3.171 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=222 \[ -\frac{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{5 a^4 b \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{10 a^2 b^3 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a b^4 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{b^5 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{10 a^3 b^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^2*(a + b*x)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))
+ (10*a^2*b^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a*b^4*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a
+ b*x)) + (b^5*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Lo
g[x])/(a + b*x)

________________________________________________________________________________________

Rubi [A]  time = 0.0509013, antiderivative size = 222, 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} \[ -\frac{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{5 a^4 b \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{10 a^2 b^3 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a b^4 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{b^5 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{10 a^3 b^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^3,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^2*(a + b*x)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))
+ (10*a^2*b^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a*b^4*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a
+ b*x)) + (b^5*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Lo
g[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 )^{5/2}}{x^3} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^5}{x^3} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (10 a^2 b^8+\frac{a^5 b^5}{x^3}+\frac{5 a^4 b^6}{x^2}+\frac{10 a^3 b^7}{x}+5 a b^9 x+b^{10} x^2\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{5 a^4 b \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{10 a^2 b^3 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a b^4 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{b^5 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{10 a^3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} \log (x)}{a+b x}\\ \end{align*}

Mathematica [A]  time = 0.0220932, size = 79, normalized size = 0.36 \[ \frac{\sqrt{(a+b x)^2} \left (60 a^2 b^3 x^3+60 a^3 b^2 x^2 \log (x)-30 a^4 b x-3 a^5+15 a b^4 x^4+2 b^5 x^5\right )}{6 x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^3,x]

[Out]

(Sqrt[(a + b*x)^2]*(-3*a^5 - 30*a^4*b*x + 60*a^2*b^3*x^3 + 15*a*b^4*x^4 + 2*b^5*x^5 + 60*a^3*b^2*x^2*Log[x]))/
(6*x^2*(a + b*x))

________________________________________________________________________________________

Maple [A]  time = 0.235, size = 76, normalized size = 0.3 \begin{align*}{\frac{2\,{b}^{5}{x}^{5}+15\,a{b}^{4}{x}^{4}+60\,{a}^{3}{b}^{2}\ln \left ( x \right ){x}^{2}+60\,{a}^{2}{b}^{3}{x}^{3}-30\,{a}^{4}bx-3\,{a}^{5}}{6\, \left ( bx+a \right ) ^{5}{x}^{2}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{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)^(5/2)/x^3,x)

[Out]

1/6*((b*x+a)^2)^(5/2)*(2*b^5*x^5+15*a*b^4*x^4+60*a^3*b^2*ln(x)*x^2+60*a^2*b^3*x^3-30*a^4*b*x-3*a^5)/(b*x+a)^5/
x^2

________________________________________________________________________________________

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)^(5/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.7083, size = 132, normalized size = 0.59 \begin{align*} \frac{2 \, b^{5} x^{5} + 15 \, a b^{4} x^{4} + 60 \, a^{2} b^{3} x^{3} + 60 \, a^{3} b^{2} x^{2} \log \left (x\right ) - 30 \, a^{4} b x - 3 \, a^{5}}{6 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^3,x, algorithm="fricas")

[Out]

1/6*(2*b^5*x^5 + 15*a*b^4*x^4 + 60*a^2*b^3*x^3 + 60*a^3*b^2*x^2*log(x) - 30*a^4*b*x - 3*a^5)/x^2

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}{x^{3}}\, 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)**(5/2)/x**3,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x**3, x)

________________________________________________________________________________________

Giac [A]  time = 1.36792, size = 123, normalized size = 0.55 \begin{align*} \frac{1}{3} \, b^{5} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a b^{4} x^{2} \mathrm{sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} x \mathrm{sgn}\left (b x + a\right ) + 10 \, a^{3} b^{2} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{10 \, a^{4} b x \mathrm{sgn}\left (b x + a\right ) + a^{5} \mathrm{sgn}\left (b x + a\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^3,x, algorithm="giac")

[Out]

1/3*b^5*x^3*sgn(b*x + a) + 5/2*a*b^4*x^2*sgn(b*x + a) + 10*a^2*b^3*x*sgn(b*x + a) + 10*a^3*b^2*log(abs(x))*sgn
(b*x + a) - 1/2*(10*a^4*b*x*sgn(b*x + a) + a^5*sgn(b*x + a))/x^2