\(\int \frac {x^4}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [200]

Optimal result

Integrand size = 24, antiderivative size = 171 \[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {4 a}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 171, 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} \[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {3 a^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 a^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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 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^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*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.83 \[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\frac {b x \left (3 \sqrt {a^2} b^7 x^7+3 a^3 b^4 x^4 \sqrt {(a+b x)^2}-3 a^2 b^5 x^5 \sqrt {(a+b x)^2}+3 a b^6 x^6 \sqrt {(a+b x)^2}+2 a^5 b^2 x^2 \left (26 \sqrt {a^2}-11 \sqrt {(a+b x)^2}\right )+6 a^6 b x \left (7 \sqrt {a^2}-5 \sqrt {(a+b x)^2}\right )+a^4 b^3 x^3 \left (25 \sqrt {a^2}-3 \sqrt {(a+b x)^2}\right )+12 a^7 \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{a^4 (a+b x)^3 \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )}+12 \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )-12 \log \left (b^5 \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right )}{12 b^5} \]

[In]

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

[Out]

((b*x*(3*Sqrt[a^2]*b^7*x^7 + 3*a^3*b^4*x^4*Sqrt[(a + b*x)^2] - 3*a^2*b^5*x^5*Sqrt[(a + b*x)^2] + 3*a*b^6*x^6*S
qrt[(a + b*x)^2] + 2*a^5*b^2*x^2*(26*Sqrt[a^2] - 11*Sqrt[(a + b*x)^2]) + 6*a^6*b*x*(7*Sqrt[a^2] - 5*Sqrt[(a +
b*x)^2]) + a^4*b^3*x^3*(25*Sqrt[a^2] - 3*Sqrt[(a + b*x)^2]) + 12*a^7*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(a^4*(a
+ b*x)^3*(a^2 + a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2])) + 12*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2]] - 12*Log[
b^5*(Sqrt[a^2] + b*x - Sqrt[(a + b*x)^2])])/(12*b^5)

Maple [A] (verified)

Time = 2.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.49

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {4 a \,x^{3}}{b^{2}}+\frac {9 a^{2} x^{2}}{b^{3}}+\frac {22 a^{3} x}{3 b^{4}}+\frac {25 a^{4}}{12 b^{5}}\right )}{\left (b x +a \right )^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(83\)
default	\(\frac {\left (12 \ln \left (b x +a \right ) b^{4} x^{4}+48 \ln \left (b x +a \right ) a \,b^{3} x^{3}+72 \ln \left (b x +a \right ) a^{2} b^{2} x^{2}+48 a \,b^{3} x^{3}+48 \ln \left (b x +a \right ) a^{3} b x +108 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )+88 a^{3} b x +25 a^{4}\right ) \left (b x +a \right )}{12 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(123\)

[In]

int(x^4/(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)^5*(4*a*x^3/b^2+9*a^2*x^2/b^3+22/3*a^3*x/b^4+25/12*a^4/b^5)+((b*x+a)^2)^(1/2)/(b*x+a)
/b^5*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.74 \[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4} + 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.54 \[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} + \frac {\log \left (b x + a\right )}{b^{5}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.43 \[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {48 \, a b^{2} x^{3} + 108 \, a^{2} b x^{2} + 88 \, a^{3} x + \frac {25 \, a^{4}}{b}}{12 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

log(abs(b*x + a))/(b^5*sgn(b*x + a)) + 1/12*(48*a*b^2*x^3 + 108*a^2*b*x^2 + 88*a^3*x + 25*a^4/b)/((b*x + a)^4*
b^4*sgn(b*x + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

[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)