\(\int \frac {(a c+(b c+a d) x+b d x^2)^2}{(a+b x)^7} \, dx\) [1779]

Optimal result

Integrand size = 29, antiderivative size = 65 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=-\frac {(b c-a d)^2}{4 b^3 (a+b x)^4}-\frac {2 d (b c-a d)}{3 b^3 (a+b x)^3}-\frac {d^2}{2 b^3 (a+b x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, 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} \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=-\frac {2 d (b c-a d)}{3 b^3 (a+b x)^3}-\frac {(b c-a d)^2}{4 b^3 (a+b x)^4}-\frac {d^2}{2 b^3 (a+b x)^2} \]

[In]

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

[Out]

-1/4*(b*c - a*d)^2/(b^3*(a + b*x)^4) - (2*d*(b*c - a*d))/(3*b^3*(a + b*x)^3) - d^2/(2*b^3*(a + b*x)^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 \frac {(c+d x)^2}{(a+b x)^5} \, dx \\ & = \int \left (\frac {(b c-a d)^2}{b^2 (a+b x)^5}+\frac {2 d (b c-a d)}{b^2 (a+b x)^4}+\frac {d^2}{b^2 (a+b x)^3}\right ) \, dx \\ & = -\frac {(b c-a d)^2}{4 b^3 (a+b x)^4}-\frac {2 d (b c-a d)}{3 b^3 (a+b x)^3}-\frac {d^2}{2 b^3 (a+b x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=-\frac {a^2 d^2+2 a b d (c+2 d x)+b^2 \left (3 c^2+8 c d x+6 d^2 x^2\right )}{12 b^3 (a+b x)^4} \]

[In]

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

[Out]

-1/12*(a^2*d^2 + 2*a*b*d*(c + 2*d*x) + b^2*(3*c^2 + 8*c*d*x + 6*d^2*x^2))/(b^3*(a + b*x)^4)

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95

method	result	size
gosper	\(-\frac {6 d^{2} x^{2} b^{2}+4 x a b \,d^{2}+8 x \,b^{2} c d +a^{2} d^{2}+2 a b c d +3 b^{2} c^{2}}{12 b^{3} \left (b x +a \right )^{4}}\)	\(62\)
risch	\(\frac {-\frac {d^{2} x^{2}}{2 b}-\frac {d \left (a d +2 b c \right ) x}{3 b^{2}}-\frac {a^{2} d^{2}+2 a b c d +3 b^{2} c^{2}}{12 b^{3}}}{\left (b x +a \right )^{4}}\)	\(63\)
parallelrisch	\(\frac {-6 b^{3} d^{2} x^{2}-4 a \,b^{2} d^{2} x -8 b^{3} c d x -a^{2} b \,d^{2}-2 a \,b^{2} c d -3 c^{2} b^{3}}{12 b^{4} \left (b x +a \right )^{4}}\)	\(68\)
default	\(\frac {2 \left (a d -b c \right ) d}{3 b^{3} \left (b x +a \right )^{3}}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{4 b^{3} \left (b x +a \right )^{4}}-\frac {d^{2}}{2 b^{3} \left (b x +a \right )^{2}}\)	\(71\)
norman	\(\frac {\frac {a^{2} \left (-a^{2} b^{3} d^{2}-2 a \,b^{4} c d -3 b^{5} c^{2}\right )}{12 b^{6}}-\frac {b \,d^{2} x^{4}}{2}+\frac {2 \left (-2 a \,b^{3} d^{2}-b^{4} c d \right ) x^{3}}{3 b^{3}}+\frac {\left (-5 a^{2} b^{3} d^{2}-6 a \,b^{4} c d -b^{5} c^{2}\right ) x^{2}}{4 b^{4}}+\frac {a \left (-a^{2} b^{3} d^{2}-2 a \,b^{4} c d -b^{5} c^{2}\right ) x}{2 b^{5}}}{\left (b x +a \right )^{6}}\)	\(151\)

[In]

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

[Out]

-1/12/b^3*(6*b^2*d^2*x^2+4*a*b*d^2*x+8*b^2*c*d*x+a^2*d^2+2*a*b*c*d+3*b^2*c^2)/(b*x+a)^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.51 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=-\frac {6 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 2 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \]

[In]

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

[Out]

-1/12*(6*b^2*d^2*x^2 + 3*b^2*c^2 + 2*a*b*c*d + a^2*d^2 + 4*(2*b^2*c*d + a*b*d^2)*x)/(b^7*x^4 + 4*a*b^6*x^3 + 6
*a^2*b^5*x^2 + 4*a^3*b^4*x + a^4*b^3)

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=\frac {- a^{2} d^{2} - 2 a b c d - 3 b^{2} c^{2} - 6 b^{2} d^{2} x^{2} + x \left (- 4 a b d^{2} - 8 b^{2} c d\right )}{12 a^{4} b^{3} + 48 a^{3} b^{4} x + 72 a^{2} b^{5} x^{2} + 48 a b^{6} x^{3} + 12 b^{7} x^{4}} \]

[In]

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

[Out]

(-a**2*d**2 - 2*a*b*c*d - 3*b**2*c**2 - 6*b**2*d**2*x**2 + x*(-4*a*b*d**2 - 8*b**2*c*d))/(12*a**4*b**3 + 48*a*
*3*b**4*x + 72*a**2*b**5*x**2 + 48*a*b**6*x**3 + 12*b**7*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.51 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=-\frac {6 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 2 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \]

[In]

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

[Out]

-1/12*(6*b^2*d^2*x^2 + 3*b^2*c^2 + 2*a*b*c*d + a^2*d^2 + 4*(2*b^2*c*d + a*b*d^2)*x)/(b^7*x^4 + 4*a*b^6*x^3 + 6
*a^2*b^5*x^2 + 4*a^3*b^4*x + a^4*b^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=-\frac {6 \, b^{2} d^{2} x^{2} + 8 \, b^{2} c d x + 4 \, a b d^{2} x + 3 \, b^{2} c^{2} + 2 \, a b c d + a^{2} d^{2}}{12 \, {\left (b x + a\right )}^{4} b^{3}} \]

[In]

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

[Out]

-1/12*(6*b^2*d^2*x^2 + 8*b^2*c*d*x + 4*a*b*d^2*x + 3*b^2*c^2 + 2*a*b*c*d + a^2*d^2)/((b*x + a)^4*b^3)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^7} \, dx=-\frac {\frac {a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2}{12\,b^3}+\frac {d^2\,x^2}{2\,b}+\frac {d\,x\,\left (a\,d+2\,b\,c\right )}{3\,b^2}}{a^4+4\,a^3\,b\,x+6\,a^2\,b^2\,x^2+4\,a\,b^3\,x^3+b^4\,x^4} \]

[In]

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

[Out]

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