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

Optimal. Leaf size=59 \[ -\frac {2 \left (a e^2+c d^2\right )}{3 e^3 (d+e x)^{3/2}}+\frac {2 c \sqrt {d+e x}}{e^3}+\frac {4 c d}{e^3 \sqrt {d+e x}} \]

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Rubi [A]  time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {697} \begin {gather*} -\frac {2 \left (a e^2+c d^2\right )}{3 e^3 (d+e x)^{3/2}}+\frac {2 c \sqrt {d+e x}}{e^3}+\frac {4 c d}{e^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)/(d + e*x)^(5/2),x]

[Out]

(-2*(c*d^2 + a*e^2))/(3*e^3*(d + e*x)^(3/2)) + (4*c*d)/(e^3*Sqrt[d + e*x]) + (2*c*Sqrt[d + e*x])/e^3

Rule 697

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 44, normalized size = 0.75 \begin {gather*} \frac {2 \left (c \left (8 d^2+12 d e x+3 e^2 x^2\right )-a e^2\right )}{3 e^3 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(a*e^2) + 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 = 48, normalized size = 0.81 \begin {gather*} \frac {2 \left (-a e^2-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[(a + c*x^2)/(d + e*x)^(5/2),x]

[Out]

(2*(-(c*d^2) - a*e^2 + 6*c*d*(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 = 61, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (3 \, c e^{2} x^{2} + 12 \, c d e x + 8 \, c d^{2} - a e^{2}\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+a)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

2/3*(3*c*e^2*x^2 + 12*c*d*e*x + 8*c*d^2 - a*e^2)*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 = 48, normalized size = 0.81 \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} - a e^{2}\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+a)/(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 - a*e^2)*e^(-3)/(x*e + d)^(3/2)

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maple [A]  time = 0.05, size = 40, normalized size = 0.68 \begin {gather*} -\frac {2 \left (-3 c \,e^{2} x^{2}-12 c d e x +a \,e^{2}-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+a)/(e*x+d)^(5/2),x)

[Out]

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

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maxima [A]  time = 1.35, size = 52, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\frac {3 \, \sqrt {e x + d} c}{e^{2}} + \frac {6 \, {\left (e x + d\right )} c d - c d^{2} - a e^{2}}{{\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+a)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 44, normalized size = 0.75 \begin {gather*} \frac {6\,c\,{\left (d+e\,x\right )}^2-2\,a\,e^2-2\,c\,d^2+12\,c\,d\,\left (d+e\,x\right )}{3\,e^3\,{\left (d+e\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)/(d + e*x)^(5/2),x)

[Out]

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

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sympy [A]  time = 1.20, size = 168, normalized size = 2.85 \begin {gather*} \begin {cases} - \frac {2 a e^{2}}{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 {a x + \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+a)/(e*x+d)**(5/2),x)

[Out]

Piecewise((-2*a*e**2/(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*sqrt(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)), ((a*x + c*x**3/3)/d**(5/2), True))

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