3.609 \(\int x^4 (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\)

Optimal. Leaf size=255 \[ \frac {b^5 x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )}+\frac {5 a b^4 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {a^5 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )} \]

[Out]

1/5*a^5*x^5*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/7*a^4*b*x^7*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+10/9*a^3*b^2*x^9*((b*x^2
+a)^2)^(1/2)/(b*x^2+a)+10/11*a^2*b^3*x^11*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/13*a*b^4*x^13*((b*x^2+a)^2)^(1/2)/(b
*x^2+a)+1/15*b^5*x^15*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]  time = 0.06, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1112, 270} \[ \frac {b^5 x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )}+\frac {5 a b^4 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {a^5 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(a^5*x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*(a + b*x^2)) + (5*a^4*b*x^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*(
a + b*x^2)) + (10*a^3*b^2*x^9*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*(a + b*x^2)) + (10*a^2*b^3*x^11*Sqrt[a^2 + 2
*a*b*x^2 + b^2*x^4])/(11*(a + b*x^2)) + (5*a*b^4*x^13*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*(a + b*x^2)) + (b^5
*x^15*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*(a + b*x^2))

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]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^4 \left (a b+b^2 x^2\right )^5 \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^5 b^5 x^4+5 a^4 b^6 x^6+10 a^3 b^7 x^8+10 a^2 b^8 x^{10}+5 a b^9 x^{12}+b^{10} x^{14}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {a^5 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {5 a b^4 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {b^5 x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \[ \frac {x^5 \sqrt {\left (a+b x^2\right )^2} \left (9009 a^5+32175 a^4 b x^2+50050 a^3 b^2 x^4+40950 a^2 b^3 x^6+17325 a b^4 x^8+3003 b^5 x^{10}\right )}{45045 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^5*Sqrt[(a + b*x^2)^2]*(9009*a^5 + 32175*a^4*b*x^2 + 50050*a^3*b^2*x^4 + 40950*a^2*b^3*x^6 + 17325*a*b^4*x^8
+ 3003*b^5*x^10))/(45045*(a + b*x^2))

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fricas [A]  time = 1.11, size = 57, normalized size = 0.22 \[ \frac {1}{15} \, b^{5} x^{15} + \frac {5}{13} \, a b^{4} x^{13} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {10}{9} \, a^{3} b^{2} x^{9} + \frac {5}{7} \, a^{4} b x^{7} + \frac {1}{5} \, a^{5} x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/15*b^5*x^15 + 5/13*a*b^4*x^13 + 10/11*a^2*b^3*x^11 + 10/9*a^3*b^2*x^9 + 5/7*a^4*b*x^7 + 1/5*a^5*x^5

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giac [A]  time = 0.16, size = 105, normalized size = 0.41 \[ \frac {1}{15} \, b^{5} x^{15} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{13} \, a b^{4} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{11} \, a^{2} b^{3} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{9} \, a^{3} b^{2} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a^{4} b x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{5} \, a^{5} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/15*b^5*x^15*sgn(b*x^2 + a) + 5/13*a*b^4*x^13*sgn(b*x^2 + a) + 10/11*a^2*b^3*x^11*sgn(b*x^2 + a) + 10/9*a^3*b
^2*x^9*sgn(b*x^2 + a) + 5/7*a^4*b*x^7*sgn(b*x^2 + a) + 1/5*a^5*x^5*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 80, normalized size = 0.31 \[ \frac {\left (3003 b^{5} x^{10}+17325 a \,b^{4} x^{8}+40950 a^{2} b^{3} x^{6}+50050 a^{3} b^{2} x^{4}+32175 a^{4} b \,x^{2}+9009 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x^{5}}{45045 \left (b \,x^{2}+a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/45045*x^5*(3003*b^5*x^10+17325*a*b^4*x^8+40950*a^2*b^3*x^6+50050*a^3*b^2*x^4+32175*a^4*b*x^2+9009*a^5)*((b*x
^2+a)^2)^(5/2)/(b*x^2+a)^5

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maxima [A]  time = 1.33, size = 57, normalized size = 0.22 \[ \frac {1}{15} \, b^{5} x^{15} + \frac {5}{13} \, a b^{4} x^{13} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {10}{9} \, a^{3} b^{2} x^{9} + \frac {5}{7} \, a^{4} b x^{7} + \frac {1}{5} \, a^{5} x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/15*b^5*x^15 + 5/13*a*b^4*x^13 + 10/11*a^2*b^3*x^11 + 10/9*a^3*b^2*x^9 + 5/7*a^4*b*x^7 + 1/5*a^5*x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**4*((a + b*x**2)**2)**(5/2), x)

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