3.89 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=90 \[ \frac {4 c \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{315 b^3 x^5}-\frac {2 \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{63 b^2 x^6}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7} \]

[Out]

-2/9*A*(c*x^2+b*x)^(5/2)/b/x^7-2/63*(-4*A*c+9*B*b)*(c*x^2+b*x)^(5/2)/b^2/x^6+4/315*c*(-4*A*c+9*B*b)*(c*x^2+b*x
)^(5/2)/b^3/x^5

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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} \[ \frac {4 c \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{315 b^3 x^5}-\frac {2 \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{63 b^2 x^6}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^7,x]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(9*b*x^7) - (2*(9*b*B - 4*A*c)*(b*x + c*x^2)^(5/2))/(63*b^2*x^6) + (4*c*(9*b*B - 4*
A*c)*(b*x + c*x^2)^(5/2))/(315*b^3*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 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 )^{3/2}}{x^7} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}+\frac {\left (2 \left (-7 (-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^6} \, dx}{9 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}-\frac {2 (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}-\frac {(2 c (9 b B-4 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{63 b^2}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}-\frac {2 (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}+\frac {4 c (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 56, normalized size = 0.62 \[ \frac {2 (x (b+c x))^{5/2} \left (A \left (-35 b^2+20 b c x-8 c^2 x^2\right )+9 b B x (2 c x-5 b)\right )}{315 b^3 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^7,x]

[Out]

(2*(x*(b + c*x))^(5/2)*(9*b*B*x*(-5*b + 2*c*x) + A*(-35*b^2 + 20*b*c*x - 8*c^2*x^2)))/(315*b^3*x^7)

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fricas [A]  time = 0.87, size = 104, normalized size = 1.16 \[ -\frac {2 \, {\left (35 \, A b^{4} - 2 \, {\left (9 \, B b c^{3} - 4 \, A c^{4}\right )} x^{4} + {\left (9 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{3} + 3 \, {\left (24 \, B b^{3} c + A b^{2} c^{2}\right )} x^{2} + 5 \, {\left (9 \, B b^{4} + 10 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, b^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^7,x, algorithm="fricas")

[Out]

-2/315*(35*A*b^4 - 2*(9*B*b*c^3 - 4*A*c^4)*x^4 + (9*B*b^2*c^2 - 4*A*b*c^3)*x^3 + 3*(24*B*b^3*c + A*b^2*c^2)*x^
2 + 5*(9*B*b^4 + 10*A*b^3*c)*x)*sqrt(c*x^2 + b*x)/(b^3*x^5)

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giac [B]  time = 0.23, size = 371, normalized size = 4.12 \[ \frac {2 \, {\left (315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B c^{\frac {5}{2}} + 945 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b c^{2} + 420 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A c^{3} + 1260 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{2} c^{\frac {3}{2}} + 1575 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b c^{\frac {5}{2}} + 882 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{3} c + 2583 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{2} c^{2} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{4} \sqrt {c} + 2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{3} c^{\frac {3}{2}} + 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{5} + 1170 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{4} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{5} \sqrt {c} + 35 \, A b^{6}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^7,x, algorithm="giac")

[Out]

2/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*c^(5/2) + 945*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b*c^2 + 420*(
sqrt(c)*x - sqrt(c*x^2 + b*x))^6*A*c^3 + 1260*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^2*c^(3/2) + 1575*(sqrt(c)*
x - sqrt(c*x^2 + b*x))^5*A*b*c^(5/2) + 882*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^3*c + 2583*(sqrt(c)*x - sqrt(
c*x^2 + b*x))^4*A*b^2*c^2 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^4*sqrt(c) + 2310*(sqrt(c)*x - sqrt(c*x^2
+ b*x))^3*A*b^3*c^(3/2) + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^5 + 1170*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2
*A*b^4*c + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^5*sqrt(c) + 35*A*b^6)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^9

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maple [A]  time = 0.05, size = 62, normalized size = 0.69 \[ -\frac {2 \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-18 B b c \,x^{2}-20 A b c x +45 B \,b^{2} x +35 A \,b^{2}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{315 b^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^7,x)

[Out]

-2/315*(c*x+b)*(8*A*c^2*x^2-18*B*b*c*x^2-20*A*b*c*x+45*B*b^2*x+35*A*b^2)*(c*x^2+b*x)^(3/2)/b^3/x^6

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maxima [B]  time = 0.98, size = 222, normalized size = 2.47 \[ \frac {4 \, \sqrt {c x^{2} + b x} B c^{3}}{35 \, b^{2} x} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{4}}{315 \, b^{3} x} - \frac {2 \, \sqrt {c x^{2} + b x} B c^{2}}{35 \, b x^{2}} + \frac {8 \, \sqrt {c x^{2} + b x} A c^{3}}{315 \, b^{2} x^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} B c}{70 \, x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{105 \, b x^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} B b}{14 \, x^{4}} + \frac {\sqrt {c x^{2} + b x} A c}{63 \, x^{4}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{2 \, x^{5}} + \frac {\sqrt {c x^{2} + b x} A b}{9 \, x^{5}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{3 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^7,x, algorithm="maxima")

[Out]

4/35*sqrt(c*x^2 + b*x)*B*c^3/(b^2*x) - 16/315*sqrt(c*x^2 + b*x)*A*c^4/(b^3*x) - 2/35*sqrt(c*x^2 + b*x)*B*c^2/(
b*x^2) + 8/315*sqrt(c*x^2 + b*x)*A*c^3/(b^2*x^2) + 3/70*sqrt(c*x^2 + b*x)*B*c/x^3 - 2/105*sqrt(c*x^2 + b*x)*A*
c^2/(b*x^3) + 3/14*sqrt(c*x^2 + b*x)*B*b/x^4 + 1/63*sqrt(c*x^2 + b*x)*A*c/x^4 - 1/2*(c*x^2 + b*x)^(3/2)*B/x^5
+ 1/9*sqrt(c*x^2 + b*x)*A*b/x^5 - 1/3*(c*x^2 + b*x)^(3/2)*A/x^6

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mupad [B]  time = 2.40, size = 188, normalized size = 2.09 \[ \frac {8\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^2\,x^2}-\frac {20\,A\,c\,\sqrt {c\,x^2+b\,x}}{63\,x^4}-\frac {2\,B\,b\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {16\,B\,c\,\sqrt {c\,x^2+b\,x}}{35\,x^3}-\frac {2\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b\,x^3}-\frac {2\,A\,b\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {16\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^3\,x}-\frac {2\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b\,x^2}+\frac {4\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{35\,b^2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^7,x)

[Out]

(8*A*c^3*(b*x + c*x^2)^(1/2))/(315*b^2*x^2) - (20*A*c*(b*x + c*x^2)^(1/2))/(63*x^4) - (2*B*b*(b*x + c*x^2)^(1/
2))/(7*x^4) - (16*B*c*(b*x + c*x^2)^(1/2))/(35*x^3) - (2*A*c^2*(b*x + c*x^2)^(1/2))/(105*b*x^3) - (2*A*b*(b*x
+ c*x^2)^(1/2))/(9*x^5) - (16*A*c^4*(b*x + c*x^2)^(1/2))/(315*b^3*x) - (2*B*c^2*(b*x + c*x^2)^(1/2))/(35*b*x^2
) + (4*B*c^3*(b*x + c*x^2)^(1/2))/(35*b^2*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**7,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**7, x)

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