3.2.6 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{10}} \, dx\)

Optimal. Leaf size=125 \[ -\frac {16 c^2 \left (b x+c x^2\right )^{7/2} (13 b B-6 A c)}{9009 b^4 x^7}+\frac {8 c \left (b x+c x^2\right )^{7/2} (13 b B-6 A c)}{1287 b^3 x^8}-\frac {2 \left (b x+c x^2\right )^{7/2} (13 b B-6 A c)}{143 b^2 x^9}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{13 b x^{10}} \]

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Rubi [A]  time = 0.12, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {792, 658, 650} \begin {gather*} -\frac {16 c^2 \left (b x+c x^2\right )^{7/2} (13 b B-6 A c)}{9009 b^4 x^7}+\frac {8 c \left (b x+c x^2\right )^{7/2} (13 b B-6 A c)}{1287 b^3 x^8}-\frac {2 \left (b x+c x^2\right )^{7/2} (13 b B-6 A c)}{143 b^2 x^9}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{13 b x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^10,x]

[Out]

(-2*A*(b*x + c*x^2)^(7/2))/(13*b*x^10) - (2*(13*b*B - 6*A*c)*(b*x + c*x^2)^(7/2))/(143*b^2*x^9) + (8*c*(13*b*B
- 6*A*c)*(b*x + c*x^2)^(7/2))/(1287*b^3*x^8) - (16*c^2*(13*b*B - 6*A*c)*(b*x + c*x^2)^(7/2))/(9009*b^4*x^7)

Rule 650

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

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(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, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 792

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

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{10}} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{13 b x^{10}}+\frac {\left (2 \left (-10 (-b B+A c)+\frac {7}{2} (-b B+2 A c)\right )\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^9} \, dx}{13 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{13 b x^{10}}-\frac {2 (13 b B-6 A c) \left (b x+c x^2\right )^{7/2}}{143 b^2 x^9}-\frac {(4 c (13 b B-6 A c)) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^8} \, dx}{143 b^2}\\ &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{13 b x^{10}}-\frac {2 (13 b B-6 A c) \left (b x+c x^2\right )^{7/2}}{143 b^2 x^9}+\frac {8 c (13 b B-6 A c) \left (b x+c x^2\right )^{7/2}}{1287 b^3 x^8}+\frac {\left (8 c^2 (13 b B-6 A c)\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^7} \, dx}{1287 b^3}\\ &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{13 b x^{10}}-\frac {2 (13 b B-6 A c) \left (b x+c x^2\right )^{7/2}}{143 b^2 x^9}+\frac {8 c (13 b B-6 A c) \left (b x+c x^2\right )^{7/2}}{1287 b^3 x^8}-\frac {16 c^2 (13 b B-6 A c) \left (b x+c x^2\right )^{7/2}}{9009 b^4 x^7}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 86, normalized size = 0.69 \begin {gather*} -\frac {2 (b+c x)^3 \sqrt {x (b+c x)} \left (3 A \left (231 b^3-126 b^2 c x+56 b c^2 x^2-16 c^3 x^3\right )+13 b B x \left (63 b^2-28 b c x+8 c^2 x^2\right )\right )}{9009 b^4 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^10,x]

[Out]

(-2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(13*b*B*x*(63*b^2 - 28*b*c*x + 8*c^2*x^2) + 3*A*(231*b^3 - 126*b^2*c*x + 56*
b*c^2*x^2 - 16*c^3*x^3)))/(9009*b^4*x^7)

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IntegrateAlgebraic [A]  time = 0.50, size = 156, normalized size = 1.25 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-693 A b^6-1701 A b^5 c x-1113 A b^4 c^2 x^2-15 A b^3 c^3 x^3+18 A b^2 c^4 x^4-24 A b c^5 x^5+48 A c^6 x^6-819 b^6 B x-2093 b^5 B c x^2-1469 b^4 B c^2 x^3-39 b^3 B c^3 x^4+52 b^2 B c^4 x^5-104 b B c^5 x^6\right )}{9009 b^4 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^10,x]

[Out]

(2*Sqrt[b*x + c*x^2]*(-693*A*b^6 - 819*b^6*B*x - 1701*A*b^5*c*x - 2093*b^5*B*c*x^2 - 1113*A*b^4*c^2*x^2 - 1469
*b^4*B*c^2*x^3 - 15*A*b^3*c^3*x^3 - 39*b^3*B*c^3*x^4 + 18*A*b^2*c^4*x^4 + 52*b^2*B*c^4*x^5 - 24*A*b*c^5*x^5 -
104*b*B*c^5*x^6 + 48*A*c^6*x^6))/(9009*b^4*x^7)

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fricas [A]  time = 0.40, size = 153, normalized size = 1.22 \begin {gather*} -\frac {2 \, {\left (693 \, A b^{6} + 8 \, {\left (13 \, B b c^{5} - 6 \, A c^{6}\right )} x^{6} - 4 \, {\left (13 \, B b^{2} c^{4} - 6 \, A b c^{5}\right )} x^{5} + 3 \, {\left (13 \, B b^{3} c^{3} - 6 \, A b^{2} c^{4}\right )} x^{4} + {\left (1469 \, B b^{4} c^{2} + 15 \, A b^{3} c^{3}\right )} x^{3} + 7 \, {\left (299 \, B b^{5} c + 159 \, A b^{4} c^{2}\right )} x^{2} + 63 \, {\left (13 \, B b^{6} + 27 \, A b^{5} c\right )} x\right )} \sqrt {c x^{2} + b x}}{9009 \, b^{4} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^10,x, algorithm="fricas")

[Out]

-2/9009*(693*A*b^6 + 8*(13*B*b*c^5 - 6*A*c^6)*x^6 - 4*(13*B*b^2*c^4 - 6*A*b*c^5)*x^5 + 3*(13*B*b^3*c^3 - 6*A*b
^2*c^4)*x^4 + (1469*B*b^4*c^2 + 15*A*b^3*c^3)*x^3 + 7*(299*B*b^5*c + 159*A*b^4*c^2)*x^2 + 63*(13*B*b^6 + 27*A*
b^5*c)*x)*sqrt(c*x^2 + b*x)/(b^4*x^7)

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giac [B]  time = 0.24, size = 551, normalized size = 4.41 \begin {gather*} \frac {2 \, {\left (12012 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{10} B c^{4} + 63063 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9} B b c^{\frac {7}{2}} + 18018 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9} A c^{\frac {9}{2}} + 153153 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} B b^{2} c^{3} + 108108 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} A b c^{4} + 219219 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B b^{3} c^{\frac {5}{2}} + 297297 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} A b^{2} c^{\frac {7}{2}} + 199485 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b^{4} c^{2} + 485199 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A b^{3} c^{3} + 117117 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{5} c^{\frac {3}{2}} + 513513 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b^{4} c^{\frac {5}{2}} + 43043 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{6} c + 363363 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{5} c^{2} + 9009 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{7} \sqrt {c} + 171171 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{6} c^{\frac {3}{2}} + 819 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{8} + 51597 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{7} c + 9009 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{8} \sqrt {c} + 693 \, A b^{9}\right )}}{9009 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{13}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^10,x, algorithm="giac")

[Out]

2/9009*(12012*(sqrt(c)*x - sqrt(c*x^2 + b*x))^10*B*c^4 + 63063*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*B*b*c^(7/2) +
18018*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*A*c^(9/2) + 153153*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*b^2*c^3 + 1081
08*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*b*c^4 + 219219*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b^3*c^(5/2) + 297297
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*A*b^2*c^(7/2) + 199485*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^4*c^2 + 485199
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*A*b^3*c^3 + 117117*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^5*c^(3/2) + 513513
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b^4*c^(5/2) + 43043*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^6*c + 363363*(s
qrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^5*c^2 + 9009*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^7*sqrt(c) + 171171*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^3*A*b^6*c^(3/2) + 819*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^8 + 51597*(sqrt(c)*x -
sqrt(c*x^2 + b*x))^2*A*b^7*c + 9009*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^8*sqrt(c) + 693*A*b^9)/(sqrt(c)*x - s
qrt(c*x^2 + b*x))^13

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maple [A]  time = 0.05, size = 86, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-48 A \,c^{3} x^{3}+104 B b \,c^{2} x^{3}+168 A b \,c^{2} x^{2}-364 B \,b^{2} c \,x^{2}-378 A \,b^{2} c x +819 B \,b^{3} x +693 A \,b^{3}\right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{9009 b^{4} x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^10,x)

[Out]

-2/9009*(c*x+b)*(-48*A*c^3*x^3+104*B*b*c^2*x^3+168*A*b*c^2*x^2-364*B*b^2*c*x^2-378*A*b^2*c*x+819*B*b^3*x+693*A
*b^3)*(c*x^2+b*x)^(5/2)/x^9/b^4

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maxima [B]  time = 0.94, size = 350, normalized size = 2.80 \begin {gather*} -\frac {16 \, \sqrt {c x^{2} + b x} B c^{5}}{693 \, b^{3} x} + \frac {32 \, \sqrt {c x^{2} + b x} A c^{6}}{3003 \, b^{4} x} + \frac {8 \, \sqrt {c x^{2} + b x} B c^{4}}{693 \, b^{2} x^{2}} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{5}}{3003 \, b^{3} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} B c^{3}}{231 \, b x^{3}} + \frac {4 \, \sqrt {c x^{2} + b x} A c^{4}}{1001 \, b^{2} x^{3}} + \frac {5 \, \sqrt {c x^{2} + b x} B c^{2}}{693 \, x^{4}} - \frac {10 \, \sqrt {c x^{2} + b x} A c^{3}}{3003 \, b x^{4}} - \frac {5 \, \sqrt {c x^{2} + b x} B b c}{792 \, x^{5}} + \frac {5 \, \sqrt {c x^{2} + b x} A c^{2}}{1716 \, x^{5}} - \frac {5 \, \sqrt {c x^{2} + b x} B b^{2}}{88 \, x^{6}} - \frac {3 \, \sqrt {c x^{2} + b x} A b c}{1144 \, x^{6}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{24 \, x^{7}} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{2}}{104 \, x^{7}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{3 \, x^{8}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{8 \, x^{8}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{4 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^10,x, algorithm="maxima")

[Out]

-16/693*sqrt(c*x^2 + b*x)*B*c^5/(b^3*x) + 32/3003*sqrt(c*x^2 + b*x)*A*c^6/(b^4*x) + 8/693*sqrt(c*x^2 + b*x)*B*
c^4/(b^2*x^2) - 16/3003*sqrt(c*x^2 + b*x)*A*c^5/(b^3*x^2) - 2/231*sqrt(c*x^2 + b*x)*B*c^3/(b*x^3) + 4/1001*sqr
t(c*x^2 + b*x)*A*c^4/(b^2*x^3) + 5/693*sqrt(c*x^2 + b*x)*B*c^2/x^4 - 10/3003*sqrt(c*x^2 + b*x)*A*c^3/(b*x^4) -
5/792*sqrt(c*x^2 + b*x)*B*b*c/x^5 + 5/1716*sqrt(c*x^2 + b*x)*A*c^2/x^5 - 5/88*sqrt(c*x^2 + b*x)*B*b^2/x^6 - 3
/1144*sqrt(c*x^2 + b*x)*A*b*c/x^6 + 5/24*(c*x^2 + b*x)^(3/2)*B*b/x^7 - 3/104*sqrt(c*x^2 + b*x)*A*b^2/x^7 - 1/3
*(c*x^2 + b*x)^(5/2)*B/x^8 + 1/8*(c*x^2 + b*x)^(3/2)*A*b/x^8 - 1/4*(c*x^2 + b*x)^(5/2)*A/x^9

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mupad [B]  time = 4.09, size = 280, normalized size = 2.24 \begin {gather*} \frac {4\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{1001\,b^2\,x^3}-\frac {106\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{429\,x^5}-\frac {2\,B\,b^2\,\sqrt {c\,x^2+b\,x}}{11\,x^6}-\frac {226\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{693\,x^4}-\frac {10\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{3003\,b\,x^4}-\frac {2\,A\,b^2\,\sqrt {c\,x^2+b\,x}}{13\,x^7}-\frac {16\,A\,c^5\,\sqrt {c\,x^2+b\,x}}{3003\,b^3\,x^2}+\frac {32\,A\,c^6\,\sqrt {c\,x^2+b\,x}}{3003\,b^4\,x}-\frac {2\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{231\,b\,x^3}+\frac {8\,B\,c^4\,\sqrt {c\,x^2+b\,x}}{693\,b^2\,x^2}-\frac {16\,B\,c^5\,\sqrt {c\,x^2+b\,x}}{693\,b^3\,x}-\frac {54\,A\,b\,c\,\sqrt {c\,x^2+b\,x}}{143\,x^6}-\frac {46\,B\,b\,c\,\sqrt {c\,x^2+b\,x}}{99\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^10,x)

[Out]

(4*A*c^4*(b*x + c*x^2)^(1/2))/(1001*b^2*x^3) - (106*A*c^2*(b*x + c*x^2)^(1/2))/(429*x^5) - (2*B*b^2*(b*x + c*x
^2)^(1/2))/(11*x^6) - (226*B*c^2*(b*x + c*x^2)^(1/2))/(693*x^4) - (10*A*c^3*(b*x + c*x^2)^(1/2))/(3003*b*x^4)
- (2*A*b^2*(b*x + c*x^2)^(1/2))/(13*x^7) - (16*A*c^5*(b*x + c*x^2)^(1/2))/(3003*b^3*x^2) + (32*A*c^6*(b*x + c*
x^2)^(1/2))/(3003*b^4*x) - (2*B*c^3*(b*x + c*x^2)^(1/2))/(231*b*x^3) + (8*B*c^4*(b*x + c*x^2)^(1/2))/(693*b^2*
x^2) - (16*B*c^5*(b*x + c*x^2)^(1/2))/(693*b^3*x) - (54*A*b*c*(b*x + c*x^2)^(1/2))/(143*x^6) - (46*B*b*c*(b*x
+ c*x^2)^(1/2))/(99*x^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{10}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**10,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**10, x)

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