\(\int \frac {x^6}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [198]

Optimal result

Integrand size = 24, antiderivative size = 244 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {20 a^3}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^6}{4 b^7 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^5}{b^7 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 a^4}{2 b^7 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a x (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 a^2 (a+b x) \log (a+b x)}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {15 a^2 (a+b x) \log (a+b x)}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 a x (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x)}{2 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^6}{4 b^7 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^5}{b^7 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 a^4}{2 b^7 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {20 a^3}{b^7 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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 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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.26 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\frac {b x \left (-a^2 b^5 x^5 \sqrt {(a+b x)^2}+a b^6 x^6 \sqrt {(a+b x)^2}+\sqrt {a^2} b^5 x^5 \left (-2 a^2+b^2 x^2\right )+10 a^5 b^2 x^2 \left (26 \sqrt {a^2}-11 \sqrt {(a+b x)^2}\right )+30 a^6 b x \left (7 \sqrt {a^2}-5 \sqrt {(a+b x)^2}\right )+5 a^4 b^3 x^3 \left (25 \sqrt {a^2}-3 \sqrt {(a+b x)^2}\right )+60 a^7 \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )+3 a^3 b^4 x^4 \left (4 \sqrt {a^2}+\sqrt {(a+b x)^2}\right )\right )}{(a+b x)^3 \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )}-120 a^4 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )}{4 a^2 b^7} \]

[In]

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

[Out]

((b*x*(-(a^2*b^5*x^5*Sqrt[(a + b*x)^2]) + a*b^6*x^6*Sqrt[(a + b*x)^2] + Sqrt[a^2]*b^5*x^5*(-2*a^2 + b^2*x^2) +
10*a^5*b^2*x^2*(26*Sqrt[a^2] - 11*Sqrt[(a + b*x)^2]) + 30*a^6*b*x*(7*Sqrt[a^2] - 5*Sqrt[(a + b*x)^2]) + 5*a^4
*b^3*x^3*(25*Sqrt[a^2] - 3*Sqrt[(a + b*x)^2]) + 60*a^7*(Sqrt[a^2] - Sqrt[(a + b*x)^2]) + 3*a^3*b^4*x^4*(4*Sqrt
[a^2] + Sqrt[(a + b*x)^2])))/((a + b*x)^3*(a^2 + a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2])) - 120*a^4*ArcTanh[(b*x)
/(Sqrt[a^2] - Sqrt[(a + b*x)^2])])/(4*a^2*b^7)

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.48

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{2} b \,x^{2}-5 a x \right )}{\left (b x +a \right ) b^{6}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (20 a^{3} b^{2} x^{3}+\frac {105 a^{4} b \,x^{2}}{2}+47 a^{5} x +\frac {57 a^{6}}{4 b}\right )}{\left (b x +a \right )^{5} b^{6}}+\frac {15 \sqrt {\left (b x +a \right )^{2}}\, a^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{7}}\)	\(118\)
default	\(\frac {\left (2 b^{6} x^{6}+60 \ln \left (b x +a \right ) a^{2} b^{4} x^{4}-12 a \,x^{5} b^{5}+240 \ln \left (b x +a \right ) a^{3} b^{3} x^{3}-68 a^{2} x^{4} b^{4}+360 \ln \left (b x +a \right ) a^{4} b^{2} x^{2}-32 a^{3} x^{3} b^{3}+240 \ln \left (b x +a \right ) a^{5} b x +132 a^{4} x^{2} b^{2}+60 \ln \left (b x +a \right ) a^{6}+168 a^{5} x b +57 a^{6}\right ) \left (b x +a \right )}{4 b^{7} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(158\)

[In]

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

[Out]

((b*x+a)^2)^(1/2)/(b*x+a)*(1/2*b*x^2-5*a*x)/b^6+((b*x+a)^2)^(1/2)/(b*x+a)^5*(20*a^3*b^2*x^3+105/2*a^4*b*x^2+47
*a^5*x+57/4/b*a^6)/b^6+15*((b*x+a)^2)^(1/2)/(b*x+a)/b^7*a^2*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.66 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {2 \, b^{6} x^{6} - 12 \, a b^{5} x^{5} - 68 \, a^{2} b^{4} x^{4} - 32 \, a^{3} b^{3} x^{3} + 132 \, a^{4} b^{2} x^{2} + 168 \, a^{5} b x + 57 \, a^{6} + 60 \, {\left (a^{2} b^{4} x^{4} + 4 \, a^{3} b^{3} x^{3} + 6 \, a^{4} b^{2} x^{2} + 4 \, a^{5} b x + a^{6}\right )} \log \left (b x + a\right )}{4 \, {\left (b^{11} x^{4} + 4 \, a b^{10} x^{3} + 6 \, a^{2} b^{9} x^{2} + 4 \, a^{3} b^{8} x + a^{4} b^{7}\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.52 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {2 \, b^{6} x^{6} - 12 \, a b^{5} x^{5} - 68 \, a^{2} b^{4} x^{4} - 32 \, a^{3} b^{3} x^{3} + 132 \, a^{4} b^{2} x^{2} + 168 \, a^{5} b x + 57 \, a^{6}}{4 \, {\left (b^{11} x^{4} + 4 \, a b^{10} x^{3} + 6 \, a^{2} b^{9} x^{2} + 4 \, a^{3} b^{8} x + a^{4} b^{7}\right )}} + \frac {15 \, a^{2} \log \left (b x + a\right )}{b^{7}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.45 \[ \int \frac {x^6}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {15 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{7} \mathrm {sgn}\left (b x + a\right )} + \frac {b^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a b^{4} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{10}} + \frac {80 \, a^{3} b^{3} x^{3} + 210 \, a^{4} b^{2} x^{2} + 188 \, a^{5} b x + 57 \, a^{6}}{4 \, {\left (b x + a\right )}^{4} b^{7} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

15*a^2*log(abs(b*x + a))/(b^7*sgn(b*x + a)) + 1/2*(b^5*x^2*sgn(b*x + a) - 10*a*b^4*x*sgn(b*x + a))/b^10 + 1/4*
(80*a^3*b^3*x^3 + 210*a^4*b^2*x^2 + 188*a^5*b*x + 57*a^6)/((b*x + a)^4*b^7*sgn(b*x + a))

Mupad [F(-1)]

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

[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)