3.1.78 \(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{15}} \, dx\)

Optimal. Leaf size=248 \[ \frac {b^5 x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}-\frac {2 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5 \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 x^{14} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 x^{11} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^8 \left (a+b x^3\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 248, 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} \begin {gather*} -\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 x^{14} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 x^{11} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^8 \left (a+b x^3\right )}-\frac {2 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5 \left (a+b x^3\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac {b^5 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^15,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(14*x^14*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(11*x
^11*(a + b*x^3)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*x^8*(a + b*x^3)) - (2*a^2*b^3*Sqrt[a^2 + 2*a
*b*x^3 + b^2*x^6])/(x^5*(a + b*x^3)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*x^2*(a + b*x^3)) + (b^5*x*
Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(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^{15}} \, 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^{15}} \, 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 (b^{10}+\frac {a^5 b^5}{x^{15}}+\frac {5 a^4 b^6}{x^{12}}+\frac {10 a^3 b^7}{x^9}+\frac {10 a^2 b^8}{x^6}+\frac {5 a b^9}{x^3}\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}}{14 x^{14} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 x^{11} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^8 \left (a+b x^3\right )}-\frac {2 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5 \left (a+b x^3\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^2 \left (a+b x^3\right )}+\frac {b^5 x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \begin {gather*} -\frac {\sqrt {\left (a+b x^3\right )^2} \left (22 a^5+140 a^4 b x^3+385 a^3 b^2 x^6+616 a^2 b^3 x^9+770 a b^4 x^{12}-308 b^5 x^{15}\right )}{308 x^{14} \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^15,x]

[Out]

-1/308*(Sqrt[(a + b*x^3)^2]*(22*a^5 + 140*a^4*b*x^3 + 385*a^3*b^2*x^6 + 616*a^2*b^3*x^9 + 770*a*b^4*x^12 - 308
*b^5*x^15))/(x^14*(a + b*x^3))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 19.62, size = 83, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (-22 a^5-140 a^4 b x^3-385 a^3 b^2 x^6-616 a^2 b^3 x^9-770 a b^4 x^{12}+308 b^5 x^{15}\right )}{308 x^{14} \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^15,x]

[Out]

(Sqrt[(a + b*x^3)^2]*(-22*a^5 - 140*a^4*b*x^3 - 385*a^3*b^2*x^6 - 616*a^2*b^3*x^9 - 770*a*b^4*x^12 + 308*b^5*x
^15))/(308*x^14*(a + b*x^3))

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 59, normalized size = 0.24 \begin {gather*} \frac {308 \, b^{5} x^{15} - 770 \, a b^{4} x^{12} - 616 \, a^{2} b^{3} x^{9} - 385 \, a^{3} b^{2} x^{6} - 140 \, a^{4} b x^{3} - 22 \, a^{5}}{308 \, x^{14}} \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^15,x, algorithm="fricas")

[Out]

1/308*(308*b^5*x^15 - 770*a*b^4*x^12 - 616*a^2*b^3*x^9 - 385*a^3*b^2*x^6 - 140*a^4*b*x^3 - 22*a^5)/x^14

________________________________________________________________________________________

giac [A]  time = 0.35, size = 105, normalized size = 0.42 \begin {gather*} b^{5} x \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {770 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 616 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 385 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 140 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 22 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{308 \, x^{14}} \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^15,x, algorithm="giac")

[Out]

b^5*x*sgn(b*x^3 + a) - 1/308*(770*a*b^4*x^12*sgn(b*x^3 + a) + 616*a^2*b^3*x^9*sgn(b*x^3 + a) + 385*a^3*b^2*x^6
*sgn(b*x^3 + a) + 140*a^4*b*x^3*sgn(b*x^3 + a) + 22*a^5*sgn(b*x^3 + a))/x^14

________________________________________________________________________________________

maple [A]  time = 0.01, size = 80, normalized size = 0.32 \begin {gather*} -\frac {\left (-308 b^{5} x^{15}+770 a \,b^{4} x^{12}+616 a^{2} b^{3} x^{9}+385 a^{3} b^{2} x^{6}+140 a^{4} b \,x^{3}+22 a^{5}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}}}{308 \left (b \,x^{3}+a \right )^{5} x^{14}} \end {gather*}

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^15,x)

[Out]

-1/308*(-308*b^5*x^15+770*a*b^4*x^12+616*a^2*b^3*x^9+385*a^3*b^2*x^6+140*a^4*b*x^3+22*a^5)*((b*x^3+a)^2)^(5/2)
/x^14/(b*x^3+a)^5

________________________________________________________________________________________

maxima [A]  time = 0.99, size = 59, normalized size = 0.24 \begin {gather*} \frac {308 \, b^{5} x^{15} - 770 \, a b^{4} x^{12} - 616 \, a^{2} b^{3} x^{9} - 385 \, a^{3} b^{2} x^{6} - 140 \, a^{4} b x^{3} - 22 \, a^{5}}{308 \, x^{14}} \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^15,x, algorithm="maxima")

[Out]

1/308*(308*b^5*x^15 - 770*a*b^4*x^12 - 616*a^2*b^3*x^9 - 385*a^3*b^2*x^6 - 140*a^4*b*x^3 - 22*a^5)/x^14

________________________________________________________________________________________

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^{15}} \,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^15,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^15, x)

________________________________________________________________________________________

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^{15}}\, 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**15,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**15, x)

________________________________________________________________________________________