3.244 \(\int \frac{x^{7/2} (A+B x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=108 \[ \frac{2 x^{3/2} (4 b B-A c)}{3 b c^2 \sqrt{b x+c x^2}}+\frac{4 \sqrt{x} (4 b B-A c)}{3 c^3 \sqrt{b x+c x^2}}-\frac{2 x^{7/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(b*B - A*c)*x^(7/2))/(3*b*c*(b*x + c*x^2)^(3/2)) + (4*(4*b*B - A*c)*Sqrt[x])/(3*c^3*Sqrt[b*x + c*x^2]) + (
2*(4*b*B - A*c)*x^(3/2))/(3*b*c^2*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.0896188, antiderivative size = 108, 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} \[ \frac{2 x^{3/2} (4 b B-A c)}{3 b c^2 \sqrt{b x+c x^2}}+\frac{4 \sqrt{x} (4 b B-A c)}{3 c^3 \sqrt{b x+c x^2}}-\frac{2 x^{7/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b*B - A*c)*x^(7/2))/(3*b*c*(b*x + c*x^2)^(3/2)) + (4*(4*b*B - A*c)*Sqrt[x])/(3*c^3*Sqrt[b*x + c*x^2]) + (
2*(4*b*B - A*c)*x^(3/2))/(3*b*c^2*Sqrt[b*x + c*x^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^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (b B-A c) x^{7/2}}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac{\left (2 \left (\frac{7}{2} (-b B+A c)-\frac{3}{2} (-b B+2 A c)\right )\right ) \int \frac{x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b c}\\ &=-\frac{2 (b B-A c) x^{7/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac{2 (4 b B-A c) x^{3/2}}{3 b c^2 \sqrt{b x+c x^2}}-\frac{(2 (4 b B-A c)) \int \frac{x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=-\frac{2 (b B-A c) x^{7/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac{4 (4 b B-A c) \sqrt{x}}{3 c^3 \sqrt{b x+c x^2}}+\frac{2 (4 b B-A c) x^{3/2}}{3 b c^2 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.037193, size = 53, normalized size = 0.49 \[ \frac{2 x^{3/2} \left (-2 b c (A-6 B x)+3 c^2 x (B x-A)+8 b^2 B\right )}{3 c^3 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.004, size = 59, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -3\,B{c}^{2}{x}^{2}+3\,A{c}^{2}x-12\,Bbcx+2\,Abc-8\,{b}^{2}B \right ) }{3\,{c}^{3}}{x}^{{\frac{5}{2}}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x)

[Out]

-2/3*(c*x+b)*(-3*B*c^2*x^2+3*A*c^2*x-12*B*b*c*x+2*A*b*c-8*B*b^2)*x^(5/2)/c^3/(c*x^2+b*x)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )} x^{\frac{7}{2}}}{{\left (c x^{2} + b x\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*x^(7/2)/(c*x^2 + b*x)^(5/2), x)

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Fricas [A]  time = 1.86424, size = 169, normalized size = 1.56 \begin{align*} \frac{2 \,{\left (3 \, B c^{2} x^{2} + 8 \, B b^{2} - 2 \, A b c + 3 \,{\left (4 \, B b c - A c^{2}\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{3 \,{\left (c^{5} x^{3} + 2 \, b c^{4} x^{2} + b^{2} c^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.2064, size = 96, normalized size = 0.89 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{c x + b} B + \frac{6 \,{\left (c x + b\right )} B b - B b^{2} - 3 \,{\left (c x + b\right )} A c + A b c}{{\left (c x + b\right )}^{\frac{3}{2}}}\right )}}{3 \, c^{3}} - \frac{4 \,{\left (4 \, B b - A c\right )}}{3 \, \sqrt{b} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

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