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

Optimal. Leaf size=158 \[ -\frac {3 a^2}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^3}{4 b^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.13, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \begin {gather*} \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 {3 a^2}{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 {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 {gather*}

Antiderivative was successfully verified.

[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 43

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 646

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 1111

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*} \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.03, size = 81, normalized size = 0.51 \begin {gather*} \frac {-5 a^3-4 a^2 b x^2+4 a b^2 x^4-6 a \left (a+b x^2\right )^2 \log \left (a+b x^2\right )+2 b^3 x^6}{4 b^4 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[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])

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IntegrateAlgebraic [B]  time = 1.26, size = 1386, normalized size = 8.77 \begin {gather*} \frac {-4 \sqrt {b^2} x^{10}-12 a \tanh ^{-1}\left (\frac {\sqrt {b^2 x^4+2 a b x^2+a^2}-\sqrt {b^2} x^2}{a}\right ) x^8-\frac {10 a \sqrt {b^2} x^8}{b}+4 \sqrt {b^2 x^4+2 a b x^2+a^2} x^8+\frac {4 a^2 \left (b^2\right )^{3/2} x^6}{b^4}+\frac {12 a \left (b^2\right )^{3/2} \sqrt {b^2 x^4+2 a b x^2+a^2} \tanh ^{-1}\left (\frac {\sqrt {b^2 x^4+2 a b x^2+a^2}-\sqrt {b^2} x^2}{a}\right ) x^6}{b^4}-\frac {24 a^2 \tanh ^{-1}\left (\frac {\sqrt {b^2 x^4+2 a b x^2+a^2}-\sqrt {b^2} x^2}{a}\right ) x^6}{b}+\frac {6 a \sqrt {b^2 x^4+2 a b x^2+a^2} x^6}{b}+\frac {12 a^2 \sqrt {b^2} \sqrt {b^2 x^4+2 a b x^2+a^2} \tanh ^{-1}\left (\frac {\sqrt {b^2 x^4+2 a b x^2+a^2}-\sqrt {b^2} x^2}{a}\right ) x^4}{b^3}-\frac {12 a^3 \tanh ^{-1}\left (\frac {\sqrt {b^2 x^4+2 a b x^2+a^2}-\sqrt {b^2} x^2}{a}\right ) x^4}{b^2}+\frac {16 a^3 \sqrt {b^2} x^4}{b^3}-\frac {10 a^2 \sqrt {b^2 x^4+2 a b x^2+a^2} x^4}{b^2}+\frac {8 a^4 \sqrt {b^2} x^2}{b^4}-\frac {6 a^3 \sqrt {b^2 x^4+2 a b x^2+a^2} x^2}{b^3}-\frac {2 a^4 \sqrt {b^2 x^4+2 a b x^2+a^2}}{b^4}}{\left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right )^2 \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right )^2}+\frac {\frac {6 a b \log \left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^8}{\sqrt {b^2}}+\frac {6 a b \log \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^8}{\sqrt {b^2}}-\frac {6 a \sqrt {b^2 x^4+2 a b x^2+a^2} \log \left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^6}{b}+\frac {12 a^2 \log \left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^6}{\sqrt {b^2}}-\frac {6 a \sqrt {b^2 x^4+2 a b x^2+a^2} \log \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^6}{b}+\frac {12 a^2 \log \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^6}{\sqrt {b^2}}-\frac {6 a^2 \sqrt {b^2 x^4+2 a b x^2+a^2} \log \left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^4}{b^2}+\frac {6 a^3 \log \left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^4}{b \sqrt {b^2}}-\frac {6 a^2 \sqrt {b^2 x^4+2 a b x^2+a^2} \log \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^4}{b^2}+\frac {6 a^3 \log \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right ) x^4}{b \sqrt {b^2}}-\frac {8 a^3 x^4}{b \sqrt {b^2}}+\frac {8 a^3 \sqrt {b^2 x^4+2 a b x^2+a^2} x^2}{b^3}-\frac {8 a^4 x^2}{\left (b^2\right )^{3/2}}-\frac {2 a^5}{b^3 \sqrt {b^2}}}{\left (-\sqrt {b^2} x^2-a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right )^2 \left (-\sqrt {b^2} x^2+a+\sqrt {b^2 x^4+2 a b x^2+a^2}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4*Sqrt[b^2]*x^2)/b^4 + (16*a^3*Sqrt[b^2]*x^4)/b^3 + (4*a^2*(b^2)^(3/2)*x^6)/b^4 - (10*a*Sqrt[b^2]*x^8)/b
- 4*Sqrt[b^2]*x^10 - (2*a^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^4 - (6*a^3*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
)/b^3 - (10*a^2*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^2 + (6*a*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b + 4*x^8
*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4] - (12*a^3*x^4*ArcTanh[(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/a]
)/b^2 - (24*a^2*x^6*ArcTanh[(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/a])/b - 12*a*x^8*ArcTanh[(-(S
qrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/a] + (12*a^2*Sqrt[b^2]*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Ar
cTanh[(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/a])/b^3 + (12*a*(b^2)^(3/2)*x^6*Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4]*ArcTanh[(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/a])/b^4)/((-a - Sqrt[b^2]*x^2 + Sqrt
[a^2 + 2*a*b*x^2 + b^2*x^4])^2*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^2) + ((-2*a^5)/(b^3*Sqrt[
b^2]) - (8*a^4*x^2)/(b^2)^(3/2) - (8*a^3*x^4)/(b*Sqrt[b^2]) + (8*a^3*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^3
+ (6*a^3*x^4*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(b*Sqrt[b^2]) + (12*a^2*x^6*Log[-a - S
qrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] + (6*a*b*x^8*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4]])/Sqrt[b^2] - (6*a^2*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4]])/b^2 - (6*a*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4]])/b + (6*a^3*x^4*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(b*Sqrt[b^2]) +
(12*a^2*x^6*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] + (6*a*b*x^8*Log[a - Sqrt[b^2]
*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/Sqrt[b^2] - (6*a^2*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sqrt[b
^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/b^2 - (6*a*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sqrt[b^2]*x
^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/b)/((-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^2*(a - Sqrt[
b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^2)

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fricas [A]  time = 1.49, size = 91, normalized size = 0.58 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.25, size = 83, normalized size = 0.53 \begin {gather*} \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 {6 \, a^{2} b x^{2} + 5 \, a^{3}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*(6*a^2*b*x^2 + 5*a^3)/((b*
x^2 + a)^2*b^4*sgn(b*x^2 + a))

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maple [A]  time = 0.02, size = 103, normalized size = 0.65 \begin {gather*} -\frac {\left (-2 b^{3} x^{6}+6 a \,b^{2} x^{4} \ln \left (b \,x^{2}+a \right )-4 a \,b^{2} x^{4}+12 a^{2} b \,x^{2} \ln \left (b \,x^{2}+a \right )+4 a^{2} b \,x^{2}+6 a^{3} \ln \left (b \,x^{2}+a \right )+5 a^{3}\right ) \left (b \,x^{2}+a \right )}{4 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-2*b^3*x^6+6*ln(b*x^2+a)*x^4*a*b^2-4*a*b^2*x^4+12*ln(b*x^2+a)*x^2*a^2*b+4*a^2*b*x^2+6*ln(b*x^2+a)*a^3+5*
a^3)*(b*x^2+a)/b^4/((b*x^2+a)^2)^(3/2)

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maxima [A]  time = 1.36, size = 66, normalized size = 0.42 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^7}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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