3.2.57 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{x^{11}} \, dx\)

Optimal. Leaf size=231 \[ -\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 231, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^11,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(10*x^10*(a + b*x)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*x^9*(a +
b*x)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^8*(a + b*x)) - (10*a^2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/(7*x^7*(a + b*x)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*x^6*(a + b*x)) - (b^5*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(5*x^5*(a + b*x))

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])

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]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{11}} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^{11}} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5 b^5}{x^{11}}+\frac {5 a^4 b^6}{x^{10}}+\frac {10 a^3 b^7}{x^9}+\frac {10 a^2 b^8}{x^8}+\frac {5 a b^9}{x^7}+\frac {b^{10}}{x^6}\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}}{10 x^{10} (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^8 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 77, normalized size = 0.33 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (126 a^5+700 a^4 b x+1575 a^3 b^2 x^2+1800 a^2 b^3 x^3+1050 a b^4 x^4+252 b^5 x^5\right )}{1260 x^{10} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^11,x]

[Out]

-1/1260*(Sqrt[(a + b*x)^2]*(126*a^5 + 700*a^4*b*x + 1575*a^3*b^2*x^2 + 1800*a^2*b^3*x^3 + 1050*a*b^4*x^4 + 252
*b^5*x^5))/(x^10*(a + b*x))

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IntegrateAlgebraic [B]  time = 1.67, size = 608, normalized size = 2.63 \begin {gather*} \frac {128 b^9 \sqrt {a^2+2 a b x+b^2 x^2} \left (-126 a^{14} b-1834 a^{13} b^2 x-12411 a^{12} b^3 x^2-51759 a^{11} b^4 x^3-148626 a^{10} b^5 x^4-310878 a^9 b^6 x^5-488502 a^8 b^7 x^6-585858 a^7 b^8 x^7-538902 a^6 b^9 x^8-378378 a^5 b^{10} x^9-199627 a^4 b^{11} x^{10}-76743 a^3 b^{12} x^{11}-20322 a^2 b^{13} x^{12}-3318 a b^{14} x^{13}-252 b^{15} x^{14}\right )+128 \sqrt {b^2} b^9 \left (126 a^{15}+1960 a^{14} b x+14245 a^{13} b^2 x^2+64170 a^{12} b^3 x^3+200385 a^{11} b^4 x^4+459504 a^{10} b^5 x^5+799380 a^9 b^6 x^6+1074360 a^8 b^7 x^7+1124760 a^7 b^8 x^8+917280 a^6 b^9 x^9+578005 a^5 b^{10} x^{10}+276370 a^4 b^{11} x^{11}+97065 a^3 b^{12} x^{12}+23640 a^2 b^{13} x^{13}+3570 a b^{14} x^{14}+252 b^{15} x^{15}\right )}{315 \sqrt {b^2} x^{10} \sqrt {a^2+2 a b x+b^2 x^2} \left (-512 a^9 b^9-4608 a^8 b^{10} x-18432 a^7 b^{11} x^2-43008 a^6 b^{12} x^3-64512 a^5 b^{13} x^4-64512 a^4 b^{14} x^5-43008 a^3 b^{15} x^6-18432 a^2 b^{16} x^7-4608 a b^{17} x^8-512 b^{18} x^9\right )+315 x^{10} \left (512 a^{10} b^{10}+5120 a^9 b^{11} x+23040 a^8 b^{12} x^2+61440 a^7 b^{13} x^3+107520 a^6 b^{14} x^4+129024 a^5 b^{15} x^5+107520 a^4 b^{16} x^6+61440 a^3 b^{17} x^7+23040 a^2 b^{18} x^8+5120 a b^{19} x^9+512 b^{20} x^{10}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^11,x]

[Out]

(128*b^9*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-126*a^14*b - 1834*a^13*b^2*x - 12411*a^12*b^3*x^2 - 51759*a^11*b^4*x^
3 - 148626*a^10*b^5*x^4 - 310878*a^9*b^6*x^5 - 488502*a^8*b^7*x^6 - 585858*a^7*b^8*x^7 - 538902*a^6*b^9*x^8 -
378378*a^5*b^10*x^9 - 199627*a^4*b^11*x^10 - 76743*a^3*b^12*x^11 - 20322*a^2*b^13*x^12 - 3318*a*b^14*x^13 - 25
2*b^15*x^14) + 128*b^9*Sqrt[b^2]*(126*a^15 + 1960*a^14*b*x + 14245*a^13*b^2*x^2 + 64170*a^12*b^3*x^3 + 200385*
a^11*b^4*x^4 + 459504*a^10*b^5*x^5 + 799380*a^9*b^6*x^6 + 1074360*a^8*b^7*x^7 + 1124760*a^7*b^8*x^8 + 917280*a
^6*b^9*x^9 + 578005*a^5*b^10*x^10 + 276370*a^4*b^11*x^11 + 97065*a^3*b^12*x^12 + 23640*a^2*b^13*x^13 + 3570*a*
b^14*x^14 + 252*b^15*x^15))/(315*Sqrt[b^2]*x^10*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-512*a^9*b^9 - 4608*a^8*b^10*x
- 18432*a^7*b^11*x^2 - 43008*a^6*b^12*x^3 - 64512*a^5*b^13*x^4 - 64512*a^4*b^14*x^5 - 43008*a^3*b^15*x^6 - 184
32*a^2*b^16*x^7 - 4608*a*b^17*x^8 - 512*b^18*x^9) + 315*x^10*(512*a^10*b^10 + 5120*a^9*b^11*x + 23040*a^8*b^12
*x^2 + 61440*a^7*b^13*x^3 + 107520*a^6*b^14*x^4 + 129024*a^5*b^15*x^5 + 107520*a^4*b^16*x^6 + 61440*a^3*b^17*x
^7 + 23040*a^2*b^18*x^8 + 5120*a*b^19*x^9 + 512*b^20*x^10))

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fricas [A]  time = 0.39, size = 57, normalized size = 0.25 \begin {gather*} -\frac {252 \, b^{5} x^{5} + 1050 \, a b^{4} x^{4} + 1800 \, a^{2} b^{3} x^{3} + 1575 \, a^{3} b^{2} x^{2} + 700 \, a^{4} b x + 126 \, a^{5}}{1260 \, x^{10}} \end {gather*}

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^11,x, algorithm="fricas")

[Out]

-1/1260*(252*b^5*x^5 + 1050*a*b^4*x^4 + 1800*a^2*b^3*x^3 + 1575*a^3*b^2*x^2 + 700*a^4*b*x + 126*a^5)/x^10

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giac [A]  time = 0.16, size = 108, normalized size = 0.47 \begin {gather*} -\frac {b^{10} \mathrm {sgn}\left (b x + a\right )}{1260 \, a^{5}} - \frac {252 \, b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 1050 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 1800 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1575 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 700 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 126 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{1260 \, x^{10}} \end {gather*}

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^11,x, algorithm="giac")

[Out]

-1/1260*b^10*sgn(b*x + a)/a^5 - 1/1260*(252*b^5*x^5*sgn(b*x + a) + 1050*a*b^4*x^4*sgn(b*x + a) + 1800*a^2*b^3*
x^3*sgn(b*x + a) + 1575*a^3*b^2*x^2*sgn(b*x + a) + 700*a^4*b*x*sgn(b*x + a) + 126*a^5*sgn(b*x + a))/x^10

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maple [A]  time = 0.05, size = 74, normalized size = 0.32 \begin {gather*} -\frac {\left (252 b^{5} x^{5}+1050 a \,b^{4} x^{4}+1800 a^{2} b^{3} x^{3}+1575 a^{3} b^{2} x^{2}+700 a^{4} b x +126 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{1260 \left (b x +a \right )^{5} x^{10}} \end {gather*}

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^11,x)

[Out]

-1/1260*(252*b^5*x^5+1050*a*b^4*x^4+1800*a^2*b^3*x^3+1575*a^3*b^2*x^2+700*a^4*b*x+126*a^5)*((b*x+a)^2)^(5/2)/x
^10/(b*x+a)^5

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maxima [B]  time = 1.67, size = 312, normalized size = 1.35 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{10}}{6 \, a^{10}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{9}}{6 \, a^{9} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{8}}{6 \, a^{10} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{7}}{6 \, a^{9} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{6}}{6 \, a^{8} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{5}}{6 \, a^{7} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{4}}{6 \, a^{6} x^{6}} + \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{1260 \, a^{5} x^{7}} - \frac {29 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{180 \, a^{4} x^{8}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{90 \, a^{3} x^{9}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{10 \, a^{2} x^{10}} \end {gather*}

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^11,x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^10/a^10 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^9/(a^9*x) - 1/6*(b^2*x^2
+ 2*a*b*x + a^2)^(7/2)*b^8/(a^10*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^7/(a^9*x^3) - 1/6*(b^2*x^2 + 2*
a*b*x + a^2)^(7/2)*b^6/(a^8*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^5/(a^7*x^5) - 1/6*(b^2*x^2 + 2*a*b*x
+ a^2)^(7/2)*b^4/(a^6*x^6) + 209/1260*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^3/(a^5*x^7) - 29/180*(b^2*x^2 + 2*a*b*
x + a^2)^(7/2)*b^2/(a^4*x^8) + 13/90*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^9) - 1/10*(b^2*x^2 + 2*a*b*x + a
^2)^(7/2)/(a^2*x^10)

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mupad [B]  time = 0.20, size = 207, normalized size = 0.90 \begin {gather*} -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{10\,x^{10}\,\left (a+b\,x\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^8\,\left (a+b\,x\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^11,x)

[Out]

- (a^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(10*x^10*(a + b*x)) - (b^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a
+ b*x)) - (10*a^2*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (5*a^3*b^2*(a^2 + b^2*x^2 + 2*a*b*x
)^(1/2))/(4*x^8*(a + b*x)) - (5*a*b^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (5*a^4*b*(a^2 + b^2
*x^2 + 2*a*b*x)^(1/2))/(9*x^9*(a + b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{11}}\, dx \end {gather*}

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**11,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x**11, x)

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