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

Optimal. Leaf size=296 \[ \frac {20 x+37}{217 (2 x+1)^{3/2} \left (5 x^2+3 x+2\right )}-\frac {4680}{10633 \sqrt {2 x+1}}-\frac {820}{4557 (2 x+1)^{3/2}}-\frac {5 \sqrt {\frac {1}{434} \left (2632525 \sqrt {35}-12504542\right )} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{10633}+\frac {5 \sqrt {\frac {1}{434} \left (2632525 \sqrt {35}-12504542\right )} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{10633}+\frac {5 \sqrt {\frac {2}{217} \left (12504542+2632525 \sqrt {35}\right )} \tan ^{-1}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{10633}-\frac {5 \sqrt {\frac {2}{217} \left (12504542+2632525 \sqrt {35}\right )} \tan ^{-1}\left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )}{10633} \]

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Rubi [A]  time = 0.44, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {740, 828, 826, 1169, 634, 618, 204, 628} \begin {gather*} \frac {20 x+37}{217 (2 x+1)^{3/2} \left (5 x^2+3 x+2\right )}-\frac {4680}{10633 \sqrt {2 x+1}}-\frac {820}{4557 (2 x+1)^{3/2}}-\frac {5 \sqrt {\frac {1}{434} \left (2632525 \sqrt {35}-12504542\right )} \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{10633}+\frac {5 \sqrt {\frac {1}{434} \left (2632525 \sqrt {35}-12504542\right )} \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )}{10633}+\frac {5 \sqrt {\frac {2}{217} \left (12504542+2632525 \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 )}{10633}-\frac {5 \sqrt {\frac {2}{217} \left (12504542+2632525 \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 )}{10633} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-820/(4557*(1 + 2*x)^(3/2)) - 4680/(10633*Sqrt[1 + 2*x]) + (37 + 20*x)/(217*(1 + 2*x)^(3/2)*(2 + 3*x + 5*x^2))
+ (5*Sqrt[(2*(12504542 + 2632525*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*
(-2 + Sqrt[35])]])/10633 - (5*Sqrt[(2*(12504542 + 2632525*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10
*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/10633 - (5*Sqrt[(-12504542 + 2632525*Sqrt[35])/434]*Log[Sqrt[35] -
Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10633 + (5*Sqrt[(-12504542 + 2632525*Sqrt[35])/434]*Log[
Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10633

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 740

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

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 828

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 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}{(1+2 x)^{5/2} \left (2+3 x+5 x^2\right )^2} \, dx &=\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac {1}{217} \int \frac {255+100 x}{(1+2 x)^{5/2} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac {820}{4557 (1+2 x)^{3/2}}+\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac {\int \frac {145-2050 x}{(1+2 x)^{3/2} \left (2+3 x+5 x^2\right )} \, dx}{1519}\\ &=-\frac {820}{4557 (1+2 x)^{3/2}}-\frac {4680}{10633 \sqrt {1+2 x}}+\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac {\int \frac {-8345-11700 x}{\sqrt {1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{10633}\\ &=-\frac {820}{4557 (1+2 x)^{3/2}}-\frac {4680}{10633 \sqrt {1+2 x}}+\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {-4990-11700 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt {1+2 x}\right )}{10633}\\ &=-\frac {820}{4557 (1+2 x)^{3/2}}-\frac {4680}{10633 \sqrt {1+2 x}}+\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {-998 \sqrt {10 \left (2+\sqrt {35}\right )}-\left (-4990+2340 \sqrt {35}\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 )}{10633 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\operatorname {Subst}\left (\int \frac {-998 \sqrt {10 \left (2+\sqrt {35}\right )}+\left (-4990+2340 \sqrt {35}\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 )}{10633 \sqrt {14 \left (2+\sqrt {35}\right )}}\\ &=-\frac {820}{4557 (1+2 x)^{3/2}}-\frac {4680}{10633 \sqrt {1+2 x}}+\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac {\left (5 \left (499-234 \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 )}{10633 \sqrt {14 \left (2+\sqrt {35}\right )}}-\frac {\left (5 \left (499-234 \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 )}{10633 \sqrt {14 \left (2+\sqrt {35}\right )}}-\frac {\left (8190+499 \sqrt {35}\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 )}{74431}-\frac {\left (8190+499 \sqrt {35}\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 )}{74431}\\ &=-\frac {820}{4557 (1+2 x)^{3/2}}-\frac {4680}{10633 \sqrt {1+2 x}}+\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}-\frac {5 \sqrt {-\frac {6252271}{217}+\frac {376075 \sqrt {35}}{62}} \log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{10633}+\frac {5 \sqrt {-\frac {6252271}{217}+\frac {376075 \sqrt {35}}{62}} \log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{10633}+\frac {\left (2 \left (8190+499 \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 )}{74431}+\frac {\left (2 \left (8190+499 \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 )}{74431}\\ &=-\frac {820}{4557 (1+2 x)^{3/2}}-\frac {4680}{10633 \sqrt {1+2 x}}+\frac {37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac {5 \sqrt {\frac {2}{7 \left (-2+\sqrt {35}\right )}} \left (499+234 \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 )}{10633}-\frac {5 \sqrt {\frac {2}{7 \left (-2+\sqrt {35}\right )}} \left (499+234 \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 )}{10633}-\frac {5 \sqrt {-\frac {6252271}{217}+\frac {376075 \sqrt {35}}{62}} \log \left (\sqrt {35}-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{10633}+\frac {5 \sqrt {-\frac {6252271}{217}+\frac {376075 \sqrt {35}}{62}} \log \left (\sqrt {35}+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}+5 (1+2 x)\right )}{10633}\\ \end {align*}

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Mathematica [C]  time = 0.52, size = 176, normalized size = 0.59 \begin {gather*} \frac {1}{217} \left (\frac {20 x+37}{(2 x+1)^{3/2} \left (5 x^2+3 x+2\right )}-\frac {4680}{49 \sqrt {2 x+1}}-\frac {820}{21 (2 x+1)^{3/2}}+\frac {2 i \sqrt {5} \left (\sqrt {2-i \sqrt {31}} \left (9188 \sqrt {31}+15469 i\right ) \tanh ^{-1}\left (\frac {\sqrt {10 x+5}}{\sqrt {2-i \sqrt {31}}}\right )+\left (-9188 \sqrt {31}+15469 i\right ) \sqrt {2+i \sqrt {31}} \tanh ^{-1}\left (\frac {\sqrt {10 x+5}}{\sqrt {2+i \sqrt {31}}}\right )\right )}{10633}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-820/(21*(1 + 2*x)^(3/2)) - 4680/(49*Sqrt[1 + 2*x]) + (37 + 20*x)/((1 + 2*x)^(3/2)*(2 + 3*x + 5*x^2)) + ((2*I
)/10633)*Sqrt[5]*(Sqrt[2 - I*Sqrt[31]]*(15469*I + 9188*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]]
+ (15469*I - 9188*Sqrt[31])*Sqrt[2 + I*Sqrt[31]]*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/217

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IntegrateAlgebraic [C]  time = 1.77, size = 167, normalized size = 0.56 \begin {gather*} -\frac {4 \left (17550 (2 x+1)^3-6865 (2 x+1)^2+17360 (2 x+1)+6076\right )}{31899 (2 x+1)^{3/2} \left (5 (2 x+1)^2-4 (2 x+1)+7\right )}-\frac {10 \sqrt {\frac {1}{217} \left (12504542-1667459 i \sqrt {31}\right )} \tan ^{-1}\left (\sqrt {-\frac {2}{7}-\frac {i \sqrt {31}}{7}} \sqrt {2 x+1}\right )}{10633}-\frac {10 \sqrt {\frac {1}{217} \left (12504542+1667459 i \sqrt {31}\right )} \tan ^{-1}\left (\sqrt {-\frac {2}{7}+\frac {i \sqrt {31}}{7}} \sqrt {2 x+1}\right )}{10633} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(6076 + 17360*(1 + 2*x) - 6865*(1 + 2*x)^2 + 17550*(1 + 2*x)^3))/(31899*(1 + 2*x)^(3/2)*(7 - 4*(1 + 2*x) +
5*(1 + 2*x)^2)) - (10*Sqrt[(12504542 - (1667459*I)*Sqrt[31])/217]*ArcTan[Sqrt[-2/7 - (I/7)*Sqrt[31]]*Sqrt[1 +
2*x]])/10633 - (10*Sqrt[(12504542 + (1667459*I)*Sqrt[31])/217]*ArcTan[Sqrt[-2/7 + (I/7)*Sqrt[31]]*Sqrt[1 + 2*
x]])/10633

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fricas [B]  time = 0.45, size = 653, normalized size = 2.21

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10765101069366070602*(620294748*161637035^(1/4)*sqrt(4298)*sqrt(217)*sqrt(35)*(20*x^4 + 32*x^3 + 25*x^2 + 1
1*x + 2)*sqrt(12504542*sqrt(35) + 92138375)*arctan(1/55152316249116723757744225*161637035^(3/4)*sqrt(4298)*sqr
t(1535)*sqrt(217)*sqrt(149)*sqrt(161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*sqrt(31))*s
qrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375) + 7775911578470*x + 777591157847*sqrt(35) + 3887955789235)*sqr
t(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) - 1/58486518937462105*161637035^(3/4)*sqrt(4298)*sqrt(21
7)*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt
(31)) + 620294748*161637035^(1/4)*sqrt(4298)*sqrt(217)*sqrt(35)*(20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2)*sqrt(125
04542*sqrt(35) + 92138375)*arctan(1/4729311118361759062226567293750*161637035^(3/4)*sqrt(4298)*sqrt(217)*sqrt(
149)*sqrt(-11286950937500*161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*sqrt(31))*sqrt(2*x
+ 1)*sqrt(12504542*sqrt(35) + 92138375) + 87766332480529071315625000*x + 8776633248052907131562500*sqrt(35) +
43883166240264535657812500)*sqrt(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) - 1/58486518937462105*16
1637035^(3/4)*sqrt(4298)*sqrt(217)*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) - 1/
31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 3*161637035^(1/4)*sqrt(4298)*sqrt(217)*(12504542*sqrt(35)*sqrt(31)*(20
*x^4 + 32*x^3 + 25*x^2 + 11*x + 2) - 92138375*sqrt(31)*(20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2))*sqrt(12504542*sq
rt(35) + 92138375)*log(11286950937500/53789*161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*
sqrt(31))*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375) + 1631678084376528125000*x + 163167808437652812500*
sqrt(35) + 815839042188264062500) - 3*161637035^(1/4)*sqrt(4298)*sqrt(217)*(12504542*sqrt(35)*sqrt(31)*(20*x^4
+ 32*x^3 + 25*x^2 + 11*x + 2) - 92138375*sqrt(31)*(20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2))*sqrt(12504542*sqrt(3
5) + 92138375)*log(-11286950937500/53789*161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*sqr
t(31))*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375) + 1631678084376528125000*x + 163167808437652812500*sqr
t(35) + 815839042188264062500) + 337474562505598*(140400*x^3 + 183140*x^2 + 112560*x + 34121)*sqrt(2*x + 1))/(
20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2)

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giac [B]  time = 0.99, size = 638, normalized size = 2.16

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7914248230*sqrt(31)*(24570*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - 117*sqrt(31)
*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) - 234*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) - 49140*(7/5)^(3/4)*sq
rt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 244510*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) - 489020*(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)) - 1/7914248230*sqrt(31)*(24570*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-
140*sqrt(35) + 2450) - 117*sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) - 234*(7/5)^(3/4)*(140*sqrt(35) +
2450)^(3/2) - 49140*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 244510*sqrt(31)*(7/5)^(1/4)*sqr
t(-140*sqrt(35) + 2450) - 489020*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450))*arctan(-5/7*(7/5)^(3/4)*((7/5)^(1/4)*s
qrt(1/35*sqrt(35) + 1/2) - sqrt(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2)) - 1/15828496460*sqrt(31)*(117*sqrt(31)*(
7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 24570*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35
) + 49140*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - 234*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3
/2) + 244510*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) + 489020*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450))*l
og(2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(35) + 1/2) + 2*x + sqrt(7/5) + 1) + 1/15828496460*sqrt(31)*(117*
sqrt(31)*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 24570*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqr
t(35) - 35) + 49140*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - 234*(7/5)^(3/4)*(-140*sqrt(35)
+ 2450)^(3/2) + 244510*sqrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) + 489020*(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) - 4/10633*(890*(2*x
+ 1)^(3/2) + 233*sqrt(2*x + 1))/(5*(2*x + 1)^2 - 8*x + 3) - 16/1029*(48*x + 31)/(2*x + 1)^(3/2)

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maple [B]  time = 0.36, size = 660, normalized size = 2.23 \begin {gather*} \frac {24175 \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 )}{329623 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {45940 \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 )}{2307361 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {9980 \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 )}{74431 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {24175 \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 )}{329623 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {45940 \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 )}{2307361 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}-\frac {9980 \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 )}{74431 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {4835 \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 )}{659246}-\frac {22970 \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 )}{2307361}-\frac {4835 \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 )}{659246}+\frac {22970 \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 )}{2307361}-\frac {16 \left (\frac {89 \left (2 x +1\right )^{\frac {3}{2}}}{62}+\frac {233 \sqrt {2 x +1}}{620}\right )}{343 \left (-\frac {8 x}{5}+\left (2 x +1\right )^{2}+\frac {3}{5}\right )}-\frac {16}{147 \left (2 x +1\right )^{\frac {3}{2}}}-\frac {128}{343 \sqrt {2 x +1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/343*(89/62*(2*x+1)^(3/2)+233/620*(2*x+1)^(1/2))/(-8/5*x+(2*x+1)^2+3/5)-4835/659246*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)+22970/2307361*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)+241
75/329623/(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))-45940/2307361/(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))-
9980/74431/(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))+4835/659246*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)-22970/2307361*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)+24175/329623/(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))-45940/2307361/(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))-9980/74431/(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))-16/147/(2*x+1)^(3/2)-128/343/(2*x+1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.17, size = 226, normalized size = 0.76 \begin {gather*} -\frac {\frac {128\,x}{147}-\frac {5492\,{\left (2\,x+1\right )}^2}{31899}+\frac {4680\,{\left (2\,x+1\right )}^3}{10633}+\frac {144}{245}}{\frac {7\,{\left (2\,x+1\right )}^{3/2}}{5}-\frac {4\,{\left (2\,x+1\right )}^{5/2}}{5}+{\left (2\,x+1\right )}^{7/2}}+\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {\sqrt {217}\,\sqrt {-12504542-\sqrt {31}\,1667459{}\mathrm {i}}\,\sqrt {2\,x+1}\,6884992{}\mathrm {i}}{1900211000023\,\left (-\frac {63259306496}{271458714289}+\frac {\sqrt {31}\,3435611008{}\mathrm {i}}{271458714289}\right )}-\frac {13769984\,\sqrt {31}\,\sqrt {217}\,\sqrt {-12504542-\sqrt {31}\,1667459{}\mathrm {i}}\,\sqrt {2\,x+1}}{58906541000713\,\left (-\frac {63259306496}{271458714289}+\frac {\sqrt {31}\,3435611008{}\mathrm {i}}{271458714289}\right )}\right )\,\sqrt {-12504542-\sqrt {31}\,1667459{}\mathrm {i}}\,10{}\mathrm {i}}{2307361}-\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {\sqrt {217}\,\sqrt {-12504542+\sqrt {31}\,1667459{}\mathrm {i}}\,\sqrt {2\,x+1}\,6884992{}\mathrm {i}}{1900211000023\,\left (\frac {63259306496}{271458714289}+\frac {\sqrt {31}\,3435611008{}\mathrm {i}}{271458714289}\right )}+\frac {13769984\,\sqrt {31}\,\sqrt {217}\,\sqrt {-12504542+\sqrt {31}\,1667459{}\mathrm {i}}\,\sqrt {2\,x+1}}{58906541000713\,\left (\frac {63259306496}{271458714289}+\frac {\sqrt {31}\,3435611008{}\mathrm {i}}{271458714289}\right )}\right )\,\sqrt {-12504542+\sqrt {31}\,1667459{}\mathrm {i}}\,10{}\mathrm {i}}{2307361} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(217^(1/2)*atan((217^(1/2)*(- 31^(1/2)*1667459i - 12504542)^(1/2)*(2*x + 1)^(1/2)*6884992i)/(1900211000023*((3
1^(1/2)*3435611008i)/271458714289 - 63259306496/271458714289)) - (13769984*31^(1/2)*217^(1/2)*(- 31^(1/2)*1667
459i - 12504542)^(1/2)*(2*x + 1)^(1/2))/(58906541000713*((31^(1/2)*3435611008i)/271458714289 - 63259306496/271
458714289)))*(- 31^(1/2)*1667459i - 12504542)^(1/2)*10i)/2307361 - ((128*x)/147 - (5492*(2*x + 1)^2)/31899 + (
4680*(2*x + 1)^3)/10633 + 144/245)/((7*(2*x + 1)^(3/2))/5 - (4*(2*x + 1)^(5/2))/5 + (2*x + 1)^(7/2)) - (217^(1
/2)*atan((217^(1/2)*(31^(1/2)*1667459i - 12504542)^(1/2)*(2*x + 1)^(1/2)*6884992i)/(1900211000023*((31^(1/2)*3
435611008i)/271458714289 + 63259306496/271458714289)) + (13769984*31^(1/2)*217^(1/2)*(31^(1/2)*1667459i - 1250
4542)^(1/2)*(2*x + 1)^(1/2))/(58906541000713*((31^(1/2)*3435611008i)/271458714289 + 63259306496/271458714289))
)*(31^(1/2)*1667459i - 12504542)^(1/2)*10i)/2307361

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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