\(\int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [181]

Optimal result

Integrand size = 24, antiderivative size = 182 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {a^3 x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^3 (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^4 (a+b x)}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^4 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 182, 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} \[ \int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {x^4 (a+b x)}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^3 (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^4 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[x^4/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

-((a^3*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (a^2*x^2*(a + b*x))/(2*b^3*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) - (a*x^3*(a + b*x))/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x^4*(a + b*x))/(4*b*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]) + (a^4*(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 (a b+b^2 x\right ) \int \frac {x^4}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (a b+b^2 x\right ) \int \left (-\frac {a^3}{b^5}+\frac {a^2 x}{b^4}-\frac {a x^2}{b^3}+\frac {x^3}{b^2}+\frac {a^4}{b^5 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {a^3 x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^3 (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^4 (a+b x)}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^4 (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 = 0.80 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.75 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\frac {b x \left (12 a^3-6 a^2 b x+4 a b^2 x^2-3 b^3 x^3\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}}-24 a^4 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )}{12 b^5} \]

[In]

Integrate[x^4/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((b*x*(12*a^3 - 6*a^2*b*x + 4*a*b^2*x^2 - 3*b^3*x^3)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(a^2
+ a*b*x - Sqrt[a^2]*Sqrt[(a + b*x)^2]) - 24*a^4*ArcTanh[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])])/(12*b^5)

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.37

method	result	size
default	\(\frac {\left (b x +a \right ) \left (3 b^{4} x^{4}-4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )-12 a^{3} b x \right )}{12 \sqrt {\left (b x +a \right )^{2}}\, b^{5}}\)	\(67\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{4} b^{3} x^{4}-\frac {1}{3} a \,b^{2} x^{3}+\frac {1}{2} a^{2} b \,x^{2}-a^{3} x \right )}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{4} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(84\)

[In]

int(x^4/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.29 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, b^{4} x^{4} - 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 12 \, a^{3} b x + 12 \, a^{4} \log \left (b x + a\right )}{12 \, b^{5}} \]

[In]

integrate(x^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.13 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.15 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\begin {cases} \frac {a^{4} \left (\frac {a}{b} + x\right ) \log {\left (\frac {a}{b} + x \right )}}{b^{4} \sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} + \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {25 a^{3}}{12 b^{5}} + \frac {13 a^{2} x}{12 b^{4}} - \frac {7 a x^{2}}{12 b^{3}} + \frac {x^{3}}{4 b^{2}}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {a^{8} \sqrt {a^{2} + 2 a b x} - \frac {4 a^{6} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {6 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {4 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}}{16 a^{5} b^{5}} & \text {for}\: a b \neq 0 \\\frac {x^{5}}{5 \sqrt {a^{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**4/((b*x+a)**2)**(1/2),x)

[Out]

Piecewise((a**4*(a/b + x)*log(a/b + x)/(b**4*sqrt(b**2*(a/b + x)**2)) + sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-25*
a**3/(12*b**5) + 13*a**2*x/(12*b**4) - 7*a*x**2/(12*b**3) + x**3/(4*b**2)), Ne(b**2, 0)), ((a**8*sqrt(a**2 + 2
*a*b*x) - 4*a**6*(a**2 + 2*a*b*x)**(3/2)/3 + 6*a**4*(a**2 + 2*a*b*x)**(5/2)/5 - 4*a**2*(a**2 + 2*a*b*x)**(7/2)
/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(16*a**5*b**5), Ne(a*b, 0)), (x**5/(5*sqrt(a**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.63 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} x^{3}}{4 \, b^{2}} + \frac {13 \, a^{2} x^{2}}{12 \, b^{3}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a x^{2}}{12 \, b^{3}} - \frac {13 \, a^{3} x}{6 \, b^{4}} + \frac {a^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}}{6 \, b^{5}} \]

[In]

integrate(x^4/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*x^3/b^2 + 13/12*a^2*x^2/b^3 - 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*x^2/b^3 -
13/6*a^3*x/b^4 + a^4*log(x + a/b)/b^5 + 7/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3/b^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.46 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a^{4} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{b^{5}} + \frac {3 \, b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) - 12 \, a^{3} x \mathrm {sgn}\left (b x + a\right )}{12 \, b^{4}} \]

[In]

integrate(x^4/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^4/((a + b*x)^2)^(1/2),x)

[Out]

int(x^4/((a + b*x)^2)^(1/2), x)