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]

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])

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Rubi [A]  time = 0.0341693, antiderivative size = 67, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [C]  time = 0.0193607, 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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Maple [A]  time = 0.006, size = 47, normalized size = 0.7 \begin{align*} -{\frac{9}{20} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{2\,\sqrt{55}}{15125}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) }+{\frac{343}{44}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{162}{25}\sqrt{1-2\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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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Maxima [A]  time = 1.791, size = 86, normalized size = 1.28 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 1.70726, size = 185, normalized size = 2.76 \begin{align*} \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 )}} \end{align*}

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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Sympy [A]  time = 37.9039, size = 102, normalized size = 1.52 \begin{align*} - \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}} \end{align*}

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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Giac [A]  time = 2.03867, size = 90, normalized size = 1.34 \begin{align*} -\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}} \end{align*}

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)