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

Optimal. Leaf size=64 \[ \frac {2 (2 c d-b e)}{e^3 \sqrt {d+e x}}-\frac {2 d (c d-b e)}{3 e^3 (d+e x)^{3/2}}+\frac {2 c \sqrt {d+e x}}{e^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {698} \begin {gather*} \frac {2 (2 c d-b e)}{e^3 \sqrt {d+e x}}-\frac {2 d (c d-b e)}{3 e^3 (d+e x)^{3/2}}+\frac {2 c \sqrt {d+e x}}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)/(d + e*x)^(5/2),x]

[Out]

(-2*d*(c*d - b*e))/(3*e^3*(d + e*x)^(3/2)) + (2*(2*c*d - b*e))/(e^3*Sqrt[d + e*x]) + (2*c*Sqrt[d + e*x])/e^3

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 50, normalized size = 0.78 \begin {gather*} \frac {2 \left (c \left (8 d^2+12 d e x+3 e^2 x^2\right )-b e (2 d+3 e x)\right )}{3 e^3 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(b*e*(2*d + 3*e*x)) + c*(8*d^2 + 12*d*e*x + 3*e^2*x^2)))/(3*e^3*(d + e*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.04, size = 55, normalized size = 0.86 \begin {gather*} \frac {2 \left (-3 b e (d+e x)+b d e-c d^2+6 c d (d+e x)+3 c (d+e x)^2\right )}{3 e^3 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 69, normalized size = 1.08 \begin {gather*} \frac {2 \, {\left (3 \, c e^{2} x^{2} + 8 \, c d^{2} - 2 \, b d e + 3 \, {\left (4 \, c d e - b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*c*e^2*x^2 + 8*c*d^2 - 2*b*d*e + 3*(4*c*d*e - b*e^2)*x)*sqrt(e*x + d)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)

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giac [A]  time = 0.18, size = 59, normalized size = 0.92 \begin {gather*} 2 \, \sqrt {x e + d} c e^{\left (-3\right )} + \frac {2 \, {\left (6 \, {\left (x e + d\right )} c d - c d^{2} - 3 \, {\left (x e + d\right )} b e + b d e\right )} e^{\left (-3\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x*e + d)*c*e^(-3) + 2/3*(6*(x*e + d)*c*d - c*d^2 - 3*(x*e + d)*b*e + b*d*e)*e^(-3)/(x*e + d)^(3/2)

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maple [A]  time = 0.04, size = 47, normalized size = 0.73 \begin {gather*} -\frac {2 \left (-3 c \,e^{2} x^{2}+3 b \,e^{2} x -12 c d e x +2 b d e -8 c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-3*c*e^2*x^2+3*b*e^2*x-12*c*d*e*x+2*b*d*e-8*c*d^2)/(e*x+d)^(3/2)/e^3

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maxima [A]  time = 1.30, size = 58, normalized size = 0.91 \begin {gather*} \frac {2 \, {\left (\frac {3 \, \sqrt {e x + d} c}{e^{2}} - \frac {c d^{2} - b d e - 3 \, {\left (2 \, c d - b e\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{2}}\right )}}{3 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 52, normalized size = 0.81 \begin {gather*} \frac {6\,c\,{\left (d+e\,x\right )}^2-2\,c\,d^2-6\,b\,e\,\left (d+e\,x\right )+12\,c\,d\,\left (d+e\,x\right )+2\,b\,d\,e}{3\,e^3\,{\left (d+e\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)/(d + e*x)^(5/2),x)

[Out]

(6*c*(d + e*x)^2 - 2*c*d^2 - 6*b*e*(d + e*x) + 12*c*d*(d + e*x) + 2*b*d*e)/(3*e^3*(d + e*x)^(3/2))

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sympy [A]  time = 1.35, size = 211, normalized size = 3.30 \begin {gather*} \begin {cases} - \frac {4 b d e}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} - \frac {6 b e^{2} x}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {16 c d^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {24 c d e x}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {6 c e^{2} x^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {\frac {b x^{2}}{2} + \frac {c x^{3}}{3}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-4*b*d*e/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) - 6*b*e**2*x/(3*d*e**3*sqrt(d + e*x) + 3
*e**4*x*sqrt(d + e*x)) + 16*c*d**2/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) + 24*c*d*e*x/(3*d*e**3*sq
rt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) + 6*c*e**2*x**2/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)), Ne(e,
0)), ((b*x**2/2 + c*x**3/3)/d**(5/2), True))

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