\(\int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx\) [551]

Optimal result

Integrand size = 27, antiderivative size = 50 \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {g (e f+d g) x}{e^2}-\frac {(f+g x)^2}{2 e}-\frac {(e f+d g)^2 \log (d-e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {813, 45} \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {(d g+e f)^2 \log (d-e x)}{e^3}-\frac {g x (d g+e f)}{e^2}-\frac {(f+g x)^2}{2 e} \]

[In]

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

[Out]

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

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 813

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^m*(
f + g*x)^(p + 1)*(a/f + (c/g)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m}, x] && EqQ[c*f^2 + a*g^2, 0] && (Integer
Q[p] || (GtQ[a, 0] && GtQ[f, 0] && EqQ[p, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \frac {(f+g x)^2}{d-e x} \, dx \\ & = \int \left (-\frac {g (e f+d g)}{e^2}+\frac {(e f+d g)^2}{e^2 (d-e x)}-\frac {g (f+g x)}{e}\right ) \, dx \\ & = -\frac {g (e f+d g) x}{e^2}-\frac {(f+g x)^2}{2 e}-\frac {(e f+d g)^2 \log (d-e x)}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e g x (4 e f+2 d g+e g x)+2 (e f+d g)^2 \log (d-e x)}{2 e^3} \]

[In]

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

[Out]

-1/2*(e*g*x*(4*e*f + 2*d*g + e*g*x) + 2*(e*f + d*g)^2*Log[d - e*x])/e^3

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.18

method	result	size
default	\(-\frac {g \left (\frac {1}{2} e g \,x^{2}+d g x +2 e f x \right )}{e^{2}}+\frac {\left (-d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(59\)
norman	\(-\frac {g^{2} x^{2}}{2 e}-\frac {g \left (d g +2 e f \right ) x}{e^{2}}-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(61\)
risch	\(-\frac {g^{2} x^{2}}{2 e}-\frac {g^{2} d x}{e^{2}}-\frac {2 g f x}{e}-\frac {\ln \left (-e x +d \right ) d^{2} g^{2}}{e^{3}}-\frac {2 \ln \left (-e x +d \right ) d f g}{e^{2}}-\frac {\ln \left (-e x +d \right ) f^{2}}{e}\)	\(79\)
parallelrisch	\(-\frac {g^{2} x^{2} e^{2}+2 \ln \left (e x -d \right ) d^{2} g^{2}+4 \ln \left (e x -d \right ) d e f g +2 \ln \left (e x -d \right ) e^{2} f^{2}+2 x d e \,g^{2}+4 x \,e^{2} f g}{2 e^{3}}\)	\(79\)

[In]

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

[Out]

-g/e^2*(1/2*e*g*x^2+d*g*x+2*e*f*x)+(-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 = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28 \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e^{2} g^{2} x^{2} + 2 \, {\left (2 \, e^{2} f g + d e g^{2}\right )} x + 2 \, {\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, e^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=- x \left (\frac {d g^{2}}{e^{2}} + \frac {2 f g}{e}\right ) - \frac {g^{2} x^{2}}{2 e} - \frac {\left (d g + e f\right )^{2} \log {\left (- d + e x \right )}}{e^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e g^{2} x^{2} + 2 \, {\left (2 \, e f g + d g^{2}\right )} x}{2 \, e^{2}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e g^{2} x^{2} + 4 \, e f g x + 2 \, d g^{2} x}{2 \, e^{2}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.93 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx=-x\,\left (\frac {d\,g^2}{e^2}+\frac {2\,f\,g}{e}\right )-\frac {\ln \left (e\,x-d\right )\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{e^3}-\frac {g^2\,x^2}{2\,e} \]

[In]

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

[Out]

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