3.691 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^2} \, dx\)

Optimal. Leaf size=294 \[ \frac{5 a^3 b x \sqrt{a^2+2 a b x+b^2 x^2} (a B+2 A b)}{a+b x}+\frac{5 a^2 b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{5 a b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2} (2 a B+A b)}{3 (a+b x)}+\frac{b^4 x^4 \sqrt{a^2+2 a b x+b^2 x^2} (5 a B+A b)}{4 (a+b x)}+\frac{a^4 \log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+5 A b)}{a+b x}-\frac{a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^5 B x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

[Out]

-((a^5*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))) + (5*a^3*b*(2*A*b + a*B)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/(a + b*x) + (5*a^2*b^2*(A*b + a*B)*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a*b^3*(A*b + 2*a*B)*x^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (b^4*(A*b + 5*a*B)*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(a +
b*x)) + (b^5*B*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (a^4*(5*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 = 0.123289, antiderivative size = 294, 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} \[ \frac{5 a^3 b x \sqrt{a^2+2 a b x+b^2 x^2} (a B+2 A b)}{a+b x}+\frac{5 a^2 b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{5 a b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2} (2 a B+A b)}{3 (a+b x)}+\frac{b^4 x^4 \sqrt{a^2+2 a b x+b^2 x^2} (5 a B+A b)}{4 (a+b x)}+\frac{a^4 \log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+5 A b)}{a+b x}-\frac{a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b^5 B x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^5*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))) + (5*a^3*b*(2*A*b + a*B)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/(a + b*x) + (5*a^2*b^2*(A*b + a*B)*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a*b^3*(A*b + 2*a*B)*x^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (b^4*(A*b + 5*a*B)*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(a +
b*x)) + (b^5*B*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (a^4*(5*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 )^{5/2}}{x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^5 (A+B x)}{x^2} \, 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^3 b^6 (2 A b+a B)+\frac{a^5 A b^5}{x^2}+\frac{a^4 b^5 (5 A b+a B)}{x}+10 a^2 b^7 (A b+a B) x+5 a b^8 (A b+2 a B) x^2+b^9 (A b+5 a B) x^3+b^{10} B x^4\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac{a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{5 a^3 b (2 A b+a B) x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a^2 b^2 (A b+a B) x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a b^3 (A b+2 a B) x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{b^4 (A b+5 a B) x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{b^5 B x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac{a^4 (5 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 = 0.0481426, size = 128, normalized size = 0.44 \[ \frac{\sqrt{(a+b x)^2} \left (300 a^3 b^2 x^2 (2 A+B x)+100 a^2 b^3 x^3 (3 A+2 B x)+60 a^4 x \log (x) (a B+5 A b)-60 a^5 A+300 a^4 b B x^2+25 a b^4 x^4 (4 A+3 B x)+3 b^5 x^5 (5 A+4 B x)\right )}{60 x (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(-60*a^5*A + 300*a^4*b*B*x^2 + 300*a^3*b^2*x^2*(2*A + B*x) + 100*a^2*b^3*x^3*(3*A + 2*B*x)
+ 25*a*b^4*x^4*(4*A + 3*B*x) + 3*b^5*x^5*(5*A + 4*B*x) + 60*a^4*(5*A*b + a*B)*x*Log[x]))/(60*x*(a + b*x))

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Maple [A]  time = 0.013, size = 144, normalized size = 0.5 \begin{align*}{\frac{12\,B{b}^{5}{x}^{6}+15\,A{x}^{5}{b}^{5}+75\,B{x}^{5}a{b}^{4}+100\,A{x}^{4}a{b}^{4}+200\,B{x}^{4}{a}^{2}{b}^{3}+300\,A{x}^{3}{a}^{2}{b}^{3}+300\,B{x}^{3}{a}^{3}{b}^{2}+300\,A\ln \left ( x \right ) x{a}^{4}b+600\,A{x}^{2}{a}^{3}{b}^{2}+60\,B\ln \left ( x \right ) x{a}^{5}+300\,B{x}^{2}{a}^{4}b-60\,A{a}^{5}}{60\, \left ( bx+a \right ) ^{5}x} \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*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^2,x)

[Out]

1/60*((b*x+a)^2)^(5/2)*(12*B*b^5*x^6+15*A*x^5*b^5+75*B*x^5*a*b^4+100*A*x^4*a*b^4+200*B*x^4*a^2*b^3+300*A*x^3*a
^2*b^3+300*B*x^3*a^3*b^2+300*A*ln(x)*x*a^4*b+600*A*x^2*a^3*b^2+60*B*ln(x)*x*a^5+300*B*x^2*a^4*b-60*A*a^5)/(b*x
+a)^5/x

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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)^(5/2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.26308, size = 269, normalized size = 0.91 \begin{align*} \frac{12 \, B b^{5} x^{6} - 60 \, A a^{5} + 15 \,{\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 100 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 300 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 300 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 60 \,{\left (B a^{5} + 5 \, A a^{4} b\right )} x \log \left (x\right )}{60 \, 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)^(5/2)/x^2,x, algorithm="fricas")

[Out]

1/60*(12*B*b^5*x^6 - 60*A*a^5 + 15*(5*B*a*b^4 + A*b^5)*x^5 + 100*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 300*(B*a^3*b^2
+ A*a^2*b^3)*x^3 + 300*(B*a^4*b + 2*A*a^3*b^2)*x^2 + 60*(B*a^5 + 5*A*a^4*b)*x*log(x))/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{5}{2}}}{x^{2}}\, 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)**(5/2)/x**2,x)

[Out]

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

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Giac [A]  time = 1.17555, size = 258, normalized size = 0.88 \begin{align*} \frac{1}{5} \, B b^{5} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{4} \, B a b^{4} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, A b^{5} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, B a^{2} b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, A a b^{4} x^{3} \mathrm{sgn}\left (b x + a\right ) + 5 \, B a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, A a^{2} b^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, B a^{4} b x \mathrm{sgn}\left (b x + a\right ) + 10 \, A a^{3} b^{2} x \mathrm{sgn}\left (b x + a\right ) - \frac{A a^{5} \mathrm{sgn}\left (b x + a\right )}{x} +{\left (B a^{5} \mathrm{sgn}\left (b x + a\right ) + 5 \, A a^{4} b \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\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)^(5/2)/x^2,x, algorithm="giac")

[Out]

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