\(\int \frac {5-x}{(3+2 x)^{5/2} (2+5 x+3 x^2)^2} \, dx\) [2564]

Optimal result

Integrand size = 27, antiderivative size = 98 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {262}{15 (3+2 x)^{3/2}}-\frac {686}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-10 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {936}{25} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]

[Out]

-262/15/(3+2*x)^(3/2)-3/5*(37+47*x)/(3+2*x)^(3/2)/(3*x^2+5*x+2)-10*arctanh((3+2*x)^(1/2))+936/125*arctanh(1/5*
15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-686/25/(3+2*x)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {836, 842, 840, 1180, 213} \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=-10 \text {arctanh}\left (\sqrt {2 x+3}\right )+\frac {936}{25} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )-\frac {3 (47 x+37)}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}-\frac {686}{25 \sqrt {2 x+3}}-\frac {262}{15 (2 x+3)^{3/2}} \]

[In]

Int[(5 - x)/((3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)^2),x]

[Out]

-262/(15*(3 + 2*x)^(3/2)) - 686/(25*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)) - 1
0*ArcTanh[Sqrt[3 + 2*x]] + (936*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 836

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 840

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 842

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e
*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(d +
e*x)^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac {1}{5} \int \frac {730+705 x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx \\ & = -\frac {262}{15 (3+2 x)^{3/2}}-\frac {3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac {1}{25} \int \frac {2090+1965 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx \\ & = -\frac {262}{15 (3+2 x)^{3/2}}-\frac {686}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac {1}{125} \int \frac {5770+5145 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )} \, dx \\ & = -\frac {262}{15 (3+2 x)^{3/2}}-\frac {686}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac {2}{125} \text {Subst}\left (\int \frac {-3895+5145 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt {3+2 x}\right ) \\ & = -\frac {262}{15 (3+2 x)^{3/2}}-\frac {686}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+30 \text {Subst}\left (\int \frac {1}{-3+3 x^2} \, dx,x,\sqrt {3+2 x}\right )-\frac {2808}{25} \text {Subst}\left (\int \frac {1}{-5+3 x^2} \, dx,x,\sqrt {3+2 x}\right ) \\ & = -\frac {262}{15 (3+2 x)^{3/2}}-\frac {686}{25 \sqrt {3+2 x}}-\frac {3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-10 \tanh ^{-1}\left (\sqrt {3+2 x}\right )+\frac {936}{25} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.84 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=\frac {-16633-47767 x-43032 x^2-12348 x^3}{75 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-10 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {936}{25} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]

[In]

Integrate[(5 - x)/((3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)^2),x]

[Out]

(-16633 - 47767*x - 43032*x^2 - 12348*x^3)/(75*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)) - 10*ArcTanh[Sqrt[3 + 2*x]]
+ (936*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82

method	result	size
risch	\(-\frac {12348 x^{3}+43032 x^{2}+47767 x +16633}{75 \left (3+2 x \right )^{\frac {3}{2}} \left (3 x^{2}+5 x +2\right )}-5 \ln \left (\sqrt {3+2 x}+1\right )+\frac {936 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{125}+5 \ln \left (\sqrt {3+2 x}-1\right )\)	\(80\)
trager	\(-\frac {12348 x^{3}+43032 x^{2}+47767 x +16633}{75 \left (3+2 x \right )^{\frac {3}{2}} \left (3 x^{2}+5 x +2\right )}+\frac {468 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{125}-5 \ln \left (\frac {\sqrt {3+2 x}+2+x}{1+x}\right )\)	\(101\)
derivativedivides	\(-\frac {6}{\sqrt {3+2 x}+1}-5 \ln \left (\sqrt {3+2 x}+1\right )-\frac {6}{\sqrt {3+2 x}-1}+5 \ln \left (\sqrt {3+2 x}-1\right )-\frac {104}{75 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {1624}{125 \sqrt {3+2 x}}-\frac {306 \sqrt {3+2 x}}{125 \left (\frac {4}{3}+2 x \right )}+\frac {936 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{125}\)	\(104\)
default	\(-\frac {6}{\sqrt {3+2 x}+1}-5 \ln \left (\sqrt {3+2 x}+1\right )-\frac {6}{\sqrt {3+2 x}-1}+5 \ln \left (\sqrt {3+2 x}-1\right )-\frac {104}{75 \left (3+2 x \right )^{\frac {3}{2}}}-\frac {1624}{125 \sqrt {3+2 x}}-\frac {306 \sqrt {3+2 x}}{125 \left (\frac {4}{3}+2 x \right )}+\frac {936 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{125}\)	\(104\)
pseudoelliptic	\(\frac {16848 \sqrt {3+2 x}\, \sqrt {15}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \left (x +\frac {3}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )+11250 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \left (x +\frac {3}{2}\right ) \ln \left (\sqrt {3+2 x}-1\right )-11250 \sqrt {3+2 x}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \left (x +\frac {3}{2}\right ) \ln \left (\sqrt {3+2 x}+1\right )-61740 x^{3}-215160 x^{2}-238835 x -83165}{\left (3+2 x \right )^{\frac {3}{2}} \left (1125 x^{2}+1875 x +750\right )}\)	\(126\)

[In]

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

[Out]

-1/75*(12348*x^3+43032*x^2+47767*x+16633)/(3+2*x)^(3/2)/(3*x^2+5*x+2)-5*ln((3+2*x)^(1/2)+1)+936/125*arctanh(1/
5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+5*ln((3+2*x)^(1/2)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (73) = 146\).

Time = 0.31 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.72 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=\frac {1404 \, \sqrt {5} \sqrt {3} {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 1875 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) + 1875 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 5 \, {\left (12348 \, x^{3} + 43032 \, x^{2} + 47767 \, x + 16633\right )} \sqrt {2 \, x + 3}}{375 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )}} \]

[In]

integrate((5-x)/(3+2*x)^(5/2)/(3*x^2+5*x+2)^2,x, algorithm="fricas")

[Out]

1/375*(1404*sqrt(5)*sqrt(3)*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*log((sqrt(5)*sqrt(3)*sqrt(2*x + 3) + 3*x +
7)/(3*x + 2)) - 1875*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*log(sqrt(2*x + 3) + 1) + 1875*(12*x^4 + 56*x^3 + 9
5*x^2 + 69*x + 18)*log(sqrt(2*x + 3) - 1) - 5*(12348*x^3 + 43032*x^2 + 47767*x + 16633)*sqrt(2*x + 3))/(12*x^4
+ 56*x^3 + 95*x^2 + 69*x + 18)

Sympy [A] (verification not implemented)

Time = 46.72 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.41 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=- \frac {2187 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{625} + \frac {1836 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{25} + 5 \log {\left (\sqrt {2 x + 3} - 1 \right )} - 5 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {6}{\sqrt {2 x + 3} + 1} - \frac {6}{\sqrt {2 x + 3} - 1} - \frac {1624}{125 \sqrt {2 x + 3}} - \frac {104}{75 \left (2 x + 3\right )^{\frac {3}{2}}} \]

[In]

integrate((5-x)/(3+2*x)**(5/2)/(3*x**2+5*x+2)**2,x)

[Out]

-2187*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(15)/3))/625 + 1836*Piecewise((sqrt(
15)*(-log(sqrt(15)*sqrt(2*x + 3)/5 - 1)/4 + log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/4 - 1/(4*(sqrt(15)*sqrt(2*x + 3)
/5 + 1)) - 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 - 1)))/75, (sqrt(2*x + 3) > -sqrt(15)/3) & (sqrt(2*x + 3) < sqrt(15)
/3)))/25 + 5*log(sqrt(2*x + 3) - 1) - 5*log(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) - 1)
- 1624/(125*sqrt(2*x + 3)) - 104/(75*(2*x + 3)**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.18 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {468}{125} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {2 \, {\left (3087 \, {\left (2 \, x + 3\right )}^{3} - 6267 \, {\left (2 \, x + 3\right )}^{2} + 4040 \, x + 6320\right )}}{75 \, {\left (3 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 8 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 5 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}\right )}} - 5 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 5 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]

[In]

integrate((5-x)/(3+2*x)^(5/2)/(3*x^2+5*x+2)^2,x, algorithm="maxima")

[Out]

-468/125*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/75*(3087*(2*x + 3)^3 - 6
267*(2*x + 3)^2 + 4040*x + 6320)/(3*(2*x + 3)^(7/2) - 8*(2*x + 3)^(5/2) + 5*(2*x + 3)^(3/2)) - 5*log(sqrt(2*x
+ 3) + 1) + 5*log(sqrt(2*x + 3) - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.18 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {468}{125} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {6 \, {\left (903 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 1403 \, \sqrt {2 \, x + 3}\right )}}{125 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - \frac {16 \, {\left (609 \, x + 946\right )}}{375 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}} - 5 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 5 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]

[In]

integrate((5-x)/(3+2*x)^(5/2)/(3*x^2+5*x+2)^2,x, algorithm="giac")

[Out]

-468/125*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 6/125*(903*(2*x +
3)^(3/2) - 1403*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 16/375*(609*x + 946)/(2*x + 3)^(3/2) - 5*log(sqr
t(2*x + 3) + 1) + 5*log(abs(sqrt(2*x + 3) - 1))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.84 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx=\frac {936\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{125}-10\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {\frac {1616\,x}{45}-\frac {4178\,{\left (2\,x+3\right )}^2}{75}+\frac {686\,{\left (2\,x+3\right )}^3}{25}+\frac {2528}{45}}{\frac {5\,{\left (2\,x+3\right )}^{3/2}}{3}-\frac {8\,{\left (2\,x+3\right )}^{5/2}}{3}+{\left (2\,x+3\right )}^{7/2}} \]

[In]

int(-(x - 5)/((2*x + 3)^(5/2)*(5*x + 3*x^2 + 2)^2),x)

[Out]

(936*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/125 - 10*atanh((2*x + 3)^(1/2)) - ((1616*x)/45 - (4178*(2*x
+ 3)^2)/75 + (686*(2*x + 3)^3)/25 + 2528/45)/((5*(2*x + 3)^(3/2))/3 - (8*(2*x + 3)^(5/2))/3 + (2*x + 3)^(7/2)
)