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

Optimal. Leaf size=96 \[ -\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}-\frac {63 (1-2 x)^{5/2}}{550 (5 x+3)}-\frac {21}{275} (1-2 x)^{3/2}-\frac {63}{125} \sqrt {1-2 x}+\frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {78, 47, 50, 63, 206} \begin {gather*} -\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}-\frac {63 (1-2 x)^{5/2}}{550 (5 x+3)}-\frac {21}{275} (1-2 x)^{3/2}-\frac {63}{125} \sqrt {1-2 x}+\frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-63*Sqrt[1 - 2*x])/125 - (21*(1 - 2*x)^(3/2))/275 - (1 - 2*x)^(7/2)/(110*(3 + 5*x)^2) - (63*(1 - 2*x)^(5/2))/
(550*(3 + 5*x)) + (63*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

Rule 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}+\frac {63}{110} \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {63}{110} \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {63}{50} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=-\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {693}{250} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {693}{250} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 48, normalized size = 0.50 \begin {gather*} -\frac {(1-2 x)^{7/2} \left (36 (5 x+3)^2 \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {5}{11} (2 x-1)\right )+121\right )}{13310 (5 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/13310*((1 - 2*x)^(7/2)*(121 + 36*(3 + 5*x)^2*Hypergeometric2F1[2, 7/2, 9/2, (-5*(-1 + 2*x))/11]))/(3 + 5*x)
^2

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IntegrateAlgebraic [A]  time = 0.18, size = 81, normalized size = 0.84 \begin {gather*} \frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-\frac {\left (100 (1-2 x)^3+840 (1-2 x)^2-5775 (1-2 x)+7623\right ) \sqrt {1-2 x}}{125 (5 (1-2 x)-11)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/125*((7623 - 5775*(1 - 2*x) + 840*(1 - 2*x)^2 + 100*(1 - 2*x)^3)*Sqrt[1 - 2*x])/(-11 + 5*(1 - 2*x))^2 + (63
*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

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fricas [A]  time = 1.53, size = 86, normalized size = 0.90 \begin {gather*} \frac {63 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 5 \, {\left (400 \, x^{3} - 2280 \, x^{2} - 3795 \, x - 1394\right )} \sqrt {-2 \, x + 1}}{1250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1250*(63*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) +
5*(400*x^3 - 2280*x^2 - 3795*x - 1394)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.08, size = 86, normalized size = 0.90 \begin {gather*} -\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{2500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*
x + 1))) - 256/625*sqrt(-2*x + 1) + 11/2500*(285*(-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.69 \begin {gather*} \frac {63 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{625}-\frac {4 \left (-2 x +1\right )^{\frac {3}{2}}}{125}-\frac {256 \sqrt {-2 x +1}}{625}-\frac {44 \left (-\frac {57 \left (-2 x +1\right )^{\frac {3}{2}}}{20}+\frac {649 \sqrt {-2 x +1}}{100}\right )}{25 \left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(-2*x+1)^(3/2)-256/625*(-2*x+1)^(1/2)-44/25*(-57/20*(-2*x+1)^(3/2)+649/100*(-2*x+1)^(1/2))/(-10*x-6)^2+
63/625*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.28, size = 92, normalized size = 0.96 \begin {gather*} -\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) -
256/625*sqrt(-2*x + 1) + 11/625*(285*(-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.08, size = 74, normalized size = 0.77 \begin {gather*} -\frac {256\,\sqrt {1-2\,x}}{625}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {\frac {7139\,\sqrt {1-2\,x}}{15625}-\frac {627\,{\left (1-2\,x\right )}^{3/2}}{3125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}-\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,63{}\mathrm {i}}{625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*63i)/625 - (256*(1 - 2*x)^(1/2))/625 - (4*(1 - 2*x)^(3/2))/
125 - ((7139*(1 - 2*x)^(1/2))/15625 - (627*(1 - 2*x)^(3/2))/3125)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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

[Out]

Timed out

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