3.1981 \(\int \frac{(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=121 \[ -\frac{(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}+\frac{11}{75} (1-2 x)^{5/2} (3 x+2)^2+\frac{188 (1-2 x)^{3/2}}{9375}-\frac{2 (1-2 x)^{5/2} (2850 x+6191)}{65625}+\frac{2068 \sqrt{1-2 x}}{15625}-\frac{2068 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{15625} \]

[Out]

(2068*Sqrt[1 - 2*x])/15625 + (188*(1 - 2*x)^(3/2))/9375 + (11*(1 - 2*x)^(5/2)*(2 + 3*x)^2)/75 - ((1 - 2*x)^(5/
2)*(2 + 3*x)^3)/(5*(3 + 5*x)) - (2*(1 - 2*x)^(5/2)*(6191 + 2850*x))/65625 - (2068*Sqrt[11/5]*ArcTanh[Sqrt[5/11
]*Sqrt[1 - 2*x]])/15625

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Rubi [A]  time = 0.0396276, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}+\frac{11}{75} (1-2 x)^{5/2} (3 x+2)^2+\frac{188 (1-2 x)^{3/2}}{9375}-\frac{2 (1-2 x)^{5/2} (2850 x+6191)}{65625}+\frac{2068 \sqrt{1-2 x}}{15625}-\frac{2068 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{15625} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2068*Sqrt[1 - 2*x])/15625 + (188*(1 - 2*x)^(3/2))/9375 + (11*(1 - 2*x)^(5/2)*(2 + 3*x)^2)/75 - ((1 - 2*x)^(5/
2)*(2 + 3*x)^3)/(5*(3 + 5*x)) - (2*(1 - 2*x)^(5/2)*(6191 + 2850*x))/65625 - (2068*Sqrt[11/5]*ArcTanh[Sqrt[5/11
]*Sqrt[1 - 2*x]])/15625

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)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx &=-\frac{(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}+\frac{1}{5} \int \frac{(-1-33 x) (1-2 x)^{3/2} (2+3 x)^2}{3+5 x} \, dx\\ &=\frac{11}{75} (1-2 x)^{5/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}-\frac{1}{225} \int \frac{(-306-228 x) (1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx\\ &=\frac{11}{75} (1-2 x)^{5/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}-\frac{2 (1-2 x)^{5/2} (6191+2850 x)}{65625}+\frac{94}{625} \int \frac{(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=\frac{188 (1-2 x)^{3/2}}{9375}+\frac{11}{75} (1-2 x)^{5/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}-\frac{2 (1-2 x)^{5/2} (6191+2850 x)}{65625}+\frac{1034 \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx}{3125}\\ &=\frac{2068 \sqrt{1-2 x}}{15625}+\frac{188 (1-2 x)^{3/2}}{9375}+\frac{11}{75} (1-2 x)^{5/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}-\frac{2 (1-2 x)^{5/2} (6191+2850 x)}{65625}+\frac{11374 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{15625}\\ &=\frac{2068 \sqrt{1-2 x}}{15625}+\frac{188 (1-2 x)^{3/2}}{9375}+\frac{11}{75} (1-2 x)^{5/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}-\frac{2 (1-2 x)^{5/2} (6191+2850 x)}{65625}-\frac{11374 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{15625}\\ &=\frac{2068 \sqrt{1-2 x}}{15625}+\frac{188 (1-2 x)^{3/2}}{9375}+\frac{11}{75} (1-2 x)^{5/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}-\frac{2 (1-2 x)^{5/2} (6191+2850 x)}{65625}-\frac{2068 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{15625}\\ \end{align*}

Mathematica [A]  time = 0.042904, size = 73, normalized size = 0.6 \[ \frac{\frac{5 \sqrt{1-2 x} \left (1575000 x^5+427500 x^4-1858950 x^3+152105 x^2+680930 x+16794\right )}{5 x+3}-43428 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{1640625} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(16794 + 680930*x + 152105*x^2 - 1858950*x^3 + 427500*x^4 + 1575000*x^5))/(3 + 5*x) - 43428*
Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1640625

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Maple [A]  time = 0.01, size = 81, normalized size = 0.7 \begin{align*}{\frac{3}{50} \left ( 1-2\,x \right ) ^{{\frac{9}{2}}}}-{\frac{351}{1750} \left ( 1-2\,x \right ) ^{{\frac{7}{2}}}}+{\frac{18}{3125} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{194}{9375} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{418}{3125}\sqrt{1-2\,x}}+{\frac{242}{78125}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{2068\,\sqrt{55}}{78125}{\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)^(5/2)*(2+3*x)^3/(3+5*x)^2,x)

[Out]

3/50*(1-2*x)^(9/2)-351/1750*(1-2*x)^(7/2)+18/3125*(1-2*x)^(5/2)+194/9375*(1-2*x)^(3/2)+418/3125*(1-2*x)^(1/2)+
242/78125*(1-2*x)^(1/2)/(-2*x-6/5)-2068/78125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 2.5741, size = 132, normalized size = 1.09 \begin{align*} \frac{3}{50} \,{\left (-2 \, x + 1\right )}^{\frac{9}{2}} - \frac{351}{1750} \,{\left (-2 \, x + 1\right )}^{\frac{7}{2}} + \frac{18}{3125} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + \frac{194}{9375} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1034}{78125} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{418}{3125} \, \sqrt{-2 \, x + 1} - \frac{121 \, \sqrt{-2 \, x + 1}}{15625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/50*(-2*x + 1)^(9/2) - 351/1750*(-2*x + 1)^(7/2) + 18/3125*(-2*x + 1)^(5/2) + 194/9375*(-2*x + 1)^(3/2) + 103
4/78125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 418/3125*sqrt(-2*x + 1) -
121/15625*sqrt(-2*x + 1)/(5*x + 3)

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Fricas [A]  time = 1.42323, size = 279, normalized size = 2.31 \begin{align*} \frac{21714 \, \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 (1575000 \, x^{5} + 427500 \, x^{4} - 1858950 \, x^{3} + 152105 \, x^{2} + 680930 \, x + 16794\right )} \sqrt{-2 \, x + 1}}{1640625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1640625*(21714*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 5*(15
75000*x^5 + 427500*x^4 - 1858950*x^3 + 152105*x^2 + 680930*x + 16794)*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)**(5/2)*(2+3*x)**3/(3+5*x)**2,x)

[Out]

Timed out

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Giac [A]  time = 2.31568, size = 165, normalized size = 1.36 \begin{align*} \frac{3}{50} \,{\left (2 \, x - 1\right )}^{4} \sqrt{-2 \, x + 1} + \frac{351}{1750} \,{\left (2 \, x - 1\right )}^{3} \sqrt{-2 \, x + 1} + \frac{18}{3125} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} + \frac{194}{9375} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1034}{78125} \, \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{418}{3125} \, \sqrt{-2 \, x + 1} - \frac{121 \, \sqrt{-2 \, x + 1}}{15625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/50*(2*x - 1)^4*sqrt(-2*x + 1) + 351/1750*(2*x - 1)^3*sqrt(-2*x + 1) + 18/3125*(2*x - 1)^2*sqrt(-2*x + 1) + 1
94/9375*(-2*x + 1)^(3/2) + 1034/78125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt
(-2*x + 1))) + 418/3125*sqrt(-2*x + 1) - 121/15625*sqrt(-2*x + 1)/(5*x + 3)