3.1910 \(\int \frac{(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=108 \[ -\frac{(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}+\frac{27}{175} (1-2 x)^{3/2} (3 x+2)^2-\frac{6}{625} (1-2 x)^{3/2} (9 x+29)+\frac{192 \sqrt{1-2 x}}{3125}-\frac{192 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3125} \]

[Out]

(192*Sqrt[1 - 2*x])/3125 + (27*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/175 - ((1 - 2*x)^(3/2)*(2 + 3*x)^3)/(5*(3 + 5*x))
- (6*(1 - 2*x)^(3/2)*(29 + 9*x))/625 - (192*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3125

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Rubi [A]  time = 0.0329507, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {97, 153, 147, 50, 63, 206} \[ -\frac{(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}+\frac{27}{175} (1-2 x)^{3/2} (3 x+2)^2-\frac{6}{625} (1-2 x)^{3/2} (9 x+29)+\frac{192 \sqrt{1-2 x}}{3125}-\frac{192 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3125} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(192*Sqrt[1 - 2*x])/3125 + (27*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/175 - ((1 - 2*x)^(3/2)*(2 + 3*x)^3)/(5*(3 + 5*x))
- (6*(1 - 2*x)^(3/2)*(29 + 9*x))/625 - (192*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3125

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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{(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx &=-\frac{(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}+\frac{1}{5} \int \frac{(3-27 x) \sqrt{1-2 x} (2+3 x)^2}{3+5 x} \, dx\\ &=\frac{27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac{1}{175} \int \frac{(-210-126 x) \sqrt{1-2 x} (2+3 x)}{3+5 x} \, dx\\ &=\frac{27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac{6}{625} (1-2 x)^{3/2} (29+9 x)+\frac{96}{625} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=\frac{192 \sqrt{1-2 x}}{3125}+\frac{27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac{6}{625} (1-2 x)^{3/2} (29+9 x)+\frac{1056 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{3125}\\ &=\frac{192 \sqrt{1-2 x}}{3125}+\frac{27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac{6}{625} (1-2 x)^{3/2} (29+9 x)-\frac{1056 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{3125}\\ &=\frac{192 \sqrt{1-2 x}}{3125}+\frac{27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac{6}{625} (1-2 x)^{3/2} (29+9 x)-\frac{192 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3125}\\ \end{align*}

Mathematica [A]  time = 0.0396299, size = 68, normalized size = 0.63 \[ \frac{-\frac{5 \sqrt{1-2 x} \left (67500 x^4+62100 x^3-57165 x^2-27640 x+8738\right )}{5 x+3}-1344 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{109375} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*Sqrt[1 - 2*x]*(8738 - 27640*x - 57165*x^2 + 62100*x^3 + 67500*x^4))/(3 + 5*x) - 1344*Sqrt[55]*ArcTanh[Sqr
t[5/11]*Sqrt[1 - 2*x]])/109375

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Maple [A]  time = 0.008, size = 72, normalized size = 0.7 \begin{align*}{\frac{27}{350} \left ( 1-2\,x \right ) ^{{\frac{7}{2}}}}-{\frac{351}{1250} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{6}{625} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{194}{3125}\sqrt{1-2\,x}}+{\frac{22}{15625}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{192\,\sqrt{55}}{15625}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/350*(1-2*x)^(7/2)-351/1250*(1-2*x)^(5/2)+6/625*(1-2*x)^(3/2)+194/3125*(1-2*x)^(1/2)+22/15625*(1-2*x)^(1/2)/
(-2*x-6/5)-192/15625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.53724, size = 120, normalized size = 1.11 \begin{align*} \frac{27}{350} \,{\left (-2 \, x + 1\right )}^{\frac{7}{2}} - \frac{351}{1250} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + \frac{6}{625} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{96}{15625} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{194}{3125} \, \sqrt{-2 \, x + 1} - \frac{11 \, \sqrt{-2 \, x + 1}}{3125 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/350*(-2*x + 1)^(7/2) - 351/1250*(-2*x + 1)^(5/2) + 6/625*(-2*x + 1)^(3/2) + 96/15625*sqrt(55)*log(-(sqrt(55
) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 194/3125*sqrt(-2*x + 1) - 11/3125*sqrt(-2*x + 1)/(5*x +
3)

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Fricas [A]  time = 1.56556, size = 248, normalized size = 2.3 \begin{align*} \frac{672 \, \sqrt{11} \sqrt{5}{\left (5 \, x + 3\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 5 \,{\left (67500 \, x^{4} + 62100 \, x^{3} - 57165 \, x^{2} - 27640 \, x + 8738\right )} \sqrt{-2 \, x + 1}}{109375 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/109375*(672*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 5*(67500
*x^4 + 62100*x^3 - 57165*x^2 - 27640*x + 8738)*sqrt(-2*x + 1))/(5*x + 3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 3.09979, size = 143, normalized size = 1.32 \begin{align*} -\frac{27}{350} \,{\left (2 \, x - 1\right )}^{3} \sqrt{-2 \, x + 1} - \frac{351}{1250} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} + \frac{6}{625} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{96}{15625} \, \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{194}{3125} \, \sqrt{-2 \, x + 1} - \frac{11 \, \sqrt{-2 \, x + 1}}{3125 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/350*(2*x - 1)^3*sqrt(-2*x + 1) - 351/1250*(2*x - 1)^2*sqrt(-2*x + 1) + 6/625*(-2*x + 1)^(3/2) + 96/15625*s
qrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 194/3125*sqrt(-2*x + 1)
- 11/3125*sqrt(-2*x + 1)/(5*x + 3)