3.585 \(\int x^{13} (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\)

Optimal. Leaf size=255 \[ \frac {b^5 x^{24} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 \left (a+b x^2\right )}+\frac {5 a b^4 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {a^5 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )} \]

[Out]

1/14*a^5*x^14*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/16*a^4*b*x^16*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/9*a^3*b^2*x^18*((b
*x^2+a)^2)^(1/2)/(b*x^2+a)+1/2*a^2*b^3*x^20*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/22*a*b^4*x^22*((b*x^2+a)^2)^(1/2)/
(b*x^2+a)+1/24*b^5*x^24*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]  time = 0.16, antiderivative size = 255, 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} \[ \frac {b^5 x^{24} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 \left (a+b x^2\right )}+\frac {5 a b^4 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 \left (a+b x^2\right )}+\frac {a^5 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^13*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(a^5*x^14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(14*(a + b*x^2)) + (5*a^4*b*x^16*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(
16*(a + b*x^2)) + (5*a^3*b^2*x^18*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*(a + b*x^2)) + (a^2*b^3*x^20*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])/(2*(a + b*x^2)) + (5*a*b^4*x^22*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(22*(a + b*x^2)) + (b^5
*x^24*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(24*(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^{13} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^6 \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} \operatorname {Subst}\left (\int x^6 \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} \operatorname {Subst}\left (\int \left (a^5 b^5 x^6+5 a^4 b^6 x^7+10 a^3 b^7 x^8+10 a^2 b^8 x^9+5 a b^9 x^{10}+b^{10} x^{11}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {a^5 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}+\frac {b^5 x^{24} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 83, normalized size = 0.33 \[ \frac {x^{14} \sqrt {\left (a+b x^2\right )^2} \left (792 a^5+3465 a^4 b x^2+6160 a^3 b^2 x^4+5544 a^2 b^3 x^6+2520 a b^4 x^8+462 b^5 x^{10}\right )}{11088 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^13*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^14*Sqrt[(a + b*x^2)^2]*(792*a^5 + 3465*a^4*b*x^2 + 6160*a^3*b^2*x^4 + 5544*a^2*b^3*x^6 + 2520*a*b^4*x^8 + 4
62*b^5*x^10))/(11088*(a + b*x^2))

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fricas [A]  time = 0.65, size = 57, normalized size = 0.22 \[ \frac {1}{24} \, b^{5} x^{24} + \frac {5}{22} \, a b^{4} x^{22} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {5}{9} \, a^{3} b^{2} x^{18} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{14} \, a^{5} x^{14} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/24*b^5*x^24 + 5/22*a*b^4*x^22 + 1/2*a^2*b^3*x^20 + 5/9*a^3*b^2*x^18 + 5/16*a^4*b*x^16 + 1/14*a^5*x^14

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giac [A]  time = 0.16, size = 105, normalized size = 0.41 \[ \frac {1}{24} \, b^{5} x^{24} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{22} \, a b^{4} x^{22} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{2} \, a^{2} b^{3} x^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{9} \, a^{3} b^{2} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{16} \, a^{4} b x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{14} \, a^{5} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/24*b^5*x^24*sgn(b*x^2 + a) + 5/22*a*b^4*x^22*sgn(b*x^2 + a) + 1/2*a^2*b^3*x^20*sgn(b*x^2 + a) + 5/9*a^3*b^2*
x^18*sgn(b*x^2 + a) + 5/16*a^4*b*x^16*sgn(b*x^2 + a) + 1/14*a^5*x^14*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 80, normalized size = 0.31 \[ \frac {\left (462 b^{5} x^{10}+2520 a \,b^{4} x^{8}+5544 a^{2} b^{3} x^{6}+6160 a^{3} b^{2} x^{4}+3465 a^{4} b \,x^{2}+792 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x^{14}}{11088 \left (b \,x^{2}+a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/11088*x^14*(462*b^5*x^10+2520*a*b^4*x^8+5544*a^2*b^3*x^6+6160*a^3*b^2*x^4+3465*a^4*b*x^2+792*a^5)*((b*x^2+a)
^2)^(5/2)/(b*x^2+a)^5

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maxima [A]  time = 1.32, size = 57, normalized size = 0.22 \[ \frac {1}{24} \, b^{5} x^{24} + \frac {5}{22} \, a b^{4} x^{22} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {5}{9} \, a^{3} b^{2} x^{18} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{14} \, a^{5} x^{14} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/24*b^5*x^24 + 5/22*a*b^4*x^22 + 1/2*a^2*b^3*x^20 + 5/9*a^3*b^2*x^18 + 5/16*a^4*b*x^16 + 1/14*a^5*x^14

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^{13}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^13*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{13} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**13*((a + b*x**2)**2)**(5/2), x)

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