\(\int x^8 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\) [572]

Optimal result

Integrand size = 26, antiderivative size = 167 \[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {a^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {3 a^2 b x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {3 a b^2 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {b^3 x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1126, 276} \[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {3 a b^2 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {3 a^2 b x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {b^3 x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )}+\frac {a^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )} \]

[In]

Int[x^8*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

Rule 276

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 1126

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*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^8 \left (a b+b^2 x^2\right )^3 \, dx}{b^2 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^3 b^3 x^8+3 a^2 b^4 x^{10}+3 a b^5 x^{12}+b^6 x^{14}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )} \\ & = \frac {a^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {3 a^2 b x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {3 a b^2 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {b^3 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] (verified)

Time = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {x^9 \sqrt {\left (a+b x^2\right )^2} \left (715 a^3+1755 a^2 b x^2+1485 a b^2 x^4+429 b^3 x^6\right )}{6435 \left (a+b x^2\right )} \]

[In]

Integrate[x^8*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x^9*Sqrt[(a + b*x^2)^2]*(715*a^3 + 1755*a^2*b*x^2 + 1485*a*b^2*x^4 + 429*b^3*x^6))/(6435*(a + b*x^2))

Maple [A] (verified)

Time = 2.69 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.35

method	result	size
gosper	\(\frac {x^{9} \left (429 b^{3} x^{6}+1485 b^{2} x^{4} a +1755 a^{2} b \,x^{2}+715 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{6435 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
default	\(\frac {x^{9} \left (429 b^{3} x^{6}+1485 b^{2} x^{4} a +1755 a^{2} b \,x^{2}+715 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{6435 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
risch	\(\frac {a^{3} x^{9} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{9 b \,x^{2}+9 a}+\frac {3 a^{2} b \,x^{11} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{11 \left (b \,x^{2}+a \right )}+\frac {3 a \,b^{2} x^{13} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{13 \left (b \,x^{2}+a \right )}+\frac {b^{3} x^{15} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{15 b \,x^{2}+15 a}\)	\(116\)

[In]

int(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/6435*x^9*(429*b^3*x^6+1485*a*b^2*x^4+1755*a^2*b*x^2+715*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{15} \, b^{3} x^{15} + \frac {3}{13} \, a b^{2} x^{13} + \frac {3}{11} \, a^{2} b x^{11} + \frac {1}{9} \, a^{3} x^{9} \]

[In]

integrate(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/15*b^3*x^15 + 3/13*a*b^2*x^13 + 3/11*a^2*b*x^11 + 1/9*a^3*x^9

Sympy [F]

\[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int x^{8} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate(x**8*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**8*((a + b*x**2)**2)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{15} \, b^{3} x^{15} + \frac {3}{13} \, a b^{2} x^{13} + \frac {3}{11} \, a^{2} b x^{11} + \frac {1}{9} \, a^{3} x^{9} \]

[In]

integrate(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/15*b^3*x^15 + 3/13*a*b^2*x^13 + 3/11*a^2*b*x^11 + 1/9*a^3*x^9

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.40 \[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{15} \, b^{3} x^{15} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{13} \, a b^{2} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{11} \, a^{2} b x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{9} \, a^{3} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/15*b^3*x^15*sgn(b*x^2 + a) + 3/13*a*b^2*x^13*sgn(b*x^2 + a) + 3/11*a^2*b*x^11*sgn(b*x^2 + a) + 1/9*a^3*x^9*s
gn(b*x^2 + a)

Mupad [F(-1)]

Timed out. \[ \int x^8 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int x^8\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \]

[In]

int(x^8*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int(x^8*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)