\(\int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx\) [549]

Optimal result

Integrand size = 29, antiderivative size = 109 \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {d (e f+2 d g) (3 e f+2 d g) x}{e^2}-\frac {\left (e^2 f^2+6 d e f g+4 d^2 g^2\right ) x^2}{2 e}-\frac {1}{3} g (2 e f+3 d g) x^3-\frac {1}{4} e g^2 x^4-\frac {4 d^2 (e f+d g)^2 \log (d-e x)}{e^3} \]

[Out]

-d*(2*d*g+e*f)*(2*d*g+3*e*f)*x/e^2-1/2*(4*d^2*g^2+6*d*e*f*g+e^2*f^2)*x^2/e-1/3*g*(3*d*g+2*e*f)*x^3-1/4*e*g^2*x
^4-4*d^2*(d*g+e*f)^2*ln(-e*x+d)/e^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, 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} \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {4 d^2 (d g+e f)^2 \log (d-e x)}{e^3}-\frac {x^2 \left (4 d^2 g^2+6 d e f g+e^2 f^2\right )}{2 e}-\frac {d x (2 d g+e f) (2 d g+3 e f)}{e^2}-\frac {1}{3} g x^3 (3 d g+2 e f)-\frac {1}{4} e g^2 x^4 \]

[In]

Int[((d + e*x)^3*(f + g*x)^2)/(d^2 - e^2*x^2),x]

[Out]

-((d*(e*f + 2*d*g)*(3*e*f + 2*d*g)*x)/e^2) - ((e^2*f^2 + 6*d*e*f*g + 4*d^2*g^2)*x^2)/(2*e) - (g*(2*e*f + 3*d*g
)*x^3)/3 - (e*g^2*x^4)/4 - (4*d^2*(e*f + d*g)^2*Log[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)^2 (f+g x)^2}{d-e x} \, dx \\ & = \int \left (\frac {d (-3 e f-2 d g) (e f+2 d g)}{e^2}-\frac {\left (e^2 f^2+6 d e f g+4 d^2 g^2\right ) x}{e}-g (2 e f+3 d g) x^2-e g^2 x^3-\frac {4 d^2 (e f+d g)^2}{e^2 (-d+e x)}\right ) \, dx \\ & = -\frac {d (e f+2 d g) (3 e f+2 d g) x}{e^2}-\frac {\left (e^2 f^2+6 d e f g+4 d^2 g^2\right ) x^2}{2 e}-\frac {1}{3} g (2 e f+3 d g) x^3-\frac {1}{4} e g^2 x^4-\frac {4 d^2 (e f+d g)^2 \log (d-e x)}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e x \left (48 d^3 g^2+24 d^2 e g (4 f+g x)+12 d e^2 \left (3 f^2+3 f g x+g^2 x^2\right )+e^3 x \left (6 f^2+8 f g x+3 g^2 x^2\right )\right )+48 d^2 (e f+d g)^2 \log (d-e x)}{12 e^3} \]

[In]

Integrate[((d + e*x)^3*(f + g*x)^2)/(d^2 - e^2*x^2),x]

[Out]

-1/12*(e*x*(48*d^3*g^2 + 24*d^2*e*g*(4*f + g*x) + 12*d*e^2*(3*f^2 + 3*f*g*x + g^2*x^2) + e^3*x*(6*f^2 + 8*f*g*
x + 3*g^2*x^2)) + 48*d^2*(e*f + d*g)^2*Log[d - e*x])/e^3

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.12

method	result	size
norman	\(-\frac {e \,g^{2} x^{4}}{4}-\frac {g \left (3 d g +2 e f \right ) x^{3}}{3}-\frac {\left (4 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right ) x^{2}}{2 e}-\frac {d \left (4 d^{2} g^{2}+8 d e f g +3 e^{2} f^{2}\right ) x}{e^{2}}-\frac {4 d^{2} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(122\)
risch	\(-\frac {e \,g^{2} x^{4}}{4}-x^{3} d \,g^{2}-\frac {2 e \,x^{3} f g}{3}-\frac {2 x^{2} g^{2} d^{2}}{e}-3 x^{2} f g d -\frac {e \,x^{2} f^{2}}{2}-\frac {4 d^{3} g^{2} x}{e^{2}}-\frac {8 d^{2} f g x}{e}-3 d \,f^{2} x -\frac {4 d^{4} \ln \left (-e x +d \right ) g^{2}}{e^{3}}-\frac {8 d^{3} \ln \left (-e x +d \right ) f g}{e^{2}}-\frac {4 d^{2} \ln \left (-e x +d \right ) f^{2}}{e}\)	\(142\)
default	\(-\frac {\frac {g^{2} e^{3} x^{4}}{4}+\frac {\left (\left (2 d g +e f \right ) e^{2} g +e g \left (d e g +e^{2} f \right )\right ) x^{3}}{3}+\frac {\left (\left (2 d g +e f \right ) \left (d e g +e^{2} f \right )+e g \left (2 d^{2} g +3 d e f \right )\right ) x^{2}}{2}+\left (2 d g +e f \right ) \left (2 d^{2} g +3 d e f \right ) x}{e^{2}}-\frac {4 d^{2} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(149\)
parallelrisch	\(-\frac {3 g^{2} e^{4} x^{4}+12 x^{3} d \,e^{3} g^{2}+8 x^{3} e^{4} f g +24 x^{2} d^{2} e^{2} g^{2}+36 x^{2} d \,e^{3} f g +6 x^{2} e^{4} f^{2}+48 \ln \left (e x -d \right ) d^{4} g^{2}+96 \ln \left (e x -d \right ) d^{3} e f g +48 \ln \left (e x -d \right ) d^{2} e^{2} f^{2}+48 x \,d^{3} e \,g^{2}+96 x \,d^{2} e^{2} f g +36 x d \,e^{3} f^{2}}{12 e^{3}}\)	\(158\)

[In]

int((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2),x,method=_RETURNVERBOSE)

[Out]

-1/4*e*g^2*x^4-1/3*g*(3*d*g+2*e*f)*x^3-1/2*(4*d^2*g^2+6*d*e*f*g+e^2*f^2)*x^2/e-d*(4*d^2*g^2+8*d*e*f*g+3*e^2*f^
2)/e^2*x-4*d^2*(d^2*g^2+2*d*e*f*g+e^2*f^2)/e^3*ln(-e*x+d)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {3 \, e^{4} g^{2} x^{4} + 4 \, {\left (2 \, e^{4} f g + 3 \, d e^{3} g^{2}\right )} x^{3} + 6 \, {\left (e^{4} f^{2} + 6 \, d e^{3} f g + 4 \, d^{2} e^{2} g^{2}\right )} x^{2} + 12 \, {\left (3 \, d e^{3} f^{2} + 8 \, d^{2} e^{2} f g + 4 \, d^{3} e g^{2}\right )} x + 48 \, {\left (d^{2} e^{2} f^{2} + 2 \, d^{3} e f g + d^{4} g^{2}\right )} \log \left (e x - d\right )}{12 \, e^{3}} \]

[In]

integrate((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2),x, algorithm="fricas")

[Out]

-1/12*(3*e^4*g^2*x^4 + 4*(2*e^4*f*g + 3*d*e^3*g^2)*x^3 + 6*(e^4*f^2 + 6*d*e^3*f*g + 4*d^2*e^2*g^2)*x^2 + 12*(3
*d*e^3*f^2 + 8*d^2*e^2*f*g + 4*d^3*e*g^2)*x + 48*(d^2*e^2*f^2 + 2*d^3*e*f*g + d^4*g^2)*log(e*x - d))/e^3

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=- \frac {4 d^{2} \left (d g + e f\right )^{2} \log {\left (- d + e x \right )}}{e^{3}} - \frac {e g^{2} x^{4}}{4} - x^{3} \left (d g^{2} + \frac {2 e f g}{3}\right ) - x^{2} \cdot \left (\frac {2 d^{2} g^{2}}{e} + 3 d f g + \frac {e f^{2}}{2}\right ) - x \left (\frac {4 d^{3} g^{2}}{e^{2}} + \frac {8 d^{2} f g}{e} + 3 d f^{2}\right ) \]

[In]

integrate((e*x+d)**3*(g*x+f)**2/(-e**2*x**2+d**2),x)

[Out]

-4*d**2*(d*g + e*f)**2*log(-d + e*x)/e**3 - e*g**2*x**4/4 - x**3*(d*g**2 + 2*e*f*g/3) - x**2*(2*d**2*g**2/e +
3*d*f*g + e*f**2/2) - x*(4*d**3*g**2/e**2 + 8*d**2*f*g/e + 3*d*f**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.27 \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {3 \, e^{3} g^{2} x^{4} + 4 \, {\left (2 \, e^{3} f g + 3 \, d e^{2} g^{2}\right )} x^{3} + 6 \, {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 4 \, d^{2} e g^{2}\right )} x^{2} + 12 \, {\left (3 \, d e^{2} f^{2} + 8 \, d^{2} e f g + 4 \, d^{3} g^{2}\right )} x}{12 \, e^{2}} - \frac {4 \, {\left (d^{2} e^{2} f^{2} + 2 \, d^{3} e f g + d^{4} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

[In]

integrate((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.37 \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {4 \, {\left (d^{2} e^{2} f^{2} + 2 \, d^{3} e f g + d^{4} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {3 \, e^{5} g^{2} x^{4} + 8 \, e^{5} f g x^{3} + 12 \, d e^{4} g^{2} x^{3} + 6 \, e^{5} f^{2} x^{2} + 36 \, d e^{4} f g x^{2} + 24 \, d^{2} e^{3} g^{2} x^{2} + 36 \, d e^{4} f^{2} x + 96 \, d^{2} e^{3} f g x + 48 \, d^{3} e^{2} g^{2} x}{12 \, e^{4}} \]

[In]

integrate((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2),x, algorithm="giac")

[Out]

-4*(d^2*e^2*f^2 + 2*d^3*e*f*g + d^4*g^2)*log(abs(e*x - d))/e^3 - 1/12*(3*e^5*g^2*x^4 + 8*e^5*f*g*x^3 + 12*d*e^
4*g^2*x^3 + 6*e^5*f^2*x^2 + 36*d*e^4*f*g*x^2 + 24*d^2*e^3*g^2*x^2 + 36*d*e^4*f^2*x + 96*d^2*e^3*f*g*x + 48*d^3
*e^2*g^2*x)/e^4

Mupad [B] (verification not implemented)

Time = 12.02 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.81 \[ \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx=-x^3\,\left (\frac {2\,g\,\left (d\,g+e\,f\right )}{3}+\frac {d\,g^2}{3}\right )-x^2\,\left (\frac {d^2\,g^2+4\,d\,e\,f\,g+e^2\,f^2}{2\,e}+\frac {d\,\left (2\,g\,\left (d\,g+e\,f\right )+d\,g^2\right )}{2\,e}\right )-x\,\left (\frac {d\,\left (\frac {d^2\,g^2+4\,d\,e\,f\,g+e^2\,f^2}{e}+\frac {d\,\left (2\,g\,\left (d\,g+e\,f\right )+d\,g^2\right )}{e}\right )}{e}+\frac {2\,d\,f\,\left (d\,g+e\,f\right )}{e}\right )-\frac {\ln \left (e\,x-d\right )\,\left (4\,d^4\,g^2+8\,d^3\,e\,f\,g+4\,d^2\,e^2\,f^2\right )}{e^3}-\frac {e\,g^2\,x^4}{4} \]

[In]

int(((f + g*x)^2*(d + e*x)^3)/(d^2 - e^2*x^2),x)

[Out]

- x^3*((2*g*(d*g + e*f))/3 + (d*g^2)/3) - x^2*((d^2*g^2 + e^2*f^2 + 4*d*e*f*g)/(2*e) + (d*(2*g*(d*g + e*f) + d
*g^2))/(2*e)) - x*((d*((d^2*g^2 + e^2*f^2 + 4*d*e*f*g)/e + (d*(2*g*(d*g + e*f) + d*g^2))/e))/e + (2*d*f*(d*g +
e*f))/e) - (log(e*x - d)*(4*d^4*g^2 + 4*d^2*e^2*f^2 + 8*d^3*e*f*g))/e^3 - (e*g^2*x^4)/4