3.242 \(\int \frac {x^{11/2} (A+B x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=180 \[ \frac {32 b^2 \sqrt {x} (8 b B-5 A c)}{15 c^5 \sqrt {b x+c x^2}}+\frac {16 b x^{3/2} (8 b B-5 A c)}{15 c^4 \sqrt {b x+c x^2}}-\frac {4 x^{5/2} (8 b B-5 A c)}{15 c^3 \sqrt {b x+c x^2}}+\frac {2 x^{7/2} (8 b B-5 A c)}{15 b c^2 \sqrt {b x+c x^2}}-\frac {2 x^{11/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

-2/3*(-A*c+B*b)*x^(11/2)/b/c/(c*x^2+b*x)^(3/2)+16/15*b*(-5*A*c+8*B*b)*x^(3/2)/c^4/(c*x^2+b*x)^(1/2)-4/15*(-5*A
*c+8*B*b)*x^(5/2)/c^3/(c*x^2+b*x)^(1/2)+2/15*(-5*A*c+8*B*b)*x^(7/2)/b/c^2/(c*x^2+b*x)^(1/2)+32/15*b^2*(-5*A*c+
8*B*b)*x^(1/2)/c^5/(c*x^2+b*x)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {788, 656, 648} \[ \frac {32 b^2 \sqrt {x} (8 b B-5 A c)}{15 c^5 \sqrt {b x+c x^2}}+\frac {2 x^{7/2} (8 b B-5 A c)}{15 b c^2 \sqrt {b x+c x^2}}-\frac {4 x^{5/2} (8 b B-5 A c)}{15 c^3 \sqrt {b x+c x^2}}+\frac {16 b x^{3/2} (8 b B-5 A c)}{15 c^4 \sqrt {b x+c x^2}}-\frac {2 x^{11/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b*B - A*c)*x^(11/2))/(3*b*c*(b*x + c*x^2)^(3/2)) + (32*b^2*(8*b*B - 5*A*c)*Sqrt[x])/(15*c^5*Sqrt[b*x + c*
x^2]) + (16*b*(8*b*B - 5*A*c)*x^(3/2))/(15*c^4*Sqrt[b*x + c*x^2]) - (4*(8*b*B - 5*A*c)*x^(5/2))/(15*c^3*Sqrt[b
*x + c*x^2]) + (2*(8*b*B - 5*A*c)*x^(7/2))/(15*b*c^2*Sqrt[b*x + c*x^2])

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), 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 + p, 0]

Rule 656

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

Rule 788

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

Rubi steps

\begin {align*} \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (b B-A c) x^{11/2}}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {1}{3} \left (\frac {5 A}{b}-\frac {8 B}{c}\right ) \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac {2 (b B-A c) x^{11/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {2 (8 b B-5 A c) x^{7/2}}{15 b c^2 \sqrt {b x+c x^2}}-\frac {(2 (8 b B-5 A c)) \int \frac {x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 c^2}\\ &=-\frac {2 (b B-A c) x^{11/2}}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {4 (8 b B-5 A c) x^{5/2}}{15 c^3 \sqrt {b x+c x^2}}+\frac {2 (8 b B-5 A c) x^{7/2}}{15 b c^2 \sqrt {b x+c x^2}}+\frac {(8 b (8 b B-5 A c)) \int \frac {x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{15 c^3}\\ &=-\frac {2 (b B-A c) x^{11/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {16 b (8 b B-5 A c) x^{3/2}}{15 c^4 \sqrt {b x+c x^2}}-\frac {4 (8 b B-5 A c) x^{5/2}}{15 c^3 \sqrt {b x+c x^2}}+\frac {2 (8 b B-5 A c) x^{7/2}}{15 b c^2 \sqrt {b x+c x^2}}-\frac {\left (16 b^2 (8 b B-5 A c)\right ) \int \frac {x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{15 c^4}\\ &=-\frac {2 (b B-A c) x^{11/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {32 b^2 (8 b B-5 A c) \sqrt {x}}{15 c^5 \sqrt {b x+c x^2}}+\frac {16 b (8 b B-5 A c) x^{3/2}}{15 c^4 \sqrt {b x+c x^2}}-\frac {4 (8 b B-5 A c) x^{5/2}}{15 c^3 \sqrt {b x+c x^2}}+\frac {2 (8 b B-5 A c) x^{7/2}}{15 b c^2 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 93, normalized size = 0.52 \[ \frac {2 x^{3/2} \left (b^3 (192 B c x-80 A c)+24 b^2 c^2 x (2 B x-5 A)-2 b c^3 x^2 (15 A+4 B x)+c^4 x^3 (5 A+3 B x)+128 b^4 B\right )}{15 c^5 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(2*x^(3/2)*(128*b^4*B + 24*b^2*c^2*x*(-5*A + 2*B*x) + c^4*x^3*(5*A + 3*B*x) - 2*b*c^3*x^2*(15*A + 4*B*x) + b^3
*(-80*A*c + 192*B*c*x)))/(15*c^5*(x*(b + c*x))^(3/2))

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fricas [A]  time = 0.69, size = 127, normalized size = 0.71 \[ \frac {2 \, {\left (3 \, B c^{4} x^{4} + 128 \, B b^{4} - 80 \, A b^{3} c - {\left (8 \, B b c^{3} - 5 \, A c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 24 \, {\left (8 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{15 \, {\left (c^{7} x^{3} + 2 \, b c^{6} x^{2} + b^{2} c^{5} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*B*c^4*x^4 + 128*B*b^4 - 80*A*b^3*c - (8*B*b*c^3 - 5*A*c^4)*x^3 + 6*(8*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 24*
(8*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^7*x^3 + 2*b*c^6*x^2 + b^2*c^5*x)

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giac [A]  time = 0.21, size = 147, normalized size = 0.82 \[ -\frac {32 \, {\left (8 \, B b^{3} - 5 \, A b^{2} c\right )}}{15 \, \sqrt {b} c^{5}} + \frac {2 \, {\left (12 \, {\left (c x + b\right )} B b^{3} - B b^{4} - 9 \, {\left (c x + b\right )} A b^{2} c + A b^{3} c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{5}} + \frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} B c^{20} - 20 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c^{20} + 90 \, \sqrt {c x + b} B b^{2} c^{20} + 5 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{21} - 45 \, \sqrt {c x + b} A b c^{21}\right )}}{15 \, c^{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

-32/15*(8*B*b^3 - 5*A*b^2*c)/(sqrt(b)*c^5) + 2/3*(12*(c*x + b)*B*b^3 - B*b^4 - 9*(c*x + b)*A*b^2*c + A*b^3*c)/
((c*x + b)^(3/2)*c^5) + 2/15*(3*(c*x + b)^(5/2)*B*c^20 - 20*(c*x + b)^(3/2)*B*b*c^20 + 90*sqrt(c*x + b)*B*b^2*
c^20 + 5*(c*x + b)^(3/2)*A*c^21 - 45*sqrt(c*x + b)*A*b*c^21)/c^25

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maple [A]  time = 0.05, size = 107, normalized size = 0.59 \[ -\frac {2 \left (c x +b \right ) \left (-3 B \,x^{4} c^{4}-5 A \,c^{4} x^{3}+8 B b \,c^{3} x^{3}+30 A b \,c^{3} x^{2}-48 B \,b^{2} c^{2} x^{2}+120 A \,b^{2} c^{2} x -192 B \,b^{3} c x +80 A \,b^{3} c -128 b^{4} B \right ) x^{\frac {5}{2}}}{15 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x)

[Out]

-2/15*(c*x+b)*(-3*B*c^4*x^4-5*A*c^4*x^3+8*B*b*c^3*x^3+30*A*b*c^3*x^2-48*B*b^2*c^2*x^2+120*A*b^2*c^2*x-192*B*b^
3*c*x+80*A*b^3*c-128*B*b^4)*x^(5/2)/c^5/(c*x^2+b*x)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left ({\left (3 \, B c^{3} x^{2} + B b c^{2} x - 2 \, B b^{2} c\right )} x^{4} - {\left (6 \, B b^{3} + {\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} + {\left (12 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} x^{3}\right )} \sqrt {c x + b}}{15 \, {\left (c^{6} x^{4} + 3 \, b c^{5} x^{3} + 3 \, b^{2} c^{4} x^{2} + b^{3} c^{3} x\right )}} - \int -\frac {2 \, {\left (4 \, B b^{4} + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + {\left (13 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x + b} x^{3}}{5 \, {\left (c^{7} x^{6} + 4 \, b c^{6} x^{5} + 6 \, b^{2} c^{5} x^{4} + 4 \, b^{3} c^{4} x^{3} + b^{4} c^{3} x^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*B*c^3*x^2 + B*b*c^2*x - 2*B*b^2*c)*x^4 - (6*B*b^3 + (6*B*b*c^2 - 5*A*c^3)*x^2 + (12*B*b^2*c - 5*A*b*c
^2)*x)*x^3)*sqrt(c*x + b)/(c^6*x^4 + 3*b*c^5*x^3 + 3*b^2*c^4*x^2 + b^3*c^3*x) - integrate(-2/5*(4*B*b^4 + (9*B
*b^2*c^2 - 5*A*b*c^3)*x^2 + (13*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(c*x + b)*x^3/(c^7*x^6 + 4*b*c^6*x^5 + 6*b^2*c^5
*x^4 + 4*b^3*c^4*x^3 + b^4*c^3*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{11/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x)

[Out]

int((x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(11/2)*(B*x+A)/(c*x**2+b*x)**(5/2),x)

[Out]

Timed out

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