3.2155 \(\int \frac {(3+5 x)^2}{(1-2 x)^{5/2} (2+3 x)^4} \, dx\)

Optimal. Leaf size=121 \[ \frac {1415}{7203 \sqrt {1-2 x}}-\frac {1415}{6174 \sqrt {1-2 x} (3 x+2)}-\frac {283}{882 \sqrt {1-2 x} (3 x+2)^2}-\frac {1091}{882 \sqrt {1-2 x} (3 x+2)^3}+\frac {121}{42 (1-2 x)^{3/2} (3 x+2)^3}-\frac {1415 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2401 \sqrt {21}} \]

[Out]

121/42/(1-2*x)^(3/2)/(2+3*x)^3-1415/50421*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+1415/7203/(1-2*x)^(1/2)
-1091/882/(2+3*x)^3/(1-2*x)^(1/2)-283/882/(2+3*x)^2/(1-2*x)^(1/2)-1415/6174/(2+3*x)/(1-2*x)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 128, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {89, 78, 51, 63, 206} \[ -\frac {1415 \sqrt {1-2 x}}{4802 (3 x+2)}-\frac {1415 \sqrt {1-2 x}}{2058 (3 x+2)^2}+\frac {566}{441 \sqrt {1-2 x} (3 x+2)^2}-\frac {1091}{882 \sqrt {1-2 x} (3 x+2)^3}+\frac {121}{42 (1-2 x)^{3/2} (3 x+2)^3}-\frac {1415 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2401 \sqrt {21}} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^2/((1 - 2*x)^(5/2)*(2 + 3*x)^4),x]

[Out]

121/(42*(1 - 2*x)^(3/2)*(2 + 3*x)^3) - 1091/(882*Sqrt[1 - 2*x]*(2 + 3*x)^3) + 566/(441*Sqrt[1 - 2*x]*(2 + 3*x)
^2) - (1415*Sqrt[1 - 2*x])/(2058*(2 + 3*x)^2) - (1415*Sqrt[1 - 2*x])/(4802*(2 + 3*x)) - (1415*ArcTanh[Sqrt[3/7
]*Sqrt[1 - 2*x]])/(2401*Sqrt[21])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 206

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

Rubi steps

\begin {align*} \int \frac {(3+5 x)^2}{(1-2 x)^{5/2} (2+3 x)^4} \, dx &=\frac {121}{42 (1-2 x)^{3/2} (2+3 x)^3}-\frac {1}{42} \int \frac {-741+525 x}{(1-2 x)^{3/2} (2+3 x)^4} \, dx\\ &=\frac {121}{42 (1-2 x)^{3/2} (2+3 x)^3}-\frac {1091}{882 \sqrt {1-2 x} (2+3 x)^3}+\frac {283}{63} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\\ &=\frac {121}{42 (1-2 x)^{3/2} (2+3 x)^3}-\frac {1091}{882 \sqrt {1-2 x} (2+3 x)^3}+\frac {566}{441 \sqrt {1-2 x} (2+3 x)^2}+\frac {1415}{147} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^3} \, dx\\ &=\frac {121}{42 (1-2 x)^{3/2} (2+3 x)^3}-\frac {1091}{882 \sqrt {1-2 x} (2+3 x)^3}+\frac {566}{441 \sqrt {1-2 x} (2+3 x)^2}-\frac {1415 \sqrt {1-2 x}}{2058 (2+3 x)^2}+\frac {1415}{686} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {121}{42 (1-2 x)^{3/2} (2+3 x)^3}-\frac {1091}{882 \sqrt {1-2 x} (2+3 x)^3}+\frac {566}{441 \sqrt {1-2 x} (2+3 x)^2}-\frac {1415 \sqrt {1-2 x}}{2058 (2+3 x)^2}-\frac {1415 \sqrt {1-2 x}}{4802 (2+3 x)}+\frac {1415 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx}{4802}\\ &=\frac {121}{42 (1-2 x)^{3/2} (2+3 x)^3}-\frac {1091}{882 \sqrt {1-2 x} (2+3 x)^3}+\frac {566}{441 \sqrt {1-2 x} (2+3 x)^2}-\frac {1415 \sqrt {1-2 x}}{2058 (2+3 x)^2}-\frac {1415 \sqrt {1-2 x}}{4802 (2+3 x)}-\frac {1415 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{4802}\\ &=\frac {121}{42 (1-2 x)^{3/2} (2+3 x)^3}-\frac {1091}{882 \sqrt {1-2 x} (2+3 x)^3}+\frac {566}{441 \sqrt {1-2 x} (2+3 x)^2}-\frac {1415 \sqrt {1-2 x}}{2058 (2+3 x)^2}-\frac {1415 \sqrt {1-2 x}}{4802 (2+3 x)}-\frac {1415 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2401 \sqrt {21}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 59, normalized size = 0.49 \[ -\frac {2264 (2 x-1) (3 x+2)^3 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {3}{7}-\frac {6 x}{7}\right )-49 (1091 x+725)}{21609 (1-2 x)^{3/2} (3 x+2)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^2/((1 - 2*x)^(5/2)*(2 + 3*x)^4),x]

[Out]

-1/21609*(-49*(725 + 1091*x) + 2264*(-1 + 2*x)*(2 + 3*x)^3*Hypergeometric2F1[-1/2, 3, 1/2, 3/7 - (6*x)/7])/((1
- 2*x)^(3/2)*(2 + 3*x)^3)

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fricas [A]  time = 0.81, size = 114, normalized size = 0.94 \[ \frac {1415 \, \sqrt {21} {\left (108 \, x^{5} + 108 \, x^{4} - 45 \, x^{3} - 58 \, x^{2} + 4 \, x + 8\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 7 \, {\left (152820 \, x^{4} + 169800 \, x^{3} - 26319 \, x^{2} - 83655 \, x - 23872\right )} \sqrt {-2 \, x + 1}}{100842 \, {\left (108 \, x^{5} + 108 \, x^{4} - 45 \, x^{3} - 58 \, x^{2} + 4 \, x + 8\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)^2/(1-2*x)^(5/2)/(2+3*x)^4,x, algorithm="fricas")

[Out]

1/100842*(1415*sqrt(21)*(108*x^5 + 108*x^4 - 45*x^3 - 58*x^2 + 4*x + 8)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5
)/(3*x + 2)) - 7*(152820*x^4 + 169800*x^3 - 26319*x^2 - 83655*x - 23872)*sqrt(-2*x + 1))/(108*x^5 + 108*x^4 -
45*x^3 - 58*x^2 + 4*x + 8)

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giac [A]  time = 1.29, size = 95, normalized size = 0.79 \[ \frac {1415}{100842} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {38205 \, {\left (2 \, x - 1\right )}^{4} + 237720 \, {\left (2 \, x - 1\right )}^{3} + 457611 \, {\left (2 \, x - 1\right )}^{2} + 375144 \, x - 353584}{7203 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)^2/(1-2*x)^(5/2)/(2+3*x)^4,x, algorithm="giac")

[Out]

1415/100842*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/7203*(3820
5*(2*x - 1)^4 + 237720*(2*x - 1)^3 + 457611*(2*x - 1)^2 + 375144*x - 353584)/(3*(-2*x + 1)^(3/2) - 7*sqrt(-2*x
+ 1))^3

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maple [A]  time = 0.02, size = 75, normalized size = 0.62 \[ -\frac {1415 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{50421}+\frac {484}{7203 \left (-2 x +1\right )^{\frac {3}{2}}}+\frac {2728}{16807 \sqrt {-2 x +1}}+\frac {\frac {15489 \left (-2 x +1\right )^{\frac {5}{2}}}{16807}-\frac {1420 \left (-2 x +1\right )^{\frac {3}{2}}}{343}+\frac {1595 \sqrt {-2 x +1}}{343}}{\left (-6 x -4\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x+3)^2/(-2*x+1)^(5/2)/(3*x+2)^4,x)

[Out]

484/7203/(-2*x+1)^(3/2)+2728/16807/(-2*x+1)^(1/2)+108/16807*(1721/12*(-2*x+1)^(5/2)-17395/27*(-2*x+1)^(3/2)+78
155/108*(-2*x+1)^(1/2))/(-6*x-4)^3-1415/50421*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.19, size = 110, normalized size = 0.91 \[ \frac {1415}{100842} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {38205 \, {\left (2 \, x - 1\right )}^{4} + 237720 \, {\left (2 \, x - 1\right )}^{3} + 457611 \, {\left (2 \, x - 1\right )}^{2} + 375144 \, x - 353584}{7203 \, {\left (27 \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - 189 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + 441 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 343 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)^2/(1-2*x)^(5/2)/(2+3*x)^4,x, algorithm="maxima")

[Out]

1415/100842*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/7203*(38205*(2*x -
1)^4 + 237720*(2*x - 1)^3 + 457611*(2*x - 1)^2 + 375144*x - 353584)/(27*(-2*x + 1)^(9/2) - 189*(-2*x + 1)^(7/2
) + 441*(-2*x + 1)^(5/2) - 343*(-2*x + 1)^(3/2))

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mupad [B]  time = 1.22, size = 92, normalized size = 0.76 \[ -\frac {\frac {2552\,x}{1323}+\frac {3113\,{\left (2\,x-1\right )}^2}{1323}+\frac {11320\,{\left (2\,x-1\right )}^3}{9261}+\frac {1415\,{\left (2\,x-1\right )}^4}{7203}-\frac {7216}{3969}}{\frac {343\,{\left (1-2\,x\right )}^{3/2}}{27}-\frac {49\,{\left (1-2\,x\right )}^{5/2}}{3}+7\,{\left (1-2\,x\right )}^{7/2}-{\left (1-2\,x\right )}^{9/2}}-\frac {1415\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{50421} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 3)^2/((1 - 2*x)^(5/2)*(3*x + 2)^4),x)

[Out]

- ((2552*x)/1323 + (3113*(2*x - 1)^2)/1323 + (11320*(2*x - 1)^3)/9261 + (1415*(2*x - 1)^4)/7203 - 7216/3969)/(
(343*(1 - 2*x)^(3/2))/27 - (49*(1 - 2*x)^(5/2))/3 + 7*(1 - 2*x)^(7/2) - (1 - 2*x)^(9/2)) - (1415*21^(1/2)*atan
h((21^(1/2)*(1 - 2*x)^(1/2))/7))/50421

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)**2/(1-2*x)**(5/2)/(2+3*x)**4,x)

[Out]

Timed out

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