3.41.6 \(\int \frac {-4-40 x-100 x^2+x^5+10 x^6+25 x^7+e^{\frac {4 x^2}{1+5 x}} (8 x^6+20 x^7)}{x^5+10 x^6+25 x^7} \, dx\)

Optimal. Leaf size=24 \[ -\frac {e^9}{2}+e^{\frac {4 x}{5+\frac {1}{x}}}+\frac {1}{x^4}+x \]

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Rubi [A]  time = 0.74, antiderivative size = 19, normalized size of antiderivative = 0.79, number of steps used = 15, number of rules used = 6, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1594, 27, 6742, 44, 43, 6706} \begin {gather*} \frac {1}{x^4}+e^{\frac {4 x^2}{5 x+1}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 40*x - 100*x^2 + x^5 + 10*x^6 + 25*x^7 + E^((4*x^2)/(1 + 5*x))*(8*x^6 + 20*x^7))/(x^5 + 10*x^6 + 25*
x^7),x]

[Out]

E^((4*x^2)/(1 + 5*x)) + x^(-4) + x

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4-40 x-100 x^2+x^5+10 x^6+25 x^7+e^{\frac {4 x^2}{1+5 x}} \left (8 x^6+20 x^7\right )}{x^5 \left (1+10 x+25 x^2\right )} \, dx\\ &=\int \frac {-4-40 x-100 x^2+x^5+10 x^6+25 x^7+e^{\frac {4 x^2}{1+5 x}} \left (8 x^6+20 x^7\right )}{x^5 (1+5 x)^2} \, dx\\ &=\int \left (\frac {1}{(1+5 x)^2}-\frac {4}{x^5 (1+5 x)^2}-\frac {40}{x^4 (1+5 x)^2}-\frac {100}{x^3 (1+5 x)^2}+\frac {10 x}{(1+5 x)^2}+\frac {25 x^2}{(1+5 x)^2}+\frac {4 e^{\frac {4 x^2}{1+5 x}} x (2+5 x)}{(1+5 x)^2}\right ) \, dx\\ &=-\frac {1}{5 (1+5 x)}-4 \int \frac {1}{x^5 (1+5 x)^2} \, dx+4 \int \frac {e^{\frac {4 x^2}{1+5 x}} x (2+5 x)}{(1+5 x)^2} \, dx+10 \int \frac {x}{(1+5 x)^2} \, dx+25 \int \frac {x^2}{(1+5 x)^2} \, dx-40 \int \frac {1}{x^4 (1+5 x)^2} \, dx-100 \int \frac {1}{x^3 (1+5 x)^2} \, dx\\ &=e^{\frac {4 x^2}{1+5 x}}-\frac {1}{5 (1+5 x)}-4 \int \left (\frac {1}{x^5}-\frac {10}{x^4}+\frac {75}{x^3}-\frac {500}{x^2}+\frac {3125}{x}-\frac {3125}{(1+5 x)^2}-\frac {15625}{1+5 x}\right ) \, dx+10 \int \left (-\frac {1}{5 (1+5 x)^2}+\frac {1}{5 (1+5 x)}\right ) \, dx+25 \int \left (\frac {1}{25}+\frac {1}{25 (1+5 x)^2}-\frac {2}{25 (1+5 x)}\right ) \, dx-40 \int \left (\frac {1}{x^4}-\frac {10}{x^3}+\frac {75}{x^2}-\frac {500}{x}+\frac {625}{(1+5 x)^2}+\frac {2500}{1+5 x}\right ) \, dx-100 \int \left (\frac {1}{x^3}-\frac {10}{x^2}+\frac {75}{x}-\frac {125}{(1+5 x)^2}-\frac {375}{1+5 x}\right ) \, dx\\ &=e^{\frac {4 x^2}{1+5 x}}+\frac {1}{x^4}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 19, normalized size = 0.79 \begin {gather*} e^{\frac {4 x^2}{1+5 x}}+\frac {1}{x^4}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 40*x - 100*x^2 + x^5 + 10*x^6 + 25*x^7 + E^((4*x^2)/(1 + 5*x))*(8*x^6 + 20*x^7))/(x^5 + 10*x^6
+ 25*x^7),x]

[Out]

E^((4*x^2)/(1 + 5*x)) + x^(-4) + x

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fricas [A]  time = 0.48, size = 26, normalized size = 1.08 \begin {gather*} \frac {x^{5} + x^{4} e^{\left (\frac {4 \, x^{2}}{5 \, x + 1}\right )} + 1}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^7+8*x^6)*exp(4*x^2/(1+5*x))+25*x^7+10*x^6+x^5-100*x^2-40*x-4)/(25*x^7+10*x^6+x^5),x, algorith
m="fricas")

[Out]

(x^5 + x^4*e^(4*x^2/(5*x + 1)) + 1)/x^4

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giac [A]  time = 0.14, size = 26, normalized size = 1.08 \begin {gather*} \frac {x^{5} + x^{4} e^{\left (\frac {4 \, x^{2}}{5 \, x + 1}\right )} + 1}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^7+8*x^6)*exp(4*x^2/(1+5*x))+25*x^7+10*x^6+x^5-100*x^2-40*x-4)/(25*x^7+10*x^6+x^5),x, algorith
m="giac")

[Out]

(x^5 + x^4*e^(4*x^2/(5*x + 1)) + 1)/x^4

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maple [A]  time = 0.24, size = 19, normalized size = 0.79

method	result	size
risch	\(\frac {1}{x^{4}}+x +{\mathrm e}^{\frac {4 x^{2}}{1+5 x}}\)	\(19\)
norman	\(\frac {1+x^{4} {\mathrm e}^{\frac {4 x^{2}}{1+5 x}}+x^{5}+5 x +5 x^{6}+5 x^{5} {\mathrm e}^{\frac {4 x^{2}}{1+5 x}}}{x^{4} \left (1+5 x \right )}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x^7+8*x^6)*exp(4*x^2/(1+5*x))+25*x^7+10*x^6+x^5-100*x^2-40*x-4)/(25*x^7+10*x^6+x^5),x,method=_RETURNV
ERBOSE)

[Out]

1/x^4+x+exp(4*x^2/(1+5*x))

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maxima [B]  time = 0.43, size = 101, normalized size = 4.21 \begin {gather*} x - \frac {37500 \, x^{4} + 3750 \, x^{3} - 250 \, x^{2} + 25 \, x - 3}{3 \, {\left (5 \, x^{5} + x^{4}\right )}} + \frac {40 \, {\left (1500 \, x^{3} + 150 \, x^{2} - 10 \, x + 1\right )}}{3 \, {\left (5 \, x^{4} + x^{3}\right )}} - \frac {50 \, {\left (150 \, x^{2} + 15 \, x - 1\right )}}{5 \, x^{3} + x^{2}} + e^{\left (\frac {4}{5} \, x + \frac {4}{25 \, {\left (5 \, x + 1\right )}} - \frac {4}{25}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^7+8*x^6)*exp(4*x^2/(1+5*x))+25*x^7+10*x^6+x^5-100*x^2-40*x-4)/(25*x^7+10*x^6+x^5),x, algorith
m="maxima")

[Out]

x - 1/3*(37500*x^4 + 3750*x^3 - 250*x^2 + 25*x - 3)/(5*x^5 + x^4) + 40/3*(1500*x^3 + 150*x^2 - 10*x + 1)/(5*x^
4 + x^3) - 50*(150*x^2 + 15*x - 1)/(5*x^3 + x^2) + e^(4/5*x + 4/25/(5*x + 1) - 4/25)

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mupad [B]  time = 2.70, size = 18, normalized size = 0.75 \begin {gather*} x+{\mathrm {e}}^{\frac {4\,x^2}{5\,x+1}}+\frac {1}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((4*x^2)/(5*x + 1))*(8*x^6 + 20*x^7) - 40*x - 100*x^2 + x^5 + 10*x^6 + 25*x^7 - 4)/(x^5 + 10*x^6 + 25*
x^7),x)

[Out]

x + exp((4*x^2)/(5*x + 1)) + 1/x^4

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sympy [A]  time = 0.18, size = 17, normalized size = 0.71 \begin {gather*} x + e^{\frac {4 x^{2}}{5 x + 1}} + \frac {1}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x**7+8*x**6)*exp(4*x**2/(1+5*x))+25*x**7+10*x**6+x**5-100*x**2-40*x-4)/(25*x**7+10*x**6+x**5),x
)

[Out]

x + exp(4*x**2/(5*x + 1)) + x**(-4)

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