3.2.5 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^9} \, dx\)

Optimal. Leaf size=90 \[ \frac {4 c \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{693 b^3 x^7}-\frac {2 \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{99 b^2 x^8}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9} \]

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Rubi [A]  time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {4 c \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{693 b^3 x^7}-\frac {2 \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{99 b^2 x^8}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^9,x]

[Out]

(-2*A*(b*x + c*x^2)^(7/2))/(11*b*x^9) - (2*(11*b*B - 4*A*c)*(b*x + c*x^2)^(7/2))/(99*b^2*x^8) + (4*c*(11*b*B -
4*A*c)*(b*x + c*x^2)^(7/2))/(693*b^3*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^9} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9}+\frac {\left (2 \left (-9 (-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^8} \, dx}{11 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9}-\frac {2 (11 b B-4 A c) \left (b x+c x^2\right )^{7/2}}{99 b^2 x^8}-\frac {(2 c (11 b B-4 A c)) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^7} \, dx}{99 b^2}\\ &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9}-\frac {2 (11 b B-4 A c) \left (b x+c x^2\right )^{7/2}}{99 b^2 x^8}+\frac {4 c (11 b B-4 A c) \left (b x+c x^2\right )^{7/2}}{693 b^3 x^7}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.70 \begin {gather*} \frac {2 (b+c x)^3 \sqrt {x (b+c x)} \left (A \left (-63 b^2+28 b c x-8 c^2 x^2\right )+11 b B x (2 c x-7 b)\right )}{693 b^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^9,x]

[Out]

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(11*b*B*x*(-7*b + 2*c*x) + A*(-63*b^2 + 28*b*c*x - 8*c^2*x^2)))/(693*b^3*x^6)

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IntegrateAlgebraic [A]  time = 0.47, size = 132, normalized size = 1.47 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (63 A b^5+161 A b^4 c x+113 A b^3 c^2 x^2+3 A b^2 c^3 x^3-4 A b c^4 x^4+8 A c^5 x^5+77 b^5 B x+209 b^4 B c x^2+165 b^3 B c^2 x^3+11 b^2 B c^3 x^4-22 b B c^4 x^5\right )}{693 b^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^9,x]

[Out]

(-2*Sqrt[b*x + c*x^2]*(63*A*b^5 + 77*b^5*B*x + 161*A*b^4*c*x + 209*b^4*B*c*x^2 + 113*A*b^3*c^2*x^2 + 165*b^3*B
*c^2*x^3 + 3*A*b^2*c^3*x^3 + 11*b^2*B*c^3*x^4 - 4*A*b*c^4*x^4 - 22*b*B*c^4*x^5 + 8*A*c^5*x^5))/(693*b^3*x^6)

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fricas [A]  time = 0.41, size = 127, normalized size = 1.41 \begin {gather*} -\frac {2 \, {\left (63 \, A b^{5} - 2 \, {\left (11 \, B b c^{4} - 4 \, A c^{5}\right )} x^{5} + {\left (11 \, B b^{2} c^{3} - 4 \, A b c^{4}\right )} x^{4} + 3 \, {\left (55 \, B b^{3} c^{2} + A b^{2} c^{3}\right )} x^{3} + {\left (209 \, B b^{4} c + 113 \, A b^{3} c^{2}\right )} x^{2} + 7 \, {\left (11 \, B b^{5} + 23 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x}}{693 \, b^{3} x^{6}} \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^9,x, algorithm="fricas")

[Out]

-2/693*(63*A*b^5 - 2*(11*B*b*c^4 - 4*A*c^5)*x^5 + (11*B*b^2*c^3 - 4*A*b*c^4)*x^4 + 3*(55*B*b^3*c^2 + A*b^2*c^3
)*x^3 + (209*B*b^4*c + 113*A*b^3*c^2)*x^2 + 7*(11*B*b^5 + 23*A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^3*x^6)

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giac [B]  time = 0.22, size = 491, normalized size = 5.46 \begin {gather*} \frac {2 \, {\left (693 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9} B c^{\frac {7}{2}} + 3003 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} B b c^{3} + 924 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} A c^{4} + 6237 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B b^{2} c^{\frac {5}{2}} + 4851 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} A b c^{\frac {7}{2}} + 7623 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b^{3} c^{2} + 11781 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A b^{2} c^{3} + 5775 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{4} c^{\frac {3}{2}} + 16863 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b^{3} c^{\frac {5}{2}} + 2673 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{5} c + 15345 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{4} c^{2} + 693 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{6} \sqrt {c} + 9009 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{5} c^{\frac {3}{2}} + 77 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{7} + 3311 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{6} c + 693 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{7} \sqrt {c} + 63 \, A b^{8}\right )}}{693 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{11}} \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^9,x, algorithm="giac")

[Out]

2/693*(693*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*B*c^(7/2) + 3003*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*b*c^3 + 924*
(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*c^4 + 6237*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b^2*c^(5/2) + 4851*(sqrt(c)
*x - sqrt(c*x^2 + b*x))^7*A*b*c^(7/2) + 7623*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^3*c^2 + 11781*(sqrt(c)*x -
sqrt(c*x^2 + b*x))^6*A*b^2*c^3 + 5775*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^4*c^(3/2) + 16863*(sqrt(c)*x - sqr
t(c*x^2 + b*x))^5*A*b^3*c^(5/2) + 2673*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^5*c + 15345*(sqrt(c)*x - sqrt(c*x
^2 + b*x))^4*A*b^4*c^2 + 693*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^6*sqrt(c) + 9009*(sqrt(c)*x - sqrt(c*x^2 +
b*x))^3*A*b^5*c^(3/2) + 77*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^7 + 3311*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*
b^6*c + 693*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^7*sqrt(c) + 63*A*b^8)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^11

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maple [A]  time = 0.05, size = 62, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-22 B b c \,x^{2}-28 A b c x +77 B \,b^{2} x +63 A \,b^{2}\right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{693 b^{3} x^{8}} \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^9,x)

[Out]

-2/693*(c*x+b)*(8*A*c^2*x^2-22*B*b*c*x^2-28*A*b*c*x+77*B*b^2*x+63*A*b^2)*(c*x^2+b*x)^(5/2)/b^3/x^8

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maxima [B]  time = 0.97, size = 304, normalized size = 3.38 \begin {gather*} \frac {4 \, \sqrt {c x^{2} + b x} B c^{4}}{63 \, b^{2} x} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{5}}{693 \, b^{3} x} - \frac {2 \, \sqrt {c x^{2} + b x} B c^{3}}{63 \, b x^{2}} + \frac {8 \, \sqrt {c x^{2} + b x} A c^{4}}{693 \, b^{2} x^{2}} + \frac {\sqrt {c x^{2} + b x} B c^{2}}{42 \, x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{3}}{231 \, b x^{3}} - \frac {5 \, \sqrt {c x^{2} + b x} B b c}{252 \, x^{4}} + \frac {5 \, \sqrt {c x^{2} + b x} A c^{2}}{693 \, x^{4}} - \frac {5 \, \sqrt {c x^{2} + b x} B b^{2}}{36 \, x^{5}} - \frac {5 \, \sqrt {c x^{2} + b x} A b c}{792 \, x^{5}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{12 \, x^{6}} - \frac {5 \, \sqrt {c x^{2} + b x} A b^{2}}{88 \, x^{6}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{2 \, x^{7}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{24 \, x^{7}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{3 \, x^{8}} \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^9,x, algorithm="maxima")

[Out]

4/63*sqrt(c*x^2 + b*x)*B*c^4/(b^2*x) - 16/693*sqrt(c*x^2 + b*x)*A*c^5/(b^3*x) - 2/63*sqrt(c*x^2 + b*x)*B*c^3/(
b*x^2) + 8/693*sqrt(c*x^2 + b*x)*A*c^4/(b^2*x^2) + 1/42*sqrt(c*x^2 + b*x)*B*c^2/x^3 - 2/231*sqrt(c*x^2 + b*x)*
A*c^3/(b*x^3) - 5/252*sqrt(c*x^2 + b*x)*B*b*c/x^4 + 5/693*sqrt(c*x^2 + b*x)*A*c^2/x^4 - 5/36*sqrt(c*x^2 + b*x)
*B*b^2/x^5 - 5/792*sqrt(c*x^2 + b*x)*A*b*c/x^5 + 5/12*(c*x^2 + b*x)^(3/2)*B*b/x^6 - 5/88*sqrt(c*x^2 + b*x)*A*b
^2/x^6 - 1/2*(c*x^2 + b*x)^(5/2)*B/x^7 + 5/24*(c*x^2 + b*x)^(3/2)*A*b/x^7 - 1/3*(c*x^2 + b*x)^(5/2)*A/x^8

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mupad [B]  time = 3.46, size = 234, normalized size = 2.60 \begin {gather*} \frac {8\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{693\,b^2\,x^2}-\frac {226\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{693\,x^4}-\frac {2\,B\,b^2\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {10\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{21\,x^3}-\frac {2\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{231\,b\,x^3}-\frac {2\,A\,b^2\,\sqrt {c\,x^2+b\,x}}{11\,x^6}-\frac {16\,A\,c^5\,\sqrt {c\,x^2+b\,x}}{693\,b^3\,x}-\frac {2\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{63\,b\,x^2}+\frac {4\,B\,c^4\,\sqrt {c\,x^2+b\,x}}{63\,b^2\,x}-\frac {46\,A\,b\,c\,\sqrt {c\,x^2+b\,x}}{99\,x^5}-\frac {38\,B\,b\,c\,\sqrt {c\,x^2+b\,x}}{63\,x^4} \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^9,x)

[Out]

(8*A*c^4*(b*x + c*x^2)^(1/2))/(693*b^2*x^2) - (226*A*c^2*(b*x + c*x^2)^(1/2))/(693*x^4) - (2*B*b^2*(b*x + c*x^
2)^(1/2))/(9*x^5) - (10*B*c^2*(b*x + c*x^2)^(1/2))/(21*x^3) - (2*A*c^3*(b*x + c*x^2)^(1/2))/(231*b*x^3) - (2*A
*b^2*(b*x + c*x^2)^(1/2))/(11*x^6) - (16*A*c^5*(b*x + c*x^2)^(1/2))/(693*b^3*x) - (2*B*c^3*(b*x + c*x^2)^(1/2)
)/(63*b*x^2) + (4*B*c^4*(b*x + c*x^2)^(1/2))/(63*b^2*x) - (46*A*b*c*(b*x + c*x^2)^(1/2))/(99*x^5) - (38*B*b*c*
(b*x + c*x^2)^(1/2))/(63*x^4)

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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^{9}}\, 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**9,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**9, x)

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