\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^2} \, dx\) [672]

Optimal result

Integrand size = 29, antiderivative size = 200 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {3 a b (A b+a B) x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^2 (A b+3 a B) x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^3 B x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {784, 77} \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\frac {3 a b x \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac {b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{2 (a+b x)}+\frac {a^2 \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{a+b x}+\frac {b^3 B x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)} \]

[In]

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

[Out]

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

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 784

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 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 (A+B x)}{x^2} \, 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 b^4 (A b+a B)+\frac {a^3 A b^3}{x^2}+\frac {a^2 b^3 (3 A b+a B)}{x}+b^5 (A b+3 a B) x+b^6 B x^2\right ) \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = -\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {3 a b (A b+a B) x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^2 (A b+3 a B) x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^3 B x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.44 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (-6 a^3 A+18 a^2 b B x^2+9 a b^2 x^2 (2 A+B x)+b^3 x^3 (3 A+2 B x)+6 a^2 (3 A b+a B) x \log (x)\right )}{6 x (a+b x)} \]

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(-6*a^3*A + 18*a^2*b*B*x^2 + 9*a*b^2*x^2*(2*A + B*x) + b^3*x^3*(3*A + 2*B*x) + 6*a^2*(3*A*b
+ a*B)*x*Log[x]))/(6*x*(a + b*x))

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.48

method	result	size
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (2 x^{4} B \,b^{3}+3 A \,b^{3} x^{3}+9 B a \,b^{2} x^{3}+18 A \ln \left (x \right ) x \,a^{2} b +18 A a \,b^{2} x^{2}+6 B \,a^{3} \ln \left (x \right ) x +18 B \,a^{2} b \,x^{2}-6 A \,a^{3}\right )}{6 \left (b x +a \right )^{3} x}\)	\(96\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b \left (\frac {1}{3} B \,b^{2} x^{3}+\frac {1}{2} A \,b^{2} x^{2}+\frac {3}{2} B a b \,x^{2}+3 a A b x +3 a^{2} B x \right )}{b x +a}-\frac {a^{3} A \sqrt {\left (b x +a \right )^{2}}}{x \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A \,a^{2} b +B \,a^{3}\right ) \ln \left (x \right )}{b x +a}\)	\(117\)

[In]

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

[Out]

1/6*((b*x+a)^2)^(3/2)*(2*x^4*B*b^3+3*A*b^3*x^3+9*B*a*b^2*x^3+18*A*ln(x)*x*a^2*b+18*A*a*b^2*x^2+6*B*a^3*ln(x)*x
+18*B*a^2*b*x^2-6*A*a^3)/(b*x+a)^3/x

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\frac {2 \, B b^{3} x^{4} - 6 \, A a^{3} + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x \log \left (x\right )}{6 \, x} \]

[In]

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

[Out]

1/6*(2*B*b^3*x^4 - 6*A*a^3 + 3*(3*B*a*b^2 + A*b^3)*x^3 + 18*(B*a^2*b + A*a*b^2)*x^2 + 6*(B*a^3 + 3*A*a^2*b)*x*
log(x))/x

Sympy [F]

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

[In]

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

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**2, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (141) = 282\).

Time = 0.21 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) + 3 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{2} b \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - 3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{2} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a b x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2} x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} + \frac {9}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a b + \frac {1}{3} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{x} \]

[In]

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

[Out]

(-1)^(2*b^2*x + 2*a*b)*B*a^3*log(2*b^2*x + 2*a*b) + 3*(-1)^(2*b^2*x + 2*a*b)*A*a^2*b*log(2*b^2*x + 2*a*b) - (-
1)^(2*a*b*x + 2*a^2)*B*a^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) - 3*(-1)^(2*a*b*x + 2*a^2)*A*a^2*b*log(2*a*b*x/a
bs(x) + 2*a^2/abs(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*b*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^2*x
+ 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^2 + 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a*b + 1/3*(b^2*x^2 + 2*a*b*x +
a^2)^(3/2)*B - (b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^2} \, dx=\frac {1}{3} \, B b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, B a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 3 \, A a b^{2} x \mathrm {sgn}\left (b x + a\right ) - \frac {A a^{3} \mathrm {sgn}\left (b x + a\right )}{x} + {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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