3.88 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{25}} \, dx\)

Optimal. Leaf size=128 \[ -\frac{b^2 \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{504 a^3 x^{18}}+\frac{b \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{84 a^2 x^{21}}-\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{24 a x^{24}} \]

[Out]

-((a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(24*a*x^24) + (b*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6
])/(84*a^2*x^21) - (b^2*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(504*a^3*x^18)

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Rubi [A]  time = 0.0566844, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1355, 266, 45, 37} \[ -\frac{b^2 \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{504 a^3 x^{18}}+\frac{b \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{84 a^2 x^{21}}-\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{24 a x^{24}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^25,x]

[Out]

-((a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(24*a*x^24) + (b*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6
])/(84*a^2*x^21) - (b^2*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(504*a^3*x^18)

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 266

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 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{25}} \, dx &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int \frac{\left (a b+b^2 x^3\right )^5}{x^{25}} \, dx}{b^4 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^5}{x^9} \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )}\\ &=-\frac{\left (a+b x^3\right )^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{24 a x^{24}}-\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^5}{x^8} \, dx,x,x^3\right )}{12 a b^3 \left (a b+b^2 x^3\right )}\\ &=-\frac{\left (a+b x^3\right )^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{24 a x^{24}}+\frac{b \left (a+b x^3\right )^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{84 a^2 x^{21}}+\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^5}{x^7} \, dx,x,x^3\right )}{84 a^2 b^2 \left (a b+b^2 x^3\right )}\\ &=-\frac{\left (a+b x^3\right )^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{24 a x^{24}}+\frac{b \left (a+b x^3\right )^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{84 a^2 x^{21}}-\frac{b^2 \left (a+b x^3\right )^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{504 a^3 x^{18}}\\ \end{align*}

Mathematica [A]  time = 0.0196098, size = 83, normalized size = 0.65 \[ -\frac{\sqrt{\left (a+b x^3\right )^2} \left (336 a^2 b^3 x^9+280 a^3 b^2 x^6+120 a^4 b x^3+21 a^5+210 a b^4 x^{12}+56 b^5 x^{15}\right )}{504 x^{24} \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^25,x]

[Out]

-(Sqrt[(a + b*x^3)^2]*(21*a^5 + 120*a^4*b*x^3 + 280*a^3*b^2*x^6 + 336*a^2*b^3*x^9 + 210*a*b^4*x^12 + 56*b^5*x^
15))/(504*x^24*(a + b*x^3))

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Maple [A]  time = 0.007, size = 80, normalized size = 0.6 \begin{align*} -{\frac{56\,{b}^{5}{x}^{15}+210\,a{b}^{4}{x}^{12}+336\,{a}^{2}{b}^{3}{x}^{9}+280\,{a}^{3}{b}^{2}{x}^{6}+120\,{a}^{4}b{x}^{3}+21\,{a}^{5}}{504\,{x}^{24} \left ( b{x}^{3}+a \right ) ^{5}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^25,x)

[Out]

-1/504*(56*b^5*x^15+210*a*b^4*x^12+336*a^2*b^3*x^9+280*a^3*b^2*x^6+120*a^4*b*x^3+21*a^5)*((b*x^3+a)^2)^(5/2)/x
^24/(b*x^3+a)^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^25,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73209, size = 142, normalized size = 1.11 \begin{align*} -\frac{56 \, b^{5} x^{15} + 210 \, a b^{4} x^{12} + 336 \, a^{2} b^{3} x^{9} + 280 \, a^{3} b^{2} x^{6} + 120 \, a^{4} b x^{3} + 21 \, a^{5}}{504 \, x^{24}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^25,x, algorithm="fricas")

[Out]

-1/504*(56*b^5*x^15 + 210*a*b^4*x^12 + 336*a^2*b^3*x^9 + 280*a^3*b^2*x^6 + 120*a^4*b*x^3 + 21*a^5)/x^24

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac{5}{2}}}{x^{25}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x**25,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**25, x)

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Giac [A]  time = 1.11714, size = 144, normalized size = 1.12 \begin{align*} -\frac{56 \, b^{5} x^{15} \mathrm{sgn}\left (b x^{3} + a\right ) + 210 \, a b^{4} x^{12} \mathrm{sgn}\left (b x^{3} + a\right ) + 336 \, a^{2} b^{3} x^{9} \mathrm{sgn}\left (b x^{3} + a\right ) + 280 \, a^{3} b^{2} x^{6} \mathrm{sgn}\left (b x^{3} + a\right ) + 120 \, a^{4} b x^{3} \mathrm{sgn}\left (b x^{3} + a\right ) + 21 \, a^{5} \mathrm{sgn}\left (b x^{3} + a\right )}{504 \, x^{24}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^25,x, algorithm="giac")

[Out]

-1/504*(56*b^5*x^15*sgn(b*x^3 + a) + 210*a*b^4*x^12*sgn(b*x^3 + a) + 336*a^2*b^3*x^9*sgn(b*x^3 + a) + 280*a^3*
b^2*x^6*sgn(b*x^3 + a) + 120*a^4*b*x^3*sgn(b*x^3 + a) + 21*a^5*sgn(b*x^3 + a))/x^24