3.2.90 \(\int \frac {x^4}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [190]

Optimal. Leaf size=172 \[ \frac {4 a^3}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 a^2 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 172, 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, 45} \begin {gather*} \frac {6 a^2 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a^3}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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

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Mathematica [A]
time = 0.02, size = 83, normalized size = 0.48 \begin {gather*} \frac {7 a^4+2 a^3 b x-11 a^2 b^2 x^2-4 a b^3 x^3+b^4 x^4+12 a^2 (a+b x)^2 \log (a+b x)}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(7*a^4 + 2*a^3*b*x - 11*a^2*b^2*x^2 - 4*a*b^3*x^3 + b^4*x^4 + 12*a^2*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x
)*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.52, size = 101, normalized size = 0.59

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{2} b \,x^{2}-3 a x \right )}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (4 a^{3} x +\frac {7 a^{4}}{2 b}\right )}{\left (b x +a \right )^{3} b^{4}}+\frac {6 \sqrt {\left (b x +a \right )^{2}}\, a^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(98\)
default	\(\frac {\left (b^{4} x^{4}+12 \ln \left (b x +a \right ) a^{2} b^{2} x^{2}-4 a \,b^{3} x^{3}+24 \ln \left (b x +a \right ) a^{3} b x -11 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )+2 a^{3} b x +7 a^{4}\right ) \left (b x +a \right )}{2 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(b^4*x^4+12*ln(b*x+a)*a^2*b^2*x^2-4*a*b^3*x^3+24*ln(b*x+a)*a^3*b*x-11*a^2*b^2*x^2+12*a^4*ln(b*x+a)+2*a^3*b
*x+7*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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Maxima [A]
time = 0.28, size = 131, normalized size = 0.76 \begin {gather*} \frac {x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, a x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {6 \, a^{2} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {5 \, a^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {12 \, a^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, a^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*a*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 6*a^2*log(x + a/
b)/b^5 - 5*a^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) + 12*a^3*x/(b^6*(x + a/b)^2) + 23/2*a^4/(b^7*(x + a/b)^2)

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Fricas [A]
time = 1.56, size = 95, normalized size = 0.55 \begin {gather*} \frac {b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4} + 12 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4 + 12*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*log(b*x +
a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(x**4/((a + b*x)**2)**(3/2), x)

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Giac [A]
time = 0.82, size = 89, normalized size = 0.52 \begin {gather*} \frac {6 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, a b^{2} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{6}} + \frac {8 \, a^{3} b x + 7 \, a^{4}}{2 \, {\left (b x + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

6*a^2*log(abs(b*x + a))/(b^5*sgn(b*x + a)) + 1/2*(b^3*x^2*sgn(b*x + a) - 6*a*b^2*x*sgn(b*x + a))/b^6 + 1/2*(8*
a^3*b*x + 7*a^4)/((b*x + a)^2*b^5*sgn(b*x + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(x^4/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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