\(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{x} \, dx\) [154]

Optimal result

Integrand size = 24, antiderivative size = 143 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x} \, dx=\frac {3 a^2 b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {3 a b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 143, 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} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x} \, dx=\frac {3 a^2 b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {3 a b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^3 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

[In]

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

[Out]

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

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 {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3}{x} \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (3 a^2 b^4+\frac {a^3 b^3}{x}+3 a b^5 x+b^6 x^2\right ) \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {3 a^2 b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {3 a b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x} \, dx=\frac {1}{2} \left (\frac {b x \left (18 a^2+9 a b x+2 b^2 x^2\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{-3 a^2-3 a b x+3 \sqrt {a^2} \sqrt {(a+b x)^2}}-2 a^3 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )-2 \left (a^2\right )^{3/2} \log (x)+\left (a^2\right )^{3/2} \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )+\left (a^2\right )^{3/2} \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right ) \]

[In]

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

[Out]

((b*x*(18*a^2 + 9*a*b*x + 2*b^2*x^2)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(-3*a^2 - 3*a*b*x +
3*Sqrt[a^2]*Sqrt[(a + b*x)^2]) - 2*a^3*ArcTanh[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])] - 2*(a^2)^(3/2)*Log[x] +
(a^2)^(3/2)*Log[Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2]] + (a^2)^(3/2)*Log[Sqrt[a^2] + b*x - Sqrt[(a + b*x)^2]])/
2

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.36

method	result	size
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (2 b^{3} x^{3}+9 a \,b^{2} x^{2}+6 a^{3} \ln \left (x \right )+18 a^{2} b x \right )}{6 \left (b x +a \right )^{3}}\)	\(51\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b \left (\frac {1}{3} b^{2} x^{3}+\frac {3}{2} a b \,x^{2}+3 a^{2} x \right )}{b x +a}+\frac {a^{3} \ln \left (x \right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(64\)

[In]

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

[Out]

1/6*((b*x+a)^2)^(3/2)*(2*b^3*x^3+9*a*b^2*x^2+6*a^3*ln(x)+18*a^2*b*x)/(b*x+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} + \frac {3}{2} \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3} \log \left (x\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x}\, dx \]

[In]

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

[Out]

Integral(((a + b*x)**2)**(3/2)/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} a^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a b x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} + \frac {1}{3} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

(-1)^(2*b^2*x + 2*a*b)*a^3*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*a^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x)
) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*b*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2 + 1/3*(b^2*x^2 + 2*a*b*x +
a^2)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.39 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b x \mathrm {sgn}\left (b x + a\right ) + a^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/x,x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/x, x)