Integrand size = 40, antiderivative size = 158 \[ \int \frac {\sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{(5+2 x)^4} \, dx=-\frac {(44378877-7400779 x) \sqrt {3-x+2 x^2}}{5971968}-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}+\frac {158527 \left (3-x+2 x^2\right )^{3/2}}{82944 (5+2 x)^2}-\frac {6467659 \left (3-x+2 x^2\right )^{3/2}}{5971968 (5+2 x)}-\frac {10939 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{256 \sqrt {2}}+\frac {170114729 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{3981312 \sqrt {2}} \]
-3667/1728*(2*x^2-x+3)^(3/2)/(5+2*x)^3+158527/82944*(2*x^2-x+3)^(3/2)/(5+2 *x)^2-6467659/5971968*(2*x^2-x+3)^(3/2)/(5+2*x)-10939/512*arcsinh(1/23*(1- 4*x)*23^(1/2))*2^(1/2)+170114729/7962624*arctanh(1/24*(17-22*x)*2^(1/2)/(2 *x^2-x+3)^(1/2))*2^(1/2)-1/5971968*(44378877-7400779*x)*(2*x^2-x+3)^(1/2)
Time = 0.68 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{(5+2 x)^4} \, dx=\frac {\frac {12 \sqrt {3-x+2 x^2} \left (-327735797-329667508 x-97682900 x^2-5453568 x^3+414720 x^4\right )}{(5+2 x)^3}-170114729 \sqrt {2} \text {arctanh}\left (\frac {1}{6} \left (5+2 x-\sqrt {6-2 x+4 x^2}\right )\right )-85061664 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{3981312} \]
((12*Sqrt[3 - x + 2*x^2]*(-327735797 - 329667508*x - 97682900*x^2 - 545356 8*x^3 + 414720*x^4))/(5 + 2*x)^3 - 170114729*Sqrt[2]*ArcTanh[(5 + 2*x - Sq rt[6 - 2*x + 4*x^2])/6] - 85061664*Sqrt[2]*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/3981312
Time = 0.54 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.275, Rules used = {2181, 27, 2181, 2181, 1231, 27, 1269, 1090, 222, 1154, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt {2 x^2-x+3} \left (5 x^4-x^3+3 x^2+x+2\right )}{(2 x+5)^4} \, dx\) |
| \(\Big \downarrow \) 2181 |
| \(\displaystyle -\frac {1}{216} \int \frac {3 \sqrt {2 x^2-x+3} \left (-2880 x^3+7776 x^2-21168 x+12007\right )}{16 (2 x+5)^3}dx-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sqrt {2 x^2-x+3} \left (-2880 x^3+7776 x^2-21168 x+12007\right )}{(2 x+5)^3}dx}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 2181 |
| \(\displaystyle \frac {\frac {1}{144} \int \frac {\sqrt {2 x^2-x+3} \left (207360 x^2-1712380 x+890709\right )}{(2 x+5)^2}dx+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 2181 |
| \(\displaystyle \frac {\frac {1}{144} \left (-\frac {1}{72} \int \frac {(22099149-59206232 x) \sqrt {2 x^2-x+3}}{2 x+5}dx-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {\frac {1}{144} \left (\frac {1}{72} \left (\frac {1}{32} \int -\frac {576 (28345289-56707776 x)}{(2 x+5) \sqrt {2 x^2-x+3}}dx-2 (44378877-7400779 x) \sqrt {2 x^2-x+3}\right )-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{144} \left (\frac {1}{72} \left (-18 \int \frac {28345289-56707776 x}{(2 x+5) \sqrt {2 x^2-x+3}}dx-2 \sqrt {2 x^2-x+3} (44378877-7400779 x)\right )-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {\frac {1}{144} \left (\frac {1}{72} \left (-18 \left (170114729 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-28353888 \int \frac {1}{\sqrt {2 x^2-x+3}}dx\right )-2 \sqrt {2 x^2-x+3} (44378877-7400779 x)\right )-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {\frac {1}{144} \left (\frac {1}{72} \left (-18 \left (170114729 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-14176944 \sqrt {\frac {2}{23}} \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)\right )-2 \sqrt {2 x^2-x+3} (44378877-7400779 x)\right )-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {\frac {1}{144} \left (\frac {1}{72} \left (-18 \left (170114729 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-14176944 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )\right )-2 \sqrt {2 x^2-x+3} (44378877-7400779 x)\right )-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {\frac {1}{144} \left (\frac {1}{72} \left (-18 \left (-340229458 \int \frac {1}{288-\frac {(17-22 x)^2}{2 x^2-x+3}}d\frac {17-22 x}{\sqrt {2 x^2-x+3}}-14176944 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )\right )-2 \sqrt {2 x^2-x+3} (44378877-7400779 x)\right )-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{144} \left (\frac {1}{72} \left (-18 \left (-14176944 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )-\frac {170114729 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{6 \sqrt {2}}\right )-2 \sqrt {2 x^2-x+3} (44378877-7400779 x)\right )-\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{36 (2 x+5)}\right )+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{72 (2 x+5)^2}}{1152}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}\) |
(-3667*(3 - x + 2*x^2)^(3/2))/(1728*(5 + 2*x)^3) + ((158527*(3 - x + 2*x^2 )^(3/2))/(72*(5 + 2*x)^2) + ((-6467659*(3 - x + 2*x^2)^(3/2))/(36*(5 + 2*x )) + (-2*(44378877 - 7400779*x)*Sqrt[3 - x + 2*x^2] - 18*(-14176944*Sqrt[2 ]*ArcSinh[(-1 + 4*x)/Sqrt[23]] - (170114729*ArcTanh[(17 - 22*x)/(12*Sqrt[2 ]*Sqrt[3 - x + 2*x^2])])/(6*Sqrt[2])))/72)/144)/1152
3.4.29.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ (c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c ^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x ] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] || !R ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (Integer Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = Polynomi alRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*x^2) ^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Qx + c*d*R*(m + 1) - b*e*R*(m + p + 2) - c*e*R *(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]
Timed out.
hanged
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{(5+2 x)^4} \, dx=\frac {170123328 \, \sqrt {2} {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 170114729 \, \sqrt {2} {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )} \log \left (\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} - 1060 \, x^{2} + 1036 \, x - 1153}{4 \, x^{2} + 20 \, x + 25}\right ) + 48 \, {\left (414720 \, x^{4} - 5453568 \, x^{3} - 97682900 \, x^{2} - 329667508 \, x - 327735797\right )} \sqrt {2 \, x^{2} - x + 3}}{15925248 \, {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )}} \]
1/15925248*(170123328*sqrt(2)*(8*x^3 + 60*x^2 + 150*x + 125)*log(-4*sqrt(2 )*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 170114729*sqrt(2)* (8*x^3 + 60*x^2 + 150*x + 125)*log((24*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) - 1060*x^2 + 1036*x - 1153)/(4*x^2 + 20*x + 25)) + 48*(414720*x^4 - 5 453568*x^3 - 97682900*x^2 - 329667508*x - 327735797)*sqrt(2*x^2 - x + 3))/ (8*x^3 + 60*x^2 + 150*x + 125)
\[ \int \frac {\sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{(5+2 x)^4} \, dx=\int \frac {\sqrt {2 x^{2} - x + 3} \cdot \left (5 x^{4} - x^{3} + 3 x^{2} + x + 2\right )}{\left (2 x + 5\right )^{4}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{(5+2 x)^4} \, dx=\frac {5}{32} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {10939}{512} \, \sqrt {2} \operatorname {arsinh}\left (\frac {4}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) - \frac {170114729}{7962624} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) - \frac {693775}{165888} \, \sqrt {2 \, x^{2} - x + 3} - \frac {3667 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{1728 \, {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )}} + \frac {158527 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{82944 \, {\left (4 \, x^{2} + 20 \, x + 25\right )}} - \frac {6467659 \, \sqrt {2 \, x^{2} - x + 3}}{331776 \, {\left (2 \, x + 5\right )}} \]
5/32*sqrt(2*x^2 - x + 3)*x + 10939/512*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1 /23*sqrt(23)) - 170114729/7962624*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) - 693775/165888*sqrt(2*x^2 - x + 3) - 3667/1728*(2*x^2 - x + 3)^(3/2)/(8*x^3 + 60*x^2 + 150*x + 125) + 158527/8 2944*(2*x^2 - x + 3)^(3/2)/(4*x^2 + 20*x + 25) - 6467659/331776*sqrt(2*x^2 - x + 3)/(2*x + 5)
Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (127) = 254\).
Time = 0.30 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.92 \[ \int \frac {\sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{(5+2 x)^4} \, dx=\frac {1}{128} \, \sqrt {2 \, x^{2} - x + 3} {\left (20 \, x - 413\right )} - \frac {10939}{512} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {170114729}{7962624} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x + \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) - \frac {170114729}{7962624} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x - 11 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) - \frac {\sqrt {2} {\left (575810908 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{5} + 9206213116 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{4} + 9688786604 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{3} - 73157325092 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} + 49481952947 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} - 20269228621\right )}}{663552 \, {\left (2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} + 10 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} - 11\right )}^{3}} \]
1/128*sqrt(2*x^2 - x + 3)*(20*x - 413) - 10939/512*sqrt(2)*log(-2*sqrt(2)* (sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 170114729/7962624*sqrt(2)*log(abs (-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) - 170114729/7962624*sqrt (2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2*x^2 - x + 3))) - 1/663552 *sqrt(2)*(575810908*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^5 + 92062131 16*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^4 + 9688786604*sqrt(2)*(sqrt(2)*x - s qrt(2*x^2 - x + 3))^3 - 73157325092*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 49481952947*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) - 20269228621)/(2*(s qrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 10*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) - 11)^3
Timed out. \[ \int \frac {\sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{(5+2 x)^4} \, dx=\int \frac {\sqrt {2\,x^2-x+3}\,\left (5\,x^4-x^3+3\,x^2+x+2\right )}{{\left (2\,x+5\right )}^4} \,d x \]