\(\int \frac {(a c+(b c+a d) x+b d x^2)^3}{(a+b x)^2} \, dx\) [1788]

Optimal result

Integrand size = 29, antiderivative size = 38 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=-\frac {(b c-a d) (c+d x)^4}{4 d^2}+\frac {b (c+d x)^5}{5 d^2} \]

[Out]

-1/4*(-a*d+b*c)*(d*x+c)^4/d^2+1/5*b*(d*x+c)^5/d^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {640, 45} \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=\frac {b (c+d x)^5}{5 d^2}-\frac {(c+d x)^4 (b c-a d)}{4 d^2} \]

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^2,x]

[Out]

-1/4*((b*c - a*d)*(c + d*x)^4)/d^2 + (b*(c + d*x)^5)/(5*d^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int (a+b x) (c+d x)^3 \, dx \\ & = \int \left (\frac {(-b c+a d) (c+d x)^3}{d}+\frac {b (c+d x)^4}{d}\right ) \, dx \\ & = -\frac {(b c-a d) (c+d x)^4}{4 d^2}+\frac {b (c+d x)^5}{5 d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=a c^3 x+\frac {1}{2} c^2 (b c+3 a d) x^2+c d (b c+a d) x^3+\frac {1}{4} d^2 (3 b c+a d) x^4+\frac {1}{5} b d^3 x^5 \]

[In]

Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^2,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(72\) vs. \(2(34)=68\).

Time = 2.58 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.92

method	result	size
risch	\(\frac {1}{5} b \,d^{3} x^{5}+\frac {1}{4} x^{4} a \,d^{3}+\frac {3}{4} x^{4} b c \,d^{2}+a c \,d^{2} x^{3}+b \,c^{2} d \,x^{3}+\frac {3}{2} x^{2} a \,c^{2} d +\frac {1}{2} b \,c^{3} x^{2}+a \,c^{3} x\)	\(73\)
parallelrisch	\(\frac {1}{5} b \,d^{3} x^{5}+\frac {1}{4} x^{4} a \,d^{3}+\frac {3}{4} x^{4} b c \,d^{2}+a c \,d^{2} x^{3}+b \,c^{2} d \,x^{3}+\frac {3}{2} x^{2} a \,c^{2} d +\frac {1}{2} b \,c^{3} x^{2}+a \,c^{3} x\)	\(73\)
gosper	\(\frac {x \left (4 b \,d^{3} x^{4}+5 x^{3} a \,d^{3}+15 x^{3} b c \,d^{2}+20 a c \,d^{2} x^{2}+20 b \,c^{2} d \,x^{2}+30 x a \,c^{2} d +10 x \,c^{3} b +20 c^{3} a \right )}{20}\)	\(74\)
default	\(\frac {b \,d^{3} x^{5}}{5}+\frac {\left (2 b c \,d^{2}+d^{2} \left (a d +b c \right )\right ) x^{4}}{4}+\frac {\left (b \,c^{2} d +2 c d \left (a d +b c \right )+a c \,d^{2}\right ) x^{3}}{3}+\frac {\left (c^{2} \left (a d +b c \right )+2 a \,c^{2} d \right ) x^{2}}{2}+a \,c^{3} x\)	\(94\)
norman	\(\frac {\left (\frac {9}{20} a b \,d^{3}+\frac {3}{4} b^{2} c \,d^{2}\right ) x^{5}+\left (\frac {3}{2} a^{2} c^{2} d +\frac {3}{2} a b \,c^{3}\right ) x^{2}+\left (\frac {1}{4} a^{2} d^{3}+\frac {7}{4} a b c \,d^{2}+b^{2} c^{2} d \right ) x^{4}+\left (c \,a^{2} d^{2}+\frac {5}{2} a b \,c^{2} d +\frac {1}{2} b^{2} c^{3}\right ) x^{3}+a^{2} c^{3} x +\frac {b^{2} d^{3} x^{6}}{5}}{b x +a}\)	\(129\)

[In]

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

[Out]

1/5*b*d^3*x^5+1/4*x^4*a*d^3+3/4*x^4*b*c*d^2+a*c*d^2*x^3+b*c^2*d*x^3+3/2*x^2*a*c^2*d+1/2*b*c^3*x^2+a*c^3*x

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.82 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=\frac {1}{5} \, b d^{3} x^{5} + a c^{3} x + \frac {1}{4} \, {\left (3 \, b c d^{2} + a d^{3}\right )} x^{4} + {\left (b c^{2} d + a c d^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2} \]

[In]

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

[Out]

1/5*b*d^3*x^5 + a*c^3*x + 1/4*(3*b*c*d^2 + a*d^3)*x^4 + (b*c^2*d + a*c*d^2)*x^3 + 1/2*(b*c^3 + 3*a*c^2*d)*x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (32) = 64\).

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.92 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=a c^{3} x + \frac {b d^{3} x^{5}}{5} + x^{4} \left (\frac {a d^{3}}{4} + \frac {3 b c d^{2}}{4}\right ) + x^{3} \left (a c d^{2} + b c^{2} d\right ) + x^{2} \cdot \left (\frac {3 a c^{2} d}{2} + \frac {b c^{3}}{2}\right ) \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.82 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=\frac {1}{5} \, b d^{3} x^{5} + a c^{3} x + \frac {1}{4} \, {\left (3 \, b c d^{2} + a d^{3}\right )} x^{4} + {\left (b c^{2} d + a c d^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2} \]

[In]

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

[Out]

1/5*b*d^3*x^5 + a*c^3*x + 1/4*(3*b*c*d^2 + a*d^3)*x^4 + (b*c^2*d + a*c*d^2)*x^3 + 1/2*(b*c^3 + 3*a*c^2*d)*x^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (34) = 68\).

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 4.08 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=\frac {{\left (\frac {10 \, b^{3} c^{3}}{{\left (b x + a\right )}^{3}} + \frac {20 \, b^{2} c^{2} d}{{\left (b x + a\right )}^{2}} - \frac {30 \, a b^{2} c^{2} d}{{\left (b x + a\right )}^{3}} + \frac {15 \, b c d^{2}}{b x + a} - \frac {40 \, a b c d^{2}}{{\left (b x + a\right )}^{2}} + \frac {30 \, a^{2} b c d^{2}}{{\left (b x + a\right )}^{3}} - \frac {15 \, a d^{3}}{b x + a} + \frac {20 \, a^{2} d^{3}}{{\left (b x + a\right )}^{2}} - \frac {10 \, a^{3} d^{3}}{{\left (b x + a\right )}^{3}} + 4 \, d^{3}\right )} {\left (b x + a\right )}^{5}}{20 \, b^{4}} \]

[In]

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

[Out]

1/20*(10*b^3*c^3/(b*x + a)^3 + 20*b^2*c^2*d/(b*x + a)^2 - 30*a*b^2*c^2*d/(b*x + a)^3 + 15*b*c*d^2/(b*x + a) -
40*a*b*c*d^2/(b*x + a)^2 + 30*a^2*b*c*d^2/(b*x + a)^3 - 15*a*d^3/(b*x + a) + 20*a^2*d^3/(b*x + a)^2 - 10*a^3*d
^3/(b*x + a)^3 + 4*d^3)*(b*x + a)^5/b^4

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^2} \, dx=x^2\,\left (\frac {b\,c^3}{2}+\frac {3\,a\,d\,c^2}{2}\right )+x^4\,\left (\frac {a\,d^3}{4}+\frac {3\,b\,c\,d^2}{4}\right )+\frac {b\,d^3\,x^5}{5}+a\,c^3\,x+c\,d\,x^3\,\left (a\,d+b\,c\right ) \]

[In]

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

[Out]

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