3.6.90 \(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx\)

Optimal. Leaf size=114 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{4 x^4 (a+b x)}-\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 114, 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} \begin {gather*} -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{4 x^4 (a+b x)}-\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^6,x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.43 \begin {gather*} -\frac {\sqrt {(a+b x)^2} (3 a (4 A+5 B x)+5 b x (3 A+4 B x))}{60 x^5 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^6,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(5*b*x*(3*A + 4*B*x) + 3*a*(4*A + 5*B*x)))/(x^5*(a + b*x))

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IntegrateAlgebraic [B]  time = 1.52, size = 448, normalized size = 3.93 \begin {gather*} \frac {4 b^4 \sqrt {a^2+2 a b x+b^2 x^2} \left (-12 a^5 A b-15 a^5 b B x-63 a^4 A b^2 x-80 a^4 b^2 B x^2-132 a^3 A b^3 x^2-170 a^3 b^3 B x^3-138 a^2 A b^4 x^3-180 a^2 b^4 B x^4-72 a A b^5 x^4-95 a b^5 B x^5-15 A b^6 x^5-20 b^6 B x^6\right )+4 \sqrt {b^2} b^4 \left (12 a^6 A+15 a^6 B x+75 a^5 A b x+95 a^5 b B x^2+195 a^4 A b^2 x^2+250 a^4 b^2 B x^3+270 a^3 A b^3 x^3+350 a^3 b^3 B x^4+210 a^2 A b^4 x^4+275 a^2 b^4 B x^5+87 a A b^5 x^5+115 a b^5 B x^6+15 A b^6 x^6+20 b^6 B x^7\right )}{15 \sqrt {b^2} x^5 \sqrt {a^2+2 a b x+b^2 x^2} \left (-16 a^4 b^4-64 a^3 b^5 x-96 a^2 b^6 x^2-64 a b^7 x^3-16 b^8 x^4\right )+15 x^5 \left (16 a^5 b^5+80 a^4 b^6 x+160 a^3 b^7 x^2+160 a^2 b^8 x^3+80 a b^9 x^4+16 b^{10} x^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^6,x]

[Out]

(4*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-12*a^5*A*b - 63*a^4*A*b^2*x - 15*a^5*b*B*x - 132*a^3*A*b^3*x^2 - 80*a^4
*b^2*B*x^2 - 138*a^2*A*b^4*x^3 - 170*a^3*b^3*B*x^3 - 72*a*A*b^5*x^4 - 180*a^2*b^4*B*x^4 - 15*A*b^6*x^5 - 95*a*
b^5*B*x^5 - 20*b^6*B*x^6) + 4*b^4*Sqrt[b^2]*(12*a^6*A + 75*a^5*A*b*x + 15*a^6*B*x + 195*a^4*A*b^2*x^2 + 95*a^5
*b*B*x^2 + 270*a^3*A*b^3*x^3 + 250*a^4*b^2*B*x^3 + 210*a^2*A*b^4*x^4 + 350*a^3*b^3*B*x^4 + 87*a*A*b^5*x^5 + 27
5*a^2*b^4*B*x^5 + 15*A*b^6*x^6 + 115*a*b^5*B*x^6 + 20*b^6*B*x^7))/(15*Sqrt[b^2]*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x
^2]*(-16*a^4*b^4 - 64*a^3*b^5*x - 96*a^2*b^6*x^2 - 64*a*b^7*x^3 - 16*b^8*x^4) + 15*x^5*(16*a^5*b^5 + 80*a^4*b^
6*x + 160*a^3*b^7*x^2 + 160*a^2*b^8*x^3 + 80*a*b^9*x^4 + 16*b^10*x^5))

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fricas [A]  time = 0.40, size = 27, normalized size = 0.24 \begin {gather*} -\frac {20 \, B b x^{2} + 12 \, A a + 15 \, {\left (B a + A b\right )} x}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="fricas")

[Out]

-1/60*(20*B*b*x^2 + 12*A*a + 15*(B*a + A*b)*x)/x^5

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giac [A]  time = 0.22, size = 77, normalized size = 0.68 \begin {gather*} -\frac {{\left (5 \, B a b^{4} - 3 \, A b^{5}\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, a^{4}} - \frac {20 \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, B a x \mathrm {sgn}\left (b x + a\right ) + 15 \, A b x \mathrm {sgn}\left (b x + a\right ) + 12 \, A a \mathrm {sgn}\left (b x + a\right )}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/60*(5*B*a*b^4 - 3*A*b^5)*sgn(b*x + a)/a^4 - 1/60*(20*B*b*x^2*sgn(b*x + a) + 15*B*a*x*sgn(b*x + a) + 15*A*b*
x*sgn(b*x + a) + 12*A*a*sgn(b*x + a))/x^5

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maple [A]  time = 0.05, size = 44, normalized size = 0.39 \begin {gather*} -\frac {\left (20 B b \,x^{2}+15 A b x +15 B a x +12 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{60 \left (b x +a \right ) x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x)

[Out]

-1/60*(20*B*b*x^2+15*A*b*x+15*B*a*x+12*A*a)*((b*x+a)^2)^(1/2)/x^5/(b*x+a)

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maxima [B]  time = 0.65, size = 315, normalized size = 2.76 \begin {gather*} \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{4}}{2 \, a^{4}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{5}}{2 \, a^{5}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{3}}{2 \, a^{3} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{4}}{2 \, a^{4} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2}}{2 \, a^{4} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{2 \, a^{5} x^{2}} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b}{12 \, a^{3} x^{3}} - \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2}}{20 \, a^{4} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B}{4 \, a^{2} x^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b}{20 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{5 \, a^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="maxima")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^4/a^4 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^5/a^5 + 1/2*sqrt(b^2*x^2 +
2*a*b*x + a^2)*B*b^3/(a^3*x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^4/(a^4*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2
)^(3/2)*B*b^2/(a^4*x^2) + 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/(a^5*x^2) + 5/12*(b^2*x^2 + 2*a*b*x + a^2)
^(3/2)*B*b/(a^3*x^3) - 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^2/(a^4*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3
/2)*B/(a^2*x^4) + 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b/(a^3*x^4) - 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A/(
a^2*x^5)

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mupad [B]  time = 1.12, size = 43, normalized size = 0.38 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (12\,A\,a+15\,A\,b\,x+15\,B\,a\,x+20\,B\,b\,x^2\right )}{60\,x^5\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^6,x)

[Out]

-(((a + b*x)^2)^(1/2)*(12*A*a + 15*A*b*x + 15*B*a*x + 20*B*b*x^2))/(60*x^5*(a + b*x))

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sympy [A]  time = 0.43, size = 31, normalized size = 0.27 \begin {gather*} \frac {- 12 A a - 20 B b x^{2} + x \left (- 15 A b - 15 B a\right )}{60 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**6,x)

[Out]

(-12*A*a - 20*B*b*x**2 + x*(-15*A*b - 15*B*a))/(60*x**5)

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