3.620 \(\int x^m (a+b x^4)^2 \, dx\)

Optimal. Leaf size=43 \[ \frac {a^2 x^{m+1}}{m+1}+\frac {2 a b x^{m+5}}{m+5}+\frac {b^2 x^{m+9}}{m+9} \]

[Out]

a^2*x^(1+m)/(1+m)+2*a*b*x^(5+m)/(5+m)+b^2*x^(9+m)/(9+m)

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {270} \[ \frac {a^2 x^{m+1}}{m+1}+\frac {2 a b x^{m+5}}{m+5}+\frac {b^2 x^{m+9}}{m+9} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x^4)^2,x]

[Out]

(a^2*x^(1 + m))/(1 + m) + (2*a*b*x^(5 + m))/(5 + m) + (b^2*x^(9 + m))/(9 + m)

Rule 270

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

Rubi steps

\begin {align*} \int x^m \left (a+b x^4\right )^2 \, dx &=\int \left (a^2 x^m+2 a b x^{4+m}+b^2 x^{8+m}\right ) \, dx\\ &=\frac {a^2 x^{1+m}}{1+m}+\frac {2 a b x^{5+m}}{5+m}+\frac {b^2 x^{9+m}}{9+m}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 40, normalized size = 0.93 \[ x^{m+1} \left (\frac {a^2}{m+1}+\frac {2 a b x^4}{m+5}+\frac {b^2 x^8}{m+9}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x^4)^2,x]

[Out]

x^(1 + m)*(a^2/(1 + m) + (2*a*b*x^4)/(5 + m) + (b^2*x^8)/(9 + m))

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fricas [A]  time = 0.64, size = 85, normalized size = 1.98 \[ \frac {{\left ({\left (b^{2} m^{2} + 6 \, b^{2} m + 5 \, b^{2}\right )} x^{9} + 2 \, {\left (a b m^{2} + 10 \, a b m + 9 \, a b\right )} x^{5} + {\left (a^{2} m^{2} + 14 \, a^{2} m + 45 \, a^{2}\right )} x\right )} x^{m}}{m^{3} + 15 \, m^{2} + 59 \, m + 45} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^2*m^2 + 6*b^2*m + 5*b^2)*x^9 + 2*(a*b*m^2 + 10*a*b*m + 9*a*b)*x^5 + (a^2*m^2 + 14*a^2*m + 45*a^2)*x)*x^m/(
m^3 + 15*m^2 + 59*m + 45)

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giac [B]  time = 0.16, size = 117, normalized size = 2.72 \[ \frac {b^{2} m^{2} x^{9} x^{m} + 6 \, b^{2} m x^{9} x^{m} + 5 \, b^{2} x^{9} x^{m} + 2 \, a b m^{2} x^{5} x^{m} + 20 \, a b m x^{5} x^{m} + 18 \, a b x^{5} x^{m} + a^{2} m^{2} x x^{m} + 14 \, a^{2} m x x^{m} + 45 \, a^{2} x x^{m}}{m^{3} + 15 \, m^{2} + 59 \, m + 45} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*m^2*x^9*x^m + 6*b^2*m*x^9*x^m + 5*b^2*x^9*x^m + 2*a*b*m^2*x^5*x^m + 20*a*b*m*x^5*x^m + 18*a*b*x^5*x^m + a
^2*m^2*x*x^m + 14*a^2*m*x*x^m + 45*a^2*x*x^m)/(m^3 + 15*m^2 + 59*m + 45)

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maple [B]  time = 0.01, size = 93, normalized size = 2.16 \[ \frac {\left (b^{2} m^{2} x^{8}+6 b^{2} m \,x^{8}+5 b^{2} x^{8}+2 a b \,m^{2} x^{4}+20 a b m \,x^{4}+18 a b \,x^{4}+a^{2} m^{2}+14 a^{2} m +45 a^{2}\right ) x^{m +1}}{\left (m +9\right ) \left (m +5\right ) \left (m +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(m+1)*(b^2*m^2*x^8+6*b^2*m*x^8+5*b^2*x^8+2*a*b*m^2*x^4+20*a*b*m*x^4+18*a*b*x^4+a^2*m^2+14*a^2*m+45*a^2)/(m+9
)/(m+5)/(m+1)

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maxima [A]  time = 1.32, size = 43, normalized size = 1.00 \[ \frac {b^{2} x^{m + 9}}{m + 9} + \frac {2 \, a b x^{m + 5}}{m + 5} + \frac {a^{2} x^{m + 1}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x^(m + 9)/(m + 9) + 2*a*b*x^(m + 5)/(m + 5) + a^2*x^(m + 1)/(m + 1)

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mupad [B]  time = 1.32, size = 93, normalized size = 2.16 \[ x^m\,\left (\frac {a^2\,x\,\left (m^2+14\,m+45\right )}{m^3+15\,m^2+59\,m+45}+\frac {b^2\,x^9\,\left (m^2+6\,m+5\right )}{m^3+15\,m^2+59\,m+45}+\frac {2\,a\,b\,x^5\,\left (m^2+10\,m+9\right )}{m^3+15\,m^2+59\,m+45}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a + b*x^4)^2,x)

[Out]

x^m*((a^2*x*(14*m + m^2 + 45))/(59*m + 15*m^2 + m^3 + 45) + (b^2*x^9*(6*m + m^2 + 5))/(59*m + 15*m^2 + m^3 + 4
5) + (2*a*b*x^5*(10*m + m^2 + 9))/(59*m + 15*m^2 + m^3 + 45))

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sympy [A]  time = 3.94, size = 309, normalized size = 7.19 \[ \begin {cases} - \frac {a^{2}}{8 x^{8}} - \frac {a b}{2 x^{4}} + b^{2} \log {\relax (x )} & \text {for}\: m = -9 \\- \frac {a^{2}}{4 x^{4}} + 2 a b \log {\relax (x )} + \frac {b^{2} x^{4}}{4} & \text {for}\: m = -5 \\a^{2} \log {\relax (x )} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{8}}{8} & \text {for}\: m = -1 \\\frac {a^{2} m^{2} x x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {14 a^{2} m x x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {45 a^{2} x x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {2 a b m^{2} x^{5} x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {20 a b m x^{5} x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {18 a b x^{5} x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {b^{2} m^{2} x^{9} x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {6 b^{2} m x^{9} x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} + \frac {5 b^{2} x^{9} x^{m}}{m^{3} + 15 m^{2} + 59 m + 45} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2/(8*x**8) - a*b/(2*x**4) + b**2*log(x), Eq(m, -9)), (-a**2/(4*x**4) + 2*a*b*log(x) + b**2*x**4
/4, Eq(m, -5)), (a**2*log(x) + a*b*x**4/2 + b**2*x**8/8, Eq(m, -1)), (a**2*m**2*x*x**m/(m**3 + 15*m**2 + 59*m
+ 45) + 14*a**2*m*x*x**m/(m**3 + 15*m**2 + 59*m + 45) + 45*a**2*x*x**m/(m**3 + 15*m**2 + 59*m + 45) + 2*a*b*m*
*2*x**5*x**m/(m**3 + 15*m**2 + 59*m + 45) + 20*a*b*m*x**5*x**m/(m**3 + 15*m**2 + 59*m + 45) + 18*a*b*x**5*x**m
/(m**3 + 15*m**2 + 59*m + 45) + b**2*m**2*x**9*x**m/(m**3 + 15*m**2 + 59*m + 45) + 6*b**2*m*x**9*x**m/(m**3 +
15*m**2 + 59*m + 45) + 5*b**2*x**9*x**m/(m**3 + 15*m**2 + 59*m + 45), True))

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