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

Optimal. Leaf size=116 \[ \frac {160}{2401 \sqrt {1-2 x}}-\frac {16}{147 (1-2 x)^{3/2} (3 x+2)}+\frac {160}{3087 (1-2 x)^{3/2}}-\frac {16}{147 (1-2 x)^{3/2} (3 x+2)^2}+\frac {1}{63 (1-2 x)^{3/2} (3 x+2)^3}-\frac {160 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2401} \]

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Rubi [A]  time = 0.03, antiderivative size = 130, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 206} \begin {gather*} -\frac {240 \sqrt {1-2 x}}{2401 (3 x+2)}-\frac {80 \sqrt {1-2 x}}{343 (3 x+2)^2}+\frac {64}{147 \sqrt {1-2 x} (3 x+2)^2}+\frac {64}{441 (1-2 x)^{3/2} (3 x+2)^2}+\frac {1}{63 (1-2 x)^{3/2} (3 x+2)^3}-\frac {160 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2401} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(63*(1 - 2*x)^(3/2)*(2 + 3*x)^3) + 64/(441*(1 - 2*x)^(3/2)*(2 + 3*x)^2) + 64/(147*Sqrt[1 - 2*x]*(2 + 3*x)^2)
- (80*Sqrt[1 - 2*x])/(343*(2 + 3*x)^2) - (240*Sqrt[1 - 2*x])/(2401*(2 + 3*x)) - (160*Sqrt[3/7]*ArcTanh[Sqrt[3
/7]*Sqrt[1 - 2*x]])/2401

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 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}{(1-2 x)^{5/2} (2+3 x)^4} \, dx &=\frac {1}{63 (1-2 x)^{3/2} (2+3 x)^3}+\frac {32}{21} \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3} \, dx\\ &=\frac {1}{63 (1-2 x)^{3/2} (2+3 x)^3}+\frac {64}{441 (1-2 x)^{3/2} (2+3 x)^2}+\frac {32}{21} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\\ &=\frac {1}{63 (1-2 x)^{3/2} (2+3 x)^3}+\frac {64}{441 (1-2 x)^{3/2} (2+3 x)^2}+\frac {64}{147 \sqrt {1-2 x} (2+3 x)^2}+\frac {160}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^3} \, dx\\ &=\frac {1}{63 (1-2 x)^{3/2} (2+3 x)^3}+\frac {64}{441 (1-2 x)^{3/2} (2+3 x)^2}+\frac {64}{147 \sqrt {1-2 x} (2+3 x)^2}-\frac {80 \sqrt {1-2 x}}{343 (2+3 x)^2}+\frac {240}{343} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {1}{63 (1-2 x)^{3/2} (2+3 x)^3}+\frac {64}{441 (1-2 x)^{3/2} (2+3 x)^2}+\frac {64}{147 \sqrt {1-2 x} (2+3 x)^2}-\frac {80 \sqrt {1-2 x}}{343 (2+3 x)^2}-\frac {240 \sqrt {1-2 x}}{2401 (2+3 x)}+\frac {240 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx}{2401}\\ &=\frac {1}{63 (1-2 x)^{3/2} (2+3 x)^3}+\frac {64}{441 (1-2 x)^{3/2} (2+3 x)^2}+\frac {64}{147 \sqrt {1-2 x} (2+3 x)^2}-\frac {80 \sqrt {1-2 x}}{343 (2+3 x)^2}-\frac {240 \sqrt {1-2 x}}{2401 (2+3 x)}-\frac {240 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{2401}\\ &=\frac {1}{63 (1-2 x)^{3/2} (2+3 x)^3}+\frac {64}{441 (1-2 x)^{3/2} (2+3 x)^2}+\frac {64}{147 \sqrt {1-2 x} (2+3 x)^2}-\frac {80 \sqrt {1-2 x}}{343 (2+3 x)^2}-\frac {240 \sqrt {1-2 x}}{2401 (2+3 x)}-\frac {160 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2401}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 42, normalized size = 0.36 \begin {gather*} \frac {256 \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};\frac {3}{7}-\frac {6 x}{7}\right )+\frac {343}{(3 x+2)^3}}{21609 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(343/(2 + 3*x)^3 + 256*Hypergeometric2F1[-3/2, 3, -1/2, 3/7 - (6*x)/7])/(21609*(1 - 2*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.26, size = 90, normalized size = 0.78 \begin {gather*} \frac {8 \left (1620 (1-2 x)^4-10080 (1-2 x)^3+19404 (1-2 x)^2-9408 (1-2 x)-3773\right )}{7203 (3 (1-2 x)-7)^3 (1-2 x)^{3/2}}-\frac {160 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{2401} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-3773 - 9408*(1 - 2*x) + 19404*(1 - 2*x)^2 - 10080*(1 - 2*x)^3 + 1620*(1 - 2*x)^4))/(7203*(-7 + 3*(1 - 2*x
))^3*(1 - 2*x)^(3/2)) - (160*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/2401

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fricas [A]  time = 1.51, size = 120, normalized size = 1.03 \begin {gather*} \frac {240 \, \sqrt {7} \sqrt {3} {\left (108 \, x^{5} + 108 \, x^{4} - 45 \, x^{3} - 58 \, x^{2} + 4 \, x + 8\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 7 \, {\left (25920 \, x^{4} + 28800 \, x^{3} - 4464 \, x^{2} - 11280 \, x - 2237\right )} \sqrt {-2 \, x + 1}}{50421 \, {\left (108 \, x^{5} + 108 \, x^{4} - 45 \, x^{3} - 58 \, x^{2} + 4 \, x + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50421*(240*sqrt(7)*sqrt(3)*(108*x^5 + 108*x^4 - 45*x^3 - 58*x^2 + 4*x + 8)*log((sqrt(7)*sqrt(3)*sqrt(-2*x +
1) + 3*x - 5)/(3*x + 2)) - 7*(25920*x^4 + 28800*x^3 - 4464*x^2 - 11280*x - 2237)*sqrt(-2*x + 1))/(108*x^5 + 10
8*x^4 - 45*x^3 - 58*x^2 + 4*x + 8)

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giac [A]  time = 1.24, size = 95, normalized size = 0.82 \begin {gather*} \frac {80}{16807} \, \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 {8 \, {\left (1620 \, {\left (2 \, x - 1\right )}^{4} + 10080 \, {\left (2 \, x - 1\right )}^{3} + 19404 \, {\left (2 \, x - 1\right )}^{2} + 18816 \, x - 13181\right )}}{7203 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

80/16807*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 8/7203*(1620*(2
*x - 1)^4 + 10080*(2*x - 1)^3 + 19404*(2*x - 1)^2 + 18816*x - 13181)/(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 = 0.65 \begin {gather*} -\frac {160 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{16807}+\frac {88}{7203 \left (-2 x +1\right )^{\frac {3}{2}}}+\frac {776}{16807 \sqrt {-2 x +1}}+\frac {\frac {9288 \left (-2 x +1\right )^{\frac {5}{2}}}{16807}-\frac {960 \left (-2 x +1\right )^{\frac {3}{2}}}{343}+\frac {1200 \sqrt {-2 x +1}}{343}}{\left (-6 x -4\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

88/7203/(-2*x+1)^(3/2)+776/16807/(-2*x+1)^(1/2)+648/16807*(43/3*(-2*x+1)^(5/2)-1960/27*(-2*x+1)^(3/2)+2450/27*
(-2*x+1)^(1/2))/(-6*x-4)^3-160/16807*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.22, size = 110, normalized size = 0.95 \begin {gather*} \frac {80}{16807} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {8 \, {\left (1620 \, {\left (2 \, x - 1\right )}^{4} + 10080 \, {\left (2 \, x - 1\right )}^{3} + 19404 \, {\left (2 \, x - 1\right )}^{2} + 18816 \, x - 13181\right )}}{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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

80/16807*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 8/7203*(1620*(2*x - 1)^4
+ 10080*(2*x - 1)^3 + 19404*(2*x - 1)^2 + 18816*x - 13181)/(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 = 1.20, size = 92, normalized size = 0.79 \begin {gather*} -\frac {\frac {1024\,x}{1323}+\frac {352\,{\left (2\,x-1\right )}^2}{441}+\frac {1280\,{\left (2\,x-1\right )}^3}{3087}+\frac {160\,{\left (2\,x-1\right )}^4}{2401}-\frac {2152}{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 {160\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{16807} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((1024*x)/1323 + (352*(2*x - 1)^2)/441 + (1280*(2*x - 1)^3)/3087 + (160*(2*x - 1)^4)/2401 - 2152/3969)/((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)) - (160*21^(1/2)*atanh((21
^(1/2)*(1 - 2*x)^(1/2))/7))/16807

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

[Out]

Timed out

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