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

Optimal. Leaf size=182 \[ \frac{3 a^2 A b x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{3 a A b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{A b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{a^3 A \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{B (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0618986, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {770, 80, 43} \[ \frac{3 a^2 A b x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{3 a A b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{A b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{a^3 A \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{B (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*A*b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (3*a*A*b^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a +
b*x)) + (A*b^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (B*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(4*b) + (a^3*A*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 43

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])

Rubi steps

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

Mathematica [A]  time = 0.0385336, size = 83, normalized size = 0.46 \[ \frac{\sqrt{(a+b x)^2} \left (x \left (18 a^2 b (2 A+B x)+12 a^3 B+6 a b^2 x (3 A+2 B x)+b^3 x^2 (4 A+3 B x)\right )+12 a^3 A \log (x)\right )}{12 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(x*(12*a^3*B + 18*a^2*b*(2*A + B*x) + 6*a*b^2*x*(3*A + 2*B*x) + b^3*x^2*(4*A + 3*B*x)) + 12
*a^3*A*Log[x]))/(12*(a + b*x))

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Maple [A]  time = 0.007, size = 91, normalized size = 0.5 \begin{align*}{\frac{3\,B{x}^{4}{b}^{3}+4\,A{b}^{3}{x}^{3}+12\,B{x}^{3}a{b}^{2}+18\,A{x}^{2}a{b}^{2}+18\,B{x}^{2}{a}^{2}b+12\,{a}^{3}A\ln \left ( x \right ) +36\,A{a}^{2}bx+12\,{a}^{3}Bx}{12\, \left ( bx+a \right ) ^{3}} \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,x)

[Out]

1/12*((b*x+a)^2)^(3/2)*(3*B*x^4*b^3+4*A*b^3*x^3+12*B*x^3*a*b^2+18*A*x^2*a*b^2+18*B*x^2*a^2*b+12*a^3*A*ln(x)+36
*A*a^2*b*x+12*a^3*B*x)/(b*x+a)^3

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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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58315, size = 154, normalized size = 0.85 \begin{align*} \frac{1}{4} \, B b^{3} x^{4} + A a^{3} \log \left (x\right ) + \frac{1}{3} \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + \frac{3}{2} \,{\left (B a^{2} b + A a b^{2}\right )} x^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} 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,x, algorithm="fricas")

[Out]

1/4*B*b^3*x^4 + A*a^3*log(x) + 1/3*(3*B*a*b^2 + A*b^3)*x^3 + 3/2*(B*a^2*b + A*a*b^2)*x^2 + (B*a^3 + 3*A*a^2*b)
*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}\, 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,x)

[Out]

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

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Giac [A]  time = 1.27704, size = 159, normalized size = 0.87 \begin{align*} \frac{1}{4} \, B b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + B a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, A b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, B a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, A a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + B a^{3} x \mathrm{sgn}\left (b x + a\right ) + 3 \, A a^{2} b x \mathrm{sgn}\left (b x + a\right ) + A a^{3} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\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,x, algorithm="giac")

[Out]

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