\(\int \frac {1}{(\frac {c}{a+b x})^{5/2}} \, dx\) [2824]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\frac {2 (a+b x)^3}{7 b c^2 \sqrt {\frac {c}{a+b x}}} \]

[Out]

2/7*(b*x+a)^3/b/c^2/(c/(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {253, 15, 30} \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\frac {2 (a+b x)^3}{7 b c^2 \sqrt {\frac {c}{a+b x}}} \]

[In]

Int[(c/(a + b*x))^(-5/2),x]

[Out]

(2*(a + b*x)^3)/(7*b*c^2*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 253

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (\frac {c}{x}\right )^{5/2}} \, dx,x,a+b x\right )}{b} \\ & = \frac {\text {Subst}\left (\int x^{5/2} \, dx,x,a+b x\right )}{b c^2 \sqrt {\frac {c}{a+b x}} \sqrt {a+b x}} \\ & = \frac {2 (a+b x)^3}{7 b c^2 \sqrt {\frac {c}{a+b x}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\frac {2 c}{7 b \left (\frac {c}{a+b x}\right )^{7/2}} \]

[In]

Integrate[(c/(a + b*x))^(-5/2),x]

[Out]

(2*c)/(7*b*(c/(a + b*x))^(7/2))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
gosper	\(\frac {\frac {2 b x}{7}+\frac {2 a}{7}}{b \left (\frac {c}{b x +a}\right )^{\frac {5}{2}}}\)	\(22\)
default	\(\frac {\frac {2 b x}{7}+\frac {2 a}{7}}{b \left (\frac {c}{b x +a}\right )^{\frac {5}{2}}}\)	\(22\)
risch	\(\frac {\frac {2}{7} a^{3}+\frac {6}{7} a^{2} b x +\frac {6}{7} a \,b^{2} x^{2}+\frac {2}{7} b^{3} x^{3}}{c^{2} \sqrt {\frac {c}{b x +a}}\, b}\)	\(47\)
trager	\(\frac {2 \left (b x +a \right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right ) \sqrt {\frac {c}{b x +a}}}{7 c^{3} b}\)	\(52\)

[In]

int(1/(c/(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/7*(b*x+a)/b/(c/(b*x+a))^(5/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (26) = 52\).

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\frac {2 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \sqrt {\frac {c}{b x + a}}}{7 \, b c^{3}} \]

[In]

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

[Out]

2/7*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*sqrt(c/(b*x + a))/(b*c^3)

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\begin {cases} \frac {2 a}{7 b \left (\frac {c}{a + b x}\right )^{\frac {5}{2}}} + \frac {2 x}{7 \left (\frac {c}{a + b x}\right )^{\frac {5}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\frac {c}{a}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a/(7*b*(c/(a + b*x))**(5/2)) + 2*x/(7*(c/(a + b*x))**(5/2)), Ne(b, 0)), (x/(c/a)**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\frac {2 \, c}{7 \, b \left (\frac {c}{b x + a}\right )^{\frac {7}{2}}} \]

[In]

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

[Out]

2/7*c/(b*(c/(b*x + a))^(7/2))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (26) = 52\).

Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 6.20 \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\frac {2 \, {\left (35 \, \sqrt {b c x + a c} a^{3} - \frac {35 \, {\left (3 \, \sqrt {b c x + a c} a c - {\left (b c x + a c\right )}^{\frac {3}{2}}\right )} a^{2}}{c} + \frac {7 \, {\left (15 \, \sqrt {b c x + a c} a^{2} c^{2} - 10 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a c + 3 \, {\left (b c x + a c\right )}^{\frac {5}{2}}\right )} a}{c^{2}} - \frac {35 \, \sqrt {b c x + a c} a^{3} c^{3} - 35 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a^{2} c^{2} + 21 \, {\left (b c x + a c\right )}^{\frac {5}{2}} a c - 5 \, {\left (b c x + a c\right )}^{\frac {7}{2}}}{c^{3}}\right )}}{35 \, b c^{3} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

2/35*(35*sqrt(b*c*x + a*c)*a^3 - 35*(3*sqrt(b*c*x + a*c)*a*c - (b*c*x + a*c)^(3/2))*a^2/c + 7*(15*sqrt(b*c*x +
a*c)*a^2*c^2 - 10*(b*c*x + a*c)^(3/2)*a*c + 3*(b*c*x + a*c)^(5/2))*a/c^2 - (35*sqrt(b*c*x + a*c)*a^3*c^3 - 35
*(b*c*x + a*c)^(3/2)*a^2*c^2 + 21*(b*c*x + a*c)^(5/2)*a*c - 5*(b*c*x + a*c)^(7/2))/c^3)/(b*c^3*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 6.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (\frac {c}{a+b x}\right )^{5/2}} \, dx=\frac {2\,\sqrt {\frac {c}{a+b\,x}}\,{\left (a+b\,x\right )}^4}{7\,b\,c^3} \]

[In]

int(1/(c/(a + b*x))^(5/2),x)

[Out]

(2*(c/(a + b*x))^(1/2)*(a + b*x)^4)/(7*b*c^3)