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

Optimal. Leaf size=248 \[ \frac {b^5 x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{11 \left (a+b x^2\right )^5}+\frac {5 a b^4 x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{9 \left (a+b x^2\right )^5}+\frac {10 a^2 b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{7 \left (a+b x^2\right )^5}+\frac {a^5 x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{\left (a+b x^2\right )^5}+\frac {5 a^4 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{3 \left (a+b x^2\right )^5}+\frac {2 a^3 b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{\left (a+b x^2\right )^5} \]

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Rubi [A]  time = 0.05, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1088, 194} \begin {gather*} \frac {b^5 x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{11 \left (a+b x^2\right )^5}+\frac {5 a b^4 x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{9 \left (a+b x^2\right )^5}+\frac {10 a^2 b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{7 \left (a+b x^2\right )^5}+\frac {2 a^3 b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{\left (a+b x^2\right )^5}+\frac {5 a^4 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{3 \left (a+b x^2\right )^5}+\frac {a^5 x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{\left (a+b x^2\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^5*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(a + b*x^2)^5 + (5*a^4*b*x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(3*(
a + b*x^2)^5) + (2*a^3*b^2*x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(a + b*x^2)^5 + (10*a^2*b^3*x^7*(a^2 + 2*a*b
*x^2 + b^2*x^4)^(5/2))/(7*(a + b*x^2)^5) + (5*a*b^4*x^9*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(9*(a + b*x^2)^5) +
(b^5*x^11*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(11*(a + b*x^2)^5)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 1088

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^p/(b + 2*c*x^2)^(2*p), In
t[(b + 2*c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \int \left (2 a b+2 b^2 x^2\right )^5 \, dx}{\left (2 a b+2 b^2 x^2\right )^5}\\ &=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \int \left (32 a^5 b^5+160 a^4 b^6 x^2+320 a^3 b^7 x^4+320 a^2 b^8 x^6+160 a b^9 x^8+32 b^{10} x^{10}\right ) \, dx}{\left (2 a b+2 b^2 x^2\right )^5}\\ &=\frac {a^5 x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{\left (a+b x^2\right )^5}+\frac {5 a^4 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{3 \left (a+b x^2\right )^5}+\frac {2 a^3 b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{\left (a+b x^2\right )^5}+\frac {10 a^2 b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{7 \left (a+b x^2\right )^5}+\frac {5 a b^4 x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{9 \left (a+b x^2\right )^5}+\frac {b^5 x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{11 \left (a+b x^2\right )^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 81, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (693 a^5 x+1155 a^4 b x^3+1386 a^3 b^2 x^5+990 a^2 b^3 x^7+385 a b^4 x^9+63 b^5 x^{11}\right )}{693 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(693*a^5*x + 1155*a^4*b*x^3 + 1386*a^3*b^2*x^5 + 990*a^2*b^3*x^7 + 385*a*b^4*x^9 + 63*b^5
*x^11))/(693*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 6.35, size = 81, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (693 a^5 x+1155 a^4 b x^3+1386 a^3 b^2 x^5+990 a^2 b^3 x^7+385 a b^4 x^9+63 b^5 x^{11}\right )}{693 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(693*a^5*x + 1155*a^4*b*x^3 + 1386*a^3*b^2*x^5 + 990*a^2*b^3*x^7 + 385*a*b^4*x^9 + 63*b^5
*x^11))/(693*(a + b*x^2))

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fricas [A]  time = 0.81, size = 54, normalized size = 0.22 \begin {gather*} \frac {1}{11} \, b^{5} x^{11} + \frac {5}{9} \, a b^{4} x^{9} + \frac {10}{7} \, a^{2} b^{3} x^{7} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{3} \, a^{4} b x^{3} + a^{5} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*b^5*x^11 + 5/9*a*b^4*x^9 + 10/7*a^2*b^3*x^7 + 2*a^3*b^2*x^5 + 5/3*a^4*b*x^3 + a^5*x

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giac [A]  time = 0.16, size = 102, normalized size = 0.41 \begin {gather*} \frac {1}{11} \, b^{5} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{9} \, a b^{4} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{7} \, a^{2} b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{3} b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{3} \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} x \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*b^5*x^11*sgn(b*x^2 + a) + 5/9*a*b^4*x^9*sgn(b*x^2 + a) + 10/7*a^2*b^3*x^7*sgn(b*x^2 + a) + 2*a^3*b^2*x^5*
sgn(b*x^2 + a) + 5/3*a^4*b*x^3*sgn(b*x^2 + a) + a^5*x*sgn(b*x^2 + a)

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maple [A]  time = 0.00, size = 78, normalized size = 0.31 \begin {gather*} \frac {\left (63 b^{5} x^{10}+385 a \,b^{4} x^{8}+990 a^{2} b^{3} x^{6}+1386 a^{3} b^{2} x^{4}+1155 a^{4} b \,x^{2}+693 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x}{693 \left (b \,x^{2}+a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/693*x*(63*b^5*x^10+385*a*b^4*x^8+990*a^2*b^3*x^6+1386*a^3*b^2*x^4+1155*a^4*b*x^2+693*a^5)*((b*x^2+a)^2)^(5/2
)/(b*x^2+a)^5

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maxima [A]  time = 1.32, size = 54, normalized size = 0.22 \begin {gather*} \frac {1}{11} \, b^{5} x^{11} + \frac {5}{9} \, a b^{4} x^{9} + \frac {10}{7} \, a^{2} b^{3} x^{7} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{3} \, a^{4} b x^{3} + a^{5} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*b^5*x^11 + 5/9*a*b^4*x^9 + 10/7*a^2*b^3*x^7 + 2*a^3*b^2*x^5 + 5/3*a^4*b*x^3 + a^5*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \begin {gather*} \int \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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