3.5.49 \(\int x^3 (a^2+2 a b x^2+b^2 x^4)^3 \, dx\) [449]

Optimal. Leaf size=34 \[ -\frac {a \left (a+b x^2\right )^7}{14 b^2}+\frac {\left (a+b x^2\right )^8}{16 b^2} \]

[Out]

-1/14*a*(b*x^2+a)^7/b^2+1/16*(b*x^2+a)^8/b^2

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Rubi [A]
time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 45} \begin {gather*} \frac {\left (a+b x^2\right )^8}{16 b^2}-\frac {a \left (a+b x^2\right )^7}{14 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

-1/14*(a*(a + b*x^2)^7)/b^2 + (a + b*x^2)^8/(16*b^2)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx &=\frac {\int x^3 \left (a b+b^2 x^2\right )^6 \, dx}{b^6}\\ &=\frac {\text {Subst}\left (\int x \left (a b+b^2 x\right )^6 \, dx,x,x^2\right )}{2 b^6}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {a \left (a b+b^2 x\right )^6}{b}+\frac {\left (a b+b^2 x\right )^7}{b^2}\right ) \, dx,x,x^2\right )}{2 b^6}\\ &=-\frac {a \left (a+b x^2\right )^7}{14 b^2}+\frac {\left (a+b x^2\right )^8}{16 b^2}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(77\) vs. \(2(34)=68\).
time = 0.00, size = 77, normalized size = 2.26 \begin {gather*} \frac {a^6 x^4}{4}+a^5 b x^6+\frac {15}{8} a^4 b^2 x^8+2 a^3 b^3 x^{10}+\frac {5}{4} a^2 b^4 x^{12}+\frac {3}{7} a b^5 x^{14}+\frac {b^6 x^{16}}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

(a^6*x^4)/4 + a^5*b*x^6 + (15*a^4*b^2*x^8)/8 + 2*a^3*b^3*x^10 + (5*a^2*b^4*x^12)/4 + (3*a*b^5*x^14)/7 + (b^6*x
^16)/16

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs. \(2(30)=60\).
time = 0.05, size = 68, normalized size = 2.00

method	result	size
default	\(\frac {1}{4} a^{6} x^{4}+a^{5} b \,x^{6}+\frac {15}{8} a^{4} b^{2} x^{8}+2 a^{3} b^{3} x^{10}+\frac {5}{4} a^{2} b^{4} x^{12}+\frac {3}{7} a \,b^{5} x^{14}+\frac {1}{16} b^{6} x^{16}\)	\(68\)
norman	\(\frac {1}{4} a^{6} x^{4}+a^{5} b \,x^{6}+\frac {15}{8} a^{4} b^{2} x^{8}+2 a^{3} b^{3} x^{10}+\frac {5}{4} a^{2} b^{4} x^{12}+\frac {3}{7} a \,b^{5} x^{14}+\frac {1}{16} b^{6} x^{16}\)	\(68\)
risch	\(\frac {1}{4} a^{6} x^{4}+a^{5} b \,x^{6}+\frac {15}{8} a^{4} b^{2} x^{8}+2 a^{3} b^{3} x^{10}+\frac {5}{4} a^{2} b^{4} x^{12}+\frac {3}{7} a \,b^{5} x^{14}+\frac {1}{16} b^{6} x^{16}\)	\(68\)
gosper	\(\frac {x^{4} \left (7 b^{6} x^{12}+48 a \,b^{5} x^{10}+140 a^{2} b^{4} x^{8}+224 a^{3} b^{3} x^{6}+210 a^{4} b^{2} x^{4}+112 a^{5} b \,x^{2}+28 a^{6}\right )}{112}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b^2*x^4+2*a*b*x^2+a^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/4*a^6*x^4+a^5*b*x^6+15/8*a^4*b^2*x^8+2*a^3*b^3*x^10+5/4*a^2*b^4*x^12+3/7*a*b^5*x^14+1/16*b^6*x^16

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (30) = 60\).
time = 0.29, size = 67, normalized size = 1.97 \begin {gather*} \frac {1}{16} \, b^{6} x^{16} + \frac {3}{7} \, a b^{5} x^{14} + \frac {5}{4} \, a^{2} b^{4} x^{12} + 2 \, a^{3} b^{3} x^{10} + \frac {15}{8} \, a^{4} b^{2} x^{8} + a^{5} b x^{6} + \frac {1}{4} \, a^{6} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")

[Out]

1/16*b^6*x^16 + 3/7*a*b^5*x^14 + 5/4*a^2*b^4*x^12 + 2*a^3*b^3*x^10 + 15/8*a^4*b^2*x^8 + a^5*b*x^6 + 1/4*a^6*x^
4

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (30) = 60\).
time = 0.34, size = 67, normalized size = 1.97 \begin {gather*} \frac {1}{16} \, b^{6} x^{16} + \frac {3}{7} \, a b^{5} x^{14} + \frac {5}{4} \, a^{2} b^{4} x^{12} + 2 \, a^{3} b^{3} x^{10} + \frac {15}{8} \, a^{4} b^{2} x^{8} + a^{5} b x^{6} + \frac {1}{4} \, a^{6} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")

[Out]

1/16*b^6*x^16 + 3/7*a*b^5*x^14 + 5/4*a^2*b^4*x^12 + 2*a^3*b^3*x^10 + 15/8*a^4*b^2*x^8 + a^5*b*x^6 + 1/4*a^6*x^
4

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (27) = 54\).
time = 0.02, size = 75, normalized size = 2.21 \begin {gather*} \frac {a^{6} x^{4}}{4} + a^{5} b x^{6} + \frac {15 a^{4} b^{2} x^{8}}{8} + 2 a^{3} b^{3} x^{10} + \frac {5 a^{2} b^{4} x^{12}}{4} + \frac {3 a b^{5} x^{14}}{7} + \frac {b^{6} x^{16}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b**2*x**4+2*a*b*x**2+a**2)**3,x)

[Out]

a**6*x**4/4 + a**5*b*x**6 + 15*a**4*b**2*x**8/8 + 2*a**3*b**3*x**10 + 5*a**2*b**4*x**12/4 + 3*a*b**5*x**14/7 +
b**6*x**16/16

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (30) = 60\).
time = 3.95, size = 67, normalized size = 1.97 \begin {gather*} \frac {1}{16} \, b^{6} x^{16} + \frac {3}{7} \, a b^{5} x^{14} + \frac {5}{4} \, a^{2} b^{4} x^{12} + 2 \, a^{3} b^{3} x^{10} + \frac {15}{8} \, a^{4} b^{2} x^{8} + a^{5} b x^{6} + \frac {1}{4} \, a^{6} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")

[Out]

1/16*b^6*x^16 + 3/7*a*b^5*x^14 + 5/4*a^2*b^4*x^12 + 2*a^3*b^3*x^10 + 15/8*a^4*b^2*x^8 + a^5*b*x^6 + 1/4*a^6*x^
4

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Mupad [B]
time = 0.03, size = 67, normalized size = 1.97 \begin {gather*} \frac {a^6\,x^4}{4}+a^5\,b\,x^6+\frac {15\,a^4\,b^2\,x^8}{8}+2\,a^3\,b^3\,x^{10}+\frac {5\,a^2\,b^4\,x^{12}}{4}+\frac {3\,a\,b^5\,x^{14}}{7}+\frac {b^6\,x^{16}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^3,x)

[Out]

(a^6*x^4)/4 + (b^6*x^16)/16 + a^5*b*x^6 + (3*a*b^5*x^14)/7 + (15*a^4*b^2*x^8)/8 + 2*a^3*b^3*x^10 + (5*a^2*b^4*
x^12)/4

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