3.4.60 \(\int \frac {(d+e x)^6 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\)

Optimal. Leaf size=177 \[ \frac {16 d^4 (d g+e f)^2}{e^3 (d-e x)}+\frac {32 d^3 (d g+e f) (2 d g+e f) \log (d-e x)}{e^3}+\frac {1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac {d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac {d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac {1}{2} e g x^4 (3 d g+e f)+\frac {1}{5} e^2 g^2 x^5 \]

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Rubi [A]  time = 0.23, antiderivative size = 177, 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 {1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac {d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac {d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac {16 d^4 (d g+e f)^2}{e^3 (d-e x)}+\frac {32 d^3 (d g+e f) (2 d g+e f) \log (d-e x)}{e^3}+\frac {1}{2} e g x^4 (3 d g+e f)+\frac {1}{5} e^2 g^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(17*e^2*f^2 + 64*d*e*f*g + 48*d^2*g^2)*x)/e^2 + (d*(3*e^2*f^2 + 17*d*e*f*g + 16*d^2*g^2)*x^2)/e + ((e^2*f
^2 + 12*d*e*f*g + 17*d^2*g^2)*x^3)/3 + (e*g*(e*f + 3*d*g)*x^4)/2 + (e^2*g^2*x^5)/5 + (16*d^4*(e*f + d*g)^2)/(e
^3*(d - e*x)) + (32*d^3*(e*f + d*g)*(e*f + 2*d*g)*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)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^4 (f+g x)^2}{(d-e x)^2} \, dx\\ &=\int \left (\frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right )}{e^2}+\frac {2 d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x}{e}+\left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^2+2 e g (e f+3 d g) x^3+e^2 g^2 x^4+\frac {32 d^3 (-e f-2 d g) (e f+d g)}{e^2 (d-e x)}+\frac {16 d^4 (e f+d g)^2}{e^2 (-d+e x)^2}\right ) \, dx\\ &=\frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right ) x}{e^2}+\frac {d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x^2}{e}+\frac {1}{3} \left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^3+\frac {1}{2} e g (e f+3 d g) x^4+\frac {1}{5} e^2 g^2 x^5+\frac {16 d^4 (e f+d g)^2}{e^3 (d-e x)}+\frac {32 d^3 (e f+d g) (e f+2 d g) \log (d-e x)}{e^3}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 185, normalized size = 1.05 \begin {gather*} -\frac {16 d^4 (d g+e f)^2}{e^3 (e x-d)}+\frac {1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac {d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac {d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac {32 d^3 \left (2 d^2 g^2+3 d e f g+e^2 f^2\right ) \log (d-e x)}{e^3}+\frac {1}{2} e g x^4 (3 d g+e f)+\frac {1}{5} e^2 g^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(17*e^2*f^2 + 64*d*e*f*g + 48*d^2*g^2)*x)/e^2 + (d*(3*e^2*f^2 + 17*d*e*f*g + 16*d^2*g^2)*x^2)/e + ((e^2*f
^2 + 12*d*e*f*g + 17*d^2*g^2)*x^3)/3 + (e*g*(e*f + 3*d*g)*x^4)/2 + (e^2*g^2*x^5)/5 - (16*d^4*(e*f + d*g)^2)/(e
^3*(-d + e*x)) + (32*d^3*(e^2*f^2 + 3*d*e*f*g + 2*d^2*g^2)*Log[d - e*x])/e^3

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 288, normalized size = 1.63 \begin {gather*} \frac {6 \, e^{6} g^{2} x^{6} - 480 \, d^{4} e^{2} f^{2} - 960 \, d^{5} e f g - 480 \, d^{6} g^{2} + 3 \, {\left (5 \, e^{6} f g + 13 \, d e^{5} g^{2}\right )} x^{5} + 5 \, {\left (2 \, e^{6} f^{2} + 21 \, d e^{5} f g + 25 \, d^{2} e^{4} g^{2}\right )} x^{4} + 10 \, {\left (8 \, d e^{5} f^{2} + 39 \, d^{2} e^{4} f g + 31 \, d^{3} e^{3} g^{2}\right )} x^{3} + 30 \, {\left (14 \, d^{2} e^{4} f^{2} + 47 \, d^{3} e^{3} f g + 32 \, d^{4} e^{2} g^{2}\right )} x^{2} - 30 \, {\left (17 \, d^{3} e^{3} f^{2} + 64 \, d^{4} e^{2} f g + 48 \, d^{5} e g^{2}\right )} x - 960 \, {\left (d^{4} e^{2} f^{2} + 3 \, d^{5} e f g + 2 \, d^{6} g^{2} - {\left (d^{3} e^{3} f^{2} + 3 \, d^{4} e^{2} f g + 2 \, d^{5} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{30 \, {\left (e^{4} x - d 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)^2,x, algorithm="fricas")

[Out]

1/30*(6*e^6*g^2*x^6 - 480*d^4*e^2*f^2 - 960*d^5*e*f*g - 480*d^6*g^2 + 3*(5*e^6*f*g + 13*d*e^5*g^2)*x^5 + 5*(2*
e^6*f^2 + 21*d*e^5*f*g + 25*d^2*e^4*g^2)*x^4 + 10*(8*d*e^5*f^2 + 39*d^2*e^4*f*g + 31*d^3*e^3*g^2)*x^3 + 30*(14
*d^2*e^4*f^2 + 47*d^3*e^3*f*g + 32*d^4*e^2*g^2)*x^2 - 30*(17*d^3*e^3*f^2 + 64*d^4*e^2*f*g + 48*d^5*e*g^2)*x -
960*(d^4*e^2*f^2 + 3*d^5*e*f*g + 2*d^6*g^2 - (d^3*e^3*f^2 + 3*d^4*e^2*f*g + 2*d^5*e*g^2)*x)*log(e*x - d))/(e^4
*x - d*e^3)

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giac [A]  time = 0.18, size = 327, normalized size = 1.85 \begin {gather*} 16 \, {\left (2 \, d^{5} g^{2} e^{5} + 3 \, d^{4} f g e^{6} + d^{3} f^{2} e^{7}\right )} e^{\left (-8\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) + \frac {1}{30} \, {\left (6 \, g^{2} x^{5} e^{22} + 45 \, d g^{2} x^{4} e^{21} + 170 \, d^{2} g^{2} x^{3} e^{20} + 480 \, d^{3} g^{2} x^{2} e^{19} + 1440 \, d^{4} g^{2} x e^{18} + 15 \, f g x^{4} e^{22} + 120 \, d f g x^{3} e^{21} + 510 \, d^{2} f g x^{2} e^{20} + 1920 \, d^{3} f g x e^{19} + 10 \, f^{2} x^{3} e^{22} + 90 \, d f^{2} x^{2} e^{21} + 510 \, d^{2} f^{2} x e^{20}\right )} e^{\left (-20\right )} + \frac {16 \, {\left (2 \, d^{6} g^{2} e^{6} + 3 \, d^{5} f g e^{7} + d^{4} f^{2} e^{8}\right )} e^{\left (-9\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 |}} - \frac {16 \, {\left (d^{7} g^{2} e^{5} + 2 \, d^{6} f g e^{6} + d^{5} f^{2} e^{7} + {\left (d^{6} g^{2} e^{6} + 2 \, d^{5} f g e^{7} + d^{4} f^{2} e^{8}\right )} x\right )} e^{\left (-8\right )}}{x^{2} e^{2} - d^{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)^2,x, algorithm="giac")

[Out]

16*(2*d^5*g^2*e^5 + 3*d^4*f*g*e^6 + d^3*f^2*e^7)*e^(-8)*log(abs(x^2*e^2 - d^2)) + 1/30*(6*g^2*x^5*e^22 + 45*d*
g^2*x^4*e^21 + 170*d^2*g^2*x^3*e^20 + 480*d^3*g^2*x^2*e^19 + 1440*d^4*g^2*x*e^18 + 15*f*g*x^4*e^22 + 120*d*f*g
*x^3*e^21 + 510*d^2*f*g*x^2*e^20 + 1920*d^3*f*g*x*e^19 + 10*f^2*x^3*e^22 + 90*d*f^2*x^2*e^21 + 510*d^2*f^2*x*e
^20)*e^(-20) + 16*(2*d^6*g^2*e^6 + 3*d^5*f*g*e^7 + d^4*f^2*e^8)*e^(-9)*log(abs(2*x*e^2 - 2*abs(d)*e)/abs(2*x*e
^2 + 2*abs(d)*e))/abs(d) - 16*(d^7*g^2*e^5 + 2*d^6*f*g*e^6 + d^5*f^2*e^7 + (d^6*g^2*e^6 + 2*d^5*f*g*e^7 + d^4*
f^2*e^8)*x)*e^(-8)/(x^2*e^2 - d^2)

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maple [A]  time = 0.01, size = 245, normalized size = 1.38 \begin {gather*} \frac {e^{2} g^{2} x^{5}}{5}+\frac {3 d e \,g^{2} x^{4}}{2}+\frac {e^{2} f g \,x^{4}}{2}+\frac {17 d^{2} g^{2} x^{3}}{3}+4 d e f g \,x^{3}+\frac {e^{2} f^{2} x^{3}}{3}+\frac {16 d^{3} g^{2} x^{2}}{e}+17 d^{2} f g \,x^{2}+3 d e \,f^{2} x^{2}-\frac {16 d^{6} g^{2}}{\left (e x -d \right ) e^{3}}-\frac {32 d^{5} f g}{\left (e x -d \right ) e^{2}}+\frac {64 d^{5} g^{2} \ln \left (e x -d \right )}{e^{3}}-\frac {16 d^{4} f^{2}}{\left (e x -d \right ) e}+\frac {96 d^{4} f g \ln \left (e x -d \right )}{e^{2}}+\frac {48 d^{4} g^{2} x}{e^{2}}+\frac {32 d^{3} f^{2} \ln \left (e x -d \right )}{e}+\frac {64 d^{3} f g x}{e}+17 d^{2} f^{2} x \end {gather*}

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)^2,x)

[Out]

1/5*e^2*g^2*x^5+3/2*d*e*g^2*x^4+1/2*e^2*f*g*x^4+17/3*d^2*g^2*x^3+4*d*e*f*g*x^3+1/3*e^2*f^2*x^3+16*d^3/e*g^2*x^
2+17*d^2*f*g*x^2+3*d*e*f^2*x^2+48*d^4/e^2*g^2*x+64*d^3/e*f*g*x+17*d^2*f^2*x+64*d^5/e^3*g^2*ln(e*x-d)+96*d^4/e^
2*f*g*ln(e*x-d)+32*d^3/e*f^2*ln(e*x-d)-16*d^6/e^3/(e*x-d)*g^2-32*d^5/e^2/(e*x-d)*f*g-16*d^4/e/(e*x-d)*f^2

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maxima [A]  time = 0.45, size = 218, normalized size = 1.23 \begin {gather*} -\frac {16 \, {\left (d^{4} e^{2} f^{2} + 2 \, d^{5} e f g + d^{6} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac {6 \, e^{4} g^{2} x^{5} + 15 \, {\left (e^{4} f g + 3 \, d e^{3} g^{2}\right )} x^{4} + 10 \, {\left (e^{4} f^{2} + 12 \, d e^{3} f g + 17 \, d^{2} e^{2} g^{2}\right )} x^{3} + 30 \, {\left (3 \, d e^{3} f^{2} + 17 \, d^{2} e^{2} f g + 16 \, d^{3} e g^{2}\right )} x^{2} + 30 \, {\left (17 \, d^{2} e^{2} f^{2} + 64 \, d^{3} e f g + 48 \, d^{4} g^{2}\right )} x}{30 \, e^{2}} + \frac {32 \, {\left (d^{3} e^{2} f^{2} + 3 \, d^{4} e f g + 2 \, d^{5} 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)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="maxima")

[Out]

-16*(d^4*e^2*f^2 + 2*d^5*e*f*g + d^6*g^2)/(e^4*x - d*e^3) + 1/30*(6*e^4*g^2*x^5 + 15*(e^4*f*g + 3*d*e^3*g^2)*x
^4 + 10*(e^4*f^2 + 12*d*e^3*f*g + 17*d^2*e^2*g^2)*x^3 + 30*(3*d*e^3*f^2 + 17*d^2*e^2*f*g + 16*d^3*e*g^2)*x^2 +
30*(17*d^2*e^2*f^2 + 64*d^3*e*f*g + 48*d^4*g^2)*x)/e^2 + 32*(d^3*e^2*f^2 + 3*d^4*e*f*g + 2*d^5*g^2)*log(e*x -
d)/e^3

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mupad [B]  time = 2.61, size = 565, normalized size = 3.19 \begin {gather*} x^2\,\left (\frac {2\,d\,\left (d^2\,g^2+3\,d\,e\,f\,g+e^2\,f^2\right )}{e}-\frac {d^2\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{2\,e^2}+\frac {d\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e}\right )+x^4\,\left (\frac {e\,g\,\left (2\,d\,g+e\,f\right )}{2}+\frac {d\,e\,g^2}{2}\right )+x\,\left (\frac {d^4\,g^2+8\,d^3\,e\,f\,g+6\,d^2\,e^2\,f^2}{e^2}-\frac {d^2\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e^2}+\frac {2\,d\,\left (\frac {4\,d\,\left (d^2\,g^2+3\,d\,e\,f\,g+e^2\,f^2\right )}{e}-\frac {d^2\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e^2}+\frac {2\,d\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e}\right )}{e}\right )+x^3\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{3\,e^2}-\frac {d^2\,g^2}{3}+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{3\,e}\right )+\frac {\ln \left (e\,x-d\right )\,\left (64\,d^5\,g^2+96\,d^4\,e\,f\,g+32\,d^3\,e^2\,f^2\right )}{e^3}+\frac {16\,\left (d^6\,g^2+2\,d^5\,e\,f\,g+d^4\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {e^2\,g^2\,x^5}{5} \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)^2,x)

[Out]

x^2*((2*d*(d^2*g^2 + e^2*f^2 + 3*d*e*f*g))/e - (d^2*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/(2*e^2) + (d*((e^4*f^2
+ 6*d^2*e^2*g^2 + 8*d*e^3*f*g)/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e) + x^4*((e*g*(2*d
*g + e*f))/2 + (d*e*g^2)/2) + x*((d^4*g^2 + 6*d^2*e^2*f^2 + 8*d^3*e*f*g)/e^2 - (d^2*((e^4*f^2 + 6*d^2*e^2*g^2
+ 8*d*e^3*f*g)/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e^2 + (2*d*((4*d*(d^2*g^2 + e^2*f^2
+ 3*d*e*f*g))/e - (d^2*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e^2 + (2*d*((e^4*f^2 + 6*d^2*e^2*g^2 + 8*d*e^3*f*g)
/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e))/e) + x^3*((e^4*f^2 + 6*d^2*e^2*g^2 + 8*d*e^3*
f*g)/(3*e^2) - (d^2*g^2)/3 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/(3*e)) + (log(e*x - d)*(64*d^5*g^2 + 32*d
^3*e^2*f^2 + 96*d^4*e*f*g))/e^3 + (16*(d^6*g^2 + d^4*e^2*f^2 + 2*d^5*e*f*g))/(e*(d*e^2 - e^3*x)) + (e^2*g^2*x^
5)/5

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sympy [A]  time = 1.01, size = 199, normalized size = 1.12 \begin {gather*} \frac {32 d^{3} \left (d g + e f\right ) \left (2 d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} + \frac {e^{2} g^{2} x^{5}}{5} + x^{4} \left (\frac {3 d e g^{2}}{2} + \frac {e^{2} f g}{2}\right ) + x^{3} \left (\frac {17 d^{2} g^{2}}{3} + 4 d e f g + \frac {e^{2} f^{2}}{3}\right ) + x^{2} \left (\frac {16 d^{3} g^{2}}{e} + 17 d^{2} f g + 3 d e f^{2}\right ) + x \left (\frac {48 d^{4} g^{2}}{e^{2}} + \frac {64 d^{3} f g}{e} + 17 d^{2} f^{2}\right ) + \frac {- 16 d^{6} g^{2} - 32 d^{5} e f g - 16 d^{4} e^{2} f^{2}}{- d e^{3} + e^{4} x} \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)**2,x)

[Out]

32*d**3*(d*g + e*f)*(2*d*g + e*f)*log(-d + e*x)/e**3 + e**2*g**2*x**5/5 + x**4*(3*d*e*g**2/2 + e**2*f*g/2) + x
**3*(17*d**2*g**2/3 + 4*d*e*f*g + e**2*f**2/3) + x**2*(16*d**3*g**2/e + 17*d**2*f*g + 3*d*e*f**2) + x*(48*d**4
*g**2/e**2 + 64*d**3*f*g/e + 17*d**2*f**2) + (-16*d**6*g**2 - 32*d**5*e*f*g - 16*d**4*e**2*f**2)/(-d*e**3 + e*
*4*x)

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