3.39 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=162 \[ -\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {3 a b^2 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )} \]

[Out]

-1/6*a^3*((b*x^3+a)^2)^(1/2)/x^6/(b*x^3+a)-a^2*b*((b*x^3+a)^2)^(1/2)/x^3/(b*x^3+a)+1/3*b^3*x^3*((b*x^3+a)^2)^(
1/2)/(b*x^3+a)+3*a*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 = 162, 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 = {1355, 266, 43} \[ -\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {b^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {3 a b^2 \log (x) \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(6*x^6*(a + b*x^3)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^3*(a +
b*x^3)) + (b^3*x^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(3*(a + b*x^3)) + (3*a*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6
]*Log[x])/(a + b*x^3)

Rule 43

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 266

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 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 )^{3/2}}{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 )^3}{x^7} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \operatorname {Subst}\left (\int \frac {\left (a b+b^2 x\right )^3}{x^3} \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \operatorname {Subst}\left (\int \left (b^6+\frac {a^3 b^3}{x^3}+\frac {3 a^2 b^4}{x^2}+\frac {3 a b^5}{x}\right ) \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {b^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {3 a 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 = 61, normalized size = 0.38 \[ -\frac {\sqrt {\left (a+b x^3\right )^2} \left (a^3+6 a^2 b x^3-18 a b^2 x^6 \log (x)-2 b^3 x^9\right )}{6 x^6 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(Sqrt[(a + b*x^3)^2]*(a^3 + 6*a^2*b*x^3 - 2*b^3*x^9 - 18*a*b^2*x^6*Log[x]))/(x^6*(a + b*x^3))

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fricas [A]  time = 0.86, size = 39, normalized size = 0.24 \[ \frac {2 \, b^{3} x^{9} + 18 \, a b^{2} x^{6} \log \relax (x) - 6 \, a^{2} b x^{3} - a^{3}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^7,x, algorithm="fricas")

[Out]

1/6*(2*b^3*x^9 + 18*a*b^2*x^6*log(x) - 6*a^2*b*x^3 - a^3)/x^6

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giac [A]  time = 0.34, size = 86, normalized size = 0.53 \[ \frac {1}{3} \, b^{3} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {9 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 6 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^7,x, algorithm="giac")

[Out]

1/3*b^3*x^3*sgn(b*x^3 + a) + 3*a*b^2*log(abs(x))*sgn(b*x^3 + a) - 1/6*(9*a*b^2*x^6*sgn(b*x^3 + a) + 6*a^2*b*x^
3*sgn(b*x^3 + a) + a^3*sgn(b*x^3 + a))/x^6

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maple [A]  time = 0.01, size = 60, normalized size = 0.37 \[ \frac {\left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}} \left (2 b^{3} x^{9}+18 a \,b^{2} x^{6} \ln \relax (x )-6 a^{2} b \,x^{3}-a^{3}\right )}{6 \left (b \,x^{3}+a \right )^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((b*x^3+a)^2)^(3/2)*(2*b^3*x^9+18*a*b^2*ln(x)*x^6-6*a^2*b*x^3-a^3)/(b*x^3+a)^3/x^6

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maxima [A]  time = 0.70, size = 220, normalized size = 1.36 \[ \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{3} x^{3}}{2 \, a} + \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a b^{2} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{2} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{2}}{6 \, a^{2}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b}{6 \, a x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{6 \, a^{2} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^7,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**7,x)

[Out]

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

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