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

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

[Out]

-(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^2*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
x*(a + b*x)) + (b^2*(A*b + 3*a*B)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (b^3*B*x^2*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(2*(a + b*x)) + (3*a*b*(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 = 0.0884344, 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} \[ -\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{x (a+b x)}+\frac{b^2 x \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{a+b x}+\frac{3 a b \log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}+\frac{b^3 B x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^2*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
x*(a + b*x)) + (b^2*(A*b + 3*a*B)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (b^3*B*x^2*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(2*(a + b*x)) + (3*a*b*(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^3} \, 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^3} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (b^5 (A b+3 a B)+\frac{a^3 A b^3}{x^3}+\frac{a^2 b^3 (3 A b+a B)}{x^2}+\frac{3 a b^4 (A b+a B)}{x}+b^6 B x\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}}{2 x^2 (a+b x)}-\frac{a^2 (3 A b+a B) \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^2 (A b+3 a B) x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^3 B x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{3 a b (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 = 0.0390654, size = 85, normalized size = 0.42 \[ \frac{\sqrt{(a+b x)^2} \left (-6 a^2 A b x+a^3 (-(A+2 B x))+6 a b x^2 \log (x) (a B+A b)+6 a b^2 B x^3+b^3 x^3 (2 A+B x)\right )}{2 x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 95, normalized size = 0.5 \begin{align*}{\frac{B{x}^{4}{b}^{3}+6\,A\ln \left ( x \right ){x}^{2}a{b}^{2}+2\,A{b}^{3}{x}^{3}+6\,B\ln \left ( x \right ){x}^{2}{a}^{2}b+6\,B{x}^{3}a{b}^{2}-6\,A{a}^{2}bx-2\,{a}^{3}Bx-A{a}^{3}}{2\, \left ( bx+a \right ) ^{3}{x}^{2}} \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^3,x)

[Out]

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

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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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.16492, size = 159, normalized size = 0.8 \begin{align*} \frac{B b^{3} x^{4} - A a^{3} + 2 \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 6 \,{\left (B a^{2} b + A a b^{2}\right )} x^{2} \log \left (x\right ) - 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

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

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Giac [A]  time = 1.1799, size = 158, normalized size = 0.79 \begin{align*} \frac{1}{2} \, B b^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, B a b^{2} x \mathrm{sgn}\left (b x + a\right ) + A b^{3} x \mathrm{sgn}\left (b x + a\right ) + 3 \,{\left (B a^{2} b \mathrm{sgn}\left (b x + a\right ) + A a b^{2} \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac{A a^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (B a^{3} \mathrm{sgn}\left (b x + a\right ) + 3 \, A a^{2} b \mathrm{sgn}\left (b x + a\right )\right )} x}{2 \, x^{2}} \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^3,x, algorithm="giac")

[Out]

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