3.6.93 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^3} \, dx\) [593]

Optimal. Leaf size=250 \[ -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a^2 b^3 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {b^5 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]

[Out]

-1/2*a^5*((b*x^2+a)^2)^(1/2)/x^2/(b*x^2+a)+5*a^3*b^2*x^2*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/2*a^2*b^3*x^4*((b*x^2
+a)^2)^(1/2)/(b*x^2+a)+5/6*a*b^4*x^6*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/8*b^5*x^8*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5
*a^4*b*ln(x)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]
time = 0.05, antiderivative size = 250, 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 = {1126, 272, 45} \begin {gather*} \frac {b^5 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a b^4 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {5 a^4 b \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^3,x]

[Out]

-1/2*(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^2*(a + b*x^2)) + (5*a^3*b^2*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
/(a + b*x^2) + (5*a^2*b^3*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*(a + b*x^2)) + (5*a*b^4*x^6*Sqrt[a^2 + 2*a*b
*x^2 + b^2*x^4])/(6*(a + b*x^2)) + (b^5*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*(a + b*x^2)) + (5*a^4*b*Sqrt[a
^2 + 2*a*b*x^2 + b^2*x^4]*Log[x])/(a + b*x^2)

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^3} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^5}{x^3} \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^5}{x^2} \, 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} \text {Subst}\left (\int \left (10 a^3 b^7+\frac {a^5 b^5}{x^2}+\frac {5 a^4 b^6}{x}+10 a^2 b^8 x+5 a b^9 x^2+b^{10} x^3\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a^2 b^3 x^4 \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^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {b^5 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 85, normalized size = 0.34 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (-12 a^5+120 a^3 b^2 x^4+60 a^2 b^3 x^6+20 a b^4 x^8+3 b^5 x^{10}+120 a^4 b x^2 \log (x)\right )}{24 x^2 \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^3,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(-12*a^5 + 120*a^3*b^2*x^4 + 60*a^2*b^3*x^6 + 20*a*b^4*x^8 + 3*b^5*x^10 + 120*a^4*b*x^2*L
og[x]))/(24*x^2*(a + b*x^2))

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Maple [A]
time = 0.03, size = 82, normalized size = 0.33

method	result	size
default	\(\frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} \left (3 b^{5} x^{10}+20 b^{4} a \,x^{8}+60 a^{2} b^{3} x^{6}+120 b^{2} a^{3} x^{4}+120 b \,a^{4} \ln \left (x \right ) x^{2}-12 a^{5}\right )}{24 \left (b \,x^{2}+a \right )^{5} x^{2}}\)	\(82\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{2} \left (\frac {1}{8} b^{3} x^{8}+\frac {5}{6} a \,b^{2} x^{6}+\frac {5}{2} a^{2} b \,x^{4}+5 a^{3} x^{2}\right )}{b \,x^{2}+a}-\frac {a^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 x^{2} \left (b \,x^{2}+a \right )}+\frac {5 a^{4} b \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(117\)

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^3,x,method=_RETURNVERBOSE)

[Out]

1/24*((b*x^2+a)^2)^(5/2)*(3*b^5*x^10+20*b^4*a*x^8+60*a^2*b^3*x^6+120*b^2*a^3*x^4+120*b*a^4*ln(x)*x^2-12*a^5)/(
b*x^2+a)^5/x^2

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Maxima [A]
time = 0.28, size = 56, normalized size = 0.22 \begin {gather*} \frac {1}{8} \, b^{5} x^{8} + \frac {5}{6} \, a b^{4} x^{6} + \frac {5}{2} \, a^{2} b^{3} x^{4} + 5 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b \log \left (x\right ) - \frac {a^{5}}{2 \, x^{2}} \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^3,x, algorithm="maxima")

[Out]

1/8*b^5*x^8 + 5/6*a*b^4*x^6 + 5/2*a^2*b^3*x^4 + 5*a^3*b^2*x^2 + 5*a^4*b*log(x) - 1/2*a^5/x^2

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Fricas [A]
time = 0.35, size = 61, normalized size = 0.24 \begin {gather*} \frac {3 \, b^{5} x^{10} + 20 \, a b^{4} x^{8} + 60 \, a^{2} b^{3} x^{6} + 120 \, a^{3} b^{2} x^{4} + 120 \, a^{4} b x^{2} \log \left (x\right ) - 12 \, a^{5}}{24 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

1/24*(3*b^5*x^10 + 20*a*b^4*x^8 + 60*a^2*b^3*x^6 + 120*a^3*b^2*x^4 + 120*a^4*b*x^2*log(x) - 12*a^5)/x^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{3}}\, 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**3,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**3, x)

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Giac [A]
time = 3.92, size = 125, normalized size = 0.50 \begin {gather*} \frac {1}{8} \, b^{5} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{6} \, a b^{4} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a^{4} b \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {5 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, x^{2}} \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^3,x, algorithm="giac")

[Out]

1/8*b^5*x^8*sgn(b*x^2 + a) + 5/6*a*b^4*x^6*sgn(b*x^2 + a) + 5/2*a^2*b^3*x^4*sgn(b*x^2 + a) + 5*a^3*b^2*x^2*sgn
(b*x^2 + a) + 5/2*a^4*b*log(x^2)*sgn(b*x^2 + a) - 1/2*(5*a^4*b*x^2*sgn(b*x^2 + a) + a^5*sgn(b*x^2 + a))/x^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^3} \,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^3,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^3, x)

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