\(\int \frac {1}{(c (a+b x)^3)^{5/2}} \, dx\) [2818]

Optimal result

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

[Out]

-2/13/b/c^2/(b*x+a)^5/(c*(b*x+a)^3)^(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} \[ \int \frac {1}{\left (c (a+b x)^3\right )^{5/2}} \, dx=-\frac {2}{13 b c^2 (a+b x)^5 \sqrt {c (a+b x)^3}} \]

[In]

Int[(c*(a + b*x)^3)^(-5/2),x]

[Out]

-2/(13*b*c^2*(a + b*x)^5*Sqrt[c*(a + b*x)^3])

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 (c x^3\right )^{5/2}} \, dx,x,a+b x\right )}{b} \\ & = \frac {(a+b x)^{3/2} \text {Subst}\left (\int \frac {1}{x^{15/2}} \, dx,x,a+b x\right )}{b c^2 \sqrt {c (a+b x)^3}} \\ & = -\frac {2}{13 b c^2 (a+b x)^5 \sqrt {c (a+b x)^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (c (a+b x)^3\right )^{5/2}} \, dx=-\frac {2 (a+b x)}{13 b \left (c (a+b x)^3\right )^{5/2}} \]

[In]

Integrate[(c*(a + b*x)^3)^(-5/2),x]

[Out]

(-2*(a + b*x))/(13*b*(c*(a + b*x)^3)^(5/2))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )}{13 b \left (c \left (b x +a \right )^{3}\right )^{\frac {5}{2}}}\)	\(22\)
default	\(-\frac {2 \left (b x +a \right )^{5} \left (c \left (b x +a \right )\right )^{\frac {5}{2}} c^{4}}{13 \left (c \left (b x +a \right )^{3}\right )^{\frac {5}{2}} b \left (b c x +a c \right )^{\frac {13}{2}}}\)	\(46\)
trager	\(-\frac {2 \sqrt {b^{3} c \,x^{3}+3 a \,b^{2} c \,x^{2}+3 a^{2} b c x +c \,a^{3}}}{13 c^{3} \left (b x +a \right )^{8} b}\)	\(50\)

[In]

int(1/(c*(b*x+a)^3)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/13*(b*x+a)/b/(c*(b*x+a)^3)^(5/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (26) = 52\).

Time = 0.26 (sec) , antiderivative size = 151, normalized size of antiderivative = 5.03 \[ \int \frac {1}{\left (c (a+b x)^3\right )^{5/2}} \, dx=-\frac {2 \, \sqrt {b^{3} c x^{3} + 3 \, a b^{2} c x^{2} + 3 \, a^{2} b c x + a^{3} c}}{13 \, {\left (b^{9} c^{3} x^{8} + 8 \, a b^{8} c^{3} x^{7} + 28 \, a^{2} b^{7} c^{3} x^{6} + 56 \, a^{3} b^{6} c^{3} x^{5} + 70 \, a^{4} b^{5} c^{3} x^{4} + 56 \, a^{5} b^{4} c^{3} x^{3} + 28 \, a^{6} b^{3} c^{3} x^{2} + 8 \, a^{7} b^{2} c^{3} x + a^{8} b c^{3}\right )}} \]

[In]

integrate(1/(c*(b*x+a)^3)^(5/2),x, algorithm="fricas")

[Out]

-2/13*sqrt(b^3*c*x^3 + 3*a*b^2*c*x^2 + 3*a^2*b*c*x + a^3*c)/(b^9*c^3*x^8 + 8*a*b^8*c^3*x^7 + 28*a^2*b^7*c^3*x^
6 + 56*a^3*b^6*c^3*x^5 + 70*a^4*b^5*c^3*x^4 + 56*a^5*b^4*c^3*x^3 + 28*a^6*b^3*c^3*x^2 + 8*a^7*b^2*c^3*x + a^8*
b*c^3)

Sympy [A] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (c (a+b x)^3\right )^{5/2}} \, dx=\begin {cases} - \frac {2 \left (\frac {a}{b} + x\right )}{13 \left (c \left (a + b x\right )^{3}\right )^{\frac {5}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (a^{3} c\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*(a/b + x)/(13*(c*(a + b*x)**3)**(5/2)), Ne(b, 0)), (x/(a**3*c)**(5/2), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (26) = 52\).

Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.83 \[ \int \frac {1}{\left (c (a+b x)^3\right )^{5/2}} \, dx=-\frac {2 \, \sqrt {c}}{13 \, {\left (b^{6} c^{3} x^{5} + 5 \, a b^{5} c^{3} x^{4} + 10 \, a^{2} b^{4} c^{3} x^{3} + 10 \, a^{3} b^{3} c^{3} x^{2} + 5 \, a^{4} b^{2} c^{3} x + a^{5} b c^{3}\right )} {\left (b x + a\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(c*(b*x+a)^3)^(5/2),x, algorithm="maxima")

[Out]

-2/13*sqrt(c)/((b^6*c^3*x^5 + 5*a*b^5*c^3*x^4 + 10*a^2*b^4*c^3*x^3 + 10*a^3*b^3*c^3*x^2 + 5*a^4*b^2*c^3*x + a^
5*b*c^3)*(b*x + a)^(3/2))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (c (a+b x)^3\right )^{5/2}} \, dx=-\frac {2 \, c^{4}}{13 \, {\left (b c x + a c\right )}^{\frac {13}{2}} b \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate(1/(c*(b*x+a)^3)^(5/2),x, algorithm="giac")

[Out]

-2/13*c^4/((b*c*x + a*c)^(13/2)*b*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 6.38 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (c (a+b x)^3\right )^{5/2}} \, dx=-\frac {2\,\sqrt {c\,{\left (a+b\,x\right )}^3}}{13\,b\,c^3\,{\left (a+b\,x\right )}^8} \]

[In]

int(1/(c*(a + b*x)^3)^(5/2),x)

[Out]

-(2*(c*(a + b*x)^3)^(1/2))/(13*b*c^3*(a + b*x)^8)