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

Optimal. Leaf size=244 \[ \frac {15 a^2 (a+b x) \log (a+b x)}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a x (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^6}{4 b^7 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^5}{b^7 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 a^4}{2 b^7 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {20 a^3}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 244, 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} \[ -\frac {a^6}{4 b^7 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^5}{b^7 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 a^4}{2 b^7 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {20 a^3}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a x (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 a^2 (a+b x) \log (a+b x)}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(20*a^3)/(b^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - a^6/(4*b^7*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*a^5)
/(b^7*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (15*a^4)/(2*b^7*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) -
(5*a*x*(a + b*x))/(b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x^2*(a + b*x))/(2*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
+ (15*a^2*(a + b*x)*Log[a + b*x])/(b^7*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^6}{\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^6}{\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 {5 a}{b^{11}}+\frac {x}{b^{10}}+\frac {a^6}{b^{11} (a+b x)^5}-\frac {6 a^5}{b^{11} (a+b x)^4}+\frac {15 a^4}{b^{11} (a+b x)^3}-\frac {20 a^3}{b^{11} (a+b x)^2}+\frac {15 a^2}{b^{11} (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {20 a^3}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^6}{4 b^7 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^5}{b^7 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 a^4}{2 b^7 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a x (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 a^2 (a+b x) \log (a+b x)}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 106, normalized size = 0.43 \[ \frac {57 a^6+168 a^5 b x+132 a^4 b^2 x^2-32 a^3 b^3 x^3-68 a^2 b^4 x^4+60 a^2 (a+b x)^4 \log (a+b x)-12 a b^5 x^5+2 b^6 x^6}{4 b^7 (a+b x)^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(57*a^6 + 168*a^5*b*x + 132*a^4*b^2*x^2 - 32*a^3*b^3*x^3 - 68*a^2*b^4*x^4 - 12*a*b^5*x^5 + 2*b^6*x^6 + 60*a^2*
(a + b*x)^4*Log[a + b*x])/(4*b^7*(a + b*x)^3*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.74, size = 162, normalized size = 0.66 \[ \frac {2 \, b^{6} x^{6} - 12 \, a b^{5} x^{5} - 68 \, a^{2} b^{4} x^{4} - 32 \, a^{3} b^{3} x^{3} + 132 \, a^{4} b^{2} x^{2} + 168 \, a^{5} b x + 57 \, a^{6} + 60 \, {\left (a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{3} + 6 \, a^{4} b^{2} x^{2} + 4 \, a^{5} b x + a^{6}\right )} \log \left (b x + a\right )}{4 \, {\left (b^{11} x^{4} + 4 \, a b^{10} x^{3} + 6 \, a^{2} b^{9} x^{2} + 4 \, a^{3} b^{8} x + a^{4} b^{7}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/4*(2*b^6*x^6 - 12*a*b^5*x^5 - 68*a^2*b^4*x^4 - 32*a^3*b^3*x^3 + 132*a^4*b^2*x^2 + 168*a^5*b*x + 57*a^6 + 60*
(a^2*b^4*x^4 + 4*a^3*b^3*x^3 + 6*a^4*b^2*x^2 + 4*a^5*b*x + a^6)*log(b*x + a))/(b^11*x^4 + 4*a*b^10*x^3 + 6*a^2
*b^9*x^2 + 4*a^3*b^8*x + a^4*b^7)

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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^6/(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 = 0.24, size = 158, normalized size = 0.65 \[ \frac {\left (2 b^{6} x^{6}+60 a^{2} b^{4} x^{4} \ln \left (b x +a \right )-12 a \,b^{5} x^{5}+240 a^{3} b^{3} x^{3} \ln \left (b x +a \right )-68 a^{2} b^{4} x^{4}+360 a^{4} b^{2} x^{2} \ln \left (b x +a \right )-32 a^{3} b^{3} x^{3}+240 a^{5} b x \ln \left (b x +a \right )+132 a^{4} b^{2} x^{2}+60 a^{6} \ln \left (b x +a \right )+168 a^{5} b x +57 a^{6}\right ) \left (b x +a \right )}{4 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b^6*x^6+60*ln(b*x+a)*x^4*a^2*b^4-12*x^5*a*b^5+240*ln(b*x+a)*x^3*a^3*b^3-68*a^2*x^4*b^4+360*ln(b*x+a)*x^
2*a^4*b^2-32*a^3*x^3*b^3+240*ln(b*x+a)*x*a^5*b+132*a^4*x^2*b^2+60*ln(b*x+a)*a^6+168*x*a^5*b+57*a^6)*(b*x+a)/b^
7/((b*x+a)^2)^(5/2)

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maxima [A]  time = 1.57, size = 126, normalized size = 0.52 \[ \frac {2 \, b^{6} x^{6} - 12 \, a b^{5} x^{5} - 68 \, a^{2} b^{4} x^{4} - 32 \, a^{3} b^{3} x^{3} + 132 \, a^{4} b^{2} x^{2} + 168 \, a^{5} b x + 57 \, a^{6}}{4 \, {\left (b^{11} x^{4} + 4 \, a b^{10} x^{3} + 6 \, a^{2} b^{9} x^{2} + 4 \, a^{3} b^{8} x + a^{4} b^{7}\right )}} + \frac {15 \, a^{2} \log \left (b x + a\right )}{b^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/4*(2*b^6*x^6 - 12*a*b^5*x^5 - 68*a^2*b^4*x^4 - 32*a^3*b^3*x^3 + 132*a^4*b^2*x^2 + 168*a^5*b*x + 57*a^6)/(b^1
1*x^4 + 4*a*b^10*x^3 + 6*a^2*b^9*x^2 + 4*a^3*b^8*x + a^4*b^7) + 15*a^2*log(b*x + a)/b^7

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^6}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^6/(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 \[ \int \frac {x^{6}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**6/((a + b*x)**2)**(5/2), x)

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