\(\int x^7 (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\) [588]

Optimal result

Integrand size = 26, antiderivative size = 160 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=-\frac {a^3 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^4}+\frac {3 a^2 \left (a+b x^2\right )^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 b^4}-\frac {3 a \left (a+b x^2\right )^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 b^4}+\frac {\left (a+b x^2\right )^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 b^4} \]

[Out]

-1/12*a^3*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/b^4+3/14*a^2*(b*x^2+a)^6*((b*x^2+a)^2)^(1/2)/b^4-3/16*a*(b*x^2+a)^7*
((b*x^2+a)^2)^(1/2)/b^4+1/18*(b*x^2+a)^8*((b*x^2+a)^2)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^8}{18 b^4}-\frac {3 a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^7}{16 b^4}+\frac {3 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^6}{14 b^4}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^4} \]

[In]

Int[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

-1/12*(a^3*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^4 + (3*a^2*(a + b*x^2)^6*Sqrt[a^2 + 2*a*b*x^2 + b^
2*x^4])/(14*b^4) - (3*a*(a + b*x^2)^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*b^4) + ((a + b*x^2)^8*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4])/(18*b^4)

Rule 45

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])

Rule 660

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

Rule 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int x^3 \left (a b+b^2 x\right )^5 \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (-\frac {a^3 \left (a b+b^2 x\right )^5}{b^3}+\frac {3 a^2 \left (a b+b^2 x\right )^6}{b^4}-\frac {3 a \left (a b+b^2 x\right )^7}{b^5}+\frac {\left (a b+b^2 x\right )^8}{b^6}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = -\frac {a^3 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^4}+\frac {3 a^2 \left (a+b x^2\right )^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 b^4}-\frac {3 a \left (a+b x^2\right )^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 b^4}+\frac {\left (a+b x^2\right )^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.84 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^8 \left (126 a^5+504 a^4 b x^2+840 a^3 b^2 x^4+720 a^2 b^3 x^6+315 a b^4 x^8+56 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{1008 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^8*(126*a^5 + 504*a^4*b*x^2 + 840*a^3*b^2*x^4 + 720*a^2*b^3*x^6 + 315*a*b^4*x^8 + 56*b^5*x^10)*(Sqrt[a^2]*b*
x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])))/(1008*(-a^2 - a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.41

method	result	size
pseudoelliptic	\(\frac {\left (\frac {4}{9} x^{10} b^{5}+\frac {5}{2} a \,x^{8} b^{4}+\frac {40}{7} a^{2} x^{6} b^{3}+\frac {20}{3} a^{3} x^{4} b^{2}+4 x^{2} a^{4} b +a^{5}\right ) x^{8} \operatorname {csgn}\left (b \,x^{2}+a \right )}{8}\)	\(66\)
gosper	\(\frac {x^{8} \left (56 x^{10} b^{5}+315 a \,x^{8} b^{4}+720 a^{2} x^{6} b^{3}+840 a^{3} x^{4} b^{2}+504 x^{2} a^{4} b +126 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{1008 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(\frac {x^{8} \left (56 x^{10} b^{5}+315 a \,x^{8} b^{4}+720 a^{2} x^{6} b^{3}+840 a^{3} x^{4} b^{2}+504 x^{2} a^{4} b +126 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{1008 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{5} x^{8}}{8 b \,x^{2}+8 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{10}}{2 b \,x^{2}+2 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{12}}{6 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{14}}{7 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{16}}{16 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{5} x^{18}}{18 b \,x^{2}+18 a}\)	\(178\)

[In]

int(x^7*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/8*(4/9*x^10*b^5+5/2*a*x^8*b^4+40/7*a^2*x^6*b^3+20/3*a^3*x^4*b^2+4*x^2*a^4*b+a^5)*x^8*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{18} \, b^{5} x^{18} + \frac {5}{16} \, a b^{4} x^{16} + \frac {5}{7} \, a^{2} b^{3} x^{14} + \frac {5}{6} \, a^{3} b^{2} x^{12} + \frac {1}{2} \, a^{4} b x^{10} + \frac {1}{8} \, a^{5} x^{8} \]

[In]

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/18*b^5*x^18 + 5/16*a*b^4*x^16 + 5/7*a^2*b^3*x^14 + 5/6*a^3*b^2*x^12 + 1/2*a^4*b*x^10 + 1/8*a^5*x^8

Sympy [F]

\[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{7} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate(x**7*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**7*((a + b*x**2)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{18} \, b^{5} x^{18} + \frac {5}{16} \, a b^{4} x^{16} + \frac {5}{7} \, a^{2} b^{3} x^{14} + \frac {5}{6} \, a^{3} b^{2} x^{12} + \frac {1}{2} \, a^{4} b x^{10} + \frac {1}{8} \, a^{5} x^{8} \]

[In]

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/18*b^5*x^18 + 5/16*a*b^4*x^16 + 5/7*a^2*b^3*x^14 + 5/6*a^3*b^2*x^12 + 1/2*a^4*b*x^10 + 1/8*a^5*x^8

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.66 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{18} \, b^{5} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{16} \, a b^{4} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a^{2} b^{3} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{6} \, a^{3} b^{2} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{2} \, a^{4} b x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{8} \, a^{5} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/18*b^5*x^18*sgn(b*x^2 + a) + 5/16*a*b^4*x^16*sgn(b*x^2 + a) + 5/7*a^2*b^3*x^14*sgn(b*x^2 + a) + 5/6*a^3*b^2*
x^12*sgn(b*x^2 + a) + 1/2*a^4*b*x^10*sgn(b*x^2 + a) + 1/8*a^5*x^8*sgn(b*x^2 + a)

Mupad [F(-1)]

Timed out. \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^7\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]

[In]

int(x^7*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^7*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)