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

Optimal. Leaf size=167 \[ \frac {a b^2 x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {3 a^2 b x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {b^3 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {a^3 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )} \]

[Out]

1/8*a^3*x^8*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+3/10*a^2*b*x^10*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/4*a*b^2*x^12*((b*x^2
+a)^2)^(1/2)/(b*x^2+a)+1/14*b^3*x^14*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]  time = 0.11, antiderivative size = 167, 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} \[ \frac {b^3 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {a b^2 x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {3 a^2 b x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {a^3 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*(a + b*x^2)) + (3*a^2*b*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10
*(a + b*x^2)) + (a*b^2*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*(a + b*x^2)) + (b^3*x^14*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(14*(a + b*x^2))

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 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 x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int x^3 \left (a b+b^2 x\right )^3 \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \left (a^3 b^3 x^3+3 a^2 b^4 x^4+3 a b^5 x^5+b^6 x^6\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {a^3 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {3 a^2 b x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {a b^2 x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {b^3 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.37 \[ \frac {x^8 \sqrt {\left (a+b x^2\right )^2} \left (35 a^3+84 a^2 b x^2+70 a b^2 x^4+20 b^3 x^6\right )}{280 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^8*Sqrt[(a + b*x^2)^2]*(35*a^3 + 84*a^2*b*x^2 + 70*a*b^2*x^4 + 20*b^3*x^6))/(280*(a + b*x^2))

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fricas [A]  time = 0.84, size = 35, normalized size = 0.21 \[ \frac {1}{14} \, b^{3} x^{14} + \frac {1}{4} \, a b^{2} x^{12} + \frac {3}{10} \, a^{2} b x^{10} + \frac {1}{8} \, a^{3} x^{8} \]

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]

1/14*b^3*x^14 + 1/4*a*b^2*x^12 + 3/10*a^2*b*x^10 + 1/8*a^3*x^8

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giac [A]  time = 0.16, size = 67, normalized size = 0.40 \[ \frac {1}{14} \, b^{3} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{4} \, a b^{2} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{10} \, a^{2} b x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{8} \, a^{3} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) \]

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]

1/14*b^3*x^14*sgn(b*x^2 + a) + 1/4*a*b^2*x^12*sgn(b*x^2 + a) + 3/10*a^2*b*x^10*sgn(b*x^2 + a) + 1/8*a^3*x^8*sg
n(b*x^2 + a)

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \[ \frac {\left (20 b^{3} x^{6}+70 a \,b^{2} x^{4}+84 a^{2} b \,x^{2}+35 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} x^{8}}{280 \left (b \,x^{2}+a \right )^{3}} \]

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]

1/280*x^8*(20*b^3*x^6+70*a*b^2*x^4+84*a^2*b*x^2+35*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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maxima [A]  time = 1.32, size = 35, normalized size = 0.21 \[ \frac {1}{14} \, b^{3} x^{14} + \frac {1}{4} \, a b^{2} x^{12} + \frac {3}{10} \, a^{2} b x^{10} + \frac {1}{8} \, a^{3} x^{8} \]

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]

1/14*b^3*x^14 + 1/4*a*b^2*x^12 + 3/10*a^2*b*x^10 + 1/8*a^3*x^8

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

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

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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