\(\int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx\) [1449]

Optimal result

Integrand size = 20, antiderivative size = 22 \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\frac {2 (a c+b c x)^{5/2}}{5 b c^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {21, 32} \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\frac {2 (a c+b c x)^{5/2}}{5 b c^6} \]

[In]

Int[(a + b*x)^5/(a*c + b*c*x)^(7/2),x]

[Out]

(2*(a*c + b*c*x)^(5/2))/(5*b*c^6)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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*} \text {integral}& = \frac {\int (a c+b c x)^{3/2} \, dx}{c^5} \\ & = \frac {2 (a c+b c x)^{5/2}}{5 b c^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\frac {2 (a+b x)^6}{5 b (c (a+b x))^{7/2}} \]

[In]

Integrate[(a + b*x)^5/(a*c + b*c*x)^(7/2),x]

[Out]

(2*(a + b*x)^6)/(5*b*(c*(a + b*x))^(7/2))

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {2 \left (b c x +a c \right )^{\frac {5}{2}}}{5 b \,c^{6}}\)	\(19\)
default	\(\frac {2 \left (b c x +a c \right )^{\frac {5}{2}}}{5 b \,c^{6}}\)	\(19\)
gosper	\(\frac {2 \left (b x +a \right )^{6}}{5 b \left (b c x +a c \right )^{\frac {7}{2}}}\)	\(23\)
pseudoelliptic	\(\frac {2 \left (b x +a \right )^{2} \sqrt {c \left (b x +a \right )}}{5 c^{4} b}\)	\(25\)
trager	\(\frac {2 \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \sqrt {b c x +a c}}{5 c^{4} b}\)	\(35\)
risch	\(\frac {2 \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b x +a \right )}{5 c^{3} b \sqrt {c \left (b x +a \right )}}\)	\(39\)

[In]

int((b*x+a)^5/(b*c*x+a*c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b c x + a c}}{5 \, b c^{4}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(7/2),x, algorithm="fricas")

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*c*x + a*c)/(b*c^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (19) = 38\).

Time = 0.64 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.64 \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\begin {cases} \frac {2 a^{2} \sqrt {a c + b c x}}{5 b c^{4}} + \frac {4 a x \sqrt {a c + b c x}}{5 c^{4}} + \frac {2 b x^{2} \sqrt {a c + b c x}}{5 c^{4}} & \text {for}\: b \neq 0 \\\frac {a^{5} x}{\left (a c\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a**2*sqrt(a*c + b*c*x)/(5*b*c**4) + 4*a*x*sqrt(a*c + b*c*x)/(5*c**4) + 2*b*x**2*sqrt(a*c + b*c*x)
/(5*c**4), Ne(b, 0)), (a**5*x/(a*c)**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\frac {2 \, {\left (b c x + a c\right )}^{\frac {5}{2}}}{5 \, b c^{6}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(7/2),x, algorithm="maxima")

[Out]

2/5*(b*c*x + a*c)^(5/2)/(b*c^6)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (18) = 36\).

Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.82 \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\frac {2 \, {\left (15 \, \sqrt {b c x + a c} a^{2} - \frac {10 \, {\left (3 \, \sqrt {b c x + a c} a c - {\left (b c x + a c\right )}^{\frac {3}{2}}\right )} a}{c} + \frac {15 \, \sqrt {b c x + a c} a^{2} c^{2} - 10 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a c + 3 \, {\left (b c x + a c\right )}^{\frac {5}{2}}}{c^{2}}\right )}}{15 \, b c^{4}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^(7/2),x, algorithm="giac")

[Out]

2/15*(15*sqrt(b*c*x + a*c)*a^2 - 10*(3*sqrt(b*c*x + a*c)*a*c - (b*c*x + a*c)^(3/2))*a/c + (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))/c^2)/(b*c^4)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^5}{(a c+b c x)^{7/2}} \, dx=\frac {2\,{\left (c\,\left (a+b\,x\right )\right )}^{5/2}}{5\,b\,c^6} \]

[In]

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

[Out]

(2*(c*(a + b*x))^(5/2))/(5*b*c^6)