3.6.88 \(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^4} \, dx\)

Optimal. Leaf size=75 \[ \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (A b-a B)}{2 a^2 x^2}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 a^2 x^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {769, 646, 37} \begin {gather*} \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (A b-a B)}{2 a^2 x^2}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 a^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^4,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 46, normalized size = 0.61 \begin {gather*} -\frac {\sqrt {(a+b x)^2} (a (2 A+3 B x)+3 b x (A+2 B x))}{6 x^3 (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^4,x]

[Out]

-1/6*(Sqrt[(a + b*x)^2]*(3*b*x*(A + 2*B*x) + a*(2*A + 3*B*x)))/(x^3*(a + b*x))

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IntegrateAlgebraic [B]  time = 6.41, size = 403, normalized size = 5.37 \begin {gather*} \frac {2 b^2 (a+b x)^2 (a+2 b x)^{13} \left (2 a A+3 a B x+3 A b x+6 b B x^2\right )}{3 x^3 \sqrt {a^2+2 a b x+b^2 x^2} \left (-2 a^{14} b^2-54 a^{13} b^3 x-676 a^{12} b^4 x^2-5200 a^{11} b^5 x^3-27456 a^{10} b^6 x^4-105248 a^9 b^7 x^5-302016 a^8 b^8 x^6-658944 a^7 b^9 x^7-1098240 a^6 b^{10} x^8-1391104 a^5 b^{11} x^9-1317888 a^4 b^{12} x^{10}-905216 a^3 b^{13} x^{11}-425984 a^2 b^{14} x^{12}-122880 a b^{15} x^{13}-16384 b^{16} x^{14}\right )+3 \sqrt {b^2} x^3 \left (2 a^{15} b+56 a^{14} b^2 x+730 a^{13} b^3 x^2+5876 a^{12} b^4 x^3+32656 a^{11} b^5 x^4+132704 a^{10} b^6 x^5+407264 a^9 b^7 x^6+960960 a^8 b^8 x^7+1757184 a^7 b^9 x^8+2489344 a^6 b^{10} x^9+2708992 a^5 b^{11} x^{10}+2223104 a^4 b^{12} x^{11}+1331200 a^3 b^{13} x^{12}+548864 a^2 b^{14} x^{13}+139264 a b^{15} x^{14}+16384 b^{16} x^{15}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^4,x]

[Out]

(2*b^2*(a + b*x)^2*(a + 2*b*x)^13*(2*a*A + 3*A*b*x + 3*a*B*x + 6*b*B*x^2))/(3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]*(-2*a^14*b^2 - 54*a^13*b^3*x - 676*a^12*b^4*x^2 - 5200*a^11*b^5*x^3 - 27456*a^10*b^6*x^4 - 105248*a^9*b^7*x^
5 - 302016*a^8*b^8*x^6 - 658944*a^7*b^9*x^7 - 1098240*a^6*b^10*x^8 - 1391104*a^5*b^11*x^9 - 1317888*a^4*b^12*x
^10 - 905216*a^3*b^13*x^11 - 425984*a^2*b^14*x^12 - 122880*a*b^15*x^13 - 16384*b^16*x^14) + 3*Sqrt[b^2]*x^3*(2
*a^15*b + 56*a^14*b^2*x + 730*a^13*b^3*x^2 + 5876*a^12*b^4*x^3 + 32656*a^11*b^5*x^4 + 132704*a^10*b^6*x^5 + 40
7264*a^9*b^7*x^6 + 960960*a^8*b^8*x^7 + 1757184*a^7*b^9*x^8 + 2489344*a^6*b^10*x^9 + 2708992*a^5*b^11*x^10 + 2
223104*a^4*b^12*x^11 + 1331200*a^3*b^13*x^12 + 548864*a^2*b^14*x^13 + 139264*a*b^15*x^14 + 16384*b^16*x^15))

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fricas [A]  time = 0.41, size = 27, normalized size = 0.36 \begin {gather*} -\frac {6 \, B b x^{2} + 2 \, A a + 3 \, {\left (B a + A b\right )} x}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

-1/6*(6*B*b*x^2 + 2*A*a + 3*(B*a + A*b)*x)/x^3

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giac [A]  time = 0.16, size = 77, normalized size = 1.03 \begin {gather*} -\frac {{\left (3 \, B a b^{2} - A b^{3}\right )} \mathrm {sgn}\left (b x + a\right )}{6 \, a^{2}} - \frac {6 \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a x \mathrm {sgn}\left (b x + a\right ) + 3 \, A b x \mathrm {sgn}\left (b x + a\right ) + 2 \, A a \mathrm {sgn}\left (b x + a\right )}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/6*(3*B*a*b^2 - A*b^3)*sgn(b*x + a)/a^2 - 1/6*(6*B*b*x^2*sgn(b*x + a) + 3*B*a*x*sgn(b*x + a) + 3*A*b*x*sgn(b
*x + a) + 2*A*a*sgn(b*x + a))/x^3

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maple [A]  time = 0.05, size = 44, normalized size = 0.59 \begin {gather*} -\frac {\left (6 B b \,x^{2}+3 A b x +3 B a x +2 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{6 \left (b x +a \right ) x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^4,x)

[Out]

-1/6*(6*B*b*x^2+3*A*b*x+3*B*a*x+2*A*a)*((b*x+a)^2)^(1/2)/x^3/(b*x+a)

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maxima [B]  time = 0.49, size = 195, normalized size = 2.60 \begin {gather*} \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{2}}{2 \, a^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{3}}{2 \, a^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b}{2 \, a x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2}}{2 \, a^{2} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B}{2 \, a^{2} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b}{2 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{3 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^2/a^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^3/a^3 + 1/2*sqrt(b^2*x^2 +
2*a*b*x + a^2)*B*b/(a*x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^2/(a^2*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3
/2)*B/(a^2*x^2) + 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b/(a^3*x^2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A/(a
^2*x^3)

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mupad [B]  time = 1.11, size = 43, normalized size = 0.57 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (2\,A\,a+3\,A\,b\,x+3\,B\,a\,x+6\,B\,b\,x^2\right )}{6\,x^3\,\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^4,x)

[Out]

-(((a + b*x)^2)^(1/2)*(2*A*a + 3*A*b*x + 3*B*a*x + 6*B*b*x^2))/(6*x^3*(a + b*x))

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sympy [A]  time = 0.30, size = 31, normalized size = 0.41 \begin {gather*} \frac {- 2 A a - 6 B b x^{2} + x \left (- 3 A b - 3 B a\right )}{6 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**4,x)

[Out]

(-2*A*a - 6*B*b*x**2 + x*(-3*A*b - 3*B*a))/(6*x**3)

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