3.88 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=57 \[ -\frac {2 \left (b x+c x^2\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^5}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6} \]

[Out]

-2/7*A*(c*x^2+b*x)^(5/2)/b/x^6-2/35*(-2*A*c+7*B*b)*(c*x^2+b*x)^(5/2)/b^2/x^5

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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} \[ -\frac {2 \left (b x+c x^2\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^5}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^6,x]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(7*b*x^6) - (2*(7*b*B - 2*A*c)*(b*x + c*x^2)^(5/2))/(35*b^2*x^5)

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 )^{3/2}}{x^6} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6}+\frac {\left (2 \left (-6 (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right )\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{7 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6}-\frac {2 (7 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{35 b^2 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.63 \[ -\frac {2 (x (b+c x))^{5/2} (5 A b-2 A c x+7 b B x)}{35 b^2 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^6,x]

[Out]

(-2*(x*(b + c*x))^(5/2)*(5*A*b + 7*b*B*x - 2*A*c*x))/(35*b^2*x^6)

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fricas [A]  time = 0.87, size = 78, normalized size = 1.37 \[ -\frac {2 \, {\left (5 \, A b^{3} + {\left (7 \, B b c^{2} - 2 \, A c^{3}\right )} x^{3} + {\left (14 \, B b^{2} c + A b c^{2}\right )} x^{2} + {\left (7 \, B b^{3} + 8 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{35 \, b^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x, algorithm="fricas")

[Out]

-2/35*(5*A*b^3 + (7*B*b*c^2 - 2*A*c^3)*x^3 + (14*B*b^2*c + A*b*c^2)*x^2 + (7*B*b^3 + 8*A*b^2*c)*x)*sqrt(c*x^2
+ b*x)/(b^2*x^4)

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giac [B]  time = 0.23, size = 311, normalized size = 5.46 \[ \frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B c^{2} + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b c^{\frac {3}{2}} + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A c^{\frac {5}{2}} + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{2} c + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b c^{2} + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{3} \sqrt {c} + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{2} c^{\frac {3}{2}} + 7 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{4} + 98 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{3} c + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{4} \sqrt {c} + 5 \, A b^{5}\right )}}{35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x, algorithm="giac")

[Out]

2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*c^2 + 70*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b*c^(3/2) + 35*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^5*A*c^(5/2) + 70*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^2*c + 105*(sqrt(c)*x - sqrt(
c*x^2 + b*x))^4*A*b*c^2 + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*sqrt(c) + 140*(sqrt(c)*x - sqrt(c*x^2 + b
*x))^3*A*b^2*c^(3/2) + 7*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^4 + 98*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^3*
c + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^4*sqrt(c) + 5*A*b^5)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^7

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maple [A]  time = 0.05, size = 40, normalized size = 0.70 \[ -\frac {2 \left (c x +b \right ) \left (-2 A c x +7 B b x +5 A b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{35 b^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x)

[Out]

-2/35*(c*x+b)*(-2*A*c*x+7*B*b*x+5*A*b)*(c*x^2+b*x)^(3/2)/b^2/x^5

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maxima [B]  time = 0.94, size = 176, normalized size = 3.09 \[ -\frac {2 \, \sqrt {c x^{2} + b x} B c^{2}}{5 \, b x} + \frac {4 \, \sqrt {c x^{2} + b x} A c^{3}}{35 \, b^{2} x} + \frac {\sqrt {c x^{2} + b x} B c}{5 \, x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{35 \, b x^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} B b}{5 \, x^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} A c}{70 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{x^{4}} + \frac {3 \, \sqrt {c x^{2} + b x} A b}{14 \, x^{4}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{2 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x, algorithm="maxima")

[Out]

-2/5*sqrt(c*x^2 + b*x)*B*c^2/(b*x) + 4/35*sqrt(c*x^2 + b*x)*A*c^3/(b^2*x) + 1/5*sqrt(c*x^2 + b*x)*B*c/x^2 - 2/
35*sqrt(c*x^2 + b*x)*A*c^2/(b*x^2) + 3/5*sqrt(c*x^2 + b*x)*B*b/x^3 + 3/70*sqrt(c*x^2 + b*x)*A*c/x^3 - (c*x^2 +
b*x)^(3/2)*B/x^4 + 3/14*sqrt(c*x^2 + b*x)*A*b/x^4 - 1/2*(c*x^2 + b*x)^(3/2)*A/x^5

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mupad [B]  time = 2.02, size = 142, normalized size = 2.49 \[ \frac {4\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{35\,b^2\,x}-\frac {16\,A\,c\,\sqrt {c\,x^2+b\,x}}{35\,x^3}-\frac {2\,B\,b\,\sqrt {c\,x^2+b\,x}}{5\,x^3}-\frac {4\,B\,c\,\sqrt {c\,x^2+b\,x}}{5\,x^2}-\frac {2\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b\,x^2}-\frac {2\,A\,b\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {2\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{5\,b\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^6,x)

[Out]

(4*A*c^3*(b*x + c*x^2)^(1/2))/(35*b^2*x) - (16*A*c*(b*x + c*x^2)^(1/2))/(35*x^3) - (2*B*b*(b*x + c*x^2)^(1/2))
/(5*x^3) - (4*B*c*(b*x + c*x^2)^(1/2))/(5*x^2) - (2*A*c^2*(b*x + c*x^2)^(1/2))/(35*b*x^2) - (2*A*b*(b*x + c*x^
2)^(1/2))/(7*x^4) - (2*B*c^2*(b*x + c*x^2)^(1/2))/(5*b*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**6,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**6, x)

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