3.2171 \(\int \frac{(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)} \, dx\)

Optimal. Leaf size=80 \[ \frac{27}{40} (1-2 x)^{3/2}-\frac{2889}{200} \sqrt{1-2 x}-\frac{33271}{968 \sqrt{1-2 x}}+\frac{2401}{264 (1-2 x)^{3/2}}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3025 \sqrt{55}} \]

[Out]

2401/(264*(1 - 2*x)^(3/2)) - 33271/(968*Sqrt[1 - 2*x]) - (2889*Sqrt[1 - 2*x])/200 + (27*(1 - 2*x)^(3/2))/40 -
(2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*Sqrt[55])

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Rubi [A]  time = 0.0384894, antiderivative size = 80, 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} \[ \frac{27}{40} (1-2 x)^{3/2}-\frac{2889}{200} \sqrt{1-2 x}-\frac{33271}{968 \sqrt{1-2 x}}+\frac{2401}{264 (1-2 x)^{3/2}}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3025 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^4/((1 - 2*x)^(5/2)*(3 + 5*x)),x]

[Out]

2401/(264*(1 - 2*x)^(3/2)) - 33271/(968*Sqrt[1 - 2*x]) - (2889*Sqrt[1 - 2*x])/200 + (27*(1 - 2*x)^(3/2))/40 -
(2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*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)^4}{(1-2 x)^{5/2} (3+5 x)} \, dx &=\int \left (\frac{2401}{88 (1-2 x)^{5/2}}-\frac{33271}{968 (1-2 x)^{3/2}}+\frac{621}{50 \sqrt{1-2 x}}+\frac{81 x}{20 \sqrt{1-2 x}}+\frac{1}{3025 \sqrt{1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac{2401}{264 (1-2 x)^{3/2}}-\frac{33271}{968 \sqrt{1-2 x}}-\frac{621}{50} \sqrt{1-2 x}+\frac{\int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{3025}+\frac{81}{20} \int \frac{x}{\sqrt{1-2 x}} \, dx\\ &=\frac{2401}{264 (1-2 x)^{3/2}}-\frac{33271}{968 \sqrt{1-2 x}}-\frac{621}{50} \sqrt{1-2 x}-\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{3025}+\frac{81}{20} \int \left (\frac{1}{2 \sqrt{1-2 x}}-\frac{1}{2} \sqrt{1-2 x}\right ) \, dx\\ &=\frac{2401}{264 (1-2 x)^{3/2}}-\frac{33271}{968 \sqrt{1-2 x}}-\frac{2889}{200} \sqrt{1-2 x}+\frac{27}{40} (1-2 x)^{3/2}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3025 \sqrt{55}}\\ \end{align*}

Mathematica [C]  time = 0.0239189, size = 50, normalized size = 0.62 \[ \frac{2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{5}{11} (1-2 x)\right )-33 \left (3375 x^3+31050 x^2-76545 x+24404\right )}{20625 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^4/((1 - 2*x)^(5/2)*(3 + 5*x)),x]

[Out]

(-33*(24404 - 76545*x + 31050*x^2 + 3375*x^3) + 2*Hypergeometric2F1[-3/2, 1, -1/2, (5*(1 - 2*x))/11])/(20625*(
1 - 2*x)^(3/2))

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Maple [A]  time = 0.009, size = 56, normalized size = 0.7 \begin{align*}{\frac{2401}{264} \left ( 1-2\,x \right ) ^{-{\frac{3}{2}}}}+{\frac{27}{40} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{2\,\sqrt{55}}{166375}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) }-{\frac{33271}{968}{\frac{1}{\sqrt{1-2\,x}}}}-{\frac{2889}{200}\sqrt{1-2\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x),x)

[Out]

2401/264/(1-2*x)^(3/2)+27/40*(1-2*x)^(3/2)-2/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-33271/968/(1
-2*x)^(1/2)-2889/200*(1-2*x)^(1/2)

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Maxima [A]  time = 1.98293, size = 93, normalized size = 1.16 \begin{align*} \frac{27}{40} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{166375} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{2889}{200} \, \sqrt{-2 \, x + 1} + \frac{343 \,{\left (291 \, x - 107\right )}}{1452 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x),x, algorithm="maxima")

[Out]

27/40*(-2*x + 1)^(3/2) + 1/166375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) -
2889/200*sqrt(-2*x + 1) + 343/1452*(291*x - 107)/(-2*x + 1)^(3/2)

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Fricas [A]  time = 1.45109, size = 238, normalized size = 2.98 \begin{align*} \frac{3 \, \sqrt{55}{\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \,{\left (49005 \, x^{3} + 450846 \, x^{2} - 1111431 \, x + 354344\right )} \sqrt{-2 \, x + 1}}{499125 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x),x, algorithm="fricas")

[Out]

1/499125*(3*sqrt(55)*(4*x^2 - 4*x + 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(49005*x^3 + 45
0846*x^2 - 1111431*x + 354344)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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Sympy [A]  time = 47.6126, size = 114, normalized size = 1.42 \begin{align*} \frac{27 \left (1 - 2 x\right )^{\frac{3}{2}}}{40} - \frac{2889 \sqrt{1 - 2 x}}{200} + \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 )}{3025} - \frac{33271}{968 \sqrt{1 - 2 x}} + \frac{2401}{264 \left (1 - 2 x\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)**4/(1-2*x)**(5/2)/(3+5*x),x)

[Out]

27*(1 - 2*x)**(3/2)/40 - 2889*sqrt(1 - 2*x)/200 + 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))/3025 - 33271/(968*sqrt(1 -
2*x)) + 2401/(264*(1 - 2*x)**(3/2))

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Giac [A]  time = 2.73839, size = 107, normalized size = 1.34 \begin{align*} \frac{27}{40} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1}{166375} \, \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{2889}{200} \, \sqrt{-2 \, x + 1} - \frac{343 \,{\left (291 \, x - 107\right )}}{1452 \,{\left (2 \, x - 1\right )} \sqrt{-2 \, x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x),x, algorithm="giac")

[Out]

27/40*(-2*x + 1)^(3/2) + 1/166375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*
x + 1))) - 2889/200*sqrt(-2*x + 1) - 343/1452*(291*x - 107)/((2*x - 1)*sqrt(-2*x + 1))