3.2820 \(\int (\frac {c}{a+b x})^{3/2} \, dx\)

Optimal. Leaf size=19 \[ -\frac {2 c \sqrt {\frac {c}{a+b x}}}{b} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 19, 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} \[ -\frac {2 c \sqrt {\frac {c}{a+b x}}}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x))^(3/2),x]

[Out]

(-2*c*Sqrt[c/(a + b*x)])/b

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ -\frac {2 c \sqrt {\frac {c}{a+b x}}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x))^(3/2),x]

[Out]

(-2*c*Sqrt[c/(a + b*x)])/b

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fricas [A]  time = 0.93, size = 17, normalized size = 0.89 \[ -\frac {2 \, c \sqrt {\frac {c}{b x + a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c*sqrt(c/(b*x + a))/b

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giac [A]  time = 0.20, size = 24, normalized size = 1.26 \[ -\frac {2 \, c^{2} \mathrm {sgn}\left (b x + a\right )}{\sqrt {b c x + a c} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 22, normalized size = 1.16 \[ -\frac {2 \left (b x +a \right ) \left (\frac {c}{b x +a}\right )^{\frac {3}{2}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 17, normalized size = 0.89 \[ -\frac {2 \, c \sqrt {\frac {c}{b x + a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c*sqrt(c/(b*x + a))/b

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mupad [B]  time = 1.14, size = 17, normalized size = 0.89 \[ -\frac {2\,c\,\sqrt {\frac {c}{a+b\,x}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x))^(3/2),x)

[Out]

-(2*c*(c/(a + b*x))^(1/2))/b

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sympy [A]  time = 1.23, size = 48, normalized size = 2.53 \[ \begin {cases} - \frac {2 a c^{\frac {3}{2}} \left (\frac {1}{a + b x}\right )^{\frac {3}{2}}}{b} - 2 c^{\frac {3}{2}} x \left (\frac {1}{a + b x}\right )^{\frac {3}{2}} & \text {for}\: b \neq 0 \\x \left (\frac {c}{a}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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