3.2.4 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^8} \, dx\)

Optimal. Leaf size=57 \[ -\frac {2 \left (b x+c x^2\right )^{7/2} (9 b B-2 A c)}{63 b^2 x^7}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8} \]

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Rubi [A]  time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {792, 650} \begin {gather*} -\frac {2 \left (b x+c x^2\right )^{7/2} (9 b B-2 A c)}{63 b^2 x^7}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^8,x]

[Out]

(-2*A*(b*x + c*x^2)^(7/2))/(9*b*x^8) - (2*(9*b*B - 2*A*c)*(b*x + c*x^2)^(7/2))/(63*b^2*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 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^8} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8}+\frac {\left (2 \left (-8 (-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^7} \, dx}{9 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8}-\frac {2 (9 b B-2 A c) \left (b x+c x^2\right )^{7/2}}{63 b^2 x^7}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.75 \begin {gather*} -\frac {2 (b+c x)^3 \sqrt {x (b+c x)} (7 A b-2 A c x+9 b B x)}{63 b^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^8,x]

[Out]

(-2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(7*A*b + 9*b*B*x - 2*A*c*x))/(63*b^2*x^5)

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IntegrateAlgebraic [A]  time = 0.41, size = 108, normalized size = 1.89 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-7 A b^4-19 A b^3 c x-15 A b^2 c^2 x^2-A b c^3 x^3+2 A c^4 x^4-9 b^4 B x-27 b^3 B c x^2-27 b^2 B c^2 x^3-9 b B c^3 x^4\right )}{63 b^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^8,x]

[Out]

(2*Sqrt[b*x + c*x^2]*(-7*A*b^4 - 9*b^4*B*x - 19*A*b^3*c*x - 27*b^3*B*c*x^2 - 15*A*b^2*c^2*x^2 - 27*b^2*B*c^2*x
^3 - A*b*c^3*x^3 - 9*b*B*c^3*x^4 + 2*A*c^4*x^4))/(63*b^2*x^5)

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fricas [B]  time = 0.40, size = 102, normalized size = 1.79 \begin {gather*} -\frac {2 \, {\left (7 \, A b^{4} + {\left (9 \, B b c^{3} - 2 \, A c^{4}\right )} x^{4} + {\left (27 \, B b^{2} c^{2} + A b c^{3}\right )} x^{3} + 3 \, {\left (9 \, B b^{3} c + 5 \, A b^{2} c^{2}\right )} x^{2} + {\left (9 \, B b^{4} + 19 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{63 \, b^{2} x^{5}} \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^8,x, algorithm="fricas")

[Out]

-2/63*(7*A*b^4 + (9*B*b*c^3 - 2*A*c^4)*x^4 + (27*B*b^2*c^2 + A*b*c^3)*x^3 + 3*(9*B*b^3*c + 5*A*b^2*c^2)*x^2 +
(9*B*b^4 + 19*A*b^3*c)*x)*sqrt(c*x^2 + b*x)/(b^2*x^5)

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giac [B]  time = 0.23, size = 431, normalized size = 7.56 \begin {gather*} \frac {2 \, {\left (63 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} B c^{3} + 189 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B b c^{\frac {5}{2}} + 63 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} A c^{\frac {7}{2}} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b^{2} c^{2} + 273 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A b c^{3} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{3} c^{\frac {3}{2}} + 567 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b^{2} c^{\frac {5}{2}} + 189 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{4} c + 693 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{3} c^{2} + 63 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{5} \sqrt {c} + 525 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{4} c^{\frac {3}{2}} + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{6} + 243 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{5} c + 63 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{6} \sqrt {c} + 7 \, A b^{7}\right )}}{63 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{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^8,x, algorithm="giac")

[Out]

2/63*(63*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*c^3 + 189*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b*c^(5/2) + 63*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^7*A*c^(7/2) + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^2*c^2 + 273*(sqrt(c)*x - s
qrt(c*x^2 + b*x))^6*A*b*c^3 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^3*c^(3/2) + 567*(sqrt(c)*x - sqrt(c*x^
2 + b*x))^5*A*b^2*c^(5/2) + 189*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^4*c + 693*(sqrt(c)*x - sqrt(c*x^2 + b*x)
)^4*A*b^3*c^2 + 63*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^5*sqrt(c) + 525*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b
^4*c^(3/2) + 9*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^6 + 243*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^5*c + 63*(s
qrt(c)*x - sqrt(c*x^2 + b*x))*A*b^6*sqrt(c) + 7*A*b^7)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^9

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maple [A]  time = 0.04, size = 40, normalized size = 0.70 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-2 A c x +9 B b x +7 A b \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{63 b^{2} x^{7}} \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^8,x)

[Out]

-2/63*(c*x+b)*(-2*A*c*x+9*B*b*x+7*A*b)*(c*x^2+b*x)^(5/2)/b^2/x^7

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maxima [B]  time = 1.00, size = 258, normalized size = 4.53 \begin {gather*} -\frac {2 \, \sqrt {c x^{2} + b x} B c^{3}}{7 \, b x} + \frac {4 \, \sqrt {c x^{2} + b x} A c^{4}}{63 \, b^{2} x} + \frac {\sqrt {c x^{2} + b x} B c^{2}}{7 \, x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{3}}{63 \, b x^{2}} - \frac {3 \, \sqrt {c x^{2} + b x} B b c}{28 \, x^{3}} + \frac {\sqrt {c x^{2} + b x} A c^{2}}{42 \, x^{3}} - \frac {15 \, \sqrt {c x^{2} + b x} B b^{2}}{28 \, x^{4}} - \frac {5 \, \sqrt {c x^{2} + b x} A b c}{252 \, x^{4}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{4 \, x^{5}} - \frac {5 \, \sqrt {c x^{2} + b x} A b^{2}}{36 \, x^{5}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{x^{6}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{12 \, x^{6}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{2 \, 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^8,x, algorithm="maxima")

[Out]

-2/7*sqrt(c*x^2 + b*x)*B*c^3/(b*x) + 4/63*sqrt(c*x^2 + b*x)*A*c^4/(b^2*x) + 1/7*sqrt(c*x^2 + b*x)*B*c^2/x^2 -
2/63*sqrt(c*x^2 + b*x)*A*c^3/(b*x^2) - 3/28*sqrt(c*x^2 + b*x)*B*b*c/x^3 + 1/42*sqrt(c*x^2 + b*x)*A*c^2/x^3 - 1
5/28*sqrt(c*x^2 + b*x)*B*b^2/x^4 - 5/252*sqrt(c*x^2 + b*x)*A*b*c/x^4 + 5/4*(c*x^2 + b*x)^(3/2)*B*b/x^5 - 5/36*
sqrt(c*x^2 + b*x)*A*b^2/x^5 - (c*x^2 + b*x)^(5/2)*B/x^6 + 5/12*(c*x^2 + b*x)^(3/2)*A*b/x^6 - 1/2*(c*x^2 + b*x)
^(5/2)*A/x^7

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mupad [B]  time = 2.89, size = 188, normalized size = 3.30 \begin {gather*} \frac {4\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{63\,b^2\,x}-\frac {10\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{21\,x^3}-\frac {2\,B\,b^2\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {6\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{7\,x^2}-\frac {2\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{63\,b\,x^2}-\frac {2\,A\,b^2\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {2\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{7\,b\,x}-\frac {38\,A\,b\,c\,\sqrt {c\,x^2+b\,x}}{63\,x^4}-\frac {6\,B\,b\,c\,\sqrt {c\,x^2+b\,x}}{7\,x^3} \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^8,x)

[Out]

(4*A*c^4*(b*x + c*x^2)^(1/2))/(63*b^2*x) - (10*A*c^2*(b*x + c*x^2)^(1/2))/(21*x^3) - (2*B*b^2*(b*x + c*x^2)^(1
/2))/(7*x^4) - (6*B*c^2*(b*x + c*x^2)^(1/2))/(7*x^2) - (2*A*c^3*(b*x + c*x^2)^(1/2))/(63*b*x^2) - (2*A*b^2*(b*
x + c*x^2)^(1/2))/(9*x^5) - (2*B*c^3*(b*x + c*x^2)^(1/2))/(7*b*x) - (38*A*b*c*(b*x + c*x^2)^(1/2))/(63*x^4) -
(6*B*b*c*(b*x + c*x^2)^(1/2))/(7*x^3)

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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^{8}}\, 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**8,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**8, x)

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