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

Optimal. Leaf size=141 \[ \frac{2 x^{3/2} \sqrt{b x+c x^2} (6 b B-5 A c)}{5 b c^2}-\frac{8 \sqrt{x} \sqrt{b x+c x^2} (6 b B-5 A c)}{15 c^3}+\frac{16 b \sqrt{b x+c x^2} (6 b B-5 A c)}{15 c^4 \sqrt{x}}-\frac{2 x^{7/2} (b B-A c)}{b c \sqrt{b x+c x^2}} \]

[Out]

(-2*(b*B - A*c)*x^(7/2))/(b*c*Sqrt[b*x + c*x^2]) + (16*b*(6*b*B - 5*A*c)*Sqrt[b*x + c*x^2])/(15*c^4*Sqrt[x]) -
(8*(6*b*B - 5*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2])/(15*c^3) + (2*(6*b*B - 5*A*c)*x^(3/2)*Sqrt[b*x + c*x^2])/(5*b*c
^2)

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Rubi [A]  time = 0.11096, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {788, 656, 648} \[ \frac{2 x^{3/2} \sqrt{b x+c x^2} (6 b B-5 A c)}{5 b c^2}-\frac{8 \sqrt{x} \sqrt{b x+c x^2} (6 b B-5 A c)}{15 c^3}+\frac{16 b \sqrt{b x+c x^2} (6 b B-5 A c)}{15 c^4 \sqrt{x}}-\frac{2 x^{7/2} (b B-A c)}{b c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - A*c)*x^(7/2))/(b*c*Sqrt[b*x + c*x^2]) + (16*b*(6*b*B - 5*A*c)*Sqrt[b*x + c*x^2])/(15*c^4*Sqrt[x]) -
(8*(6*b*B - 5*A*c)*Sqrt[x]*Sqrt[b*x + c*x^2])/(15*c^3) + (2*(6*b*B - 5*A*c)*x^(3/2)*Sqrt[b*x + c*x^2])/(5*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^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (b B-A c) x^{7/2}}{b c \sqrt{b x+c x^2}}-\left (\frac{5 A}{b}-\frac{6 B}{c}\right ) \int \frac{x^{5/2}}{\sqrt{b x+c x^2}} \, dx\\ &=-\frac{2 (b B-A c) x^{7/2}}{b c \sqrt{b x+c x^2}}+\frac{2 (6 b B-5 A c) x^{3/2} \sqrt{b x+c x^2}}{5 b c^2}-\frac{(4 (6 b B-5 A c)) \int \frac{x^{3/2}}{\sqrt{b x+c x^2}} \, dx}{5 c^2}\\ &=-\frac{2 (b B-A c) x^{7/2}}{b c \sqrt{b x+c x^2}}-\frac{8 (6 b B-5 A c) \sqrt{x} \sqrt{b x+c x^2}}{15 c^3}+\frac{2 (6 b B-5 A c) x^{3/2} \sqrt{b x+c x^2}}{5 b c^2}+\frac{(8 b (6 b B-5 A c)) \int \frac{\sqrt{x}}{\sqrt{b x+c x^2}} \, dx}{15 c^3}\\ &=-\frac{2 (b B-A c) x^{7/2}}{b c \sqrt{b x+c x^2}}+\frac{16 b (6 b B-5 A c) \sqrt{b x+c x^2}}{15 c^4 \sqrt{x}}-\frac{8 (6 b B-5 A c) \sqrt{x} \sqrt{b x+c x^2}}{15 c^3}+\frac{2 (6 b B-5 A c) x^{3/2} \sqrt{b x+c x^2}}{5 b c^2}\\ \end{align*}

Mathematica [A]  time = 0.0500794, size = 74, normalized size = 0.52 \[ \frac{2 \sqrt{x} \left (-8 b^2 c (5 A-3 B x)-2 b c^2 x (10 A+3 B x)+c^3 x^2 (5 A+3 B x)+48 b^3 B\right )}{15 c^4 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(48*b^3*B - 8*b^2*c*(5*A - 3*B*x) + c^3*x^2*(5*A + 3*B*x) - 2*b*c^2*x*(10*A + 3*B*x)))/(15*c^4*Sqrt
[x*(b + c*x)])

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Maple [A]  time = 0.006, size = 83, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -3\,B{c}^{3}{x}^{3}-5\,A{x}^{2}{c}^{3}+6\,B{x}^{2}b{c}^{2}+20\,Ab{c}^{2}x-24\,B{b}^{2}cx+40\,A{b}^{2}c-48\,{b}^{3}B \right ) }{15\,{c}^{4}}{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^(7/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x)

[Out]

-2/15*(c*x+b)*(-3*B*c^3*x^3-5*A*c^3*x^2+6*B*b*c^2*x^2+20*A*b*c^2*x-24*B*b^2*c*x+40*A*b^2*c-48*B*b^3)*x^(3/2)/c
^4/(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 ({\left (3 \, B c^{3} x^{2} + B b c^{2} x - 2 \, B b^{2} c\right )} x^{3} -{\left (4 \, B b^{3} +{\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} +{\left (8 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} x^{2}\right )} \sqrt{c x + b}}{15 \,{\left (c^{5} x^{3} + 2 \, b c^{4} x^{2} + b^{2} c^{3} x\right )}} - \int -\frac{4 \,{\left (2 \, B b^{4} +{\left (7 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} +{\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt{c x + b} x^{2}}{15 \,{\left (c^{6} x^{5} + 3 \, b c^{5} x^{4} + 3 \, b^{2} c^{4} x^{3} + b^{3} c^{3} x^{2}\right )}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

2/15*((3*B*c^3*x^2 + B*b*c^2*x - 2*B*b^2*c)*x^3 - (4*B*b^3 + (4*B*b*c^2 - 5*A*c^3)*x^2 + (8*B*b^2*c - 5*A*b*c^
2)*x)*x^2)*sqrt(c*x + b)/(c^5*x^3 + 2*b*c^4*x^2 + b^2*c^3*x) - integrate(-4/15*(2*B*b^4 + (7*B*b^2*c^2 - 5*A*b
*c^3)*x^2 + (9*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(c*x + b)*x^2/(c^6*x^5 + 3*b*c^5*x^4 + 3*b^2*c^4*x^3 + b^3*c^3*x^
2), x)

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Fricas [A]  time = 1.93087, size = 200, normalized size = 1.42 \begin{align*} \frac{2 \,{\left (3 \, B c^{3} x^{3} + 48 \, B b^{3} - 40 \, A b^{2} c -{\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} + 4 \,{\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{15 \,{\left (c^{5} x^{2} + b c^{4} 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)^(3/2),x, algorithm="fricas")

[Out]

2/15*(3*B*c^3*x^3 + 48*B*b^3 - 40*A*b^2*c - (6*B*b*c^2 - 5*A*c^3)*x^2 + 4*(6*B*b^2*c - 5*A*b*c^2)*x)*sqrt(c*x^
2 + b*x)*sqrt(x)/(c^5*x^2 + b*c^4*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)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.16386, size = 146, normalized size = 1.04 \begin{align*} \frac{2 \,{\left (3 \,{\left (c x + b\right )}^{\frac{5}{2}} B - 15 \,{\left (c x + b\right )}^{\frac{3}{2}} B b + 45 \, \sqrt{c x + b} B b^{2} + 5 \,{\left (c x + b\right )}^{\frac{3}{2}} A c - 30 \, \sqrt{c x + b} A b c + \frac{15 \,{\left (B b^{3} - A b^{2} c\right )}}{\sqrt{c x + b}}\right )}}{15 \, c^{4}} - \frac{16 \,{\left (6 \, B b^{3} - 5 \, A b^{2} c\right )}}{15 \, \sqrt{b} c^{4}} \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)^(3/2),x, algorithm="giac")

[Out]

2/15*(3*(c*x + b)^(5/2)*B - 15*(c*x + b)^(3/2)*B*b + 45*sqrt(c*x + b)*B*b^2 + 5*(c*x + b)^(3/2)*A*c - 30*sqrt(
c*x + b)*A*b*c + 15*(B*b^3 - A*b^2*c)/sqrt(c*x + b))/c^4 - 16/15*(6*B*b^3 - 5*A*b^2*c)/(sqrt(b)*c^4)