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

Optimal. Leaf size=131 \[ -\frac{128 c^2 (b+2 c x) (7 b B-10 A c)}{105 b^6 \sqrt{b x+c x^2}}+\frac{16 c (b+2 c x) (7 b B-10 A c)}{105 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}-\frac{2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-2*A)/(7*b*x^2*(b*x + c*x^2)^(3/2)) - (2*(7*b*B - 10*A*c))/(35*b^2*x*(b*x + c*x^2)^(3/2)) + (16*c*(7*b*B - 10
*A*c)*(b + 2*c*x))/(105*b^4*(b*x + c*x^2)^(3/2)) - (128*c^2*(7*b*B - 10*A*c)*(b + 2*c*x))/(105*b^6*Sqrt[b*x +
c*x^2])

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Rubi [A]  time = 0.11707, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {792, 658, 614, 613} \[ -\frac{128 c^2 (b+2 c x) (7 b B-10 A c)}{105 b^6 \sqrt{b x+c x^2}}+\frac{16 c (b+2 c x) (7 b B-10 A c)}{105 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}-\frac{2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(7*b*x^2*(b*x + c*x^2)^(3/2)) - (2*(7*b*B - 10*A*c))/(35*b^2*x*(b*x + c*x^2)^(3/2)) + (16*c*(7*b*B - 10
*A*c)*(b + 2*c*x))/(105*b^4*(b*x + c*x^2)^(3/2)) - (128*c^2*(7*b*B - 10*A*c)*(b + 2*c*x))/(105*b^6*Sqrt[b*x +
c*x^2])

Rule 792

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

Rule 658

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

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{A+B x}{x^2 \left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}+\frac{\left (2 \left (-2 (-b B+A c)-\frac{3}{2} (-b B+2 A c)\right )\right ) \int \frac{1}{x \left (b x+c x^2\right )^{5/2}} \, dx}{7 b}\\ &=-\frac{2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}-\frac{(8 c (7 b B-10 A c)) \int \frac{1}{\left (b x+c x^2\right )^{5/2}} \, dx}{35 b^2}\\ &=-\frac{2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}+\frac{16 c (7 b B-10 A c) (b+2 c x)}{105 b^4 \left (b x+c x^2\right )^{3/2}}+\frac{\left (64 c^2 (7 b B-10 A c)\right ) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{105 b^4}\\ &=-\frac{2 A}{7 b x^2 \left (b x+c x^2\right )^{3/2}}-\frac{2 (7 b B-10 A c)}{35 b^2 x \left (b x+c x^2\right )^{3/2}}+\frac{16 c (7 b B-10 A c) (b+2 c x)}{105 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{128 c^2 (7 b B-10 A c) (b+2 c x)}{105 b^6 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0383957, size = 123, normalized size = 0.94 \[ -\frac{2 \left (5 A \left (16 b^3 c^2 x^2-96 b^2 c^3 x^3-6 b^4 c x+3 b^5-384 b c^4 x^4-256 c^5 x^5\right )+7 b B x \left (48 b^2 c^2 x^2-8 b^3 c x+3 b^4+192 b c^3 x^3+128 c^4 x^4\right )\right )}{105 b^6 x^2 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(7*b*B*x*(3*b^4 - 8*b^3*c*x + 48*b^2*c^2*x^2 + 192*b*c^3*x^3 + 128*c^4*x^4) + 5*A*(3*b^5 - 6*b^4*c*x + 16*
b^3*c^2*x^2 - 96*b^2*c^3*x^3 - 384*b*c^4*x^4 - 256*c^5*x^5)))/(105*b^6*x^2*(x*(b + c*x))^(3/2))

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Maple [A]  time = 0.007, size = 134, normalized size = 1. \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -1280\,A{c}^{5}{x}^{5}+896\,Bb{c}^{4}{x}^{5}-1920\,Ab{c}^{4}{x}^{4}+1344\,B{b}^{2}{c}^{3}{x}^{4}-480\,A{b}^{2}{c}^{3}{x}^{3}+336\,B{b}^{3}{c}^{2}{x}^{3}+80\,A{b}^{3}{c}^{2}{x}^{2}-56\,B{b}^{4}c{x}^{2}-30\,A{b}^{4}cx+21\,B{b}^{5}x+15\,A{b}^{5} \right ) }{105\,x{b}^{6}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(c*x+b)*(-1280*A*c^5*x^5+896*B*b*c^4*x^5-1920*A*b*c^4*x^4+1344*B*b^2*c^3*x^4-480*A*b^2*c^3*x^3+336*B*b^
3*c^2*x^3+80*A*b^3*c^2*x^2-56*B*b^4*c*x^2-30*A*b^4*c*x+21*B*b^5*x+15*A*b^5)/x/b^6/(c*x^2+b*x)^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80818, size = 333, normalized size = 2.54 \begin{align*} -\frac{2 \,{\left (15 \, A b^{5} + 128 \,{\left (7 \, B b c^{4} - 10 \, A c^{5}\right )} x^{5} + 192 \,{\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 48 \,{\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{3} - 8 \,{\left (7 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} x^{2} + 3 \,{\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x}}{105 \,{\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*A*b^5 + 128*(7*B*b*c^4 - 10*A*c^5)*x^5 + 192*(7*B*b^2*c^3 - 10*A*b*c^4)*x^4 + 48*(7*B*b^3*c^2 - 10*
A*b^2*c^3)*x^3 - 8*(7*B*b^4*c - 10*A*b^3*c^2)*x^2 + 3*(7*B*b^5 - 10*A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^6*c^2*x^6
+ 2*b^7*c*x^5 + b^8*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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