3.1.76 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{13}} \, dx\) [76]

Optimal. Leaf size=252 \[ -\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 x^{12} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^6 \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {b^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]

[Out]

-1/12*a^5*((b*x^3+a)^2)^(1/2)/x^12/(b*x^3+a)-5/9*a^4*b*((b*x^3+a)^2)^(1/2)/x^9/(b*x^3+a)-5/3*a^3*b^2*((b*x^3+a
)^2)^(1/2)/x^6/(b*x^3+a)-10/3*a^2*b^3*((b*x^3+a)^2)^(1/2)/x^3/(b*x^3+a)+1/3*b^5*x^3*((b*x^3+a)^2)^(1/2)/(b*x^3
+a)+5*a*b^4*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

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Rubi [A]
time = 0.04, antiderivative size = 252, 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 = {1369, 272, 45} \begin {gather*} \frac {b^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a b^4 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 x^{12} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^6 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^13,x]

[Out]

-1/12*(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^12*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(9*
x^9*(a + b*x^3)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*x^6*(a + b*x^3)) - (10*a^2*b^3*Sqrt[a^2 + 2*
a*b*x^3 + b^2*x^6])/(3*x^3*(a + b*x^3)) + (b^5*x^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*(a + b*x^3)) + (5*a*b^4
*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*Log[x])/(a + b*x^3)

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 1369

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^{13}} \, 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^{13}} \, dx}{b^4 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^5}{x^5} \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \text {Subst}\left (\int \left (b^{10}+\frac {a^5 b^5}{x^5}+\frac {5 a^4 b^6}{x^4}+\frac {10 a^3 b^7}{x^3}+\frac {10 a^2 b^8}{x^2}+\frac {5 a b^9}{x}\right ) \, dx,x,x^3\right )}{3 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}}{12 x^{12} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^6 \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {b^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 85, normalized size = 0.34 \begin {gather*} -\frac {\sqrt {\left (a+b x^3\right )^2} \left (3 a^5+20 a^4 b x^3+60 a^3 b^2 x^6+120 a^2 b^3 x^9-12 b^5 x^{15}-180 a b^4 x^{12} \log (x)\right )}{36 x^{12} \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^13,x]

[Out]

-1/36*(Sqrt[(a + b*x^3)^2]*(3*a^5 + 20*a^4*b*x^3 + 60*a^3*b^2*x^6 + 120*a^2*b^3*x^9 - 12*b^5*x^15 - 180*a*b^4*
x^12*Log[x]))/(x^12*(a + b*x^3))

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

method	result	size
default	\(\frac {\left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}} \left (12 b^{5} x^{15}+180 b^{4} a \ln \left (x \right ) x^{12}-120 a^{2} b^{3} x^{9}-60 b^{2} a^{3} x^{6}-20 a^{4} b \,x^{3}-3 a^{5}\right )}{36 \left (b \,x^{3}+a \right )^{5} x^{12}}\)	\(82\)
risch	\(\frac {b^{5} x^{3} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{3 b \,x^{3}+3 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {10}{3} a^{2} b^{3} x^{9}-\frac {5}{3} b^{2} a^{3} x^{6}-\frac {5}{9} a^{4} b \,x^{3}-\frac {1}{12} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{12}}+\frac {5 a \,b^{4} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\)	\(119\)

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^13,x,method=_RETURNVERBOSE)

[Out]

1/36*((b*x^3+a)^2)^(5/2)*(12*b^5*x^15+180*b^4*a*ln(x)*x^12-120*a^2*b^3*x^9-60*b^2*a^3*x^6-20*a^4*b*x^3-3*a^5)/
(b*x^3+a)^5/x^12

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Maxima [A]
time = 0.29, size = 342, normalized size = 1.36 \begin {gather*} \frac {5 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{5} x^{3}}{6 \, a} + \frac {5}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a b^{4} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {5}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a b^{4} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {5 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{5} x^{3}}{12 \, a^{3}} + \frac {5}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{4} + \frac {35 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{4}}{36 \, a^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{4}}{9 \, a^{4}} - \frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{3}}{9 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{2}}{9 \, a^{4} x^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b}{36 \, a^{3} x^{9}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{12 \, a^{2} x^{12}} \end {gather*}

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^13,x, algorithm="maxima")

[Out]

5/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^5*x^3/a + 5/3*(-1)^(2*b^2*x^3 + 2*a*b)*a*b^4*log(2*b^2*x^3 + 2*a*b) - 5/
3*(-1)^(2*a*b*x^3 + 2*a^2)*a*b^4*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 5/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(
3/2)*b^5*x^3/a^3 + 5/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^4 + 35/36*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^4/a^2 +
1/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^4/a^4 - 2/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^3/(a^3*x^3) - 1/9*(b^
2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^2/(a^4*x^6) + 1/36*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b/(a^3*x^9) - 1/12*(b^2*
x^6 + 2*a*b*x^3 + a^2)^(7/2)/(a^2*x^12)

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Fricas [A]
time = 0.36, size = 61, normalized size = 0.24 \begin {gather*} \frac {12 \, b^{5} x^{15} + 180 \, a b^{4} x^{12} \log \left (x\right ) - 120 \, a^{2} b^{3} x^{9} - 60 \, a^{3} b^{2} x^{6} - 20 \, a^{4} b x^{3} - 3 \, a^{5}}{36 \, x^{12}} \end {gather*}

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^13,x, algorithm="fricas")

[Out]

1/36*(12*b^5*x^15 + 180*a*b^4*x^12*log(x) - 120*a^2*b^3*x^9 - 60*a^3*b^2*x^6 - 20*a^4*b*x^3 - 3*a^5)/x^12

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{13}}\, dx \end {gather*}

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**13,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**13, x)

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Giac [A]
time = 5.02, size = 125, normalized size = 0.50 \begin {gather*} \frac {1}{3} \, b^{5} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 5 \, a b^{4} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {125 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 120 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 60 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 20 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{36 \, x^{12}} \end {gather*}

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^13,x, algorithm="giac")

[Out]

1/3*b^5*x^3*sgn(b*x^3 + a) + 5*a*b^4*log(abs(x))*sgn(b*x^3 + a) - 1/36*(125*a*b^4*x^12*sgn(b*x^3 + a) + 120*a^
2*b^3*x^9*sgn(b*x^3 + a) + 60*a^3*b^2*x^6*sgn(b*x^3 + a) + 20*a^4*b*x^3*sgn(b*x^3 + a) + 3*a^5*sgn(b*x^3 + a))
/x^12

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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^3+b^2\,x^6\right )}^{5/2}}{x^{13}} \,d x \end {gather*}

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^13,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^13, x)

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