3.332 \(\int \frac {d+e x}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=71 \[ \frac {8 (b+2 c x) (2 c d-b e)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}} \]

[Out]

-2/3*(b*d+(-b*e+2*c*d)*x)/b^2/(c*x^2+b*x)^(3/2)+8/3*(-b*e+2*c*d)*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 71, 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} \[ \frac {8 (b+2 c x) (2 c d-b e)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(b*x + c*x^2)^(5/2),x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 67, normalized size = 0.94 \[ -\frac {2 \left (b^3 (d+3 e x)-6 b^2 c x (d-2 e x)+8 b c^2 x^2 (e x-3 d)-16 c^3 d x^3\right )}{3 b^4 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(-16*c^3*d*x^3 - 6*b^2*c*x*(d - 2*e*x) + 8*b*c^2*x^2*(-3*d + e*x) + b^3*(d + 3*e*x)))/(3*b^4*(x*(b + c*x))
^(3/2))

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fricas [A]  time = 0.96, size = 104, normalized size = 1.46 \[ -\frac {2 \, {\left (b^{3} d - 8 \, {\left (2 \, c^{3} d - b c^{2} e\right )} x^{3} - 12 \, {\left (2 \, b c^{2} d - b^{2} c e\right )} x^{2} - 3 \, {\left (2 \, b^{2} c d - b^{3} e\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(b^3*d - 8*(2*c^3*d - b*c^2*e)*x^3 - 12*(2*b*c^2*d - b^2*c*e)*x^2 - 3*(2*b^2*c*d - b^3*e)*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 = 0.25, size = 89, normalized size = 1.25 \[ \frac {2 \, {\left ({\left (4 \, x {\left (\frac {2 \, {\left (2 \, c^{3} d - b c^{2} e\right )} x}{b^{4}} + \frac {3 \, {\left (2 \, b c^{2} d - b^{2} c e\right )}}{b^{4}}\right )} + \frac {3 \, {\left (2 \, b^{2} c d - b^{3} e\right )}}{b^{4}}\right )} x - \frac {d}{b}\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((4*x*(2*(2*c^3*d - b*c^2*e)*x/b^4 + 3*(2*b*c^2*d - b^2*c*e)/b^4) + 3*(2*b^2*c*d - b^3*e)/b^4)*x - d/b)/(c
*x^2 + b*x)^(3/2)

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maple [A]  time = 0.04, size = 83, normalized size = 1.17 \[ -\frac {2 \left (c x +b \right ) \left (8 b \,c^{2} e \,x^{3}-16 c^{3} d \,x^{3}+12 b^{2} c e \,x^{2}-24 b \,c^{2} d \,x^{2}+3 b^{3} e x -6 b^{2} c d x +b^{3} d \right ) x}{3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)/(c*x^2+b*x)^(5/2),x)

[Out]

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

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maxima [B]  time = 1.37, size = 130, normalized size = 1.83 \[ -\frac {4 \, c d x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, c^{2} d x}{3 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {2 \, e x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {16 \, c e x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {2 \, d}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, c d}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {8 \, e}{3 \, \sqrt {c x^{2} + b x} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*c*d*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*c^2*d*x/(sqrt(c*x^2 + b*x)*b^4) + 2/3*e*x/((c*x^2 + b*x)^(3/2)*b)
- 16/3*c*e*x/(sqrt(c*x^2 + b*x)*b^3) - 2/3*d/((c*x^2 + b*x)^(3/2)*b) + 16/3*c*d/(sqrt(c*x^2 + b*x)*b^3) - 8/3*
e/(sqrt(c*x^2 + b*x)*b^2)

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mupad [B]  time = 0.30, size = 76, normalized size = 1.07 \[ -\frac {2\,\left (3\,e\,b^3\,x+d\,b^3+12\,e\,b^2\,c\,x^2-6\,d\,b^2\,c\,x+8\,e\,b\,c^2\,x^3-24\,d\,b\,c^2\,x^2-16\,d\,c^3\,x^3\right )}{3\,b^4\,{\left (c\,x^2+b\,x\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(b*x + c*x^2)^(5/2),x)

[Out]

-(2*(b^3*d - 16*c^3*d*x^3 + 3*b^3*e*x - 6*b^2*c*d*x - 24*b*c^2*d*x^2 + 12*b^2*c*e*x^2 + 8*b*c^2*e*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 \[ \int \frac {d + e x}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((d + e*x)/(x*(b + c*x))**(5/2), x)

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