3.172 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{5/2}}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0532879, 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} \[ -\frac{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac{5 a^4 b \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac{10 a^3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{5 a b^4 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b^5 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{10 a^2 b^3 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^4,x]

[Out]

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

Mathematica [A]  time = 0.0224764, size = 79, normalized size = 0.36 \[ \frac{\sqrt{(a+b x)^2} \left (-60 a^3 b^2 x^2+60 a^2 b^3 x^3 \log (x)-15 a^4 b x-2 a^5+30 a b^4 x^4+3 b^5 x^5\right )}{6 x^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^4,x]

[Out]

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

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Maple [A]  time = 0.225, size = 76, normalized size = 0.3 \begin{align*}{\frac{3\,{b}^{5}{x}^{5}+60\,{a}^{2}{b}^{3}\ln \left ( x \right ){x}^{3}+30\,a{b}^{4}{x}^{4}-60\,{a}^{3}{b}^{2}{x}^{2}-15\,{a}^{4}bx-2\,{a}^{5}}{6\, \left ( bx+a \right ) ^{5}{x}^{3}} \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^4,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69692, size = 132, normalized size = 0.59 \begin{align*} \frac{3 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} + 60 \, a^{2} b^{3} x^{3} \log \left (x\right ) - 60 \, a^{3} b^{2} x^{2} - 15 \, a^{4} b x - 2 \, a^{5}}{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)^(5/2)/x^4,x, algorithm="fricas")

[Out]

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

[Out]

Integral(((a + b*x)**2)**(5/2)/x**4, x)

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Giac [A]  time = 1.3254, size = 124, normalized size = 0.56 \begin{align*} \frac{1}{2} \, b^{5} x^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, a b^{4} x \mathrm{sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{60 \, a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 15 \, a^{4} b x \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{5} \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)^(5/2)/x^4,x, algorithm="giac")

[Out]

1/2*b^5*x^2*sgn(b*x + a) + 5*a*b^4*x*sgn(b*x + a) + 10*a^2*b^3*log(abs(x))*sgn(b*x + a) - 1/6*(60*a^3*b^2*x^2*
sgn(b*x + a) + 15*a^4*b*x*sgn(b*x + a) + 2*a^5*sgn(b*x + a))/x^3