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

Optimal. Leaf size=67 \[ -\frac {25}{12} (1-2 x)^{5/2}+\frac {400}{27} (1-2 x)^{3/2}-\frac {5135}{108} \sqrt {1-2 x}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}} \]

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {88, 63, 206} \begin {gather*} -\frac {25}{12} (1-2 x)^{5/2}+\frac {400}{27} (1-2 x)^{3/2}-\frac {5135}{108} \sqrt {1-2 x}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5135*Sqrt[1 - 2*x])/108 + (400*(1 - 2*x)^(3/2))/27 - (25*(1 - 2*x)^(5/2))/12 + (2*ArcTanh[Sqrt[3/7]*Sqrt[1 -
2*x]])/(27*Sqrt[21])

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -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)^3}{\sqrt {1-2 x} (2+3 x)} \, dx &=\int \left (\frac {5135}{108 \sqrt {1-2 x}}-\frac {400}{9} \sqrt {1-2 x}+\frac {125}{12} (1-2 x)^{3/2}-\frac {1}{27 \sqrt {1-2 x} (2+3 x)}\right ) \, dx\\ &=-\frac {5135}{108} \sqrt {1-2 x}+\frac {400}{27} (1-2 x)^{3/2}-\frac {25}{12} (1-2 x)^{5/2}-\frac {1}{27} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {5135}{108} \sqrt {1-2 x}+\frac {400}{27} (1-2 x)^{3/2}-\frac {25}{12} (1-2 x)^{5/2}+\frac {1}{27} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {5135}{108} \sqrt {1-2 x}+\frac {400}{27} (1-2 x)^{3/2}-\frac {25}{12} (1-2 x)^{5/2}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 51, normalized size = 0.76 \begin {gather*} \frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}}-\frac {5}{27} \sqrt {1-2 x} \left (45 x^2+115 x+188\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[1 - 2*x]*(188 + 115*x + 45*x^2))/27 + (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(27*Sqrt[21])

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IntegrateAlgebraic [A]  time = 0.06, size = 59, normalized size = 0.88 \begin {gather*} \frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{27 \sqrt {21}}-\frac {5}{108} \left (45 (1-2 x)^2-320 (1-2 x)+1027\right ) \sqrt {1-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(1027 - 320*(1 - 2*x) + 45*(1 - 2*x)^2)*Sqrt[1 - 2*x])/108 + (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(27*Sqrt
[21])

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fricas [A]  time = 1.50, size = 51, normalized size = 0.76 \begin {gather*} -\frac {5}{27} \, {\left (45 \, x^{2} + 115 \, x + 188\right )} \sqrt {-2 \, x + 1} + \frac {1}{567} \, \sqrt {21} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/27*(45*x^2 + 115*x + 188)*sqrt(-2*x + 1) + 1/567*sqrt(21)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)
)

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giac [A]  time = 1.00, size = 74, normalized size = 1.10 \begin {gather*} -\frac {25}{12} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {400}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {1}{567} \, \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 {5135}{108} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/12*(2*x - 1)^2*sqrt(-2*x + 1) + 400/27*(-2*x + 1)^(3/2) - 1/567*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(
-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 5135/108*sqrt(-2*x + 1)

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maple [A]  time = 0.01, size = 47, normalized size = 0.70 \begin {gather*} \frac {2 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{567}+\frac {400 \left (-2 x +1\right )^{\frac {3}{2}}}{27}-\frac {25 \left (-2 x +1\right )^{\frac {5}{2}}}{12}-\frac {5135 \sqrt {-2 x +1}}{108} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

400/27*(-2*x+1)^(3/2)-25/12*(-2*x+1)^(5/2)+2/567*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)-5135/108*(-2*x+
1)^(1/2)

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maxima [A]  time = 1.23, size = 64, normalized size = 0.96 \begin {gather*} -\frac {25}{12} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {400}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {1}{567} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {5135}{108} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/12*(-2*x + 1)^(5/2) + 400/27*(-2*x + 1)^(3/2) - 1/567*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21
) + 3*sqrt(-2*x + 1))) - 5135/108*sqrt(-2*x + 1)

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mupad [B]  time = 1.18, size = 48, normalized size = 0.72 \begin {gather*} \frac {400\,{\left (1-2\,x\right )}^{3/2}}{27}-\frac {5135\,\sqrt {1-2\,x}}{108}-\frac {25\,{\left (1-2\,x\right )}^{5/2}}{12}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,2{}\mathrm {i}}{567} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(400*(1 - 2*x)^(3/2))/27 - (5135*(1 - 2*x)^(1/2))/108 - (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*2i)/56
7 - (25*(1 - 2*x)^(5/2))/12

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sympy [A]  time = 36.19, size = 102, normalized size = 1.52 \begin {gather*} - \frac {25 \left (1 - 2 x\right )^{\frac {5}{2}}}{12} + \frac {400 \left (1 - 2 x\right )^{\frac {3}{2}}}{27} - \frac {5135 \sqrt {1 - 2 x}}{108} - \frac {2 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21}}{3 \sqrt {1 - 2 x}} \right )}}{21} & \text {for}\: \frac {1}{1 - 2 x} > \frac {3}{7} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21}}{3 \sqrt {1 - 2 x}} \right )}}{21} & \text {for}\: \frac {1}{1 - 2 x} < \frac {3}{7} \end {cases}\right )}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25*(1 - 2*x)**(5/2)/12 + 400*(1 - 2*x)**(3/2)/27 - 5135*sqrt(1 - 2*x)/108 - 2*Piecewise((-sqrt(21)*acoth(sqrt
(21)/(3*sqrt(1 - 2*x)))/21, 1/(1 - 2*x) > 3/7), (-sqrt(21)*atanh(sqrt(21)/(3*sqrt(1 - 2*x)))/21, 1/(1 - 2*x) <
3/7))/27

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