{9}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x+2/5*a*b*x^5+1/9*b^2*x^9

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Maxima [A]
time = 0.31, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {2}{5} \, a b x^{5} + a^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^2*x^9 + 2/5*a*b*x^5 + a^2*x

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Fricas [A]
time = 0.36, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {2}{5} \, a b x^{5} + a^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^2*x^9 + 2/5*a*b*x^5 + a^2*x

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Sympy [A]
time = 0.01, size = 22, normalized size = 0.88 \begin {gather*} a^{2} x + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{9}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x + 2*a*b*x**5/5 + b**2*x**9/9

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Giac [A]
time = 0.53, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{9} \, b^{2} x^{9} + \frac {2}{5} \, a b x^{5} + a^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^2*x^9 + 2/5*a*b*x^5 + a^2*x

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Mupad [B]
time = 0.03, size = 21, normalized size = 0.84 \begin {gather*} a^2\,x+\frac {2\,a\,b\,x^5}{5}+\frac {b^2\,x^9}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^2,x)

[Out]

a^2*x + (b^2*x^9)/9 + (2*a*b*x^5)/5

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