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

Optimal. Leaf size=109 \[ -\frac {4 d^2 (d g+e f)^2 \log (d-e x)}{e^3}-\frac {x^2 \left (4 d^2 g^2+6 d e f g+e^2 f^2\right )}{2 e}-\frac {d x (2 d g+e f) (2 d g+3 e f)}{e^2}-\frac {1}{3} g x^3 (3 d g+2 e f)-\frac {1}{4} e g^2 x^4 \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {848, 88} \begin {gather*} -\frac {x^2 \left (4 d^2 g^2+6 d e f g+e^2 f^2\right )}{2 e}-\frac {4 d^2 (d g+e f)^2 \log (d-e x)}{e^3}-\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 \end {gather*}

Antiderivative was successfully verified.

[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 88

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 848

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*x)/e)^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*} \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx &=\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]  time = 0.05, size = 103, normalized size = 0.94 \begin {gather*} -\frac {48 d^2 (d g+e f)^2 \log (d-e x)+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 )}{12 e^3} \end {gather*}

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^3 (f+g x)^2}{d^2-e^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 139, normalized size = 1.28 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [B]  time = 0.21, size = 211, normalized size = 1.94 \begin {gather*} -2 \, {\left (d^{4} g^{2} e^{3} + 2 \, d^{3} f g e^{4} + d^{2} f^{2} e^{5}\right )} e^{\left (-6\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac {1}{12} \, {\left (3 \, g^{2} x^{4} e^{9} + 12 \, d g^{2} x^{3} e^{8} + 24 \, d^{2} g^{2} x^{2} e^{7} + 48 \, d^{3} g^{2} x e^{6} + 8 \, f g x^{3} e^{9} + 36 \, d f g x^{2} e^{8} + 96 \, d^{2} f g x e^{7} + 6 \, f^{2} x^{2} e^{9} + 36 \, d f^{2} x e^{8}\right )} e^{\left (-8\right )} - \frac {2 \, {\left (d^{5} g^{2} e^{2} + 2 \, d^{4} f g e^{3} + d^{3} f^{2} e^{4}\right )} e^{\left (-5\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 )}{{\left | d \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(d^4*g^2*e^3 + 2*d^3*f*g*e^4 + d^2*f^2*e^5)*e^(-6)*log(abs(x^2*e^2 - d^2)) - 1/12*(3*g^2*x^4*e^9 + 12*d*g^2
*x^3*e^8 + 24*d^2*g^2*x^2*e^7 + 48*d^3*g^2*x*e^6 + 8*f*g*x^3*e^9 + 36*d*f*g*x^2*e^8 + 96*d^2*f*g*x*e^7 + 6*f^2
*x^2*e^9 + 36*d*f^2*x*e^8)*e^(-8) - 2*(d^5*g^2*e^2 + 2*d^4*f*g*e^3 + d^3*f^2*e^4)*e^(-5)*log(abs(2*x*e^2 - 2*a
bs(d)*e)/abs(2*x*e^2 + 2*abs(d)*e))/abs(d)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 145, normalized size = 1.33 \begin {gather*} -\frac {e \,g^{2} x^{4}}{4}-d \,g^{2} x^{3}-\frac {2 e f g \,x^{3}}{3}-\frac {2 d^{2} g^{2} x^{2}}{e}-3 d f g \,x^{2}-\frac {e \,f^{2} x^{2}}{2}-\frac {4 d^{4} g^{2} \ln \left (e x -d \right )}{e^{3}}-\frac {8 d^{3} f g \ln \left (e x -d \right )}{e^{2}}-\frac {4 d^{3} g^{2} x}{e^{2}}-\frac {4 d^{2} f^{2} \ln \left (e x -d \right )}{e}-\frac {8 d^{2} f g x}{e}-3 d \,f^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 138, normalized size = 1.27 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 2.59, size = 197, normalized size = 1.81 \begin {gather*} -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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.48, size = 109, normalized size = 1.00 \begin {gather*} - \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} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________