\(\int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx\) [2311]

Optimal result

Integrand size = 22, antiderivative size = 253 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\frac {4}{5} \sqrt {1+2 x}+\frac {1}{5} \sqrt {\frac {2}{155} \left (-178+35 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )-\frac {1}{5} \sqrt {\frac {2}{155} \left (-178+35 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )-\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right ) \]

[Out]

4/5*(1+2*x)^(1/2)+1/775*arctan((-10*(1+2*x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+10*35^(1/2))^(1/2))*(-55180+108
50*35^(1/2))^(1/2)-1/775*arctan((10*(1+2*x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+10*35^(1/2))^(1/2))*(-55180+108
50*35^(1/2))^(1/2)+1/1550*ln(5+10*x+35^(1/2)-(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))*(55180+10850*35^(1/2))^(1/2
)-1/1550*ln(5+10*x+35^(1/2)+(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))*(55180+10850*35^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {717, 840, 1183, 648, 632, 210, 642} \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\frac {1}{5} \sqrt {\frac {2}{155} \left (35 \sqrt {35}-178\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{5} \sqrt {\frac {2}{155} \left (35 \sqrt {35}-178\right )} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {4}{5} \sqrt {2 x+1}+\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )-\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right ) \]

[In]

Int[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2),x]

[Out]

(4*Sqrt[1 + 2*x])/5 + (Sqrt[(2*(-178 + 35*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/
Sqrt[10*(-2 + Sqrt[35])]])/5 - (Sqrt[(2*(-178 + 35*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1
+ 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/5 + (Sqrt[(178 + 35*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*
Sqrt[1 + 2*x] + 5*(1 + 2*x)])/5 - (Sqrt[(178 + 35*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1
+ 2*x] + 5*(1 + 2*x)])/5

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

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

Rule 717

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

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 1183

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

Rubi steps \begin{align*} \text {integral}& = \frac {4}{5} \sqrt {1+2 x}+\frac {1}{5} \int \frac {-3+8 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx \\ & = \frac {4}{5} \sqrt {1+2 x}+\frac {2}{5} \text {Subst}\left (\int \frac {-14+8 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt {1+2 x}\right ) \\ & = \frac {4}{5} \sqrt {1+2 x}+\frac {\text {Subst}\left (\int \frac {-14 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-\left (-14-8 \sqrt {\frac {7}{5}}\right ) x}{\sqrt {\frac {7}{5}}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} x+x^2} \, dx,x,\sqrt {1+2 x}\right )}{5 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\text {Subst}\left (\int \frac {-14 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}+\left (-14-8 \sqrt {\frac {7}{5}}\right ) x}{\sqrt {\frac {7}{5}}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} x+x^2} \, dx,x,\sqrt {1+2 x}\right )}{5 \sqrt {14 \left (2+\sqrt {35}\right )}} \\ & = \frac {4}{5} \sqrt {1+2 x}+\frac {1}{25} \left (4-\sqrt {35}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {\frac {7}{5}}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} x+x^2} \, dx,x,\sqrt {1+2 x}\right )+\frac {1}{25} \left (4-\sqrt {35}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {\frac {7}{5}}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} x+x^2} \, dx,x,\sqrt {1+2 x}\right )+\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \text {Subst}\left (\int \frac {-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}+2 x}{\sqrt {\frac {7}{5}}-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} x+x^2} \, dx,x,\sqrt {1+2 x}\right )-\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \text {Subst}\left (\int \frac {\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}+2 x}{\sqrt {\frac {7}{5}}+\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )} x+x^2} \, dx,x,\sqrt {1+2 x}\right ) \\ & = \frac {4}{5} \sqrt {1+2 x}+\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )-\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )-\frac {1}{25} \left (2 \left (4-\sqrt {35}\right )\right ) \text {Subst}\left (\int \frac {1}{\frac {2}{5} \left (2-\sqrt {35}\right )-x^2} \, dx,x,-\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}+2 \sqrt {1+2 x}\right )-\frac {1}{25} \left (2 \left (4-\sqrt {35}\right )\right ) \text {Subst}\left (\int \frac {1}{\frac {2}{5} \left (2-\sqrt {35}\right )-x^2} \, dx,x,\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}+2 \sqrt {1+2 x}\right ) \\ & = \frac {4}{5} \sqrt {1+2 x}+\frac {1}{5} \sqrt {\frac {2}{155} \left (-178+35 \sqrt {35}\right )} \tan ^{-1}\left (\sqrt {\frac {5}{2 \left (-2+\sqrt {35}\right )}} \left (\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-2 \sqrt {1+2 x}\right )\right )-\frac {1}{5} \sqrt {\frac {2}{155} \left (-178+35 \sqrt {35}\right )} \tan ^{-1}\left (\sqrt {\frac {5}{2 \left (-2+\sqrt {35}\right )}} \left (\sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}+2 \sqrt {1+2 x}\right )\right )+\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )-\frac {1}{5} \sqrt {\frac {1}{310} \left (178+35 \sqrt {35}\right )} \log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.45 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\frac {2}{775} \left (310 \sqrt {1+2 x}-\sqrt {155 \left (-178+19 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )-\sqrt {155 \left (-178-19 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )\right ) \]

[In]

Integrate[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2),x]

[Out]

(2*(310*Sqrt[1 + 2*x] - Sqrt[155*(-178 + (19*I)*Sqrt[31])]*ArcTan[Sqrt[(-2 - I*Sqrt[31])/7]*Sqrt[1 + 2*x]] - S
qrt[155*(-178 - (19*I)*Sqrt[31])]*ArcTan[Sqrt[(I/7)*(2*I + Sqrt[31])]*Sqrt[1 + 2*x]]))/775

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {\frac {27 \left (\sqrt {5}+\frac {10 \sqrt {7}}{27}\right ) \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}-\frac {27 \left (\sqrt {5}+\frac {10 \sqrt {7}}{27}\right ) \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}+\frac {4 \sqrt {1+2 x}\, \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}{5}+\frac {2 \left (\sqrt {5}\, \sqrt {7}-4\right ) \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right )}{5}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(273\)
derivativedivides	\(\frac {4 \sqrt {1+2 x}}{5}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}-\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(389\)
default	\(\frac {4 \sqrt {1+2 x}}{5}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}+\frac {\left (27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{1550}+\frac {2 \left (-62 \sqrt {5}\, \sqrt {7}-\frac {\left (-27 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(389\)
trager	\(\frac {4 \sqrt {1+2 x}}{5}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) \ln \left (\frac {4805 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{4} x +29605 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) x +4712 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right )+1369425 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} \sqrt {1+2 x}-9000 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right ) x +30400 \operatorname {RootOf}\left (\textit {\_Z}^{2}+24025 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2}-55180\right )-9350375 \sqrt {1+2 x}}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} x -235 x -76}\right )}{775}-\frac {\operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right ) \ln \left (-\frac {24025 x \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{5}-258385 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{3} x -23560 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{3}+44175 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} \sqrt {1+2 x}+421716 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right ) x +206112 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )+200165 \sqrt {1+2 x}}{155 \operatorname {RootOf}\left (4805 \textit {\_Z}^{4}-11036 \textit {\_Z}^{2}+8575\right )^{2} x -121 x +76}\right )}{5}\)	\(431\)
risch	\(\frac {4 \sqrt {1+2 x}}{5}+\frac {27 \ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{1550}+\frac {\ln \left (\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{155}+\frac {27 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \left (2 \sqrt {5}\, \sqrt {7}+4\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {2 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {4 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \sqrt {7}}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {27 \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{1550}-\frac {\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{155}+\frac {27 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \left (2 \sqrt {5}\, \sqrt {7}+4\right )}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {2 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{155 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {4 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \sqrt {7}}{5 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\)	\(616\)

[In]

int((1+2*x)^(3/2)/(5*x^2+3*x+2),x,method=_RETURNVERBOSE)

[Out]

2/5/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(27/620*(5^(1/2)+10/27*7^(1/2))*(10*5^(1/2)*7^(1/2)-20)^(1/2)*(2*5^(1/2)*7^(
1/2)+4)^(1/2)*ln(5^(1/2)*7^(1/2)-(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5+10*x)-27/620*(5^(1/2)+10/
27*7^(1/2))*(10*5^(1/2)*7^(1/2)-20)^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)*ln(5^(1/2)*7^(1/2)+(2*5^(1/2)*7^(1/2)+4)
^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5+10*x)+2*(1+2*x)^(1/2)*(10*5^(1/2)*7^(1/2)-20)^(1/2)+(5^(1/2)*7^(1/2)-4)*(arctan
((5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-10*(1+2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))-arctan((5^(1/2)*(2*5^(1
/2)*7^(1/2)+4)^(1/2)+10*(1+2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.75 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=-\frac {1}{1550} \, \sqrt {155} \sqrt {76 i \, \sqrt {31} + 712} \log \left (\sqrt {155} \sqrt {76 i \, \sqrt {31} + 712} {\left (2 i \, \sqrt {31} + 31\right )} + 10850 \, \sqrt {2 \, x + 1}\right ) + \frac {1}{1550} \, \sqrt {155} \sqrt {76 i \, \sqrt {31} + 712} \log \left (\sqrt {155} \sqrt {76 i \, \sqrt {31} + 712} {\left (-2 i \, \sqrt {31} - 31\right )} + 10850 \, \sqrt {2 \, x + 1}\right ) + \frac {1}{1550} \, \sqrt {155} \sqrt {-76 i \, \sqrt {31} + 712} \log \left (\sqrt {155} {\left (2 i \, \sqrt {31} - 31\right )} \sqrt {-76 i \, \sqrt {31} + 712} + 10850 \, \sqrt {2 \, x + 1}\right ) - \frac {1}{1550} \, \sqrt {155} \sqrt {-76 i \, \sqrt {31} + 712} \log \left (\sqrt {155} {\left (-2 i \, \sqrt {31} + 31\right )} \sqrt {-76 i \, \sqrt {31} + 712} + 10850 \, \sqrt {2 \, x + 1}\right ) + \frac {4}{5} \, \sqrt {2 \, x + 1} \]

[In]

integrate((1+2*x)^(3/2)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

-1/1550*sqrt(155)*sqrt(76*I*sqrt(31) + 712)*log(sqrt(155)*sqrt(76*I*sqrt(31) + 712)*(2*I*sqrt(31) + 31) + 1085
0*sqrt(2*x + 1)) + 1/1550*sqrt(155)*sqrt(76*I*sqrt(31) + 712)*log(sqrt(155)*sqrt(76*I*sqrt(31) + 712)*(-2*I*sq
rt(31) - 31) + 10850*sqrt(2*x + 1)) + 1/1550*sqrt(155)*sqrt(-76*I*sqrt(31) + 712)*log(sqrt(155)*(2*I*sqrt(31)
- 31)*sqrt(-76*I*sqrt(31) + 712) + 10850*sqrt(2*x + 1)) - 1/1550*sqrt(155)*sqrt(-76*I*sqrt(31) + 712)*log(sqrt
(155)*(-2*I*sqrt(31) + 31)*sqrt(-76*I*sqrt(31) + 712) + 10850*sqrt(2*x + 1)) + 4/5*sqrt(2*x + 1)

Sympy [F]

\[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {3}{2}}}{5 x^{2} + 3 x + 2}\, dx \]

[In]

integrate((1+2*x)**(3/2)/(5*x**2+3*x+2),x)

[Out]

Integral((2*x + 1)**(3/2)/(5*x**2 + 3*x + 2), x)

Maxima [F]

\[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\int { \frac {{\left (2 \, x + 1\right )}^{\frac {3}{2}}}{5 \, x^{2} + 3 \, x + 2} \,d x } \]

[In]

integrate((1+2*x)^(3/2)/(5*x^2+3*x+2),x, algorithm="maxima")

[Out]

integrate((2*x + 1)^(3/2)/(5*x^2 + 3*x + 2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 594 vs. \(2 (170) = 340\).

Time = 0.72 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.35 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\text {Too large to display} \]

[In]

integrate((1+2*x)^(3/2)/(5*x^2+3*x+2),x, algorithm="giac")

[Out]

1/46519375*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - sqrt(31)*(7/5)^(3
/4)*(-140*sqrt(35) + 2450)^(3/2) + 2*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/4)*sqrt(140*sqrt(3
5) + 2450)*(2*sqrt(35) - 35) - 8575*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) - 17150*(7/5)^(1/4)*sqrt(1
40*sqrt(35) + 2450))*arctan(5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35*sqrt(35) + 1/2) + sqrt(2*x + 1))/sqrt(-1/35
*sqrt(35) + 1/2)) + 1/46519375*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450)
- sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 2*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(
3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 8575*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) - 1715
0*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450))*arctan(-5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35*sqrt(35) + 1/2) - sqrt
(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2)) + 1/93038750*sqrt(31)*(sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2)
+ 210*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 420*(7/5)^(3/4)*(2*sqrt(35) + 35)*sq
rt(-140*sqrt(35) + 2450) + 2*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) - 8575*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqr
t(35) + 2450) + 17150*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450))*log(2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(3
5) + 1/2) + 2*x + sqrt(7/5) + 1) - 1/93038750*sqrt(31)*(sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 210
*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 420*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-14
0*sqrt(35) + 2450) + 2*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) - 8575*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35)
+ 2450) + 17150*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450))*log(-2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(35) +
1/2) + 2*x + sqrt(7/5) + 1) + 4/5*sqrt(2*x + 1)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.72 \[ \int \frac {(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx=\frac {4\,\sqrt {2\,x+1}}{5}-\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {178-\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}\,2432{}\mathrm {i}}{390625\,\left (-\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}+\frac {4864\,\sqrt {31}\,\sqrt {155}\,\sqrt {178-\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}}{12109375\,\left (-\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}\right )\,\sqrt {178-\sqrt {31}\,19{}\mathrm {i}}\,2{}\mathrm {i}}{775}+\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {178+\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}\,2432{}\mathrm {i}}{390625\,\left (\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}-\frac {4864\,\sqrt {31}\,\sqrt {155}\,\sqrt {178+\sqrt {31}\,19{}\mathrm {i}}\,\sqrt {2\,x+1}}{12109375\,\left (\frac {34048}{78125}+\frac {\sqrt {31}\,17024{}\mathrm {i}}{78125}\right )}\right )\,\sqrt {178+\sqrt {31}\,19{}\mathrm {i}}\,2{}\mathrm {i}}{775} \]

[In]

int((2*x + 1)^(3/2)/(3*x + 5*x^2 + 2),x)

[Out]

(4*(2*x + 1)^(1/2))/5 - (155^(1/2)*atan((155^(1/2)*(178 - 31^(1/2)*19i)^(1/2)*(2*x + 1)^(1/2)*2432i)/(390625*(
(31^(1/2)*17024i)/78125 - 34048/78125)) + (4864*31^(1/2)*155^(1/2)*(178 - 31^(1/2)*19i)^(1/2)*(2*x + 1)^(1/2))
/(12109375*((31^(1/2)*17024i)/78125 - 34048/78125)))*(178 - 31^(1/2)*19i)^(1/2)*2i)/775 + (155^(1/2)*atan((155
^(1/2)*(31^(1/2)*19i + 178)^(1/2)*(2*x + 1)^(1/2)*2432i)/(390625*((31^(1/2)*17024i)/78125 + 34048/78125)) - (4
864*31^(1/2)*155^(1/2)*(31^(1/2)*19i + 178)^(1/2)*(2*x + 1)^(1/2))/(12109375*((31^(1/2)*17024i)/78125 + 34048/
78125)))*(31^(1/2)*19i + 178)^(1/2)*2i)/775