3.6.70 \(\int \frac {(d+e x)^6 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [570]

Optimal. Leaf size=149 \[ -\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (e f+d g) (3 e f+7 d g)}{e^3 (d-e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3} \]

[Out]

-(18*d^2*g^2+12*d*e*f*g+e^2*f^2)*x/e^2-g*(3*d*g+e*f)*x^2/e-1/3*g^2*x^3+4*d^3*(d*g+e*f)^2/e^3/(-e*x+d)^2-4*d^2*
(d*g+e*f)*(7*d*g+3*e*f)/e^3/(-e*x+d)-2*d*(19*d^2*g^2+18*d*e*f*g+3*e^2*f^2)*ln(-e*x+d)/e^3

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Rubi [A]
time = 0.13, antiderivative size = 149, 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 = {862, 90} \begin {gather*} \frac {4 d^3 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (d g+e f) (7 d g+3 e f)}{e^3 (d-e x)}-\frac {x \left (18 d^2 g^2+12 d e f g+e^2 f^2\right )}{e^2}-\frac {2 d \left (19 d^2 g^2+18 d e f g+3 e^2 f^2\right ) \log (d-e x)}{e^3}-\frac {g x^2 (3 d g+e f)}{e}-\frac {g^2 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^6*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-(((e^2*f^2 + 12*d*e*f*g + 18*d^2*g^2)*x)/e^2) - (g*(e*f + 3*d*g)*x^2)/e - (g^2*x^3)/3 + (4*d^3*(e*f + d*g)^2)
/(e^3*(d - e*x)^2) - (4*d^2*(e*f + d*g)*(3*e*f + 7*d*g))/(e^3*(d - e*x)) - (2*d*(3*e^2*f^2 + 18*d*e*f*g + 19*d
^2*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*} \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^3 (f+g x)^2}{(d-e x)^3} \, dx\\ &=\int \left (\frac {-e^2 f^2-12 d e f g-18 d^2 g^2}{e^2}-\frac {2 g (e f+3 d g) x}{e}-g^2 x^2+\frac {4 d^2 (-3 e f-7 d g) (e f+d g)}{e^2 (d-e x)^2}-\frac {8 d^3 (e f+d g)^2}{e^2 (-d+e x)^3}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right )}{e^2 (-d+e x)}\right ) \, dx\\ &=-\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (e f+d g) (3 e f+7 d g)}{e^3 (d-e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 157, normalized size = 1.05 \begin {gather*} -\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}+\frac {4 d^2 \left (3 e^2 f^2+10 d e f g+7 d^2 g^2\right )}{e^3 (-d+e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^6*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-(((e^2*f^2 + 12*d*e*f*g + 18*d^2*g^2)*x)/e^2) - (g*(e*f + 3*d*g)*x^2)/e - (g^2*x^3)/3 + (4*d^3*(e*f + d*g)^2)
/(e^3*(d - e*x)^2) + (4*d^2*(3*e^2*f^2 + 10*d*e*f*g + 7*d^2*g^2))/(e^3*(-d + e*x)) - (2*d*(3*e^2*f^2 + 18*d*e*
f*g + 19*d^2*g^2)*Log[d - e*x])/e^3

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Maple [A]
time = 0.09, size = 174, normalized size = 1.17

method	result	size
default	\(-\frac {\frac {1}{3} g^{2} x^{3} e^{2}+3 d e \,g^{2} x^{2}+e^{2} f g \,x^{2}+18 d^{2} g^{2} x +12 d e f g x +e^{2} f^{2} x}{e^{2}}+\frac {4 d^{3} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )^{2}}-\frac {2 d \left (19 d^{2} g^{2}+18 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}-\frac {4 d^{2} \left (7 d^{2} g^{2}+10 d e f g +3 e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}\)	\(174\)
risch	\(-\frac {g^{2} x^{3}}{3}-\frac {3 d \,g^{2} x^{2}}{e}-f g \,x^{2}-\frac {18 d^{2} g^{2} x}{e^{2}}-\frac {12 d f g x}{e}-f^{2} x -\frac {\left (-28 d^{4} g^{2}-40 f g \,d^{3} e -12 d^{2} e^{2} f^{2}\right ) x +\frac {8 d^{3} \left (3 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right )}{e}}{e^{2} \left (-e x +d \right )^{2}}-\frac {38 d^{3} \ln \left (-e x +d \right ) g^{2}}{e^{3}}-\frac {36 d^{2} \ln \left (-e x +d \right ) f g}{e^{2}}-\frac {6 d \ln \left (-e x +d \right ) f^{2}}{e}\)	\(181\)
norman	\(\frac {\left (\frac {191}{3} d^{4} g^{2}+64 f g \,d^{3} e +14 d^{2} e^{2} f^{2}\right ) x^{3}+\left (-\frac {52}{3} g^{2} d^{2} e^{2}-12 f g d \,e^{3}-f^{2} e^{4}\right ) x^{5}+\frac {d^{2} \left (41 g^{2} d^{3} e +51 f g \,d^{2} e^{2}+16 f^{2} d \,e^{3}\right ) x^{2}}{e^{2}}-\frac {d^{4} \left (30 g^{2} d^{3} e +34 f g \,d^{2} e^{2}+8 f^{2} d \,e^{3}\right )}{e^{4}}-\frac {g^{2} e^{4} x^{7}}{3}-\frac {d^{4} \left (38 d^{2} g^{2}+36 d e f g +5 e^{2} f^{2}\right ) x}{e^{2}}-e^{3} g \left (3 d g +e f \right ) x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {2 d \left (19 d^{2} g^{2}+18 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^3,x,method=_RETURNVERBOSE)

[Out]

-1/e^2*(1/3*g^2*x^3*e^2+3*d*e*g^2*x^2+e^2*f*g*x^2+18*d^2*g^2*x+12*d*e*f*g*x+e^2*f^2*x)+4*d^3*(d^2*g^2+2*d*e*f*
g+e^2*f^2)/e^3/(-e*x+d)^2-2*d*(19*d^2*g^2+18*d*e*f*g+3*e^2*f^2)*ln(-e*x+d)/e^3-4*d^2/e^3*(7*d^2*g^2+10*d*e*f*g
+3*e^2*f^2)/(-e*x+d)

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Maxima [A]
time = 0.29, size = 182, normalized size = 1.22 \begin {gather*} -2 \, {\left (19 \, d^{3} g^{2} + 18 \, d^{2} f g e + 3 \, d f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right ) - \frac {1}{3} \, {\left (g^{2} x^{3} e^{2} + 3 \, {\left (3 \, d g^{2} e + f g e^{2}\right )} x^{2} + 3 \, {\left (18 \, d^{2} g^{2} + 12 \, d f g e + f^{2} e^{2}\right )} x\right )} e^{\left (-2\right )} - \frac {4 \, {\left (6 \, d^{5} g^{2} + 8 \, d^{4} f g e + 2 \, d^{3} f^{2} e^{2} - {\left (7 \, d^{4} g^{2} e + 10 \, d^{3} f g e^{2} + 3 \, d^{2} f^{2} e^{3}\right )} x\right )}}{x^{2} e^{5} - 2 \, d x e^{4} + d^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

-2*(19*d^3*g^2 + 18*d^2*f*g*e + 3*d*f^2*e^2)*e^(-3)*log(x*e - d) - 1/3*(g^2*x^3*e^2 + 3*(3*d*g^2*e + f*g*e^2)*
x^2 + 3*(18*d^2*g^2 + 12*d*f*g*e + f^2*e^2)*x)*e^(-2) - 4*(6*d^5*g^2 + 8*d^4*f*g*e + 2*d^3*f^2*e^2 - (7*d^4*g^
2*e + 10*d^3*f*g*e^2 + 3*d^2*f^2*e^3)*x)/(x^2*e^5 - 2*d*x*e^4 + d^2*e^3)

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Fricas [A]
time = 9.63, size = 282, normalized size = 1.89 \begin {gather*} -\frac {72 \, d^{5} g^{2} + {\left (g^{2} x^{5} + 3 \, f g x^{4} + 3 \, f^{2} x^{3}\right )} e^{5} + {\left (7 \, d g^{2} x^{4} + 30 \, d f g x^{3} - 6 \, d f^{2} x^{2}\right )} e^{4} + {\left (37 \, d^{2} g^{2} x^{3} - 69 \, d^{2} f g x^{2} - 33 \, d^{2} f^{2} x\right )} e^{3} - 3 \, {\left (33 \, d^{3} g^{2} x^{2} + 28 \, d^{3} f g x - 8 \, d^{3} f^{2}\right )} e^{2} - 6 \, {\left (5 \, d^{4} g^{2} x - 16 \, d^{4} f g\right )} e + 6 \, {\left (19 \, d^{5} g^{2} + 3 \, d f^{2} x^{2} e^{4} + 6 \, {\left (3 \, d^{2} f g x^{2} - d^{2} f^{2} x\right )} e^{3} + {\left (19 \, d^{3} g^{2} x^{2} - 36 \, d^{3} f g x + 3 \, d^{3} f^{2}\right )} e^{2} - 2 \, {\left (19 \, d^{4} g^{2} x - 9 \, d^{4} f g\right )} e\right )} \log \left (x e - d\right )}{3 \, {\left (x^{2} e^{5} - 2 \, d x e^{4} + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

-1/3*(72*d^5*g^2 + (g^2*x^5 + 3*f*g*x^4 + 3*f^2*x^3)*e^5 + (7*d*g^2*x^4 + 30*d*f*g*x^3 - 6*d*f^2*x^2)*e^4 + (3
7*d^2*g^2*x^3 - 69*d^2*f*g*x^2 - 33*d^2*f^2*x)*e^3 - 3*(33*d^3*g^2*x^2 + 28*d^3*f*g*x - 8*d^3*f^2)*e^2 - 6*(5*
d^4*g^2*x - 16*d^4*f*g)*e + 6*(19*d^5*g^2 + 3*d*f^2*x^2*e^4 + 6*(3*d^2*f*g*x^2 - d^2*f^2*x)*e^3 + (19*d^3*g^2*
x^2 - 36*d^3*f*g*x + 3*d^3*f^2)*e^2 - 2*(19*d^4*g^2*x - 9*d^4*f*g)*e)*log(x*e - d))/(x^2*e^5 - 2*d*x*e^4 + d^2
*e^3)

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Sympy [A]
time = 0.64, size = 178, normalized size = 1.19 \begin {gather*} - \frac {2 d \left (19 d^{2} g^{2} + 18 d e f g + 3 e^{2} f^{2}\right ) \log {\left (- d + e x \right )}}{e^{3}} - \frac {g^{2} x^{3}}{3} - x^{2} \cdot \left (\frac {3 d g^{2}}{e} + f g\right ) - x \left (\frac {18 d^{2} g^{2}}{e^{2}} + \frac {12 d f g}{e} + f^{2}\right ) - \frac {24 d^{5} g^{2} + 32 d^{4} e f g + 8 d^{3} e^{2} f^{2} + x \left (- 28 d^{4} e g^{2} - 40 d^{3} e^{2} f g - 12 d^{2} e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**6*(g*x+f)**2/(-e**2*x**2+d**2)**3,x)

[Out]

-2*d*(19*d**2*g**2 + 18*d*e*f*g + 3*e**2*f**2)*log(-d + e*x)/e**3 - g**2*x**3/3 - x**2*(3*d*g**2/e + f*g) - x*
(18*d**2*g**2/e**2 + 12*d*f*g/e + f**2) - (24*d**5*g**2 + 32*d**4*e*f*g + 8*d**3*e**2*f**2 + x*(-28*d**4*e*g**
2 - 40*d**3*e**2*f*g - 12*d**2*e**3*f**2))/(d**2*e**3 - 2*d*e**4*x + e**5*x**2)

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Giac [A]
time = 2.23, size = 177, normalized size = 1.19 \begin {gather*} -2 \, {\left (19 \, d^{3} g^{2} + 18 \, d^{2} f g e + 3 \, d f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) - \frac {1}{3} \, {\left (g^{2} x^{3} e^{9} + 9 \, d g^{2} x^{2} e^{8} + 54 \, d^{2} g^{2} x e^{7} + 3 \, f g x^{2} e^{9} + 36 \, d f g x e^{8} + 3 \, f^{2} x e^{9}\right )} e^{\left (-9\right )} - \frac {4 \, {\left (6 \, d^{5} g^{2} + 8 \, d^{4} f g e + 2 \, d^{3} f^{2} e^{2} - {\left (7 \, d^{4} g^{2} e + 10 \, d^{3} f g e^{2} + 3 \, d^{2} f^{2} e^{3}\right )} x\right )} e^{\left (-3\right )}}{{\left (x e - d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="giac")

[Out]

-2*(19*d^3*g^2 + 18*d^2*f*g*e + 3*d*f^2*e^2)*e^(-3)*log(abs(x*e - d)) - 1/3*(g^2*x^3*e^9 + 9*d*g^2*x^2*e^8 + 5
4*d^2*g^2*x*e^7 + 3*f*g*x^2*e^9 + 36*d*f*g*x*e^8 + 3*f^2*x*e^9)*e^(-9) - 4*(6*d^5*g^2 + 8*d^4*f*g*e + 2*d^3*f^
2*e^2 - (7*d^4*g^2*e + 10*d^3*f*g*e^2 + 3*d^2*f^2*e^3)*x)*e^(-3)/(x*e - d)^2

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Mupad [B]
time = 0.10, size = 240, normalized size = 1.61 \begin {gather*} \frac {x\,\left (28\,d^4\,g^2+40\,d^3\,e\,f\,g+12\,d^2\,e^2\,f^2\right )-\frac {8\,\left (3\,d^5\,g^2+4\,d^4\,e\,f\,g+d^3\,e^2\,f^2\right )}{e}}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-x\,\left (\frac {3\,d^2\,e\,g^2+6\,d\,e^2\,f\,g+e^3\,f^2}{e^3}+\frac {3\,d\,\left (\frac {g\,\left (3\,d\,g+2\,e\,f\right )}{e}+\frac {3\,d\,g^2}{e}\right )}{e}-\frac {3\,d^2\,g^2}{e^2}\right )-x^2\,\left (\frac {g\,\left (3\,d\,g+2\,e\,f\right )}{2\,e}+\frac {3\,d\,g^2}{2\,e}\right )-\frac {g^2\,x^3}{3}-\frac {\ln \left (e\,x-d\right )\,\left (38\,d^3\,g^2+36\,d^2\,e\,f\,g+6\,d\,e^2\,f^2\right )}{e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^2*(d + e*x)^6)/(d^2 - e^2*x^2)^3,x)

[Out]

(x*(28*d^4*g^2 + 12*d^2*e^2*f^2 + 40*d^3*e*f*g) - (8*(3*d^5*g^2 + d^3*e^2*f^2 + 4*d^4*e*f*g))/e)/(d^2*e^2 + e^
4*x^2 - 2*d*e^3*x) - x*((e^3*f^2 + 3*d^2*e*g^2 + 6*d*e^2*f*g)/e^3 + (3*d*((g*(3*d*g + 2*e*f))/e + (3*d*g^2)/e)
)/e - (3*d^2*g^2)/e^2) - x^2*((g*(3*d*g + 2*e*f))/(2*e) + (3*d*g^2)/(2*e)) - (g^2*x^3)/3 - (log(e*x - d)*(38*d
^3*g^2 + 6*d*e^2*f^2 + 36*d^2*e*f*g))/e^3

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