3.3.7 \(\int \frac {1}{x^3 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [207]

Optimal. Leaf size=278 \[ \frac {10 b^2}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{4 a^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{a^4 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2}{a^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b (a+b x)}{a^6 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 b^2 (a+b x) \log (x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 b^2 (a+b x) \log (a+b x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

10*b^2/a^6/((b*x+a)^2)^(1/2)+1/4*b^2/a^3/(b*x+a)^3/((b*x+a)^2)^(1/2)+b^2/a^4/(b*x+a)^2/((b*x+a)^2)^(1/2)+3*b^2
/a^5/(b*x+a)/((b*x+a)^2)^(1/2)+1/2*(-b*x-a)/a^5/x^2/((b*x+a)^2)^(1/2)+5*b*(b*x+a)/a^6/x/((b*x+a)^2)^(1/2)+15*b
^2*(b*x+a)*ln(x)/a^7/((b*x+a)^2)^(1/2)-15*b^2*(b*x+a)*ln(b*x+a)/a^7/((b*x+a)^2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 46} \begin {gather*} \frac {15 b^2 \log (x) (a+b x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 b^2 (a+b x) \log (a+b x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {10 b^2}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b (a+b x)}{a^6 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2}{a^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{a^4 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{4 a^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(10*b^2)/(a^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + b^2/(4*a^3*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + b^2/(a^
4*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b^2)/(a^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a + b*
x)/(2*a^5*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*b*(a + b*x))/(a^6*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (15*b^2
*(a + b*x)*Log[x])/(a^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (15*b^2*(a + b*x)*Log[a + b*x])/(a^7*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{x^3 \left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{a^5 b^5 x^3}-\frac {5}{a^6 b^4 x^2}+\frac {15}{a^7 b^3 x}-\frac {1}{a^3 b^2 (a+b x)^5}-\frac {3}{a^4 b^2 (a+b x)^4}-\frac {6}{a^5 b^2 (a+b x)^3}-\frac {10}{a^6 b^2 (a+b x)^2}-\frac {15}{a^7 b^2 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {10 b^2}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{4 a^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{a^4 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2}{a^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b (a+b x)}{a^6 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 b^2 (a+b x) \log (x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 b^2 (a+b x) \log (a+b x)}{a^7 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 121, normalized size = 0.44 \begin {gather*} \frac {a \left (-2 a^5+12 a^4 b x+125 a^3 b^2 x^2+260 a^2 b^3 x^3+210 a b^4 x^4+60 b^5 x^5\right )+60 b^2 x^2 (a+b x)^4 \log (x)-60 b^2 x^2 (a+b x)^4 \log (a+b x)}{4 a^7 x^2 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(a*(-2*a^5 + 12*a^4*b*x + 125*a^3*b^2*x^2 + 260*a^2*b^3*x^3 + 210*a*b^4*x^4 + 60*b^5*x^5) + 60*b^2*x^2*(a + b*
x)^4*Log[x] - 60*b^2*x^2*(a + b*x)^4*Log[a + b*x])/(4*a^7*x^2*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.52, size = 218, normalized size = 0.78

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {15 b^{5} x^{5}}{a^{6}}+\frac {105 b^{4} x^{4}}{2 a^{5}}+\frac {65 b^{3} x^{3}}{a^{4}}+\frac {125 b^{2} x^{2}}{4 a^{3}}+\frac {3 b x}{a^{2}}-\frac {1}{2 a}\right )}{\left (b x +a \right )^{5} x^{2}}+\frac {15 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (-x \right )}{\left (b x +a \right ) a^{7}}-\frac {15 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) a^{7}}\)	\(137\)
default	\(-\frac {\left (60 \ln \left (b x +a \right ) b^{6} x^{6}-60 \ln \left (x \right ) b^{6} x^{6}+240 \ln \left (b x +a \right ) a \,b^{5} x^{5}-240 \ln \left (x \right ) a \,b^{5} x^{5}+360 \ln \left (b x +a \right ) a^{2} b^{4} x^{4}-360 \ln \left (x \right ) a^{2} b^{4} x^{4}-60 a \,b^{5} x^{5}+240 \ln \left (b x +a \right ) a^{3} b^{3} x^{3}-240 \ln \left (x \right ) a^{3} b^{3} x^{3}-210 a^{2} b^{4} x^{4}+60 \ln \left (b x +a \right ) a^{4} b^{2} x^{2}-60 \ln \left (x \right ) a^{4} b^{2} x^{2}-260 a^{3} b^{3} x^{3}-125 a^{4} x^{2} b^{2}-12 a^{5} x b +2 a^{6}\right ) \left (b x +a \right )}{4 x^{2} a^{7} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(60*ln(b*x+a)*b^6*x^6-60*ln(x)*b^6*x^6+240*ln(b*x+a)*a*b^5*x^5-240*ln(x)*a*b^5*x^5+360*ln(b*x+a)*a^2*b^4*
x^4-360*ln(x)*a^2*b^4*x^4-60*a*b^5*x^5+240*ln(b*x+a)*a^3*b^3*x^3-240*ln(x)*a^3*b^3*x^3-210*a^2*b^4*x^4+60*ln(b
*x+a)*a^4*b^2*x^2-60*ln(x)*a^4*b^2*x^2-260*a^3*b^3*x^3-125*a^4*x^2*b^2-12*a^5*x*b+2*a^6)*(b*x+a)/x^2/a^7/((b*x
+a)^2)^(5/2)

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Maxima [A]
time = 0.28, size = 178, normalized size = 0.64 \begin {gather*} -\frac {15 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{7}} + \frac {5 \, b^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{4}} + \frac {15 \, b^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{6}} + \frac {7 \, b}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} x} - \frac {1}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{2}} + \frac {15}{2 \, a^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {1}{4 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

-15*(-1)^(2*a*b*x + 2*a^2)*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^7 + 5*b^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)
*a^4) + 15*b^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^6) + 7/2*b/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*x) - 1/2/((b^2
*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*x^2) + 15/2/(a^5*(x + a/b)^2) + 1/4/(a^3*b^2*(x + a/b)^4)

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Fricas [A]
time = 1.76, size = 218, normalized size = 0.78 \begin {gather*} \frac {60 \, a b^{5} x^{5} + 210 \, a^{2} b^{4} x^{4} + 260 \, a^{3} b^{3} x^{3} + 125 \, a^{4} b^{2} x^{2} + 12 \, a^{5} b x - 2 \, a^{6} - 60 \, {\left (b^{6} x^{6} + 4 \, a b^{5} x^{5} + 6 \, a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{3} + a^{4} b^{2} x^{2}\right )} \log \left (b x + a\right ) + 60 \, {\left (b^{6} x^{6} + 4 \, a b^{5} x^{5} + 6 \, a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{3} + a^{4} b^{2} x^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{7} b^{4} x^{6} + 4 \, a^{8} b^{3} x^{5} + 6 \, a^{9} b^{2} x^{4} + 4 \, a^{10} b x^{3} + a^{11} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/4*(60*a*b^5*x^5 + 210*a^2*b^4*x^4 + 260*a^3*b^3*x^3 + 125*a^4*b^2*x^2 + 12*a^5*b*x - 2*a^6 - 60*(b^6*x^6 + 4
*a*b^5*x^5 + 6*a^2*b^4*x^4 + 4*a^3*b^3*x^3 + a^4*b^2*x^2)*log(b*x + a) + 60*(b^6*x^6 + 4*a*b^5*x^5 + 6*a^2*b^4
*x^4 + 4*a^3*b^3*x^3 + a^4*b^2*x^2)*log(x))/(a^7*b^4*x^6 + 4*a^8*b^3*x^5 + 6*a^9*b^2*x^4 + 4*a^10*b*x^3 + a^11
*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(1/(x**3*((a + b*x)**2)**(5/2)), x)

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Giac [A]
time = 0.60, size = 121, normalized size = 0.44 \begin {gather*} -\frac {15 \, b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{7} \mathrm {sgn}\left (b x + a\right )} + \frac {15 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{7} \mathrm {sgn}\left (b x + a\right )} + \frac {60 \, a b^{5} x^{5} + 210 \, a^{2} b^{4} x^{4} + 260 \, a^{3} b^{3} x^{3} + 125 \, a^{4} b^{2} x^{2} + 12 \, a^{5} b x - 2 \, a^{6}}{4 \, {\left (b x + a\right )}^{4} a^{7} x^{2} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

-15*b^2*log(abs(b*x + a))/(a^7*sgn(b*x + a)) + 15*b^2*log(abs(x))/(a^7*sgn(b*x + a)) + 1/4*(60*a*b^5*x^5 + 210
*a^2*b^4*x^4 + 260*a^3*b^3*x^3 + 125*a^4*b^2*x^2 + 12*a^5*b*x - 2*a^6)/((b*x + a)^4*a^7*x^2*sgn(b*x + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)

[Out]

int(1/(x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)

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