\(\int \frac {(a c+(b c+a d) x+b d x^2)^2}{a+b x} \, dx\) [1773]

Optimal result

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

[Out]

-1/3*(-a*d+b*c)*(d*x+c)^3/d^2+1/4*b*(d*x+c)^4/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} \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{a+b x} \, dx=\frac {b (c+d x)^4}{4 d^2}-\frac {(c+d x)^3 (b c-a d)}{3 d^2} \]

[In]

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

[Out]

-1/3*((b*c - a*d)*(c + d*x)^3)/d^2 + (b*(c + d*x)^4)/(4*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)^2 \, dx \\ & = \int \left (\frac {(-b c+a d) (c+d x)^2}{d}+\frac {b (c+d x)^3}{d}\right ) \, dx \\ & = -\frac {(b c-a d) (c+d x)^3}{3 d^2}+\frac {b (c+d x)^4}{4 d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{a+b x} \, dx=\frac {1}{12} x \left (12 a c^2+6 c (b c+2 a d) x+4 d (2 b c+a d) x^2+3 b d^2 x^3\right ) \]

[In]

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

[Out]

(x*(12*a*c^2 + 6*c*(b*c + 2*a*d)*x + 4*d*(2*b*c + a*d)*x^2 + 3*b*d^2*x^3))/12

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {b \,d^{2} x^{4}}{4}+\left (\frac {1}{3} a \,d^{2}+\frac {2}{3} b c d \right ) x^{3}+\left (a c d +\frac {1}{2} b \,c^{2}\right ) x^{2}+a \,c^{2} x\)	\(48\)
gosper	\(\frac {x \left (3 b \,d^{2} x^{3}+4 a \,d^{2} x^{2}+8 b c d \,x^{2}+12 a d x c +6 b x \,c^{2}+12 c^{2} a \right )}{12}\)	\(50\)
risch	\(\frac {1}{4} b \,d^{2} x^{4}+\frac {1}{3} a \,d^{2} x^{3}+\frac {2}{3} b c d \,x^{3}+a c d \,x^{2}+\frac {1}{2} b \,c^{2} x^{2}+a \,c^{2} x\)	\(50\)
parallelrisch	\(\frac {1}{4} b \,d^{2} x^{4}+\frac {1}{3} a \,d^{2} x^{3}+\frac {2}{3} b c d \,x^{3}+a c d \,x^{2}+\frac {1}{2} b \,c^{2} x^{2}+a \,c^{2} x\)	\(50\)
default	\(\frac {b \,d^{2} x^{4}}{4}+\frac {\left (b c d +d \left (a d +b c \right )\right ) x^{3}}{3}+\frac {\left (c \left (a d +b c \right )+a c d \right ) x^{2}}{2}+a \,c^{2} x\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{a+b x} \, dx=\frac {1}{4} \, b d^{2} x^{4} + a c^{2} x + \frac {1}{3} \, {\left (2 \, b c d + a d^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b c^{2} + 2 \, a c d\right )} x^{2} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{a+b x} \, dx=a c^{2} x + \frac {b d^{2} x^{4}}{4} + x^{3} \left (\frac {a d^{2}}{3} + \frac {2 b c d}{3}\right ) + x^{2} \left (a c d + \frac {b c^{2}}{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{a+b x} \, dx=\frac {1}{4} \, b d^{2} x^{4} + a c^{2} x + \frac {1}{3} \, {\left (2 \, b c d + a d^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b c^{2} + 2 \, a c d\right )} x^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{a+b x} \, dx=\frac {1}{4} \, b d^{2} x^{4} + \frac {2}{3} \, b c d x^{3} + \frac {1}{3} \, a d^{2} x^{3} + \frac {1}{2} \, b c^{2} x^{2} + a c d x^{2} + a c^{2} x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{a+b x} \, dx=x^2\,\left (\frac {b\,c^2}{2}+a\,d\,c\right )+x^3\,\left (\frac {a\,d^2}{3}+\frac {2\,b\,c\,d}{3}\right )+\frac {b\,d^2\,x^4}{4}+a\,c^2\,x \]

[In]

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

[Out]

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