3.2107 \(\int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)} \, dx\)

Optimal. Leaf size=67 \[ -\frac {9}{20} (1-2 x)^{3/2}+\frac {162}{25} \sqrt {1-2 x}+\frac {343}{44 \sqrt {1-2 x}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]

[Out]

-9/20*(1-2*x)^(3/2)-2/15125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+343/44/(1-2*x)^(1/2)+162/25*(1-2*x)^
(1/2)

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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} \[ -\frac {9}{20} (1-2 x)^{3/2}+\frac {162}{25} \sqrt {1-2 x}+\frac {343}{44 \sqrt {1-2 x}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

343/(44*Sqrt[1 - 2*x]) + (162*Sqrt[1 - 2*x])/25 - (9*(1 - 2*x)^(3/2))/20 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]
])/(275*Sqrt[55])

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 {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)} \, dx &=\int \left (\frac {343}{44 (1-2 x)^{3/2}}-\frac {513}{100 \sqrt {1-2 x}}-\frac {27 x}{10 \sqrt {1-2 x}}+\frac {1}{275 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac {343}{44 \sqrt {1-2 x}}+\frac {513}{100} \sqrt {1-2 x}+\frac {1}{275} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx-\frac {27}{10} \int \frac {x}{\sqrt {1-2 x}} \, dx\\ &=\frac {343}{44 \sqrt {1-2 x}}+\frac {513}{100} \sqrt {1-2 x}-\frac {1}{275} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {27}{10} \int \left (\frac {1}{2 \sqrt {1-2 x}}-\frac {1}{2} \sqrt {1-2 x}\right ) \, dx\\ &=\frac {343}{44 \sqrt {1-2 x}}+\frac {162}{25} \sqrt {1-2 x}-\frac {9}{20} (1-2 x)^{3/2}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 45, normalized size = 0.67 \[ \frac {2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {5}{11} (1-2 x)\right )-99 \left (25 x^2+155 x-192\right )}{1375 \sqrt {1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-99*(-192 + 155*x + 25*x^2) + 2*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 - 2*x))/11])/(1375*Sqrt[1 - 2*x])

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fricas [A]  time = 1.06, size = 63, normalized size = 0.94 \[ \frac {\sqrt {55} {\left (2 \, x - 1\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (495 \, x^{2} + 3069 \, x - 3802\right )} \sqrt {-2 \, x + 1}}{15125 \, {\left (2 \, x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15125*(sqrt(55)*(2*x - 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(495*x^2 + 3069*x - 3802)*
sqrt(-2*x + 1))/(2*x - 1)

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giac [A]  time = 1.23, size = 67, normalized size = 1.00 \[ -\frac {9}{20} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{15125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {162}{25} \, \sqrt {-2 \, x + 1} + \frac {343}{44 \, \sqrt {-2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/20*(-2*x + 1)^(3/2) + 1/15125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x
+ 1))) + 162/25*sqrt(-2*x + 1) + 343/44/sqrt(-2*x + 1)

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maple [A]  time = 0.01, size = 47, normalized size = 0.70 \[ -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{15125}-\frac {9 \left (-2 x +1\right )^{\frac {3}{2}}}{20}+\frac {343}{44 \sqrt {-2 x +1}}+\frac {162 \sqrt {-2 x +1}}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/20*(-2*x+1)^(3/2)-2/15125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)+343/44/(-2*x+1)^(1/2)+162/25*(-2*x
+1)^(1/2)

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maxima [A]  time = 1.42, size = 64, normalized size = 0.96 \[ -\frac {9}{20} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{15125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {162}{25} \, \sqrt {-2 \, x + 1} + \frac {343}{44 \, \sqrt {-2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/20*(-2*x + 1)^(3/2) + 1/15125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) +
162/25*sqrt(-2*x + 1) + 343/44/sqrt(-2*x + 1)

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mupad [B]  time = 0.06, size = 48, normalized size = 0.72 \[ \frac {343}{44\,\sqrt {1-2\,x}}+\frac {162\,\sqrt {1-2\,x}}{25}-\frac {9\,{\left (1-2\,x\right )}^{3/2}}{20}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{15125} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2i)/15125 + 343/(44*(1 - 2*x)^(1/2)) + (162*(1 - 2*x)^(1/2))/
25 - (9*(1 - 2*x)^(3/2))/20

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sympy [A]  time = 62.00, size = 102, normalized size = 1.52 \[ - \frac {9 \left (1 - 2 x\right )^{\frac {3}{2}}}{20} + \frac {162 \sqrt {1 - 2 x}}{25} + \frac {2 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right )}{275} + \frac {343}{44 \sqrt {1 - 2 x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*(1 - 2*x)**(3/2)/20 + 162*sqrt(1 - 2*x)/25 + 2*Piecewise((-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))/275 + 343/(44*sqrt(1 - 2*x))

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