3.42 \(\int \frac{1}{(c+d (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=23 \[ -\frac{2}{3 b d (d (a+b x)+c)^{3/2}} \]

[Out]

-2/(3*b*d*(c + d*(a + b*x))^(3/2))

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Rubi [A]  time = 0.0108823, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {33, 32} \[ -\frac{2}{3 b d (d (a+b x)+c)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*(a + b*x))^(-5/2),x]

[Out]

-2/(3*b*d*(c + d*(a + b*x))^(3/2))

Rule 33

Int[((a_.) + (b_.)*(u_))^(m_), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(a + b*x)^m, x], x, u], x]
/; FreeQ[{a, b, m}, x] && LinearQ[u, x] && NeQ[u, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{(c+d (a+b x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{(c+d x)^{5/2}} \, dx,x,a+b x\right )}{b}\\ &=-\frac{2}{3 b d (c+d (a+b x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0304856, size = 23, normalized size = 1. \[ -\frac{2}{3 b d (d (a+b x)+c)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*(a + b*x))^(-5/2),x]

[Out]

-2/(3*b*d*(c + d*(a + b*x))^(3/2))

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Maple [A]  time = 0.002, size = 20, normalized size = 0.9 \begin{align*} -{\frac{2}{3\,bd} \left ( bdx+ad+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+d*(b*x+a))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.05801, size = 26, normalized size = 1.13 \begin{align*} -\frac{2}{3 \,{\left ({\left (b x + a\right )} d + c\right )}^{\frac{3}{2}} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-2/3/(((b*x + a)*d + c)^(3/2)*b*d)

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Fricas [B]  time = 1.57646, size = 146, normalized size = 6.35 \begin{align*} -\frac{2 \, \sqrt{b d x + a d + c}}{3 \,{\left (b^{3} d^{3} x^{2} + a^{2} b d^{3} + 2 \, a b c d^{2} + b c^{2} d + 2 \,{\left (a b^{2} d^{3} + b^{2} c d^{2}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.85587, size = 102, normalized size = 4.43 \begin{align*} \begin{cases} \frac{x}{c^{\frac{5}{2}}} & \text{for}\: b = 0 \wedge d = 0 \\\frac{x}{\left (a d + c\right )^{\frac{5}{2}}} & \text{for}\: b = 0 \\\frac{x}{c^{\frac{5}{2}}} & \text{for}\: d = 0 \\- \frac{2 \sqrt{a d + b d x + c}}{3 a^{2} b d^{3} + 6 a b^{2} d^{3} x + 6 a b c d^{2} + 3 b^{3} d^{3} x^{2} + 6 b^{2} c d^{2} x + 3 b c^{2} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*(b*x+a))**(5/2),x)

[Out]

Piecewise((x/c**(5/2), Eq(b, 0) & Eq(d, 0)), (x/(a*d + c)**(5/2), Eq(b, 0)), (x/c**(5/2), Eq(d, 0)), (-2*sqrt(
a*d + b*d*x + c)/(3*a**2*b*d**3 + 6*a*b**2*d**3*x + 6*a*b*c*d**2 + 3*b**3*d**3*x**2 + 6*b**2*c*d**2*x + 3*b*c*
*2*d), True))

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Giac [A]  time = 1.14405, size = 26, normalized size = 1.13 \begin{align*} -\frac{2}{3 \,{\left ({\left (b x + a\right )} d + c\right )}^{\frac{3}{2}} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/3/(((b*x + a)*d + c)^(3/2)*b*d)