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

Optimal. Leaf size=82 \[ -\frac {3}{35} (1-2 x)^{7/2}+\frac {2}{125} (1-2 x)^{5/2}+\frac {22}{375} (1-2 x)^{3/2}+\frac {242}{625} \sqrt {1-2 x}-\frac {242}{625} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 206} \begin {gather*} -\frac {3}{35} (1-2 x)^{7/2}+\frac {2}{125} (1-2 x)^{5/2}+\frac {22}{375} (1-2 x)^{3/2}+\frac {242}{625} \sqrt {1-2 x}-\frac {242}{625} \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),x]

[Out]

(242*Sqrt[1 - 2*x])/625 + (22*(1 - 2*x)^(3/2))/375 + (2*(1 - 2*x)^(5/2))/125 - (3*(1 - 2*x)^(7/2))/35 - (242*S
qrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/625

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 80

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

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} \, dx &=-\frac {3}{35} (1-2 x)^{7/2}+\frac {1}{5} \int \frac {(1-2 x)^{5/2}}{3+5 x} \, dx\\ &=\frac {2}{125} (1-2 x)^{5/2}-\frac {3}{35} (1-2 x)^{7/2}+\frac {11}{25} \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=\frac {22}{375} (1-2 x)^{3/2}+\frac {2}{125} (1-2 x)^{5/2}-\frac {3}{35} (1-2 x)^{7/2}+\frac {121}{125} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {242}{625} \sqrt {1-2 x}+\frac {22}{375} (1-2 x)^{3/2}+\frac {2}{125} (1-2 x)^{5/2}-\frac {3}{35} (1-2 x)^{7/2}+\frac {1331}{625} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {242}{625} \sqrt {1-2 x}+\frac {22}{375} (1-2 x)^{3/2}+\frac {2}{125} (1-2 x)^{5/2}-\frac {3}{35} (1-2 x)^{7/2}-\frac {1331}{625} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {242}{625} \sqrt {1-2 x}+\frac {22}{375} (1-2 x)^{3/2}+\frac {2}{125} (1-2 x)^{5/2}-\frac {3}{35} (1-2 x)^{7/2}-\frac {242}{625} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.68 \begin {gather*} \frac {5 \sqrt {1-2 x} \left (9000 x^3-12660 x^2+4370 x+4937\right )-5082 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{65625} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Sqrt[1 - 2*x]*(4937 + 4370*x - 12660*x^2 + 9000*x^3) - 5082*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/656
25

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IntegrateAlgebraic [A]  time = 0.06, size = 70, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {1-2 x} \left (1125 (1-2 x)^3-210 (1-2 x)^2-770 (1-2 x)-5082\right )}{13125}-\frac {242}{625} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/13125*((-5082 - 770*(1 - 2*x) - 210*(1 - 2*x)^2 + 1125*(1 - 2*x)^3)*Sqrt[1 - 2*x]) - (242*Sqrt[11/5]*ArcTan
h[Sqrt[5/11]*Sqrt[1 - 2*x]])/625

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fricas [A]  time = 1.64, size = 61, normalized size = 0.74 \begin {gather*} \frac {121}{3125} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {1}{13125} \, {\left (9000 \, x^{3} - 12660 \, x^{2} + 4370 \, x + 4937\right )} \sqrt {-2 \, x + 1} \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),x, algorithm="fricas")

[Out]

121/3125*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 1/13125*(9000*x^3 - 126
60*x^2 + 4370*x + 4937)*sqrt(-2*x + 1)

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giac [A]  time = 1.01, size = 90, normalized size = 1.10 \begin {gather*} \frac {3}{35} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {2}{125} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {22}{375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{3125} \, \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 {242}{625} \, \sqrt {-2 \, x + 1} \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),x, algorithm="giac")

[Out]

3/35*(2*x - 1)^3*sqrt(-2*x + 1) + 2/125*(2*x - 1)^2*sqrt(-2*x + 1) + 22/375*(-2*x + 1)^(3/2) + 121/3125*sqrt(5
5)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 242/625*sqrt(-2*x + 1)

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maple [A]  time = 0.01, size = 56, normalized size = 0.68 \begin {gather*} -\frac {242 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{3125}+\frac {22 \left (-2 x +1\right )^{\frac {3}{2}}}{375}+\frac {2 \left (-2 x +1\right )^{\frac {5}{2}}}{125}-\frac {3 \left (-2 x +1\right )^{\frac {7}{2}}}{35}+\frac {242 \sqrt {-2 x +1}}{625} \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),x)

[Out]

22/375*(-2*x+1)^(3/2)+2/125*(-2*x+1)^(5/2)-3/35*(-2*x+1)^(7/2)-242/3125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*
55^(1/2)+242/625*(-2*x+1)^(1/2)

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maxima [A]  time = 1.29, size = 73, normalized size = 0.89 \begin {gather*} -\frac {3}{35} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {2}{125} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {22}{375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{3125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {242}{625} \, \sqrt {-2 \, x + 1} \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),x, algorithm="maxima")

[Out]

-3/35*(-2*x + 1)^(7/2) + 2/125*(-2*x + 1)^(5/2) + 22/375*(-2*x + 1)^(3/2) + 121/3125*sqrt(55)*log(-(sqrt(55) -
5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 242/625*sqrt(-2*x + 1)

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mupad [B]  time = 0.06, size = 57, normalized size = 0.70 \begin {gather*} \frac {242\,\sqrt {1-2\,x}}{625}+\frac {22\,{\left (1-2\,x\right )}^{3/2}}{375}+\frac {2\,{\left (1-2\,x\right )}^{5/2}}{125}-\frac {3\,{\left (1-2\,x\right )}^{7/2}}{35}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{3125} \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),x)

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*242i)/3125 + (242*(1 - 2*x)^(1/2))/625 + (22*(1 - 2*x)^(3/2))
/375 + (2*(1 - 2*x)^(5/2))/125 - (3*(1 - 2*x)^(7/2))/35

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sympy [A]  time = 35.95, size = 114, normalized size = 1.39 \begin {gather*} - \frac {3 \left (1 - 2 x\right )^{\frac {7}{2}}}{35} + \frac {2 \left (1 - 2 x\right )^{\frac {5}{2}}}{125} + \frac {22 \left (1 - 2 x\right )^{\frac {3}{2}}}{375} + \frac {242 \sqrt {1 - 2 x}}{625} + \frac {2662 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right )}{625} \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),x)

[Out]

-3*(1 - 2*x)**(7/2)/35 + 2*(1 - 2*x)**(5/2)/125 + 22*(1 - 2*x)**(3/2)/375 + 242*sqrt(1 - 2*x)/625 + 2662*Piece
wise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)
/11)/55, 2*x - 1 > -11/5))/625

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