3.1.70 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^7} \, dx\) [70]

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

[Out]

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

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Rubi [A]
time = 0.05, 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^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {5 a b^4 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {10 a^3 b^2 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^7,x]

[Out]

-1/6*(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^6*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*x^
3*(a + b*x^3)) + (10*a^2*b^3*x^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*(a + b*x^3)) + (5*a*b^4*x^6*Sqrt[a^2 + 2*
a*b*x^3 + b^2*x^6])/(6*(a + b*x^3)) + (b^5*x^9*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(9*(a + b*x^3)) + (10*a^3*b^2*
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^7} \, 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^7} \, 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^3} \, 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 (10 a^2 b^8+\frac {a^5 b^5}{x^3}+\frac {5 a^4 b^6}{x^2}+\frac {10 a^3 b^7}{x}+5 a b^9 x+b^{10} x^2\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}}{6 x^6 \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {10 a^2 b^3 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 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}+\frac {b^5 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {10 a^3 b^2 \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 = 0.02, size = 85, normalized size = 0.34 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (-3 a^5-30 a^4 b x^3+60 a^2 b^3 x^9+15 a b^4 x^{12}+2 b^5 x^{15}+180 a^3 b^2 x^6 \log (x)\right )}{18 x^6 \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^7,x]

[Out]

(Sqrt[(a + b*x^3)^2]*(-3*a^5 - 30*a^4*b*x^3 + 60*a^2*b^3*x^9 + 15*a*b^4*x^12 + 2*b^5*x^15 + 180*a^3*b^2*x^6*Lo
g[x]))/(18*x^6*(a + b*x^3))

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

method	result	size
default	\(\frac {\left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}} \left (2 b^{5} x^{15}+15 b^{4} a \,x^{12}+60 a^{2} b^{3} x^{9}+180 b^{2} a^{3} \ln \left (x \right ) x^{6}-30 a^{4} b \,x^{3}-3 a^{5}\right )}{18 \left (b \,x^{3}+a \right )^{5} x^{6}}\)	\(82\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} \left (\frac {1}{9} b^{2} x^{9}+\frac {5}{6} a b \,x^{6}+\frac {10}{3} a^{2} x^{3}\right )}{b \,x^{3}+a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {5}{3} a^{4} b \,x^{3}-\frac {1}{6} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{6}}+\frac {10 a^{3} b^{2} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\)	\(119\)

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^7,x,method=_RETURNVERBOSE)

[Out]

1/18*((b*x^3+a)^2)^(5/2)*(2*b^5*x^15+15*b^4*a*x^12+60*a^2*b^3*x^9+180*b^2*a^3*ln(x)*x^6-30*a^4*b*x^3-3*a^5)/(b
*x^3+a)^5/x^6

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Maxima [A]
time = 0.29, size = 282, normalized size = 1.12 \begin {gather*} \frac {5}{3} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a b^{3} x^{3} + \frac {10}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a^{3} b^{2} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {10}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a^{3} b^{2} \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^{3} x^{3}}{6 \, a} + 5 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2} b^{2} + \frac {35}{18} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{2} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{2}}{6 \, a^{2}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{2 \, a x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{6 \, a^{2} x^{6}} \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^7,x, algorithm="maxima")

[Out]

5/3*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a*b^3*x^3 + 10/3*(-1)^(2*b^2*x^3 + 2*a*b)*a^3*b^2*log(2*b^2*x^3 + 2*a*b) -
10/3*(-1)^(2*a*b*x^3 + 2*a^2)*a^3*b^2*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 5/6*(b^2*x^6 + 2*a*b*x^3 + a
^2)^(3/2)*b^3*x^3/a + 5*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^2*b^2 + 35/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^2
+ 1/6*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^2/a^2 - 1/2*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b/(a*x^3) - 1/6*(b^2*x
^6 + 2*a*b*x^3 + a^2)^(7/2)/(a^2*x^6)

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Fricas [A]
time = 0.40, size = 61, normalized size = 0.24 \begin {gather*} \frac {2 \, b^{5} x^{15} + 15 \, a b^{4} x^{12} + 60 \, a^{2} b^{3} x^{9} + 180 \, a^{3} b^{2} x^{6} \log \left (x\right ) - 30 \, a^{4} b x^{3} - 3 \, a^{5}}{18 \, x^{6}} \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^7,x, algorithm="fricas")

[Out]

1/18*(2*b^5*x^15 + 15*a*b^4*x^12 + 60*a^2*b^3*x^9 + 180*a^3*b^2*x^6*log(x) - 30*a^4*b*x^3 - 3*a^5)/x^6

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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^{7}}\, 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**7,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**7, x)

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Giac [A]
time = 3.96, size = 126, normalized size = 0.50 \begin {gather*} \frac {1}{9} \, b^{5} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{6} \, a b^{4} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{3} \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 10 \, a^{3} b^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {30 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 10 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, x^{6}} \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^7,x, algorithm="giac")

[Out]

1/9*b^5*x^9*sgn(b*x^3 + a) + 5/6*a*b^4*x^6*sgn(b*x^3 + a) + 10/3*a^2*b^3*x^3*sgn(b*x^3 + a) + 10*a^3*b^2*log(a
bs(x))*sgn(b*x^3 + a) - 1/6*(30*a^3*b^2*x^6*sgn(b*x^3 + a) + 10*a^4*b*x^3*sgn(b*x^3 + a) + a^5*sgn(b*x^3 + a))
/x^6

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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^7} \,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^7,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^7, x)

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