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

Optimal. Leaf size=67 \[ -\frac {125}{36} (1-2 x)^{3/2}+\frac {400}{9} \sqrt {1-2 x}+\frac {1331}{28 \sqrt {1-2 x}}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{63 \sqrt {21}} \]

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 43, 63, 206} \begin {gather*} -\frac {125}{36} (1-2 x)^{3/2}+\frac {400}{9} \sqrt {1-2 x}+\frac {1331}{28 \sqrt {1-2 x}}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{63 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1331/(28*Sqrt[1 - 2*x]) + (400*Sqrt[1 - 2*x])/9 - (125*(1 - 2*x)^(3/2))/36 + (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x
]])/(63*Sqrt[21])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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}{(1-2 x)^{3/2} (2+3 x)} \, dx &=\int \left (\frac {1331}{28 (1-2 x)^{3/2}}-\frac {1225}{36 \sqrt {1-2 x}}-\frac {125 x}{6 \sqrt {1-2 x}}-\frac {1}{63 \sqrt {1-2 x} (2+3 x)}\right ) \, dx\\ &=\frac {1331}{28 \sqrt {1-2 x}}+\frac {1225}{36} \sqrt {1-2 x}-\frac {1}{63} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx-\frac {125}{6} \int \frac {x}{\sqrt {1-2 x}} \, dx\\ &=\frac {1331}{28 \sqrt {1-2 x}}+\frac {1225}{36} \sqrt {1-2 x}+\frac {1}{63} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {125}{6} \int \left (\frac {1}{2 \sqrt {1-2 x}}-\frac {1}{2} \sqrt {1-2 x}\right ) \, dx\\ &=\frac {1331}{28 \sqrt {1-2 x}}+\frac {400}{9} \sqrt {1-2 x}-\frac {125}{36} (1-2 x)^{3/2}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{63 \sqrt {21}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 45, normalized size = 0.67 \begin {gather*} \frac {-2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {3}{7}-\frac {6 x}{7}\right )-35 \left (75 x^2+405 x-478\right )}{189 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*(-478 + 405*x + 75*x^2) - 2*Hypergeometric2F1[-1/2, 1, 1/2, 3/7 - (6*x)/7])/(189*Sqrt[1 - 2*x])

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IntegrateAlgebraic [A]  time = 0.07, size = 59, normalized size = 0.88 \begin {gather*} \frac {-875 (1-2 x)^2+11200 (1-2 x)+11979}{252 \sqrt {1-2 x}}+\frac {2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{63 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11979 + 11200*(1 - 2*x) - 875*(1 - 2*x)^2)/(252*Sqrt[1 - 2*x]) + (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(63*Sqr
t[21])

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fricas [A]  time = 1.07, size = 64, normalized size = 0.96 \begin {gather*} \frac {\sqrt {21} {\left (2 \, x - 1\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (875 \, x^{2} + 4725 \, x - 5576\right )} \sqrt {-2 \, x + 1}}{1323 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1323*(sqrt(21)*(2*x - 1)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 21*(875*x^2 + 4725*x - 5576)*s
qrt(-2*x + 1))/(2*x - 1)

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giac [A]  time = 1.26, size = 67, normalized size = 1.00 \begin {gather*} -\frac {125}{36} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {1}{1323} \, \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 {400}{9} \, \sqrt {-2 \, x + 1} + \frac {1331}{28 \, \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/36*(-2*x + 1)^(3/2) - 1/1323*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x
+ 1))) + 400/9*sqrt(-2*x + 1) + 1331/28/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 )}{1323}-\frac {125 \left (-2 x +1\right )^{\frac {3}{2}}}{36}+\frac {1331}{28 \sqrt {-2 x +1}}+\frac {400 \sqrt {-2 x +1}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/36*(-2*x+1)^(3/2)+2/1323*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)+1331/28/(-2*x+1)^(1/2)+400/9*(-2*x
+1)^(1/2)

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maxima [A]  time = 1.29, size = 64, normalized size = 0.96 \begin {gather*} -\frac {125}{36} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {1}{1323} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {400}{9} \, \sqrt {-2 \, x + 1} + \frac {1331}{28 \, \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/36*(-2*x + 1)^(3/2) - 1/1323*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) +
400/9*sqrt(-2*x + 1) + 1331/28/sqrt(-2*x + 1)

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mupad [B]  time = 0.06, size = 48, normalized size = 0.72 \begin {gather*} \frac {1331}{28\,\sqrt {1-2\,x}}+\frac {400\,\sqrt {1-2\,x}}{9}-\frac {125\,{\left (1-2\,x\right )}^{3/2}}{36}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,2{}\mathrm {i}}{1323} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1331/(28*(1 - 2*x)^(1/2)) - (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*2i)/1323 + (400*(1 - 2*x)^(1/2))/9
- (125*(1 - 2*x)^(3/2))/36

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sympy [A]  time = 59.86, size = 102, normalized size = 1.52 \begin {gather*} - \frac {125 \left (1 - 2 x\right )^{\frac {3}{2}}}{36} + \frac {400 \sqrt {1 - 2 x}}{9} - \frac {2 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 < - \frac {7}{3} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 > - \frac {7}{3} \end {cases}\right )}{63} + \frac {1331}{28 \sqrt {1 - 2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125*(1 - 2*x)**(3/2)/36 + 400*sqrt(1 - 2*x)/9 - 2*Piecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*
x - 1 < -7/3), (-sqrt(21)*atanh(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 > -7/3))/63 + 1331/(28*sqrt(1 - 2*x))

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