\(\int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [182]

Optimal result

Integrand size = 24, antiderivative size = 144 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a x^2 (a+b x)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^3 (a+b x)}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 144, 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} \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {a x^2 (a+b x)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^3 (a+b x)}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[x^3/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.40 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \left (b x \left (6 a^2-3 a b x+2 b^2 x^2\right )-6 a^3 \log (a+b x)\right )}{6 b^4 \sqrt {(a+b x)^2}} \]

[In]

Integrate[x^3/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*(b*x*(6*a^2 - 3*a*b*x + 2*b^2*x^2) - 6*a^3*Log[a + b*x]))/(6*b^4*Sqrt[(a + b*x)^2])

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.39

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (-2 b^{3} x^{3}+3 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )-6 a^{2} b x \right )}{6 \sqrt {\left (b x +a \right )^{2}}\, b^{4}}\)	\(56\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{3} b^{2} x^{3}-\frac {1}{2} a b \,x^{2}+a^{2} x \right )}{\left (b x +a \right ) b^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\)	\(73\)

[In]

int(x^3/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.28 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, b^{3} x^{3} - 3 \, a b^{2} x^{2} + 6 \, a^{2} b x - 6 \, a^{3} \log \left (b x + a\right )}{6 \, b^{4}} \]

[In]

integrate(x^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.19 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\begin {cases} - \frac {a^{3} \left (\frac {a}{b} + x\right ) \log {\left (\frac {a}{b} + x \right )}}{b^{3} \sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} + \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \cdot \left (\frac {11 a^{2}}{6 b^{4}} - \frac {5 a x}{6 b^{3}} + \frac {x^{2}}{3 b^{2}}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- a^{6} \sqrt {a^{2} + 2 a b x} + a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}} - \frac {3 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4}}{4 \sqrt {a^{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3/((b*x+a)**2)**(1/2),x)

[Out]

Piecewise((-a**3*(a/b + x)*log(a/b + x)/(b**3*sqrt(b**2*(a/b + x)**2)) + sqrt(a**2 + 2*a*b*x + b**2*x**2)*(11*
a**2/(6*b**4) - 5*a*x/(6*b**3) + x**2/(3*b**2)), Ne(b**2, 0)), ((-a**6*sqrt(a**2 + 2*a*b*x) + a**4*(a**2 + 2*a
*b*x)**(3/2) - 3*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/(8*a**4*b**4), Ne(a*b, 0)), (x**4
/(4*sqrt(a**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.60 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {5 \, a x^{2}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} x^{2}}{3 \, b^{2}} + \frac {5 \, a^{2} x}{3 \, b^{3}} - \frac {a^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{3 \, b^{4}} \]

[In]

integrate(x^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.47 \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {a^{3} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{b^{4}} + \frac {2 \, b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} x \mathrm {sgn}\left (b x + a\right )}{6 \, b^{3}} \]

[In]

integrate(x^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {x^3}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

[In]

int(x^3/((a + b*x)^2)^(1/2),x)

[Out]

int(x^3/((a + b*x)^2)^(1/2), x)