3.2823 \(\int \frac {1}{(\frac {c}{a+b x})^{3/2}} \, dx\)

Optimal. Leaf size=30 \[ \frac {2 (a+b x)^2}{5 b c \sqrt {\frac {c}{a+b x}}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {247, 15, 30} \[ \frac {2 (a+b x)^2}{5 b c \sqrt {\frac {c}{a+b x}}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x))^(-3/2),x]

[Out]

(2*(a + b*x)^2)/(5*b*c*Sqrt[c/(a + b*x)])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.70 \[ \frac {2 c}{5 b \left (\frac {c}{a+b x}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x))^(-3/2),x]

[Out]

(2*c)/(5*b*(c/(a + b*x))^(5/2))

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fricas [A]  time = 0.93, size = 46, normalized size = 1.53 \[ \frac {2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {\frac {c}{b x + a}}}{5 \, b c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c/(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/5*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(c/(b*x + a))/(b*c^2)

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giac [A]  time = 0.17, size = 26, normalized size = 0.87 \[ \frac {2 \, {\left (b x + a\right )}^{2}}{5 \, b c \sqrt {\frac {c}{b x + a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c/(b*x+a))^(3/2),x, algorithm="giac")

[Out]

2/5*(b*x + a)^2/(b*c*sqrt(c/(b*x + a)))

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maple [A]  time = 0.00, size = 22, normalized size = 0.73 \[ \frac {\frac {2 b x}{5}+\frac {2 a}{5}}{\left (\frac {c}{b x +a}\right )^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/(b*x+a)*c)^(3/2),x)

[Out]

2/5*(b*x+a)/b/(1/(b*x+a)*c)^(3/2)

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maxima [A]  time = 0.50, size = 17, normalized size = 0.57 \[ \frac {2 \, c}{5 \, b \left (\frac {c}{b x + a}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c/(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/5*c/(b*(c/(b*x + a))^(5/2))

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mupad [B]  time = 1.23, size = 54, normalized size = 1.80 \[ \sqrt {\frac {c}{a+b\,x}}\,\left (\frac {6\,a^2\,x}{5\,c^2}+\frac {2\,a^3}{5\,b\,c^2}+\frac {2\,b^2\,x^3}{5\,c^2}+\frac {6\,a\,b\,x^2}{5\,c^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c/(a + b*x))^(3/2),x)

[Out]

(c/(a + b*x))^(1/2)*((6*a^2*x)/(5*c^2) + (2*a^3)/(5*b*c^2) + (2*b^2*x^3)/(5*c^2) + (6*a*b*x^2)/(5*c^2))

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sympy [A]  time = 1.69, size = 49, normalized size = 1.63 \[ \begin {cases} \frac {2 a}{5 b c^{\frac {3}{2}} \left (\frac {1}{a + b x}\right )^{\frac {3}{2}}} + \frac {2 x}{5 c^{\frac {3}{2}} \left (\frac {1}{a + b x}\right )^{\frac {3}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\frac {c}{a}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c/(b*x+a))**(3/2),x)

[Out]

Piecewise((2*a/(5*b*c**(3/2)*(1/(a + b*x))**(3/2)) + 2*x/(5*c**(3/2)*(1/(a + b*x))**(3/2)), Ne(b, 0)), (x/(c/a
)**(3/2), True))

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