3.235 \(\int \frac{x^{5/2} (A+B x)}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=106 \[ \frac{2 \sqrt{x} \sqrt{b x+c x^2} (4 b B-3 A c)}{3 b c^2}-\frac{4 \sqrt{b x+c x^2} (4 b B-3 A c)}{3 c^3 \sqrt{x}}-\frac{2 x^{5/2} (b B-A c)}{b c \sqrt{b x+c x^2}} \]

[Out]

(-2*(b*B - A*c)*x^(5/2))/(b*c*Sqrt[b*x + c*x^2]) - (4*(4*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(3*c^3*Sqrt[x]) + (2*
(4*b*B - 3*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2])/(3*b*c^2)

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Rubi [A]  time = 0.080369, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {788, 656, 648} \[ \frac{2 \sqrt{x} \sqrt{b x+c x^2} (4 b B-3 A c)}{3 b c^2}-\frac{4 \sqrt{b x+c x^2} (4 b B-3 A c)}{3 c^3 \sqrt{x}}-\frac{2 x^{5/2} (b B-A c)}{b c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*(b*B - A*c)*x^(5/2))/(b*c*Sqrt[b*x + c*x^2]) - (4*(4*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(3*c^3*Sqrt[x]) + (2*
(4*b*B - 3*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2])/(3*b*c^2)

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0344858, size = 54, normalized size = 0.51 \[ \frac{2 \sqrt{x} \left (b (6 A c-4 B c x)+c^2 x (3 A+B x)-8 b^2 B\right )}{3 c^3 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[x]*(-8*b^2*B + c^2*x*(3*A + B*x) + b*(6*A*c - 4*B*c*x)))/(3*c^3*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.005, size = 58, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( B{c}^{2}{x}^{2}+3\,A{c}^{2}x-4\,Bbcx+6\,Abc-8\,{b}^{2}B \right ) }{3\,{c}^{3}}{x}^{{\frac{3}{2}}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(c*x+b)*(B*c^2*x^2+3*A*c^2*x-4*B*b*c*x+6*A*b*c-8*B*b^2)*x^(3/2)/c^3/(c*x^2+b*x)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (B c x + B b\right )} \sqrt{c x + b} x^{2}}{3 \,{\left (c^{3} x^{2} + 2 \, b c^{2} x + b^{2} c\right )}} + \int \frac{{\left (3 \, A b c x^{2} -{\left (4 \, B b^{2} +{\left (4 \, B b c - 3 \, A c^{2}\right )} x\right )} x^{2}\right )} \sqrt{c x + b}}{3 \,{\left (c^{4} x^{4} + 3 \, b c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + b^{3} c x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/3*(B*c*x + B*b)*sqrt(c*x + b)*x^2/(c^3*x^2 + 2*b*c^2*x + b^2*c) + integrate(1/3*(3*A*b*c*x^2 - (4*B*b^2 + (4
*B*b*c - 3*A*c^2)*x)*x^2)*sqrt(c*x + b)/(c^4*x^4 + 3*b*c^3*x^3 + 3*b^2*c^2*x^2 + b^3*c*x), x)

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Fricas [A]  time = 1.98478, size = 144, normalized size = 1.36 \begin{align*} \frac{2 \,{\left (B c^{2} x^{2} - 8 \, B b^{2} + 6 \, A b c -{\left (4 \, B b c - 3 \, A c^{2}\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{3 \,{\left (c^{4} x^{2} + b c^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/3*(B*c^2*x^2 - 8*B*b^2 + 6*A*b*c - (4*B*b*c - 3*A*c^2)*x)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^4*x^2 + b*c^3*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)/(c*x**2+b*x)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14395, size = 105, normalized size = 0.99 \begin{align*} \frac{2 \,{\left ({\left (c x + b\right )}^{\frac{3}{2}} B - 6 \, \sqrt{c x + b} B b + 3 \, \sqrt{c x + b} A c - \frac{3 \,{\left (B b^{2} - A b c\right )}}{\sqrt{c x + b}}\right )}}{3 \, c^{3}} + \frac{4 \,{\left (4 \, B b^{2} - 3 \, A b c\right )}}{3 \, \sqrt{b} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

2/3*((c*x + b)^(3/2)*B - 6*sqrt(c*x + b)*B*b + 3*sqrt(c*x + b)*A*c - 3*(B*b^2 - A*b*c)/sqrt(c*x + b))/c^3 + 4/
3*(4*B*b^2 - 3*A*b*c)/(sqrt(b)*c^3)