3.157 $$\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{x^4} \, dx$$

Optimal. Leaf size=145 $-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{3 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^3 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$

[Out]

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

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Rubi [A]  time = 0.0344629, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{3 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^3 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^4,x]

[Out]

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

Rule 646

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]

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

Mathematica [A]  time = 0.0188272, size = 57, normalized size = 0.39 $-\frac{\sqrt{(a+b x)^2} \left (a \left (2 a^2+9 a b x+18 b^2 x^2\right )-6 b^3 x^3 \log (x)\right )}{6 x^3 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^4,x]

[Out]

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

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Maple [A]  time = 0.223, size = 54, normalized size = 0.4 \begin{align*}{\frac{6\,{b}^{3}\ln \left ( x \right ){x}^{3}-18\,a{b}^{2}{x}^{2}-9\,b{a}^{2}x-2\,{a}^{3}}{6\, \left ( bx+a \right ) ^{3}{x}^{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^2*x^2+2*a*b*x+a^2)^(3/2)/x^4,x)

[Out]

1/6*((b*x+a)^2)^(3/2)*(6*b^3*ln(x)*x^3-18*a*b^2*x^2-9*b*a^2*x-2*a^3)/(b*x+a)^3/x^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^2*x^2+2*a*b*x+a^2)^(3/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01924, size = 85, normalized size = 0.59 \begin{align*} \frac{6 \, b^{3} x^{3} \log \left (x\right ) - 18 \, a b^{2} x^{2} - 9 \, a^{2} b x - 2 \, a^{3}}{6 \, x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

1/6*(6*b^3*x^3*log(x) - 18*a*b^2*x^2 - 9*a^2*b*x - 2*a^3)/x^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{x^{4}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/x**4,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/x**4, x)

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Giac [A]  time = 1.26428, size = 80, normalized size = 0.55 \begin{align*} b^{3} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{18 \, a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 9 \, a^{2} b x \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{3} \mathrm{sgn}\left (b x + a\right )}{6 \, x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

b^3*log(abs(x))*sgn(b*x + a) - 1/6*(18*a*b^2*x^2*sgn(b*x + a) + 9*a^2*b*x*sgn(b*x + a) + 2*a^3*sgn(b*x + a))/x
^3