3.71 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^8} \, dx\)

Optimal. Leaf size=248 \[ \frac{b^5 x^8 \sqrt{a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac{a b^4 x^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac{10 a^3 b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac{5 a^4 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac{a^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*x^7*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*x^4*
(a + b*x^3)) - (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x*(a + b*x^3)) + (5*a^2*b^3*x^2*Sqrt[a^2 + 2*a*b*
x^3 + b^2*x^6])/(a + b*x^3) + (a*b^4*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a + b*x^3) + (b^5*x^8*Sqrt[a^2 + 2*
a*b*x^3 + b^2*x^6])/(8*(a + b*x^3))

________________________________________________________________________________________

Rubi [A]  time = 0.058232, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {1355, 270} \[ \frac{b^5 x^8 \sqrt{a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac{a b^4 x^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac{10 a^3 b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac{5 a^4 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac{a^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^8,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*x^7*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*x^4*
(a + b*x^3)) - (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x*(a + b*x^3)) + (5*a^2*b^3*x^2*Sqrt[a^2 + 2*a*b*
x^3 + b^2*x^6])/(a + b*x^3) + (a*b^4*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a + b*x^3) + (b^5*x^8*Sqrt[a^2 + 2*
a*b*x^3 + b^2*x^6])/(8*(a + b*x^3))

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 270

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

Rubi steps

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

Mathematica [A]  time = 0.0202955, size = 83, normalized size = 0.33 \[ \frac{\sqrt{\left (a+b x^3\right )^2} \left (280 a^2 b^3 x^9-560 a^3 b^2 x^6-70 a^4 b x^3-8 a^5+56 a b^4 x^{12}+7 b^5 x^{15}\right )}{56 x^7 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^8,x]

[Out]

(Sqrt[(a + b*x^3)^2]*(-8*a^5 - 70*a^4*b*x^3 - 560*a^3*b^2*x^6 + 280*a^2*b^3*x^9 + 56*a*b^4*x^12 + 7*b^5*x^15))
/(56*x^7*(a + b*x^3))

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 80, normalized size = 0.3 \begin{align*} -{\frac{-7\,{b}^{5}{x}^{15}-56\,a{b}^{4}{x}^{12}-280\,{a}^{2}{b}^{3}{x}^{9}+560\,{a}^{3}{b}^{2}{x}^{6}+70\,{a}^{4}b{x}^{3}+8\,{a}^{5}}{56\,{x}^{7} \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^8,x)

[Out]

-1/56*(-7*b^5*x^15-56*a*b^4*x^12-280*a^2*b^3*x^9+560*a^3*b^2*x^6+70*a^4*b*x^3+8*a^5)*((b*x^3+a)^2)^(5/2)/x^7/(
b*x^3+a)^5

________________________________________________________________________________________

Maxima [A]  time = 1.07002, size = 80, normalized size = 0.32 \begin{align*} \frac{7 \, b^{5} x^{15} + 56 \, a b^{4} x^{12} + 280 \, a^{2} b^{3} x^{9} - 560 \, a^{3} b^{2} x^{6} - 70 \, a^{4} b x^{3} - 8 \, a^{5}}{56 \, x^{7}} \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^8,x, algorithm="maxima")

[Out]

1/56*(7*b^5*x^15 + 56*a*b^4*x^12 + 280*a^2*b^3*x^9 - 560*a^3*b^2*x^6 - 70*a^4*b*x^3 - 8*a^5)/x^7

________________________________________________________________________________________

Fricas [A]  time = 1.77122, size = 132, normalized size = 0.53 \begin{align*} \frac{7 \, b^{5} x^{15} + 56 \, a b^{4} x^{12} + 280 \, a^{2} b^{3} x^{9} - 560 \, a^{3} b^{2} x^{6} - 70 \, a^{4} b x^{3} - 8 \, a^{5}}{56 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

1/56*(7*b^5*x^15 + 56*a*b^4*x^12 + 280*a^2*b^3*x^9 - 560*a^3*b^2*x^6 - 70*a^4*b*x^3 - 8*a^5)/x^7

________________________________________________________________________________________

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^{8}}\, 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**8,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**8, x)

________________________________________________________________________________________

Giac [A]  time = 1.10424, size = 144, normalized size = 0.58 \begin{align*} \frac{1}{8} \, b^{5} x^{8} \mathrm{sgn}\left (b x^{3} + a\right ) + a b^{4} x^{5} \mathrm{sgn}\left (b x^{3} + a\right ) + 5 \, a^{2} b^{3} x^{2} \mathrm{sgn}\left (b x^{3} + a\right ) - \frac{280 \, a^{3} b^{2} x^{6} \mathrm{sgn}\left (b x^{3} + a\right ) + 35 \, a^{4} b x^{3} \mathrm{sgn}\left (b x^{3} + a\right ) + 4 \, a^{5} \mathrm{sgn}\left (b x^{3} + a\right )}{28 \, x^{7}} \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^8,x, algorithm="giac")

[Out]

1/8*b^5*x^8*sgn(b*x^3 + a) + a*b^4*x^5*sgn(b*x^3 + a) + 5*a^2*b^3*x^2*sgn(b*x^3 + a) - 1/28*(280*a^3*b^2*x^6*s
gn(b*x^3 + a) + 35*a^4*b*x^3*sgn(b*x^3 + a) + 4*a^5*sgn(b*x^3 + a))/x^7