3.783 \(\int x^{11} (a+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=59 \[ \frac {a^2 \left (a+c x^4\right )^{5/2}}{10 c^3}+\frac {\left (a+c x^4\right )^{9/2}}{18 c^3}-\frac {a \left (a+c x^4\right )^{7/2}}{7 c^3} \]

[Out]

1/10*a^2*(c*x^4+a)^(5/2)/c^3-1/7*a*(c*x^4+a)^(7/2)/c^3+1/18*(c*x^4+a)^(9/2)/c^3

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac {a^2 \left (a+c x^4\right )^{5/2}}{10 c^3}+\frac {\left (a+c x^4\right )^{9/2}}{18 c^3}-\frac {a \left (a+c x^4\right )^{7/2}}{7 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a + c*x^4)^(3/2),x]

[Out]

(a^2*(a + c*x^4)^(5/2))/(10*c^3) - (a*(a + c*x^4)^(7/2))/(7*c^3) + (a + c*x^4)^(9/2)/(18*c^3)

Rule 43

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 266

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^{11} \left (a+c x^4\right )^{3/2} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int x^2 (a+c x)^{3/2} \, dx,x,x^4\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \left (\frac {a^2 (a+c x)^{3/2}}{c^2}-\frac {2 a (a+c x)^{5/2}}{c^2}+\frac {(a+c x)^{7/2}}{c^2}\right ) \, dx,x,x^4\right )\\ &=\frac {a^2 \left (a+c x^4\right )^{5/2}}{10 c^3}-\frac {a \left (a+c x^4\right )^{7/2}}{7 c^3}+\frac {\left (a+c x^4\right )^{9/2}}{18 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 39, normalized size = 0.66 \[ \frac {\left (a+c x^4\right )^{5/2} \left (8 a^2-20 a c x^4+35 c^2 x^8\right )}{630 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11*(a + c*x^4)^(3/2),x]

[Out]

((a + c*x^4)^(5/2)*(8*a^2 - 20*a*c*x^4 + 35*c^2*x^8))/(630*c^3)

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fricas [A]  time = 0.90, size = 57, normalized size = 0.97 \[ \frac {{\left (35 \, c^{4} x^{16} + 50 \, a c^{3} x^{12} + 3 \, a^{2} c^{2} x^{8} - 4 \, a^{3} c x^{4} + 8 \, a^{4}\right )} \sqrt {c x^{4} + a}}{630 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

1/630*(35*c^4*x^16 + 50*a*c^3*x^12 + 3*a^2*c^2*x^8 - 4*a^3*c*x^4 + 8*a^4)*sqrt(c*x^4 + a)/c^3

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giac [A]  time = 0.17, size = 43, normalized size = 0.73 \[ \frac {35 \, {\left (c x^{4} + a\right )}^{\frac {9}{2}} - 90 \, {\left (c x^{4} + a\right )}^{\frac {7}{2}} a + 63 \, {\left (c x^{4} + a\right )}^{\frac {5}{2}} a^{2}}{630 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

1/630*(35*(c*x^4 + a)^(9/2) - 90*(c*x^4 + a)^(7/2)*a + 63*(c*x^4 + a)^(5/2)*a^2)/c^3

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maple [A]  time = 0.01, size = 36, normalized size = 0.61 \[ \frac {\left (c \,x^{4}+a \right )^{\frac {5}{2}} \left (35 c^{2} x^{8}-20 a c \,x^{4}+8 a^{2}\right )}{630 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(c*x^4+a)^(3/2),x)

[Out]

1/630*(c*x^4+a)^(5/2)*(35*c^2*x^8-20*a*c*x^4+8*a^2)/c^3

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maxima [A]  time = 1.40, size = 47, normalized size = 0.80 \[ \frac {{\left (c x^{4} + a\right )}^{\frac {9}{2}}}{18 \, c^{3}} - \frac {{\left (c x^{4} + a\right )}^{\frac {7}{2}} a}{7 \, c^{3}} + \frac {{\left (c x^{4} + a\right )}^{\frac {5}{2}} a^{2}}{10 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

1/18*(c*x^4 + a)^(9/2)/c^3 - 1/7*(c*x^4 + a)^(7/2)*a/c^3 + 1/10*(c*x^4 + a)^(5/2)*a^2/c^3

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mupad [B]  time = 1.18, size = 53, normalized size = 0.90 \[ \sqrt {c\,x^4+a}\,\left (\frac {5\,a\,x^{12}}{63}+\frac {c\,x^{16}}{18}+\frac {4\,a^4}{315\,c^3}-\frac {2\,a^3\,x^4}{315\,c^2}+\frac {a^2\,x^8}{210\,c}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(a + c*x^4)^(3/2),x)

[Out]

(a + c*x^4)^(1/2)*((5*a*x^12)/63 + (c*x^16)/18 + (4*a^4)/(315*c^3) - (2*a^3*x^4)/(315*c^2) + (a^2*x^8)/(210*c)
)

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sympy [A]  time = 11.03, size = 109, normalized size = 1.85 \[ \begin {cases} \frac {4 a^{4} \sqrt {a + c x^{4}}}{315 c^{3}} - \frac {2 a^{3} x^{4} \sqrt {a + c x^{4}}}{315 c^{2}} + \frac {a^{2} x^{8} \sqrt {a + c x^{4}}}{210 c} + \frac {5 a x^{12} \sqrt {a + c x^{4}}}{63} + \frac {c x^{16} \sqrt {a + c x^{4}}}{18} & \text {for}\: c \neq 0 \\\frac {a^{\frac {3}{2}} x^{12}}{12} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11*(c*x**4+a)**(3/2),x)

[Out]

Piecewise((4*a**4*sqrt(a + c*x**4)/(315*c**3) - 2*a**3*x**4*sqrt(a + c*x**4)/(315*c**2) + a**2*x**8*sqrt(a + c
*x**4)/(210*c) + 5*a*x**12*sqrt(a + c*x**4)/63 + c*x**16*sqrt(a + c*x**4)/18, Ne(c, 0)), (a**(3/2)*x**12/12, T
rue))

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