3.6.71 \(\int \frac {(d+e x)^5 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [571]

Optimal. Leaf size=118 \[ -\frac {g (2 e f+5 d g) x}{e^2}-\frac {g^2 x^2}{2 e}+\frac {2 d^2 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d (e f+d g) (e f+3 d g)}{e^3 (d-e x)}-\frac {\left (e^2 f^2+10 d e f g+13 d^2 g^2\right ) \log (d-e x)}{e^3} \]

[Out]

-g*(5*d*g+2*e*f)*x/e^2-1/2*g^2*x^2/e+2*d^2*(d*g+e*f)^2/e^3/(-e*x+d)^2-4*d*(d*g+e*f)*(3*d*g+e*f)/e^3/(-e*x+d)-(
13*d^2*g^2+10*d*e*f*g+e^2*f^2)*ln(-e*x+d)/e^3

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Rubi [A]
time = 0.10, antiderivative size = 118, 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 {2 d^2 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {\left (13 d^2 g^2+10 d e f g+e^2 f^2\right ) \log (d-e x)}{e^3}-\frac {4 d (3 d g+e f) (d g+e f)}{e^3 (d-e x)}-\frac {g x (5 d g+2 e f)}{e^2}-\frac {g^2 x^2}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-((g*(2*e*f + 5*d*g)*x)/e^2) - (g^2*x^2)/(2*e) + (2*d^2*(e*f + d*g)^2)/(e^3*(d - e*x)^2) - (4*d*(e*f + d*g)*(e
*f + 3*d*g))/(e^3*(d - e*x)) - ((e^2*f^2 + 10*d*e*f*g + 13*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)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^2 (f+g x)^2}{(d-e x)^3} \, dx\\ &=\int \left (-\frac {g (2 e f+5 d g)}{e^2}-\frac {g^2 x}{e}+\frac {4 d (-e f-3 d g) (e f+d g)}{e^2 (d-e x)^2}-\frac {4 d^2 (e f+d g)^2}{e^2 (-d+e x)^3}+\frac {-e^2 f^2-10 d e f g-13 d^2 g^2}{e^2 (-d+e x)}\right ) \, dx\\ &=-\frac {g (2 e f+5 d g) x}{e^2}-\frac {g^2 x^2}{2 e}+\frac {2 d^2 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d (e f+d g) (e f+3 d g)}{e^3 (d-e x)}-\frac {\left (e^2 f^2+10 d e f g+13 d^2 g^2\right ) \log (d-e x)}{e^3}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 118, normalized size = 1.00 \begin {gather*} -\frac {2 e g (2 e f+5 d g) x+e^2 g^2 x^2-\frac {4 d^2 (e f+d g)^2}{(d-e x)^2}+\frac {8 d \left (e^2 f^2+4 d e f g+3 d^2 g^2\right )}{d-e x}+2 \left (e^2 f^2+10 d e f g+13 d^2 g^2\right ) \log (d-e x)}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-1/2*(2*e*g*(2*e*f + 5*d*g)*x + e^2*g^2*x^2 - (4*d^2*(e*f + d*g)^2)/(d - e*x)^2 + (8*d*(e^2*f^2 + 4*d*e*f*g +
3*d^2*g^2))/(d - e*x) + 2*(e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*Log[d - e*x])/e^3

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Maple [A]
time = 0.07, size = 133, normalized size = 1.13

method	result	size
default	\(-\frac {g \left (\frac {1}{2} e g \,x^{2}+5 d g x +2 e f x \right )}{e^{2}}-\frac {4 d \left (3 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}+\frac {2 d^{2} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )^{2}}+\frac {\left (-13 d^{2} g^{2}-10 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(133\)
risch	\(-\frac {g^{2} x^{2}}{2 e}-\frac {5 g^{2} d x}{e^{2}}-\frac {2 g f x}{e}+\frac {\left (12 d^{3} g^{2}+16 d^{2} e f g +4 d \,e^{2} f^{2}\right ) x -\frac {2 d^{2} \left (5 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right )}{e}}{e^{2} \left (-e x +d \right )^{2}}-\frac {13 \ln \left (-e x +d \right ) d^{2} g^{2}}{e^{3}}-\frac {10 \ln \left (-e x +d \right ) d f g}{e^{2}}-\frac {\ln \left (-e x +d \right ) f^{2}}{e}\)	\(150\)
norman	\(\frac {\left (22 d^{3} g^{2}+20 d^{2} e f g +4 d \,e^{2} f^{2}\right ) x^{3}-\frac {d^{4} \left (11 g^{2} d^{2} e +12 f g d \,e^{2}+2 f^{2} e^{3}\right )}{e^{4}}-\frac {e^{3} g^{2} x^{6}}{2}+\frac {d^{2} \left (31 g^{2} d^{2} e +40 f g d \,e^{2}+12 f^{2} e^{3}\right ) x^{2}}{2 e^{2}}-e^{2} g \left (5 d g +2 e f \right ) x^{5}-\frac {d^{4} g \left (13 d g +10 e f \right ) x}{e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {\left (13 d^{2} g^{2}+10 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x,method=_RETURNVERBOSE)

[Out]

-g/e^2*(1/2*e*g*x^2+5*d*g*x+2*e*f*x)-4*d/e^3*(3*d^2*g^2+4*d*e*f*g+e^2*f^2)/(-e*x+d)+2*d^2*(d^2*g^2+2*d*e*f*g+e
^2*f^2)/e^3/(-e*x+d)^2+(-13*d^2*g^2-10*d*e*f*g-e^2*f^2)/e^3*ln(-e*x+d)

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Maxima [A]
time = 0.29, size = 146, normalized size = 1.24 \begin {gather*} -{\left (13 \, d^{2} g^{2} + 10 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right ) - \frac {1}{2} \, {\left (g^{2} x^{2} e + 2 \, {\left (5 \, d g^{2} + 2 \, f g e\right )} x\right )} e^{\left (-2\right )} - \frac {2 \, {\left (5 \, d^{4} g^{2} + 6 \, d^{3} f g e + d^{2} f^{2} e^{2} - 2 \, {\left (3 \, d^{3} g^{2} e + 4 \, d^{2} f g e^{2} + d 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)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

-(13*d^2*g^2 + 10*d*f*g*e + f^2*e^2)*e^(-3)*log(x*e - d) - 1/2*(g^2*x^2*e + 2*(5*d*g^2 + 2*f*g*e)*x)*e^(-2) -
2*(5*d^4*g^2 + 6*d^3*f*g*e + d^2*f^2*e^2 - 2*(3*d^3*g^2*e + 4*d^2*f*g*e^2 + d*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 = 3.51, size = 231, normalized size = 1.96 \begin {gather*} -\frac {20 \, d^{4} g^{2} + {\left (g^{2} x^{4} + 4 \, f g x^{3}\right )} e^{4} + 8 \, {\left (d g^{2} x^{3} - d f g x^{2} - d f^{2} x\right )} e^{3} - {\left (19 \, d^{2} g^{2} x^{2} + 28 \, d^{2} f g x - 4 \, d^{2} f^{2}\right )} e^{2} - 2 \, {\left (7 \, d^{3} g^{2} x - 12 \, d^{3} f g\right )} e + 2 \, {\left (13 \, d^{4} g^{2} + f^{2} x^{2} e^{4} + 2 \, {\left (5 \, d f g x^{2} - d f^{2} x\right )} e^{3} + {\left (13 \, d^{2} g^{2} x^{2} - 20 \, d^{2} f g x + d^{2} f^{2}\right )} e^{2} - 2 \, {\left (13 \, d^{3} g^{2} x - 5 \, d^{3} f g\right )} e\right )} \log \left (x e - d\right )}{2 \, {\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)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

-1/2*(20*d^4*g^2 + (g^2*x^4 + 4*f*g*x^3)*e^4 + 8*(d*g^2*x^3 - d*f*g*x^2 - d*f^2*x)*e^3 - (19*d^2*g^2*x^2 + 28*
d^2*f*g*x - 4*d^2*f^2)*e^2 - 2*(7*d^3*g^2*x - 12*d^3*f*g)*e + 2*(13*d^4*g^2 + f^2*x^2*e^4 + 2*(5*d*f*g*x^2 - d
*f^2*x)*e^3 + (13*d^2*g^2*x^2 - 20*d^2*f*g*x + d^2*f^2)*e^2 - 2*(13*d^3*g^2*x - 5*d^3*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 = 0.57, size = 151, normalized size = 1.28 \begin {gather*} - x \left (\frac {5 d g^{2}}{e^{2}} + \frac {2 f g}{e}\right ) - \frac {10 d^{4} g^{2} + 12 d^{3} e f g + 2 d^{2} e^{2} f^{2} + x \left (- 12 d^{3} e g^{2} - 16 d^{2} e^{2} f g - 4 d e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} - \frac {g^{2} x^{2}}{2 e} - \frac {\left (13 d^{2} g^{2} + 10 d e f g + e^{2} f^{2}\right ) \log {\left (- d + e x \right )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**5*(g*x+f)**2/(-e**2*x**2+d**2)**3,x)

[Out]

-x*(5*d*g**2/e**2 + 2*f*g/e) - (10*d**4*g**2 + 12*d**3*e*f*g + 2*d**2*e**2*f**2 + x*(-12*d**3*e*g**2 - 16*d**2
*e**2*f*g - 4*d*e**3*f**2))/(d**2*e**3 - 2*d*e**4*x + e**5*x**2) - g**2*x**2/(2*e) - (13*d**2*g**2 + 10*d*e*f*
g + e**2*f**2)*log(-d + e*x)/e**3

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Giac [A]
time = 2.02, size = 138, normalized size = 1.17 \begin {gather*} -{\left (13 \, d^{2} g^{2} + 10 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) - \frac {1}{2} \, {\left (g^{2} x^{2} e^{5} + 10 \, d g^{2} x e^{4} + 4 \, f g x e^{5}\right )} e^{\left (-6\right )} - \frac {2 \, {\left (5 \, d^{4} g^{2} + 6 \, d^{3} f g e + d^{2} f^{2} e^{2} - 2 \, {\left (3 \, d^{3} g^{2} e + 4 \, d^{2} f g e^{2} + d 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)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="giac")

[Out]

-(13*d^2*g^2 + 10*d*f*g*e + f^2*e^2)*e^(-3)*log(abs(x*e - d)) - 1/2*(g^2*x^2*e^5 + 10*d*g^2*x*e^4 + 4*f*g*x*e^
5)*e^(-6) - 2*(5*d^4*g^2 + 6*d^3*f*g*e + d^2*f^2*e^2 - 2*(3*d^3*g^2*e + 4*d^2*f*g*e^2 + d*f^2*e^3)*x)*e^(-3)/(
x*e - d)^2

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Mupad [B]
time = 2.60, size = 161, normalized size = 1.36 \begin {gather*} -\frac {\frac {2\,\left (5\,d^4\,g^2+6\,d^3\,e\,f\,g+d^2\,e^2\,f^2\right )}{e}-x\,\left (12\,d^3\,g^2+16\,d^2\,e\,f\,g+4\,d\,e^2\,f^2\right )}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-x\,\left (\frac {2\,g\,\left (d\,g+e\,f\right )}{e^2}+\frac {3\,d\,g^2}{e^2}\right )-\frac {\ln \left (e\,x-d\right )\,\left (13\,d^2\,g^2+10\,d\,e\,f\,g+e^2\,f^2\right )}{e^3}-\frac {g^2\,x^2}{2\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^2*(d + e*x)^5)/(d^2 - e^2*x^2)^3,x)

[Out]

- ((2*(5*d^4*g^2 + d^2*e^2*f^2 + 6*d^3*e*f*g))/e - x*(12*d^3*g^2 + 4*d*e^2*f^2 + 16*d^2*e*f*g))/(d^2*e^2 + e^4
*x^2 - 2*d*e^3*x) - x*((2*g*(d*g + e*f))/e^2 + (3*d*g^2)/e^2) - (log(e*x - d)*(13*d^2*g^2 + e^2*f^2 + 10*d*e*f
*g))/e^3 - (g^2*x^2)/(2*e)

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