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

Optimal. Leaf size=73 \[ -\frac {2 \sqrt {x} (A c+2 b B)}{3 b c^2 \sqrt {b x+c x^2}}-\frac {2 x^{5/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {788, 648} \[ -\frac {2 \sqrt {x} (A c+2 b B)}{3 b c^2 \sqrt {b x+c x^2}}-\frac {2 x^{5/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.49 \[ -\frac {2 x^{3/2} (c (A+3 B x)+2 b B)}{3 c^2 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 56, normalized size = 0.77 \[ -\frac {2 \, {\left (3 \, B c x + 2 \, B b + A c\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (c^{4} x^{3} + 2 \, b c^{3} x^{2} + b^{2} c^{2} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*B*c*x + 2*B*b + A*c)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^4*x^3 + 2*b*c^3*x^2 + b^2*c^2*x)

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giac [A]  time = 0.19, size = 45, normalized size = 0.62 \[ -\frac {2 \, {\left (3 \, {\left (c x + b\right )} B - B b + A c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{2}} + \frac {2 \, {\left (2 \, B b + A c\right )}}{3 \, b^{\frac {3}{2}} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 38, normalized size = 0.52 \[ -\frac {2 \left (c x +b \right ) \left (3 B c x +A c +2 b B \right ) x^{\frac {5}{2}}}{3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} x^{\frac {5}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {5}{2}} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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