\(\int \frac {x^7}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [634]

Optimal result

Integrand size = 26, antiderivative size = 158 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {3 a^2}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^3}{4 b^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

-3/2*a^2/b^4/((b*x^2+a)^2)^(1/2)+1/4*a^3/b^4/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+1/2*x^2*(b*x^2+a)/b^3/((b*x^2+a)^2)
^(1/2)-3/2*a*(b*x^2+a)*ln(b*x^2+a)/b^4/((b*x^2+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 158, 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} \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {3 a^2}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^3}{4 b^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(-3*a^2)/(2*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + a^3/(4*b^4*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (
x^2*(a + b*x^2))/(2*b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (3*a*(a + b*x^2)*Log[a + b*x^2])/(2*b^4*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^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 \frac {x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = \frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \frac {x^3}{\left (a b+b^2 x\right )^3} \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \left (\frac {1}{b^6}-\frac {a^3}{b^6 (a+b x)^3}+\frac {3 a^2}{b^6 (a+b x)^2}-\frac {3 a}{b^6 (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {3 a^2}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^3}{4 b^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.51 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-5 a^3-4 a^2 b x^2+4 a b^2 x^4+2 b^3 x^6-6 a \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 b^4 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]

[In]

Integrate[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(-5*a^3 - 4*a^2*b*x^2 + 4*a*b^2*x^4 + 2*b^3*x^6 - 6*a*(a + b*x^2)^2*Log[a + b*x^2])/(4*b^4*(a + b*x^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 3.

Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.47

method	result	size
pseudoelliptic	\(-\frac {3 \left (a \left (b \,x^{2}+a \right )^{2} \ln \left (b \,x^{2}+a \right )-\frac {b^{3} x^{6}}{3}-\frac {2 b^{2} x^{4} a}{3}+\frac {2 a^{2} b \,x^{2}}{3}+\frac {5 a^{3}}{6}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right )^{2} b^{4}}\)	\(74\)
default	\(-\frac {\left (-2 b^{3} x^{6}+6 \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{2}-4 b^{2} x^{4} a +12 \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b +4 a^{2} b \,x^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{3}+5 a^{3}\right ) \left (b \,x^{2}+a \right )}{4 b^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(103\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {3 a^{2} x^{2}}{2}-\frac {5 a^{3}}{4 b}\right )}{\left (b \,x^{2}+a \right )^{3} b^{3}}-\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{4}}\)	\(105\)

[In]

int(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-3/2*(a*(b*x^2+a)^2*ln(b*x^2+a)-1/3*b^3*x^6-2/3*b^2*x^4*a+2/3*a^2*b*x^2+5/6*a^3)*csgn(b*x^2+a)/(b*x^2+a)^2/b^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.58 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {2 \, b^{3} x^{6} + 4 \, a b^{2} x^{4} - 4 \, a^{2} b x^{2} - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{4} + 2 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/4*(2*b^3*x^6 + 4*a*b^2*x^4 - 4*a^2*b*x^2 - 5*a^3 - 6*(a*b^2*x^4 + 2*a^2*b*x^2 + a^3)*log(b*x^2 + a))/(b^6*x^
4 + 2*a*b^5*x^2 + a^2*b^4)

Sympy [F]

\[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^{7}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**7/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**7/((a + b*x**2)**2)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.42 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {6 \, a^{2} b x^{2} + 5 \, a^{3}}{4 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} + \frac {x^{2}}{2 \, b^{3}} - \frac {3 \, a \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(6*a^2*b*x^2 + 5*a^3)/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4) + 1/2*x^2/b^3 - 3/2*a*log(b*x^2 + a)/b^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.58 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {x^{2}}{2 \, b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {3 \, a \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {9 \, a b^{2} x^{4} + 12 \, a^{2} b x^{2} + 4 \, a^{3}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/2*x^2/(b^3*sgn(b*x^2 + a)) - 3/2*a*log(abs(b*x^2 + a))/(b^4*sgn(b*x^2 + a)) + 1/4*(9*a*b^2*x^4 + 12*a^2*b*x^
2 + 4*a^3)/((b*x^2 + a)^2*b^4*sgn(b*x^2 + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^7}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]

[In]

int(x^7/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int(x^7/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)