3.20.92 \(\int \frac {(1+2 x)^{9/2}}{(2+3 x+5 x^2)^3} \, dx\)

Optimal. Leaf size=313 \[ -\frac {(5-4 x) (2 x+1)^{7/2}}{62 \left (5 x^2+3 x+2\right )^2}-\frac {(1143-1088 x) (2 x+1)^{3/2}}{9610 \left (5 x^2+3 x+2\right )}-\frac {1584 \sqrt {2 x+1}}{24025}+\frac {3 \sqrt {\frac {1}{310} \left (64681225 \sqrt {35}-250141922\right )} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{48050}-\frac {3 \sqrt {\frac {1}{310} \left (64681225 \sqrt {35}-250141922\right )} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{48050}-\frac {3 \sqrt {\frac {1}{310} \left (250141922+64681225 \sqrt {35}\right )} \tan ^{-1}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{24025}+\frac {3 \sqrt {\frac {1}{310} \left (250141922+64681225 \sqrt {35}\right )} \tan ^{-1}\left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{24025} \]

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Rubi [A]  time = 0.52, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {738, 818, 824, 826, 1169, 634, 618, 204, 628} \begin {gather*} -\frac {(5-4 x) (2 x+1)^{7/2}}{62 \left (5 x^2+3 x+2\right )^2}-\frac {(1143-1088 x) (2 x+1)^{3/2}}{9610 \left (5 x^2+3 x+2\right )}-\frac {1584 \sqrt {2 x+1}}{24025}+\frac {3 \sqrt {\frac {1}{310} \left (64681225 \sqrt {35}-250141922\right )} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{48050}-\frac {3 \sqrt {\frac {1}{310} \left (64681225 \sqrt {35}-250141922\right )} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{48050}-\frac {3 \sqrt {\frac {1}{310} \left (250141922+64681225 \sqrt {35}\right )} \tan ^{-1}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{24025}+\frac {3 \sqrt {\frac {1}{310} \left (250141922+64681225 \sqrt {35}\right )} \tan ^{-1}\left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{24025} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)^(9/2)/(2 + 3*x + 5*x^2)^3,x]

[Out]

(-1584*Sqrt[1 + 2*x])/24025 - ((5 - 4*x)*(1 + 2*x)^(7/2))/(62*(2 + 3*x + 5*x^2)^2) - ((1143 - 1088*x)*(1 + 2*x
)^(3/2))/(9610*(2 + 3*x + 5*x^2)) - (3*Sqrt[(250141922 + 64681225*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35]
)] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/24025 + (3*Sqrt[(250141922 + 64681225*Sqrt[35])/310]*ArcTan[
(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/24025 + (3*Sqrt[(-250141922 + 64681225
*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/48050 - (3*Sqrt[(-2501419
22 + 64681225*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/48050

Rule 204

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

Rule 618

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 628

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 634

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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 818

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

Rule 824

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

Rule 826

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 1169

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*} \int \frac {(1+2 x)^{9/2}}{\left (2+3 x+5 x^2\right )^3} \, dx &=-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac {1}{62} \int \frac {(47-4 x) (1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx\\ &=-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac {(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac {\int \frac {\sqrt {1+2 x} (-4269+1584 x)}{2+3 x+5 x^2} \, dx}{9610}\\ &=-\frac {1584 \sqrt {1+2 x}}{24025}-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac {(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac {\int \frac {-27681-44274 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{48050}\\ &=-\frac {1584 \sqrt {1+2 x}}{24025}-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac {(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {-11088-44274 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt {1+2 x}\right )}{24025}\\ &=-\frac {1584 \sqrt {1+2 x}}{24025}-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac {(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {-11088 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-\left (-11088+44274 \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 )}{48050 \sqrt {14 \left (2+\sqrt {35}\right )}}-\frac {\operatorname {Subst}\left (\int \frac {-11088 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}+\left (-11088+44274 \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 )}{48050 \sqrt {14 \left (2+\sqrt {35}\right )}}\\ &=-\frac {1584 \sqrt {1+2 x}}{24025}-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac {(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac {\left (3 \left (9240-7379 \sqrt {35}\right )\right ) \operatorname {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 )}{240250 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\left (3 \left (9240-7379 \sqrt {35}\right )\right ) \operatorname {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 )}{240250 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\left (3 \left (7379+264 \sqrt {35}\right )\right ) \operatorname {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 )}{240250}+\frac {\left (3 \left (7379+264 \sqrt {35}\right )\right ) \operatorname {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 )}{240250}\\ &=-\frac {1584 \sqrt {1+2 x}}{24025}-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac {(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}+\frac {3 \left (7379-264 \sqrt {35}\right ) \log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{48050 \sqrt {10 \left (2+\sqrt {35}\right )}}-\frac {3 \left (7379-264 \sqrt {35}\right ) \log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{48050 \sqrt {10 \left (2+\sqrt {35}\right )}}-\frac {\left (3 \left (7379+264 \sqrt {35}\right )\right ) \operatorname {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 )}{120125}-\frac {\left (3 \left (7379+264 \sqrt {35}\right )\right ) \operatorname {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 )}{120125}\\ &=-\frac {1584 \sqrt {1+2 x}}{24025}-\frac {(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac {(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac {3 \sqrt {\frac {1}{310} \left (250141922+64681225 \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 )}{24025}+\frac {3 \sqrt {\frac {1}{310} \left (250141922+64681225 \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 )}{24025}+\frac {3 \left (7379-264 \sqrt {35}\right ) \log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{48050 \sqrt {10 \left (2+\sqrt {35}\right )}}-\frac {3 \left (7379-264 \sqrt {35}\right ) \log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{48050 \sqrt {10 \left (2+\sqrt {35}\right )}}\\ \end {align*}

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Mathematica [C]  time = 0.74, size = 236, normalized size = 0.75 \begin {gather*} \frac {-\frac {7 (640 x+409) (2 x+1)^{11/2}}{5 x^2+3 x+2}+\frac {217 (20 x+37) (2 x+1)^{11/2}}{\left (5 x^2+3 x+2\right )^2}+1792 (2 x+1)^{9/2}+1932 (2 x+1)^{7/2}-2352 (2 x+1)^{5/2}-\frac {47236}{5} (2 x+1)^{3/2}-\frac {155232}{25} \sqrt {2 x+1}+\frac {294 \left (\sqrt {2-i \sqrt {31}} \left (8184-7907 i \sqrt {31}\right ) \tanh ^{-1}\left (\frac {\sqrt {10 x+5}}{\sqrt {2-i \sqrt {31}}}\right )+\sqrt {2+i \sqrt {31}} \left (8184+7907 i \sqrt {31}\right ) \tanh ^{-1}\left (\frac {\sqrt {10 x+5}}{\sqrt {2+i \sqrt {31}}}\right )\right )}{775 \sqrt {5}}}{94178} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)^(9/2)/(2 + 3*x + 5*x^2)^3,x]

[Out]

((-155232*Sqrt[1 + 2*x])/25 - (47236*(1 + 2*x)^(3/2))/5 - 2352*(1 + 2*x)^(5/2) + 1932*(1 + 2*x)^(7/2) + 1792*(
1 + 2*x)^(9/2) + (217*(1 + 2*x)^(11/2)*(37 + 20*x))/(2 + 3*x + 5*x^2)^2 - (7*(1 + 2*x)^(11/2)*(409 + 640*x))/(
2 + 3*x + 5*x^2) + (294*(Sqrt[2 - I*Sqrt[31]]*(8184 - (7907*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqr
t[31]]] + Sqrt[2 + I*Sqrt[31]]*(8184 + (7907*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/(775*
Sqrt[5]))/94178

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IntegrateAlgebraic [C]  time = 2.50, size = 167, normalized size = 0.53 \begin {gather*} -\frac {2 \sqrt {2 x+1} \left (43075 (2 x+1)^3+15332 (2 x+1)^2+14693 (2 x+1)+38808\right )}{24025 \left (5 (2 x+1)^2-4 (2 x+1)+7\right )^2}+\frac {3 \sqrt {\frac {1}{155} \left (250141922-52010281 i \sqrt {31}\right )} \tan ^{-1}\left (\sqrt {-\frac {2}{7}-\frac {i \sqrt {31}}{7}} \sqrt {2 x+1}\right )}{24025}+\frac {3 \sqrt {\frac {1}{155} \left (250141922+52010281 i \sqrt {31}\right )} \tan ^{-1}\left (\sqrt {-\frac {2}{7}+\frac {i \sqrt {31}}{7}} \sqrt {2 x+1}\right )}{24025} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + 2*x)^(9/2)/(2 + 3*x + 5*x^2)^3,x]

[Out]

(-2*Sqrt[1 + 2*x]*(38808 + 14693*(1 + 2*x) + 15332*(1 + 2*x)^2 + 43075*(1 + 2*x)^3))/(24025*(7 - 4*(1 + 2*x) +
5*(1 + 2*x)^2)^2) + (3*Sqrt[(250141922 - (52010281*I)*Sqrt[31])/155]*ArcTan[Sqrt[-2/7 - (I/7)*Sqrt[31]]*Sqrt[
1 + 2*x]])/24025 + (3*Sqrt[(250141922 + (52010281*I)*Sqrt[31])/155]*ArcTan[Sqrt[-2/7 + (I/7)*Sqrt[31]]*Sqrt[1
+ 2*x]])/24025

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fricas [B]  time = 0.46, size = 653, normalized size = 2.09

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^(9/2)/(5*x^2+3*x+2)^3,x, algorithm="fricas")

[Out]

1/44382950698994113517500*(19347824532*97578096035^(1/4)*sqrt(105602)*sqrt(155)*sqrt(35)*(25*x^4 + 30*x^3 + 29
*x^2 + 12*x + 4)*sqrt(250141922*sqrt(35) + 2263842875)*arctan(1/2160252846511970217131322639383425*97578096035
^(3/4)*sqrt(1677751)*sqrt(105602)*sqrt(7543)*sqrt(155)*sqrt(97578096035^(1/4)*sqrt(105602)*sqrt(155)*(7379*sqr
t(35)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 29796214090828850*x + 29
79621409082885*sqrt(35) + 14898107045414425)*sqrt(250141922*sqrt(35) + 2263842875)*(264*sqrt(35) - 7379) - 1/1
57326990020985410885*97578096035^(3/4)*sqrt(105602)*sqrt(155)*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 22638428
75)*(264*sqrt(35) - 7379) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 19347824532*97578096035^(1/4)*sqrt(10560
2)*sqrt(155)*sqrt(35)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(250141922*sqrt(35) + 2263842875)*arctan(1/793
8929210931490547957610699734086875*97578096035^(3/4)*sqrt(1677751)*sqrt(105602)*sqrt(155)*sqrt(-101872929375*9
7578096035^(1/4)*sqrt(105602)*sqrt(155)*(7379*sqrt(35)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*
sqrt(35) + 2263842875) + 3035427613717387271262468750*x + 303542761371738727126246875*sqrt(35) + 1517713806858
693635631234375)*sqrt(250141922*sqrt(35) + 2263842875)*(264*sqrt(35) - 7379) - 1/157326990020985410885*9757809
6035^(3/4)*sqrt(105602)*sqrt(155)*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875)*(264*sqrt(35) - 7379) -
1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 3*97578096035^(1/4)*sqrt(105602)*sqrt(155)*(250141922*sqrt(35)*sqrt(
31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 2263842875*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4))*sqrt(25
0141922*sqrt(35) + 2263842875)*log(101872929375/1677751*97578096035^(1/4)*sqrt(105602)*sqrt(155)*(7379*sqrt(35
)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 1809224142150645281250*x + 1
80922414215064528125*sqrt(35) + 904612071075322640625) - 3*97578096035^(1/4)*sqrt(105602)*sqrt(155)*(250141922
*sqrt(35)*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 2263842875*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*
x + 4))*sqrt(250141922*sqrt(35) + 2263842875)*log(-101872929375/1677751*97578096035^(1/4)*sqrt(105602)*sqrt(15
5)*(7379*sqrt(35)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 180922414215
0645281250*x + 180922414215064528125*sqrt(35) + 904612071075322640625) - 923682636815694350*(86150*x^3 + 14455
7*x^2 + 87291*x + 27977)*sqrt(2*x + 1))/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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giac [B]  time = 1.44, size = 642, normalized size = 2.05

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^(9/2)/(5*x^2+3*x+2)^3,x, algorithm="giac")

[Out]

3/1788204775000*sqrt(31)*(1549590*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - 7379*sqr
t(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 14758*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 3099180*(7/5)
^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 9055200*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) +
18110400*(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)) + 3/1788204775000*sqrt(31)*(1549590*sqrt(31)*(7/5)^(3/4)*(2*sqrt
(35) + 35)*sqrt(-140*sqrt(35) + 2450) - 7379*sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 14758*(7/5)^(
3/4)*(140*sqrt(35) + 2450)^(3/2) + 3099180*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 9055200*s
qrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) + 18110400*(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)) + 3/35764095500
00*sqrt(31)*(7379*sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 1549590*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqr
t(35) + 2450)*(2*sqrt(35) - 35) - 3099180*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) + 14758*(7/
5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 9055200*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) - 18110400*(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) - 3/3576409550000*sqrt(31)*(7379*sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 1549590*sqrt(31)*(7
/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) - 3099180*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(3
5) + 2450) + 14758*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 9055200*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) +
2450) - 18110400*(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) - 2/24025*(43075*(2*x + 1)^(7/2) + 15332*(2*x + 1)^(5/2) + 14693*(2*x + 1)^(3/2)
+ 38808*sqrt(2*x + 1))/(5*(2*x + 1)^2 - 8*x + 3)^2

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maple [B]  time = 0.25, size = 662, normalized size = 2.12 \begin {gather*} -\frac {35997 \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {2 x +1}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{744775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {23721 \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}\, \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {2 x +1}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{1489550 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {1584 \sqrt {5}\, \sqrt {7}\, \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {2 x +1}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{24025 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {35997 \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {2 x +1}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{744775 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {23721 \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}\, \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {2 x +1}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{1489550 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {1584 \sqrt {5}\, \sqrt {7}\, \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {2 x +1}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{24025 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {35997 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (10 x +\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {2 x +1}+5\right )}{7447750}+\frac {23721 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (10 x +\sqrt {5}\, \sqrt {7}-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {2 x +1}+5\right )}{2979100}+\frac {35997 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (10 x +\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {2 x +1}+5\right )}{7447750}-\frac {23721 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \ln \left (10 x +\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {2 x +1}+5\right )}{2979100}+\frac {-\frac {3446 \left (2 x +1\right )^{\frac {7}{2}}}{961}-\frac {30664 \left (2 x +1\right )^{\frac {5}{2}}}{24025}-\frac {29386 \left (2 x +1\right )^{\frac {3}{2}}}{24025}-\frac {77616 \sqrt {2 x +1}}{24025}}{\left (-8 x +5 \left (2 x +1\right )^{2}+3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x+1)^(9/2)/(5*x^2+3*x+2)^3,x)

[Out]

1600*(-1723/768800*(2*x+1)^(7/2)-3833/4805000*(2*x+1)^(5/2)-14693/19220000*(2*x+1)^(3/2)-4851/2402500*(2*x+1)^
(1/2))/(5*(2*x+1)^2-8*x+3)^2+35997/7447750*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)*ln(10*x+5^(1/2)*7^(1/2)+(2*5^(1
/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(2*x+1)^(1/2)+5)-23721/2979100*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)*ln(10*x+5^(1/2
)*7^(1/2)+(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(2*x+1)^(1/2)+5)-35997/744775/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(2*5
^(1/2)*7^(1/2)+4)*arctan((5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(2*x+1)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))
+23721/1489550/(10*5^(1/2)*7^(1/2)-20)^(1/2)*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)*7^(1/2)*arctan((5^(1/2)*(2*5^(1/2)*
7^(1/2)+4)^(1/2)+10*(2*x+1)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))+1584/24025/(10*5^(1/2)*7^(1/2)-20)^(1/2)*5^(
1/2)*7^(1/2)*arctan((5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(2*x+1)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))-3599
7/7447750*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)*ln(10*x+5^(1/2)*7^(1/2)-(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(2*x
+1)^(1/2)+5)+23721/2979100*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)*ln(10*x+5^(1/2)*7^(1/2)-(2*5^(1/2)*7^(1/2)+4)^(
1/2)*5^(1/2)*(2*x+1)^(1/2)+5)-35997/744775/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(2*5^(1/2)*7^(1/2)+4)*arctan((-5^(1/2
)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(2*x+1)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))+23721/1489550/(10*5^(1/2)*7^(1/
2)-20)^(1/2)*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)*7^(1/2)*arctan((-5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(2*x+1)^(1/
2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))+1584/24025/(10*5^(1/2)*7^(1/2)-20)^(1/2)*5^(1/2)*7^(1/2)*arctan((-5^(1/2)*(
2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(2*x+1)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x + 1\right )}^{\frac {9}{2}}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^(9/2)/(5*x^2+3*x+2)^3,x, algorithm="maxima")

[Out]

integrate((2*x + 1)^(9/2)/(5*x^2 + 3*x + 2)^3, x)

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mupad [B]  time = 1.07, size = 245, normalized size = 0.78 \begin {gather*} \frac {\frac {77616\,\sqrt {2\,x+1}}{600625}+\frac {29386\,{\left (2\,x+1\right )}^{3/2}}{600625}+\frac {30664\,{\left (2\,x+1\right )}^{5/2}}{600625}+\frac {3446\,{\left (2\,x+1\right )}^{7/2}}{24025}}{\frac {112\,x}{25}-\frac {86\,{\left (2\,x+1\right )}^2}{25}+\frac {8\,{\left (2\,x+1\right )}^3}{5}-{\left (2\,x+1\right )}^4+\frac {7}{25}}-\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-250141922-\sqrt {31}\,52010281{}\mathrm {i}}\,\sqrt {2\,x+1}\,23380272{}\mathrm {i}}{45093798828125\,\left (-\frac {1294074674928}{9018759765625}+\frac {\sqrt {31}\,43206742656{}\mathrm {i}}{9018759765625}\right )}-\frac {46760544\,\sqrt {31}\,\sqrt {155}\,\sqrt {-250141922-\sqrt {31}\,52010281{}\mathrm {i}}\,\sqrt {2\,x+1}}{1397907763671875\,\left (-\frac {1294074674928}{9018759765625}+\frac {\sqrt {31}\,43206742656{}\mathrm {i}}{9018759765625}\right )}\right )\,\sqrt {-250141922-\sqrt {31}\,52010281{}\mathrm {i}}\,3{}\mathrm {i}}{3723875}+\frac {\sqrt {155}\,\mathrm {atan}\left (\frac {\sqrt {155}\,\sqrt {-250141922+\sqrt {31}\,52010281{}\mathrm {i}}\,\sqrt {2\,x+1}\,23380272{}\mathrm {i}}{45093798828125\,\left (\frac {1294074674928}{9018759765625}+\frac {\sqrt {31}\,43206742656{}\mathrm {i}}{9018759765625}\right )}+\frac {46760544\,\sqrt {31}\,\sqrt {155}\,\sqrt {-250141922+\sqrt {31}\,52010281{}\mathrm {i}}\,\sqrt {2\,x+1}}{1397907763671875\,\left (\frac {1294074674928}{9018759765625}+\frac {\sqrt {31}\,43206742656{}\mathrm {i}}{9018759765625}\right )}\right )\,\sqrt {-250141922+\sqrt {31}\,52010281{}\mathrm {i}}\,3{}\mathrm {i}}{3723875} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 1)^(9/2)/(3*x + 5*x^2 + 2)^3,x)

[Out]

((77616*(2*x + 1)^(1/2))/600625 + (29386*(2*x + 1)^(3/2))/600625 + (30664*(2*x + 1)^(5/2))/600625 + (3446*(2*x
+ 1)^(7/2))/24025)/((112*x)/25 - (86*(2*x + 1)^2)/25 + (8*(2*x + 1)^3)/5 - (2*x + 1)^4 + 7/25) - (155^(1/2)*a
tan((155^(1/2)*(- 31^(1/2)*52010281i - 250141922)^(1/2)*(2*x + 1)^(1/2)*23380272i)/(45093798828125*((31^(1/2)*
43206742656i)/9018759765625 - 1294074674928/9018759765625)) - (46760544*31^(1/2)*155^(1/2)*(- 31^(1/2)*5201028
1i - 250141922)^(1/2)*(2*x + 1)^(1/2))/(1397907763671875*((31^(1/2)*43206742656i)/9018759765625 - 129407467492
8/9018759765625)))*(- 31^(1/2)*52010281i - 250141922)^(1/2)*3i)/3723875 + (155^(1/2)*atan((155^(1/2)*(31^(1/2)
*52010281i - 250141922)^(1/2)*(2*x + 1)^(1/2)*23380272i)/(45093798828125*((31^(1/2)*43206742656i)/901875976562
5 + 1294074674928/9018759765625)) + (46760544*31^(1/2)*155^(1/2)*(31^(1/2)*52010281i - 250141922)^(1/2)*(2*x +
1)^(1/2))/(1397907763671875*((31^(1/2)*43206742656i)/9018759765625 + 1294074674928/9018759765625)))*(31^(1/2)
*52010281i - 250141922)^(1/2)*3i)/3723875

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)**(9/2)/(5*x**2+3*x+2)**3,x)

[Out]

Timed out

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