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 \[ \frac {5 a^2 b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac {b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2} (5 a B+A b)}{4 (a+b x)}+\frac {5 a b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2} (2 a B+A b)}{3 (a+b x)}+\frac {b^5 B x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}-\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {a^4 \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (a B+5 A b)}{a+b x}+\frac {5 a^3 b x \sqrt {a^2+2 a b x+b^2 x^2} (a B+2 A b)}{a+b x} \]

[Out]

-a^5*A*((b*x+a)^2)^(1/2)/x/(b*x+a)+5*a^3*b*(2*A*b+B*a)*x*((b*x+a)^2)^(1/2)/(b*x+a)+5*a^2*b^2*(A*b+B*a)*x^2*((b
*x+a)^2)^(1/2)/(b*x+a)+5/3*a*b^3*(A*b+2*B*a)*x^3*((b*x+a)^2)^(1/2)/(b*x+a)+1/4*b^4*(A*b+5*B*a)*x^4*((b*x+a)^2)
^(1/2)/(b*x+a)+1/5*b^5*B*x^5*((b*x+a)^2)^(1/2)/(b*x+a)+a^4*(5*A*b+B*a)*ln(x)*((b*x+a)^2)^(1/2)/(b*x+a)

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Rubi [A]  time = 0.12, antiderivative size = 294, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.05, size = 128, normalized size = 0.44 \[ \frac {\sqrt {(a+b x)^2} \left (-60 a^5 A+60 a^4 x \log (x) (a B+5 A b)+300 a^4 b B x^2+300 a^3 b^2 x^2 (2 A+B x)+100 a^2 b^3 x^3 (3 A+2 B x)+25 a b^4 x^4 (4 A+3 B x)+3 b^5 x^5 (5 A+4 B x)\right )}{60 x (a+b x)} \]

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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fricas [A]  time = 0.98, size = 121, normalized size = 0.41 \[ \frac {12 \, B b^{5} x^{6} - 60 \, A a^{5} + 15 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 100 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 300 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 60 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x \log \relax (x)}{60 \, x} \]

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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giac [A]  time = 0.22, size = 191, normalized size = 0.65 \[ \frac {1}{5} \, B b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, B a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b^{5} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, B a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, A a b^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{2} b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 10 \, A a^{3} b^{2} x \mathrm {sgn}\left (b x + a\right ) - \frac {A a^{5} \mathrm {sgn}\left (b x + a\right )}{x} + {\left (B a^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{4} b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.06, size = 144, normalized size = 0.49 \[ \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (12 B \,b^{5} x^{6}+15 A \,b^{5} x^{5}+75 B a \,b^{4} x^{5}+100 A a \,b^{4} x^{4}+200 B \,a^{2} b^{3} x^{4}+300 A \,a^{2} b^{3} x^{3}+300 B \,a^{3} b^{2} x^{3}+300 A \,a^{4} b x \ln \relax (x )+600 A \,a^{3} b^{2} x^{2}+60 B \,a^{5} x \ln \relax (x )+300 B \,a^{4} b \,x^{2}-60 A \,a^{5}\right )}{60 \left (b x +a \right )^{5} x} \]

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

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maxima [A]  time = 0.59, size = 386, normalized size = 1.31 \[ \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) + 5 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{4} b \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - 5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{4} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} b x + \frac {5}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2} b^{2} x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4} + \frac {15}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3} b + \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a b x + \frac {5}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2} x + \frac {7}{12} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} + \frac {35}{12} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a b + \frac {1}{5} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{x} \]

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]

(-1)^(2*b^2*x + 2*a*b)*B*a^5*log(2*b^2*x + 2*a*b) + 5*(-1)^(2*b^2*x + 2*a*b)*A*a^4*b*log(2*b^2*x + 2*a*b) - (-
1)^(2*a*b*x + 2*a^2)*B*a^5*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) - 5*(-1)^(2*a*b*x + 2*a^2)*A*a^4*b*log(2*a*b*x/a
bs(x) + 2*a^2/abs(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^3*b*x + 5/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^2*
b^2*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^4 + 15/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^3*b + 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*B*a*b*x + 5/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^2*x + 7/12*(b^2*x^2 + 2*a*b*x + a^2)^(3
/2)*B*a^2 + 35/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a*b + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B - (b^2*x^2 + 2
*a*b*x + a^2)^(5/2)*A/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{2}}\, dx \]

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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