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

Optimal. Leaf size=249 \[ \frac {b^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a b^4 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {2 a^2 b^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

[Out]

-1/4*a^5*((b*x^3+a)^2)^(1/2)/x^4/(b*x^3+a)-5*a^4*b*((b*x^3+a)^2)^(1/2)/x/(b*x^3+a)+5*a^3*b^2*x^2*((b*x^3+a)^2)
^(1/2)/(b*x^3+a)+2*a^2*b^3*x^5*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/8*a*b^4*x^8*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/11*
b^5*x^11*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

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Rubi [A]  time = 0.06, antiderivative size = 249, normalized size of antiderivative = 1.00, 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} \[ \frac {b^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a b^4 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {2 a^2 b^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*x^4*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x*(a +
b*x^3)) + (5*a^3*b^2*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a + b*x^3) + (2*a^2*b^3*x^5*Sqrt[a^2 + 2*a*b*x^3 +
b^2*x^6])/(a + b*x^3) + (5*a*b^4*x^8*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*(a + b*x^3)) + (b^5*x^11*Sqrt[a^2 +
2*a*b*x^3 + b^2*x^6])/(11*(a + b*x^3))

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]

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]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^5} \, 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^5} \, 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^5}+\frac {5 a^4 b^6}{x^2}+10 a^3 b^7 x+10 a^2 b^8 x^4+5 a b^9 x^7+b^{10} x^{10}\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}}{4 x^4 \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {2 a^2 b^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a b^4 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {b^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \[ \frac {\sqrt {\left (a+b x^3\right )^2} \left (-22 a^5-440 a^4 b x^3+440 a^3 b^2 x^6+176 a^2 b^3 x^9+55 a b^4 x^{12}+8 b^5 x^{15}\right )}{88 x^4 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^3)^2]*(-22*a^5 - 440*a^4*b*x^3 + 440*a^3*b^2*x^6 + 176*a^2*b^3*x^9 + 55*a*b^4*x^12 + 8*b^5*x^15
))/(88*x^4*(a + b*x^3))

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fricas [A]  time = 0.86, size = 59, normalized size = 0.24 \[ \frac {8 \, b^{5} x^{15} + 55 \, a b^{4} x^{12} + 176 \, a^{2} b^{3} x^{9} + 440 \, a^{3} b^{2} x^{6} - 440 \, a^{4} b x^{3} - 22 \, a^{5}}{88 \, x^{4}} \]

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^5,x, algorithm="fricas")

[Out]

1/88*(8*b^5*x^15 + 55*a*b^4*x^12 + 176*a^2*b^3*x^9 + 440*a^3*b^2*x^6 - 440*a^4*b*x^3 - 22*a^5)/x^4

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giac [A]  time = 0.35, size = 107, normalized size = 0.43 \[ \frac {1}{11} \, b^{5} x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{8} \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) + 2 \, a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + 5 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {20 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{4 \, x^{4}} \]

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^5,x, algorithm="giac")

[Out]

1/11*b^5*x^11*sgn(b*x^3 + a) + 5/8*a*b^4*x^8*sgn(b*x^3 + a) + 2*a^2*b^3*x^5*sgn(b*x^3 + a) + 5*a^3*b^2*x^2*sgn
(b*x^3 + a) - 1/4*(20*a^4*b*x^3*sgn(b*x^3 + a) + a^5*sgn(b*x^3 + a))/x^4

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maple [A]  time = 0.01, size = 80, normalized size = 0.32 \[ -\frac {\left (-8 b^{5} x^{15}-55 a \,b^{4} x^{12}-176 a^{2} b^{3} x^{9}-440 a^{3} b^{2} x^{6}+440 a^{4} b \,x^{3}+22 a^{5}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}}}{88 \left (b \,x^{3}+a \right )^{5} x^{4}} \]

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^5,x)

[Out]

-1/88*(-8*b^5*x^15-55*a*b^4*x^12-176*a^2*b^3*x^9-440*a^3*b^2*x^6+440*a^4*b*x^3+22*a^5)*((b*x^3+a)^2)^(5/2)/x^4
/(b*x^3+a)^5

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maxima [A]  time = 0.97, size = 59, normalized size = 0.24 \[ \frac {8 \, b^{5} x^{15} + 55 \, a b^{4} x^{12} + 176 \, a^{2} b^{3} x^{9} + 440 \, a^{3} b^{2} x^{6} - 440 \, a^{4} b x^{3} - 22 \, a^{5}}{88 \, x^{4}} \]

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^5,x, algorithm="maxima")

[Out]

1/88*(8*b^5*x^15 + 55*a*b^4*x^12 + 176*a^2*b^3*x^9 + 440*a^3*b^2*x^6 - 440*a^4*b*x^3 - 22*a^5)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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**5,x)

[Out]

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

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