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

Optimal. Leaf size=70 \[ -\frac {8 (b+2 c x) (b B-2 A c)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (A b-x (b B-2 A c))}{3 b^2 \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {638, 613} \begin {gather*} -\frac {8 (b+2 c x) (b B-2 A c)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (A b-x (b B-2 A c))}{3 b^2 \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(A*b - (b*B - 2*A*c)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) - (8*(b*B - 2*A*c)*(b + 2*c*x))/(3*b^4*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 638

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 72, normalized size = 1.03 \begin {gather*} -\frac {2 \left (A \left (b^3-6 b^2 c x-24 b c^2 x^2-16 c^3 x^3\right )+b B x \left (3 b^2+12 b c x+8 c^2 x^2\right )\right )}{3 b^4 (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B*x*(3*b^2 + 12*b*c*x + 8*c^2*x^2) + A*(b^3 - 6*b^2*c*x - 24*b*c^2*x^2 - 16*c^3*x^3)))/(3*b^4*(x*(b + c
*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.38, size = 90, normalized size = 1.29 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (A b^3-6 A b^2 c x-24 A b c^2 x^2-16 A c^3 x^3+3 b^3 B x+12 b^2 B c x^2+8 b B c^2 x^3\right )}{3 b^4 x^2 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b*x + c*x^2]*(A*b^3 + 3*b^3*B*x - 6*A*b^2*c*x + 12*b^2*B*c*x^2 - 24*A*b*c^2*x^2 + 8*b*B*c^2*x^3 - 16*
A*c^3*x^3))/(3*b^4*x^2*(b + c*x)^2)

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fricas [A]  time = 0.41, size = 101, normalized size = 1.44 \begin {gather*} -\frac {2 \, {\left (A b^{3} + 8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x^{3} + 12 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} x^{2} + 3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(A*b^3 + 8*(B*b*c^2 - 2*A*c^3)*x^3 + 12*(B*b^2*c - 2*A*b*c^2)*x^2 + 3*(B*b^3 - 2*A*b^2*c)*x)*sqrt(c*x^2 +
b*x)/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2)

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giac [A]  time = 0.24, size = 82, normalized size = 1.17 \begin {gather*} -\frac {2 \, {\left ({\left (4 \, x {\left (\frac {2 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x}{b^{4}} + \frac {3 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )}}{b^{4}}\right )} + \frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )}}{b^{4}}\right )} x + \frac {A}{b}\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 83, normalized size = 1.19 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-16 A \,c^{3} x^{3}+8 B b \,c^{2} x^{3}-24 A b \,c^{2} x^{2}+12 B \,b^{2} c \,x^{2}-6 A \,b^{2} c x +3 B \,b^{3} x +A \,b^{3}\right ) x}{3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(c*x+b)*x*(-16*A*c^3*x^3+8*B*b*c^2*x^3-24*A*b*c^2*x^2+12*B*b^2*c*x^2-6*A*b^2*c*x+3*B*b^3*x+A*b^3)/b^4/(c*
x^2+b*x)^(5/2)

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maxima [B]  time = 0.57, size = 130, normalized size = 1.86 \begin {gather*} \frac {2 \, B x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {16 \, B c x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {4 \, A c x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, A c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {8 \, B}{3 \, \sqrt {c x^{2} + b x} b^{2}} - \frac {2 \, A}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, A c}{3 \, \sqrt {c x^{2} + b x} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*B*x/((c*x^2 + b*x)^(3/2)*b) - 16/3*B*c*x/(sqrt(c*x^2 + b*x)*b^3) - 4/3*A*c*x/((c*x^2 + b*x)^(3/2)*b^2) + 3
2/3*A*c^2*x/(sqrt(c*x^2 + b*x)*b^4) - 8/3*B/(sqrt(c*x^2 + b*x)*b^2) - 2/3*A/((c*x^2 + b*x)^(3/2)*b) + 16/3*A*c
/(sqrt(c*x^2 + b*x)*b^3)

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mupad [B]  time = 1.15, size = 76, normalized size = 1.09 \begin {gather*} -\frac {2\,\left (3\,B\,b^3\,x+A\,b^3+12\,B\,b^2\,c\,x^2-6\,A\,b^2\,c\,x+8\,B\,b\,c^2\,x^3-24\,A\,b\,c^2\,x^2-16\,A\,c^3\,x^3\right )}{3\,b^4\,{\left (c\,x^2+b\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(b*x + c*x^2)^(5/2),x)

[Out]

-(2*(A*b^3 - 16*A*c^3*x^3 + 3*B*b^3*x - 6*A*b^2*c*x - 24*A*b*c^2*x^2 + 12*B*b^2*c*x^2 + 8*B*b*c^2*x^3))/(3*b^4
*(b*x + c*x^2)^(3/2))

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

[Out]

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

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