\(\int \frac {(\frac {b c}{d}+b x)^5}{(c+d x)^3} \, dx\) [1007]

Optimal result

Integrand size = 20, antiderivative size = 17 \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^5 (c+d x)^3}{3 d^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, 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} \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^5 (c+d x)^3}{3 d^6} \]

[In]

Int[((b*c)/d + b*x)^5/(c + d*x)^3,x]

[Out]

(b^5*(c + d*x)^3)/(3*d^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 {b^5 \int (c+d x)^2 \, dx}{d^5} \\ & = \frac {b^5 (c+d x)^3}{3 d^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^5 (c+d x)^3}{3 d^6} \]

[In]

Integrate[((b*c)/d + b*x)^5/(c + d*x)^3,x]

[Out]

(b^5*(c + d*x)^3)/(3*d^6)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {b^{5} \left (d x +c \right )^{3}}{3 d^{6}}\)	\(16\)
gosper	\(\frac {b^{5} x \left (d^{2} x^{2}+3 c d x +3 c^{2}\right )}{3 d^{5}}\)	\(28\)
parallelrisch	\(\frac {x^{3} b^{5} d^{2}+3 x^{2} b^{5} c d +3 x \,b^{5} c^{2}}{3 d^{5}}\)	\(36\)
risch	\(\frac {b^{5} x^{3}}{3 d^{3}}+\frac {b^{5} c \,x^{2}}{d^{4}}+\frac {b^{5} c^{2} x}{d^{5}}+\frac {b^{5} c^{3}}{3 d^{6}}\)	\(46\)
norman	\(\frac {\frac {b^{5} d^{3} x^{5}}{3}+\frac {5 c^{3} b^{5} x^{2}}{2}-\frac {c^{5} b^{5}}{2 d^{2}}+\frac {5 b^{5} c \,d^{2} x^{4}}{3}+\frac {10 b^{5} c^{2} d \,x^{3}}{3}}{d^{4} \left (d x +c \right )^{2}}\)	\(70\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (15) = 30\).

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^{5} d^{2} x^{3} + 3 \, b^{5} c d x^{2} + 3 \, b^{5} c^{2} x}{3 \, d^{5}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00 \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^{5} c^{2} x}{d^{5}} + \frac {b^{5} c x^{2}}{d^{4}} + \frac {b^{5} x^{3}}{3 d^{3}} \]

[In]

integrate((b*c/d+b*x)**5/(d*x+c)**3,x)

[Out]

b**5*c**2*x/d**5 + b**5*c*x**2/d**4 + b**5*x**3/(3*d**3)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (15) = 30\).

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^{5} d^{2} x^{3} + 3 \, b^{5} c d x^{2} + 3 \, b^{5} c^{2} x}{3 \, d^{5}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (15) = 30\).

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^{5} d^{2} x^{3} + 3 \, b^{5} c d x^{2} + 3 \, b^{5} c^{2} x}{3 \, d^{5}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \frac {\left (\frac {b c}{d}+b x\right )^5}{(c+d x)^3} \, dx=\frac {b^5\,x\,\left (3\,c^2+3\,c\,d\,x+d^2\,x^2\right )}{3\,d^5} \]

[In]

int((b*x + (b*c)/d)^5/(c + d*x)^3,x)

[Out]

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