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

Optimal. Leaf size=50 \[ -\frac {(d g+e f)^2 \log (d-e x)}{e^3}-\frac {g x (d g+e f)}{e^2}-\frac {(f+g x)^2}{2 e} \]

[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

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {799, 43} \[ -\frac {g x (d g+e f)}{e^2}-\frac {(d g+e f)^2 \log (d-e x)}{e^3}-\frac {(f+g x)^2}{2 e} \]

Antiderivative was successfully verified.

[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 43

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 799

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*x)/g)^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*} \int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx &=\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*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.86 \[ -\frac {e g x (2 d g+4 e f+e g x)+2 (d g+e f)^2 \log (d-e x)}{2 e^3} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.90, size = 64, normalized size = 1.28 \[ -\frac {e^{2} g^{2} x^{2} + 2 \, {\left (2 \, e^{2} f g + d e g^{2}\right )} x + 2 \, {\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.17, size = 134, normalized size = 2.68 \[ -\frac {1}{2} \, {\left (d^{2} g^{2} e + 2 \, d f g e^{2} + f^{2} e^{3}\right )} e^{\left (-4\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac {1}{2} \, {\left (g^{2} x^{2} e^{3} + 2 \, d g^{2} x e^{2} + 4 \, f g x e^{3}\right )} e^{\left (-4\right )} - \frac {{\left (d^{3} g^{2} + 2 \, d^{2} f g e + d f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (\frac {{\left | 2 \, x e^{2} - 2 \, {\left | d \right |} e \right |}}{{\left | 2 \, x e^{2} + 2 \, {\left | d \right |} e \right |}}\right )}{2 \, {\left | d \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 82, normalized size = 1.64 \[ -\frac {g^{2} x^{2}}{2 e}-\frac {d^{2} g^{2} \ln \left (e x -d \right )}{e^{3}}-\frac {2 d f g \ln \left (e x -d \right )}{e^{2}}-\frac {d \,g^{2} x}{e^{2}}-\frac {f^{2} \ln \left (e x -d \right )}{e}-\frac {2 f g x}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 63, normalized size = 1.26 \[ -\frac {e g^{2} x^{2} + 2 \, {\left (2 \, e f g + d g^{2}\right )} x}{2 \, e^{2}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 2.61, size = 65, normalized size = 1.30 \[ -x\,\left (\frac {d\,g^2}{e^2}+\frac {2\,f\,g}{e}\right )-\frac {\ln \left (e\,x-d\right )\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{e^3}-\frac {g^2\,x^2}{2\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.29, size = 46, normalized size = 0.92 \[ - x \left (\frac {d g^{2}}{e^{2}} + \frac {2 f g}{e}\right ) - \frac {g^{2} x^{2}}{2 e} - \frac {\left (d g + e f\right )^{2} \log {\left (- d + e x \right )}}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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