3.86 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{23}} \, dx\)

Optimal. Leaf size=255 \[ -\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^{10} \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 x^{13} \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 x^{22} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 x^{19} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^{16} \left (a+b x^3\right )} \]

[Out]

-1/22*a^5*((b*x^3+a)^2)^(1/2)/x^22/(b*x^3+a)-5/19*a^4*b*((b*x^3+a)^2)^(1/2)/x^19/(b*x^3+a)-5/8*a^3*b^2*((b*x^3
+a)^2)^(1/2)/x^16/(b*x^3+a)-10/13*a^2*b^3*((b*x^3+a)^2)^(1/2)/x^13/(b*x^3+a)-1/2*a*b^4*((b*x^3+a)^2)^(1/2)/x^1
0/(b*x^3+a)-1/7*b^5*((b*x^3+a)^2)^(1/2)/x^7/(b*x^3+a)

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Rubi [A]  time = 0.06, antiderivative size = 255, 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} \[ -\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 x^{22} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 x^{19} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^{16} \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 x^{13} \left (a+b x^3\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^{10} \left (a+b x^3\right )}-\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^23,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(22*x^22*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(19*x
^19*(a + b*x^3)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*x^16*(a + b*x^3)) - (10*a^2*b^3*Sqrt[a^2 + 2
*a*b*x^3 + b^2*x^6])/(13*x^13*(a + b*x^3)) - (a*b^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*x^10*(a + b*x^3)) - (b
^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*x^7*(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^{23}} \, 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^{23}} \, 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^{23}}+\frac {5 a^4 b^6}{x^{20}}+\frac {10 a^3 b^7}{x^{17}}+\frac {10 a^2 b^8}{x^{14}}+\frac {5 a b^9}{x^{11}}+\frac {b^{10}}{x^8}\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}}{22 x^{22} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 x^{19} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^{16} \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 x^{13} \left (a+b x^3\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^{10} \left (a+b x^3\right )}-\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \[ -\frac {\sqrt {\left (a+b x^3\right )^2} \left (6916 a^5+40040 a^4 b x^3+95095 a^3 b^2 x^6+117040 a^2 b^3 x^9+76076 a b^4 x^{12}+21736 b^5 x^{15}\right )}{152152 x^{22} \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^23,x]

[Out]

-1/152152*(Sqrt[(a + b*x^3)^2]*(6916*a^5 + 40040*a^4*b*x^3 + 95095*a^3*b^2*x^6 + 117040*a^2*b^3*x^9 + 76076*a*
b^4*x^12 + 21736*b^5*x^15))/(x^22*(a + b*x^3))

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fricas [A]  time = 0.90, size = 59, normalized size = 0.23 \[ -\frac {21736 \, b^{5} x^{15} + 76076 \, a b^{4} x^{12} + 117040 \, a^{2} b^{3} x^{9} + 95095 \, a^{3} b^{2} x^{6} + 40040 \, a^{4} b x^{3} + 6916 \, a^{5}}{152152 \, x^{22}} \]

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^23,x, algorithm="fricas")

[Out]

-1/152152*(21736*b^5*x^15 + 76076*a*b^4*x^12 + 117040*a^2*b^3*x^9 + 95095*a^3*b^2*x^6 + 40040*a^4*b*x^3 + 6916
*a^5)/x^22

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giac [A]  time = 0.33, size = 107, normalized size = 0.42 \[ -\frac {21736 \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + 76076 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 117040 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 95095 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 40040 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 6916 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{152152 \, x^{22}} \]

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^23,x, algorithm="giac")

[Out]

-1/152152*(21736*b^5*x^15*sgn(b*x^3 + a) + 76076*a*b^4*x^12*sgn(b*x^3 + a) + 117040*a^2*b^3*x^9*sgn(b*x^3 + a)
+ 95095*a^3*b^2*x^6*sgn(b*x^3 + a) + 40040*a^4*b*x^3*sgn(b*x^3 + a) + 6916*a^5*sgn(b*x^3 + a))/x^22

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maple [A]  time = 0.01, size = 80, normalized size = 0.31 \[ -\frac {\left (21736 b^{5} x^{15}+76076 a \,b^{4} x^{12}+117040 a^{2} b^{3} x^{9}+95095 a^{3} b^{2} x^{6}+40040 a^{4} b \,x^{3}+6916 a^{5}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}}}{152152 \left (b \,x^{3}+a \right )^{5} x^{22}} \]

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^23,x)

[Out]

-1/152152*(21736*b^5*x^15+76076*a*b^4*x^12+117040*a^2*b^3*x^9+95095*a^3*b^2*x^6+40040*a^4*b*x^3+6916*a^5)*((b*
x^3+a)^2)^(5/2)/x^22/(b*x^3+a)^5

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maxima [A]  time = 0.98, size = 59, normalized size = 0.23 \[ -\frac {21736 \, b^{5} x^{15} + 76076 \, a b^{4} x^{12} + 117040 \, a^{2} b^{3} x^{9} + 95095 \, a^{3} b^{2} x^{6} + 40040 \, a^{4} b x^{3} + 6916 \, a^{5}}{152152 \, x^{22}} \]

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^23,x, algorithm="maxima")

[Out]

-1/152152*(21736*b^5*x^15 + 76076*a*b^4*x^12 + 117040*a^2*b^3*x^9 + 95095*a^3*b^2*x^6 + 40040*a^4*b*x^3 + 6916
*a^5)/x^22

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mupad [B]  time = 1.22, size = 231, normalized size = 0.91 \[ -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{22\,x^{22}\,\left (b\,x^3+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{7\,x^7\,\left (b\,x^3+a\right )}-\frac {a\,b^4\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{2\,x^{10}\,\left (b\,x^3+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{19\,x^{19}\,\left (b\,x^3+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{13\,x^{13}\,\left (b\,x^3+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{8\,x^{16}\,\left (b\,x^3+a\right )} \]

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^23,x)

[Out]

- (a^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(22*x^22*(a + b*x^3)) - (b^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(7*x
^7*(a + b*x^3)) - (a*b^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(2*x^10*(a + b*x^3)) - (5*a^4*b*(a^2 + b^2*x^6 + 2
*a*b*x^3)^(1/2))/(19*x^19*(a + b*x^3)) - (10*a^2*b^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(13*x^13*(a + b*x^3))
- (5*a^3*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(8*x^16*(a + b*x^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{23}}\, dx \]

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**23,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**23, x)

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