3.200 \(\int \frac {x^4}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=171 \[ -\frac {3 a^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 171, 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 = {646, 43} \[ -\frac {a^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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 646

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 )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {x^4}{\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 {a^4}{b^9 (a+b x)^5}-\frac {4 a^3}{b^9 (a+b x)^4}+\frac {6 a^2}{b^9 (a+b x)^3}-\frac {4 a}{b^9 (a+b x)^2}+\frac {1}{b^9 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {4 a}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(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 = 0.03, size = 73, normalized size = 0.43 \[ \frac {a \left (25 a^3+88 a^2 b x+108 a b^2 x^2+48 b^3 x^3\right )+12 (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(a*(25*a^3 + 88*a^2*b*x + 108*a*b^2*x^2 + 48*b^3*x^3) + 12*(a + b*x)^4*Log[a + b*x])/(12*b^5*(a + b*x)^3*Sqrt[
(a + b*x)^2])

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fricas [A]  time = 0.84, size = 127, normalized size = 0.74 \[ \frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4} + 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/12*(48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4 + 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3
*b*x + a^4)*log(b*x + a))/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.06, size = 123, normalized size = 0.72 \[ \frac {\left (12 b^{4} x^{4} \ln \left (b x +a \right )+48 a \,b^{3} x^{3} \ln \left (b x +a \right )+72 a^{2} b^{2} x^{2} \ln \left (b x +a \right )+48 a \,b^{3} x^{3}+48 a^{3} b x \ln \left (b x +a \right )+108 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )+88 a^{3} b x +25 a^{4}\right ) \left (b x +a \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/12*(12*b^4*x^4*ln(b*x+a)+48*a*b^3*x^3*ln(b*x+a)+72*a^2*b^2*x^2*ln(b*x+a)+48*a*b^3*x^3+48*a^3*b*x*ln(b*x+a)+1
08*a^2*b^2*x^2+12*a^4*ln(b*x+a)+88*a^3*b*x+25*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(5/2)

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maxima [A]  time = 1.47, size = 92, normalized size = 0.54 \[ \frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} + \frac {\log \left (b x + a\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/12*(48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6
*x + a^4*b^5) + log(b*x + a)/b^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

int(x^4/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**4/((a + b*x)**2)**(5/2), x)

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