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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0594737, antiderivative size = 255, normalized size of antiderivative = 1., 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 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]

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^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*}

Mathematica [A]  time = 0.0210503, size = 83, normalized size = 0.33 \[ \frac{x^5 \sqrt{\left (a+b x^2\right )^2} \left (40950 a^2 b^3 x^6+50050 a^3 b^2 x^4+32175 a^4 b x^2+9009 a^5+17325 a b^4 x^8+3003 b^5 x^{10}\right )}{45045 \left (a+b x^2\right )} \]

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))

________________________________________________________________________________________

Maple [A]  time = 0.164, size = 80, normalized size = 0.3 \begin{align*}{\frac{{x}^{5} \left ( 3003\,{b}^{5}{x}^{10}+17325\,a{b}^{4}{x}^{8}+40950\,{a}^{2}{b}^{3}{x}^{6}+50050\,{b}^{2}{a}^{3}{x}^{4}+32175\,{a}^{4}b{x}^{2}+9009\,{a}^{5} \right ) }{45045\, \left ( b{x}^{2}+a \right ) ^{5}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [A]  time = 1.01416, size = 77, normalized size = 0.3 \begin{align*} \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} \end{align*}

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

________________________________________________________________________________________

Fricas [A]  time = 1.32816, size = 139, normalized size = 0.55 \begin{align*} \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} \end{align*}

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [A]  time = 1.12896, size = 142, normalized size = 0.56 \begin{align*} \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 ) \end{align*}

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)