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

Optimal. Leaf size=166 \[ \frac{256 c^3 (b+2 c x) (3 b B-4 A c)}{63 b^7 \sqrt{b x+c x^2}}-\frac{32 c^2 (b+2 c x) (3 b B-4 A c)}{63 b^5 \left (b x+c x^2\right )^{3/2}}+\frac{4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac{2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}-\frac{2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}} \]

[Out]

(-2*A)/(9*b*x^3*(b*x + c*x^2)^(3/2)) - (2*(3*b*B - 4*A*c))/(21*b^2*x^2*(b*x + c*x^2)^(3/2)) + (4*c*(3*b*B - 4*
A*c))/(21*b^3*x*(b*x + c*x^2)^(3/2)) - (32*c^2*(3*b*B - 4*A*c)*(b + 2*c*x))/(63*b^5*(b*x + c*x^2)^(3/2)) + (25
6*c^3*(3*b*B - 4*A*c)*(b + 2*c*x))/(63*b^7*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.150052, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {792, 658, 614, 613} \[ \frac{256 c^3 (b+2 c x) (3 b B-4 A c)}{63 b^7 \sqrt{b x+c x^2}}-\frac{32 c^2 (b+2 c x) (3 b B-4 A c)}{63 b^5 \left (b x+c x^2\right )^{3/2}}+\frac{4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac{2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}-\frac{2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(9*b*x^3*(b*x + c*x^2)^(3/2)) - (2*(3*b*B - 4*A*c))/(21*b^2*x^2*(b*x + c*x^2)^(3/2)) + (4*c*(3*b*B - 4*
A*c))/(21*b^3*x*(b*x + c*x^2)^(3/2)) - (32*c^2*(3*b*B - 4*A*c)*(b + 2*c*x))/(63*b^5*(b*x + c*x^2)^(3/2)) + (25
6*c^3*(3*b*B - 4*A*c)*(b + 2*c*x))/(63*b^7*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^3 \left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}--\frac{\left (2 \left (-3 (-b B+A c)-\frac{3}{2} (-b B+2 A c)\right )\right ) \int \frac{1}{x^2 \left (b x+c x^2\right )^{5/2}} \, dx}{9 b}\\ &=-\frac{2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac{2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}-\frac{(10 c (3 b B-4 A c)) \int \frac{1}{x \left (b x+c x^2\right )^{5/2}} \, dx}{21 b^2}\\ &=-\frac{2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac{2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}+\frac{4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}+\frac{\left (16 c^2 (3 b B-4 A c)\right ) \int \frac{1}{\left (b x+c x^2\right )^{5/2}} \, dx}{21 b^3}\\ &=-\frac{2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac{2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}+\frac{4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac{32 c^2 (3 b B-4 A c) (b+2 c x)}{63 b^5 \left (b x+c x^2\right )^{3/2}}-\frac{\left (128 c^3 (3 b B-4 A c)\right ) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{63 b^5}\\ &=-\frac{2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}-\frac{2 (3 b B-4 A c)}{21 b^2 x^2 \left (b x+c x^2\right )^{3/2}}+\frac{4 c (3 b B-4 A c)}{21 b^3 x \left (b x+c x^2\right )^{3/2}}-\frac{32 c^2 (3 b B-4 A c) (b+2 c x)}{63 b^5 \left (b x+c x^2\right )^{3/2}}+\frac{256 c^3 (3 b B-4 A c) (b+2 c x)}{63 b^7 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0497187, size = 145, normalized size = 0.87 \[ \frac{6 b B x \left (-16 b^3 c^2 x^2+96 b^2 c^3 x^3+6 b^4 c x-3 b^5+384 b c^4 x^4+256 c^5 x^5\right )-2 A \left (24 b^4 c^2 x^2-64 b^3 c^3 x^3+384 b^2 c^4 x^4-12 b^5 c x+7 b^6+1536 b c^5 x^5+1024 c^6 x^6\right )}{63 b^7 x^3 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*B*x*(-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) - 2*A*(7*b^6 - 1
2*b^5*c*x + 24*b^4*c^2*x^2 - 64*b^3*c^3*x^3 + 384*b^2*c^4*x^4 + 1536*b*c^5*x^5 + 1024*c^6*x^6))/(63*b^7*x^3*(x
*(b + c*x))^(3/2))

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Maple [A]  time = 0.005, size = 158, normalized size = 1. \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 1024\,A{c}^{6}{x}^{6}-768\,Bb{c}^{5}{x}^{6}+1536\,Ab{c}^{5}{x}^{5}-1152\,B{b}^{2}{c}^{4}{x}^{5}+384\,A{b}^{2}{c}^{4}{x}^{4}-288\,B{b}^{3}{c}^{3}{x}^{4}-64\,A{b}^{3}{c}^{3}{x}^{3}+48\,B{b}^{4}{c}^{2}{x}^{3}+24\,A{b}^{4}{c}^{2}{x}^{2}-18\,B{b}^{5}c{x}^{2}-12\,A{b}^{5}cx+9\,{b}^{6}Bx+7\,A{b}^{6} \right ) }{63\,{x}^{2}{b}^{7}} \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^3/(c*x^2+b*x)^(5/2),x)

[Out]

-2/63*(c*x+b)*(1024*A*c^6*x^6-768*B*b*c^5*x^6+1536*A*b*c^5*x^5-1152*B*b^2*c^4*x^5+384*A*b^2*c^4*x^4-288*B*b^3*
c^3*x^4-64*A*b^3*c^3*x^3+48*B*b^4*c^2*x^3+24*A*b^4*c^2*x^2-18*B*b^5*c*x^2-12*A*b^5*c*x+9*B*b^6*x+7*A*b^6)/x^2/
b^7/(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^3/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00158, size = 374, normalized size = 2.25 \begin{align*} -\frac{2 \,{\left (7 \, A b^{6} - 256 \,{\left (3 \, B b c^{5} - 4 \, A c^{6}\right )} x^{6} - 384 \,{\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{5} - 96 \,{\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{4} + 16 \,{\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x^{3} - 6 \,{\left (3 \, B b^{5} c - 4 \, A b^{4} c^{2}\right )} x^{2} + 3 \,{\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} x\right )} \sqrt{c x^{2} + b x}}{63 \,{\left (b^{7} c^{2} x^{7} + 2 \, b^{8} c x^{6} + b^{9} x^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(7*A*b^6 - 256*(3*B*b*c^5 - 4*A*c^6)*x^6 - 384*(3*B*b^2*c^4 - 4*A*b*c^5)*x^5 - 96*(3*B*b^3*c^3 - 4*A*b^2
*c^4)*x^4 + 16*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x^3 - 6*(3*B*b^5*c - 4*A*b^4*c^2)*x^2 + 3*(3*B*b^6 - 4*A*b^5*c)*x)*
sqrt(c*x^2 + b*x)/(b^7*c^2*x^7 + 2*b^8*c*x^6 + b^9*x^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{x^{3} \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**3/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((A + B*x)/(x**3*(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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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