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

Optimal. Leaf size=200 \[ \frac{3 a b x \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{2 (a+b x)}+\frac{a^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{a+b x}-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^3 B x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

[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)

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Rubi [A]  time = 0.0857908, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {770, 76} \[ \frac{3 a b x \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{2 (a+b x)}+\frac{a^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{a+b x}-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^3 B x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

Antiderivative was successfully verified.

[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 770

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]

Rule 76

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])

Rubi steps

\begin{align*} \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^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]  time = 0.0344851, size = 89, normalized size = 0.44 \[ \frac{\sqrt{(a+b x)^2} \left (6 a^2 x \log (x) (a B+3 A b)-6 a^3 A+18 a^2 b B x^2+9 a b^2 x^2 (2 A+B x)+b^3 x^3 (3 A+2 B x)\right )}{6 x (a+b x)} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.013, size = 96, normalized size = 0.5 \begin{align*}{\frac{2\,B{x}^{4}{b}^{3}+3\,A{b}^{3}{x}^{3}+9\,B{x}^{3}a{b}^{2}+18\,A\ln \left ( x \right ) x{a}^{2}b+18\,A{x}^{2}a{b}^{2}+6\,B\ln \left ( x \right ) x{a}^{3}+18\,B{x}^{2}{a}^{2}b-6\,A{a}^{3}}{6\, \left ( bx+a \right ) ^{3}x} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.24151, size = 163, normalized size = 0.82 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.18883, size = 161, normalized size = 0.8 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))