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

Optimal. Leaf size=96 \[ \frac {16 c (b+2 c x) (5 b B-8 A c)}{15 b^5 \sqrt {b x+c x^2}}-\frac {2 (b+2 c x) (5 b B-8 A c)}{15 b^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 A}{5 b x \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {792, 614, 613} \begin {gather*} \frac {16 c (b+2 c x) (5 b B-8 A c)}{15 b^5 \sqrt {b x+c x^2}}-\frac {2 (b+2 c x) (5 b B-8 A c)}{15 b^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 A}{5 b x \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

\begin {align*} \int \frac {A+B x}{x \left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 A}{5 b x \left (b x+c x^2\right )^{3/2}}+\frac {\left (2 \left (b B-A c-\frac {3}{2} (-b B+2 A c)\right )\right ) \int \frac {1}{\left (b x+c x^2\right )^{5/2}} \, dx}{5 b}\\ &=-\frac {2 A}{5 b x \left (b x+c x^2\right )^{3/2}}-\frac {2 (5 b B-8 A c) (b+2 c x)}{15 b^3 \left (b x+c x^2\right )^{3/2}}-\frac {(8 c (5 b B-8 A c)) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{15 b^3}\\ &=-\frac {2 A}{5 b x \left (b x+c x^2\right )^{3/2}}-\frac {2 (5 b B-8 A c) (b+2 c x)}{15 b^3 \left (b x+c x^2\right )^{3/2}}+\frac {16 c (5 b B-8 A c) (b+2 c x)}{15 b^5 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 98, normalized size = 1.02 \begin {gather*} -\frac {2 \left (A \left (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\right )+5 b B x \left (b^3-6 b^2 c x-24 b c^2 x^2-16 c^3 x^3\right )\right )}{15 b^5 x (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*b*B*x*(b^3 - 6*b^2*c*x - 24*b*c^2*x^2 - 16*c^3*x^3) + A*(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)))/(15*b^5*x*(x*(b + c*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.42, size = 115, normalized size = 1.20 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-3 A b^4+8 A b^3 c x-48 A b^2 c^2 x^2-192 A b c^3 x^3-128 A c^4 x^4-5 b^4 B x+30 b^3 B c x^2+120 b^2 B c^2 x^3+80 b B c^3 x^4\right )}{15 b^5 x^3 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + c*x^2]*(-3*A*b^4 - 5*b^4*B*x + 8*A*b^3*c*x + 30*b^3*B*c*x^2 - 48*A*b^2*c^2*x^2 + 120*b^2*B*c^2*x
^3 - 192*A*b*c^3*x^3 + 80*b*B*c^3*x^4 - 128*A*c^4*x^4))/(15*b^5*x^3*(b + c*x)^2)

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fricas [A]  time = 0.41, size = 128, normalized size = 1.33 \begin {gather*} -\frac {2 \, {\left (3 \, A b^{4} - 16 \, {\left (5 \, B b c^{3} - 8 \, A c^{4}\right )} x^{4} - 24 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{3} - 6 \, {\left (5 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{2} + {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*A*b^4 - 16*(5*B*b*c^3 - 8*A*c^4)*x^4 - 24*(5*B*b^2*c^2 - 8*A*b*c^3)*x^3 - 6*(5*B*b^3*c - 8*A*b^2*c^2)
*x^2 + (5*B*b^4 - 8*A*b^3*c)*x)*sqrt(c*x^2 + b*x)/(b^5*c^2*x^5 + 2*b^6*c*x^4 + b^7*x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 107, normalized size = 1.11 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (128 A \,c^{4} x^{4}-80 B b \,c^{3} x^{4}+192 A b \,c^{3} x^{3}-120 B \,b^{2} c^{2} x^{3}+48 A \,b^{2} c^{2} x^{2}-30 B \,b^{3} c \,x^{2}-8 A \,b^{3} c x +5 b^{4} B x +3 A \,b^{4}\right )}{15 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(c*x+b)*(128*A*c^4*x^4-80*B*b*c^3*x^4+192*A*b*c^3*x^3-120*B*b^2*c^2*x^3+48*A*b^2*c^2*x^2-30*B*b^3*c*x^2-
8*A*b^3*c*x+5*B*b^4*x+3*A*b^4)/b^5/(c*x^2+b*x)^(5/2)

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maxima [B]  time = 0.71, size = 176, normalized size = 1.83 \begin {gather*} -\frac {4 \, B c x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, B c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {32 \, A c^{2} x}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} - \frac {256 \, A c^{3} x}{15 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {2 \, B}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, B c}{3 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {16 \, A c}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} - \frac {128 \, A c^{2}}{15 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {2 \, A}{5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*B*c*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*B*c^2*x/(sqrt(c*x^2 + b*x)*b^4) + 32/15*A*c^2*x/((c*x^2 + b*x)^(3/
2)*b^3) - 256/15*A*c^3*x/(sqrt(c*x^2 + b*x)*b^5) - 2/3*B/((c*x^2 + b*x)^(3/2)*b) + 16/3*B*c/(sqrt(c*x^2 + b*x)
*b^3) + 16/15*A*c/((c*x^2 + b*x)^(3/2)*b^2) - 128/15*A*c^2/(sqrt(c*x^2 + b*x)*b^4) - 2/5*A/((c*x^2 + b*x)^(3/2
)*b*x)

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mupad [B]  time = 1.25, size = 111, normalized size = 1.16 \begin {gather*} -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (5\,B\,b^4\,x+3\,A\,b^4-30\,B\,b^3\,c\,x^2-8\,A\,b^3\,c\,x-120\,B\,b^2\,c^2\,x^3+48\,A\,b^2\,c^2\,x^2-80\,B\,b\,c^3\,x^4+192\,A\,b\,c^3\,x^3+128\,A\,c^4\,x^4\right )}{15\,b^5\,x^3\,{\left (b+c\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(b*x + c*x^2)^(1/2)*(3*A*b^4 + 128*A*c^4*x^4 + 5*B*b^4*x - 8*A*b^3*c*x + 192*A*b*c^3*x^3 - 30*B*b^3*c*x^2
- 80*B*b*c^3*x^4 + 48*A*b^2*c^2*x^2 - 120*B*b^2*c^2*x^3))/(15*b^5*x^3*(b + c*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x \left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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