\(\int \frac {(d+e x)^5 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\) [559]

Optimal result

Integrand size = 29, antiderivative size = 146 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d \left (5 e^2 f^2+24 d e f g+20 d^2 g^2\right ) x}{e^2}+\frac {\left (e^2 f^2+10 d e f g+12 d^2 g^2\right ) x^2}{2 e}+\frac {1}{3} g (2 e f+5 d g) x^3+\frac {1}{4} e g^2 x^4+\frac {8 d^3 (e f+d g)^2}{e^3 (d-e x)}+\frac {4 d^2 (e f+d g) (3 e f+7 d g) \log (d-e x)}{e^3} \]

[Out]

d*(20*d^2*g^2+24*d*e*f*g+5*e^2*f^2)*x/e^2+1/2*(12*d^2*g^2+10*d*e*f*g+e^2*f^2)*x^2/e+1/3*g*(5*d*g+2*e*f)*x^3+1/
4*e*g^2*x^4+8*d^3*(d*g+e*f)^2/e^3/(-e*x+d)+4*d^2*(d*g+e*f)*(7*d*g+3*e*f)*ln(-e*x+d)/e^3

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 146, 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 = {862, 90} \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {8 d^3 (d g+e f)^2}{e^3 (d-e x)}+\frac {4 d^2 (d g+e f) (7 d g+3 e f) \log (d-e x)}{e^3}+\frac {x^2 \left (12 d^2 g^2+10 d e f g+e^2 f^2\right )}{2 e}+\frac {d x \left (20 d^2 g^2+24 d e f g+5 e^2 f^2\right )}{e^2}+\frac {1}{3} g x^3 (5 d g+2 e f)+\frac {1}{4} e g^2 x^4 \]

[In]

Int[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^2,x]

[Out]

(d*(5*e^2*f^2 + 24*d*e*f*g + 20*d^2*g^2)*x)/e^2 + ((e^2*f^2 + 10*d*e*f*g + 12*d^2*g^2)*x^2)/(2*e) + (g*(2*e*f
+ 5*d*g)*x^3)/3 + (e*g^2*x^4)/4 + (8*d^3*(e*f + d*g)^2)/(e^3*(d - e*x)) + (4*d^2*(e*f + d*g)*(3*e*f + 7*d*g)*L
og[d - e*x])/e^3

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^3 (f+g x)^2}{(d-e x)^2} \, dx \\ & = \int \left (\frac {d \left (5 e^2 f^2+24 d e f g+20 d^2 g^2\right )}{e^2}+\frac {\left (e^2 f^2+10 d e f g+12 d^2 g^2\right ) x}{e}+g (2 e f+5 d g) x^2+e g^2 x^3+\frac {4 d^2 (-3 e f-7 d g) (e f+d g)}{e^2 (d-e x)}+\frac {8 d^3 (e f+d g)^2}{e^2 (-d+e x)^2}\right ) \, dx \\ & = \frac {d \left (5 e^2 f^2+24 d e f g+20 d^2 g^2\right ) x}{e^2}+\frac {\left (e^2 f^2+10 d e f g+12 d^2 g^2\right ) x^2}{2 e}+\frac {1}{3} g (2 e f+5 d g) x^3+\frac {1}{4} e g^2 x^4+\frac {8 d^3 (e f+d g)^2}{e^3 (d-e x)}+\frac {4 d^2 (e f+d g) (3 e f+7 d g) \log (d-e x)}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d \left (5 e^2 f^2+24 d e f g+20 d^2 g^2\right ) x}{e^2}+\frac {\left (e^2 f^2+10 d e f g+12 d^2 g^2\right ) x^2}{2 e}+\frac {1}{3} g (2 e f+5 d g) x^3+\frac {1}{4} e g^2 x^4-\frac {8 d^3 (e f+d g)^2}{e^3 (-d+e x)}+\frac {4 d^2 \left (3 e^2 f^2+10 d e f g+7 d^2 g^2\right ) \log (d-e x)}{e^3} \]

[In]

Integrate[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^2,x]

[Out]

(d*(5*e^2*f^2 + 24*d*e*f*g + 20*d^2*g^2)*x)/e^2 + ((e^2*f^2 + 10*d*e*f*g + 12*d^2*g^2)*x^2)/(2*e) + (g*(2*e*f
+ 5*d*g)*x^3)/3 + (e*g^2*x^4)/4 - (8*d^3*(e*f + d*g)^2)/(e^3*(-d + e*x)) + (4*d^2*(3*e^2*f^2 + 10*d*e*f*g + 7*
d^2*g^2)*Log[d - e*x])/e^3

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\frac {1}{4} g^{2} e^{3} x^{4}+\frac {5}{3} x^{3} d \,e^{2} g^{2}+\frac {2}{3} x^{3} e^{3} f g +6 x^{2} d^{2} e \,g^{2}+5 x^{2} d \,e^{2} f g +\frac {1}{2} x^{2} e^{3} f^{2}+20 d^{3} g^{2} x +24 d^{2} e f g x +5 d \,e^{2} f^{2} x}{e^{2}}+\frac {4 d^{2} \left (7 d^{2} g^{2}+10 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {8 d^{3} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}\)	\(177\)
risch	\(\frac {e \,g^{2} x^{4}}{4}+\frac {5 x^{3} d \,g^{2}}{3}+\frac {2 e \,x^{3} f g}{3}+\frac {6 x^{2} g^{2} d^{2}}{e}+5 x^{2} f g d +\frac {e \,x^{2} f^{2}}{2}+\frac {20 d^{3} g^{2} x}{e^{2}}+\frac {24 d^{2} f g x}{e}+5 d \,f^{2} x +\frac {28 d^{4} \ln \left (-e x +d \right ) g^{2}}{e^{3}}+\frac {40 d^{3} \ln \left (-e x +d \right ) f g}{e^{2}}+\frac {12 d^{2} \ln \left (-e x +d \right ) f^{2}}{e}+\frac {8 d^{5} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {16 d^{4} f g}{e^{2} \left (-e x +d \right )}+\frac {8 d^{3} f^{2}}{e \left (-e x +d \right )}\)	\(198\)
norman	\(\frac {\left (-\frac {55}{3} d^{3} g^{2}-\frac {70}{3} d^{2} e f g -5 d \,e^{2} f^{2}\right ) x^{3}+\left (-\frac {23}{4} d^{2} g^{2} e -5 d f g \,e^{2}-\frac {1}{2} f^{2} e^{3}\right ) x^{4}+\frac {d^{3} \left (28 d^{2} g^{2}+40 d e f g +13 e^{2} f^{2}\right ) x}{e^{2}}+\frac {d^{2} \left (28 d^{4} g^{2}+42 f g e \,d^{3}+17 d^{2} e^{2} f^{2}\right )}{2 e^{3}}-\frac {e^{3} g^{2} x^{6}}{4}-\frac {e^{2} g \left (5 d g +2 e f \right ) x^{5}}{3}}{-e^{2} x^{2}+d^{2}}+\frac {4 d^{2} \left (7 d^{2} g^{2}+10 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(212\)
parallelrisch	\(\frac {3 g^{2} e^{5} x^{5}+17 x^{4} d \,e^{4} g^{2}+8 x^{4} e^{5} f g +52 x^{3} d^{2} e^{3} g^{2}+52 x^{3} d \,e^{4} f g +6 x^{3} e^{5} f^{2}+336 \ln \left (e x -d \right ) x \,d^{4} e \,g^{2}+480 \ln \left (e x -d \right ) x \,d^{3} e^{2} f g +144 \ln \left (e x -d \right ) x \,d^{2} e^{3} f^{2}+168 x^{2} d^{3} e^{2} g^{2}+228 x^{2} d^{2} e^{3} f g +54 x^{2} d \,e^{4} f^{2}-336 \ln \left (e x -d \right ) d^{5} g^{2}-480 \ln \left (e x -d \right ) d^{4} e f g -144 \ln \left (e x -d \right ) d^{3} e^{2} f^{2}-336 g^{2} d^{5}-480 f g \,d^{4} e -156 f^{2} d^{3} e^{2}}{12 e^{3} \left (e x -d \right )}\)	\(259\)

[In]

int((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/e^2*(1/4*g^2*e^3*x^4+5/3*x^3*d*e^2*g^2+2/3*x^3*e^3*f*g+6*x^2*d^2*e*g^2+5*x^2*d*e^2*f*g+1/2*x^2*e^3*f^2+20*d^
3*g^2*x+24*d^2*e*f*g*x+5*d*e^2*f^2*x)+4*d^2/e^3*(7*d^2*g^2+10*d*e*f*g+3*e^2*f^2)*ln(-e*x+d)+8*d^3*(d^2*g^2+2*d
*e*f*g+e^2*f^2)/e^3/(-e*x+d)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.72 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {3 \, e^{5} g^{2} x^{5} - 96 \, d^{3} e^{2} f^{2} - 192 \, d^{4} e f g - 96 \, d^{5} g^{2} + {\left (8 \, e^{5} f g + 17 \, d e^{4} g^{2}\right )} x^{4} + 2 \, {\left (3 \, e^{5} f^{2} + 26 \, d e^{4} f g + 26 \, d^{2} e^{3} g^{2}\right )} x^{3} + 6 \, {\left (9 \, d e^{4} f^{2} + 38 \, d^{2} e^{3} f g + 28 \, d^{3} e^{2} g^{2}\right )} x^{2} - 12 \, {\left (5 \, d^{2} e^{3} f^{2} + 24 \, d^{3} e^{2} f g + 20 \, d^{4} e g^{2}\right )} x - 48 \, {\left (3 \, d^{3} e^{2} f^{2} + 10 \, d^{4} e f g + 7 \, d^{5} g^{2} - {\left (3 \, d^{2} e^{3} f^{2} + 10 \, d^{3} e^{2} f g + 7 \, d^{4} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{12 \, {\left (e^{4} x - d e^{3}\right )}} \]

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="fricas")

[Out]

1/12*(3*e^5*g^2*x^5 - 96*d^3*e^2*f^2 - 192*d^4*e*f*g - 96*d^5*g^2 + (8*e^5*f*g + 17*d*e^4*g^2)*x^4 + 2*(3*e^5*
f^2 + 26*d*e^4*f*g + 26*d^2*e^3*g^2)*x^3 + 6*(9*d*e^4*f^2 + 38*d^2*e^3*f*g + 28*d^3*e^2*g^2)*x^2 - 12*(5*d^2*e
^3*f^2 + 24*d^3*e^2*f*g + 20*d^4*e*g^2)*x - 48*(3*d^3*e^2*f^2 + 10*d^4*e*f*g + 7*d^5*g^2 - (3*d^2*e^3*f^2 + 10
*d^3*e^2*f*g + 7*d^4*e*g^2)*x)*log(e*x - d))/(e^4*x - d*e^3)

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.11 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {4 d^{2} \left (d g + e f\right ) \left (7 d g + 3 e f\right ) \log {\left (- d + e x \right )}}{e^{3}} + \frac {e g^{2} x^{4}}{4} + x^{3} \cdot \left (\frac {5 d g^{2}}{3} + \frac {2 e f g}{3}\right ) + x^{2} \cdot \left (\frac {6 d^{2} g^{2}}{e} + 5 d f g + \frac {e f^{2}}{2}\right ) + x \left (\frac {20 d^{3} g^{2}}{e^{2}} + \frac {24 d^{2} f g}{e} + 5 d f^{2}\right ) + \frac {- 8 d^{5} g^{2} - 16 d^{4} e f g - 8 d^{3} e^{2} f^{2}}{- d e^{3} + e^{4} x} \]

[In]

integrate((e*x+d)**5*(g*x+f)**2/(-e**2*x**2+d**2)**2,x)

[Out]

4*d**2*(d*g + e*f)*(7*d*g + 3*e*f)*log(-d + e*x)/e**3 + e*g**2*x**4/4 + x**3*(5*d*g**2/3 + 2*e*f*g/3) + x**2*(
6*d**2*g**2/e + 5*d*f*g + e*f**2/2) + x*(20*d**3*g**2/e**2 + 24*d**2*f*g/e + 5*d*f**2) + (-8*d**5*g**2 - 16*d*
*4*e*f*g - 8*d**3*e**2*f**2)/(-d*e**3 + e**4*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.25 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=-\frac {8 \, {\left (d^{3} e^{2} f^{2} + 2 \, d^{4} e f g + d^{5} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac {3 \, e^{3} g^{2} x^{4} + 4 \, {\left (2 \, e^{3} f g + 5 \, d e^{2} g^{2}\right )} x^{3} + 6 \, {\left (e^{3} f^{2} + 10 \, d e^{2} f g + 12 \, d^{2} e g^{2}\right )} x^{2} + 12 \, {\left (5 \, d e^{2} f^{2} + 24 \, d^{2} e f g + 20 \, d^{3} g^{2}\right )} x}{12 \, e^{2}} + \frac {4 \, {\left (3 \, d^{2} e^{2} f^{2} + 10 \, d^{3} e f g + 7 \, d^{4} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="maxima")

[Out]

-8*(d^3*e^2*f^2 + 2*d^4*e*f*g + d^5*g^2)/(e^4*x - d*e^3) + 1/12*(3*e^3*g^2*x^4 + 4*(2*e^3*f*g + 5*d*e^2*g^2)*x
^3 + 6*(e^3*f^2 + 10*d*e^2*f*g + 12*d^2*e*g^2)*x^2 + 12*(5*d*e^2*f^2 + 24*d^2*e*f*g + 20*d^3*g^2)*x)/e^2 + 4*(
3*d^2*e^2*f^2 + 10*d^3*e*f*g + 7*d^4*g^2)*log(e*x - d)/e^3

Giac [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {4 \, {\left (3 \, d^{2} e^{2} f^{2} + 10 \, d^{3} e f g + 7 \, d^{4} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {8 \, {\left (d^{3} e^{2} f^{2} + 2 \, d^{4} e f g + d^{5} g^{2}\right )}}{{\left (e x - d\right )} e^{3}} + \frac {3 \, e^{9} g^{2} x^{4} + 8 \, e^{9} f g x^{3} + 20 \, d e^{8} g^{2} x^{3} + 6 \, e^{9} f^{2} x^{2} + 60 \, d e^{8} f g x^{2} + 72 \, d^{2} e^{7} g^{2} x^{2} + 60 \, d e^{8} f^{2} x + 288 \, d^{2} e^{7} f g x + 240 \, d^{3} e^{6} g^{2} x}{12 \, e^{8}} \]

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="giac")

[Out]

4*(3*d^2*e^2*f^2 + 10*d^3*e*f*g + 7*d^4*g^2)*log(abs(e*x - d))/e^3 - 8*(d^3*e^2*f^2 + 2*d^4*e*f*g + d^5*g^2)/(
(e*x - d)*e^3) + 1/12*(3*e^9*g^2*x^4 + 8*e^9*f*g*x^3 + 20*d*e^8*g^2*x^3 + 6*e^9*f^2*x^2 + 60*d*e^8*f*g*x^2 + 7
2*d^2*e^7*g^2*x^2 + 60*d*e^8*f^2*x + 288*d^2*e^7*f*g*x + 240*d^3*e^6*g^2*x)/e^8

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.16 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=x\,\left (\frac {d^3\,g^2+6\,d^2\,e\,f\,g+3\,d\,e^2\,f^2}{e^2}-\frac {d^2\,\left (g\,\left (3\,d\,g+2\,e\,f\right )+2\,d\,g^2\right )}{e^2}+\frac {2\,d\,\left (\frac {3\,d^2\,e\,g^2+6\,d\,e^2\,f\,g+e^3\,f^2}{e^2}-\frac {d^2\,g^2}{e}+\frac {2\,d\,\left (g\,\left (3\,d\,g+2\,e\,f\right )+2\,d\,g^2\right )}{e}\right )}{e}\right )+x^2\,\left (\frac {3\,d^2\,e\,g^2+6\,d\,e^2\,f\,g+e^3\,f^2}{2\,e^2}-\frac {d^2\,g^2}{2\,e}+\frac {d\,\left (g\,\left (3\,d\,g+2\,e\,f\right )+2\,d\,g^2\right )}{e}\right )+x^3\,\left (\frac {g\,\left (3\,d\,g+2\,e\,f\right )}{3}+\frac {2\,d\,g^2}{3}\right )+\frac {\ln \left (e\,x-d\right )\,\left (28\,d^4\,g^2+40\,d^3\,e\,f\,g+12\,d^2\,e^2\,f^2\right )}{e^3}+\frac {8\,\left (d^5\,g^2+2\,d^4\,e\,f\,g+d^3\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {e\,g^2\,x^4}{4} \]

[In]

int(((f + g*x)^2*(d + e*x)^5)/(d^2 - e^2*x^2)^2,x)

[Out]

x*((d^3*g^2 + 3*d*e^2*f^2 + 6*d^2*e*f*g)/e^2 - (d^2*(g*(3*d*g + 2*e*f) + 2*d*g^2))/e^2 + (2*d*((e^3*f^2 + 3*d^
2*e*g^2 + 6*d*e^2*f*g)/e^2 - (d^2*g^2)/e + (2*d*(g*(3*d*g + 2*e*f) + 2*d*g^2))/e))/e) + x^2*((e^3*f^2 + 3*d^2*
e*g^2 + 6*d*e^2*f*g)/(2*e^2) - (d^2*g^2)/(2*e) + (d*(g*(3*d*g + 2*e*f) + 2*d*g^2))/e) + x^3*((g*(3*d*g + 2*e*f
))/3 + (2*d*g^2)/3) + (log(e*x - d)*(28*d^4*g^2 + 12*d^2*e^2*f^2 + 40*d^3*e*f*g))/e^3 + (8*(d^5*g^2 + d^3*e^2*
f^2 + 2*d^4*e*f*g))/(e*(d*e^2 - e^3*x)) + (e*g^2*x^4)/4