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]

(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)

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Rubi [A]  time = 0.0342912, antiderivative size = 59, normalized size of antiderivative = 1., 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 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]]

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])

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*}

Mathematica [A]  time = 0.0208882, 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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Maple [A]  time = 0.007, size = 36, normalized size = 0.6 \begin{align*}{\frac{35\,{x}^{8}{c}^{2}-20\,a{x}^{4}c+8\,{a}^{2}}{630\,{c}^{3}} \left ( c{x}^{4}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

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 = 0.971273, size = 63, normalized size = 1.07 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 1.44932, size = 128, normalized size = 2.17 \begin{align*} \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}} \end{align*}

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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Sympy [A]  time = 9.61937, size = 109, normalized size = 1.85 \begin{align*} \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} \end{align*}

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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Giac [B]  time = 1.12925, size = 143, normalized size = 2.42 \begin{align*} \frac{\frac{3 \,{\left (15 \,{\left (c x^{4} + a\right )}^{\frac{7}{2}} - 42 \,{\left (c x^{4} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (c x^{4} + a\right )}^{\frac{3}{2}} a^{2}\right )} a}{c^{2}} + \frac{35 \,{\left (c x^{4} + a\right )}^{\frac{9}{2}} - 135 \,{\left (c x^{4} + a\right )}^{\frac{7}{2}} a + 189 \,{\left (c x^{4} + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (c x^{4} + a\right )}^{\frac{3}{2}} a^{3}}{c^{2}}}{630 \, c} \end{align*}

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*(3*(15*(c*x^4 + a)^(7/2) - 42*(c*x^4 + a)^(5/2)*a + 35*(c*x^4 + a)^(3/2)*a^2)*a/c^2 + (35*(c*x^4 + a)^(9
/2) - 135*(c*x^4 + a)^(7/2)*a + 189*(c*x^4 + a)^(5/2)*a^2 - 105*(c*x^4 + a)^(3/2)*a^3)/c^2)/c