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

Optimal. Leaf size=155 \[ -\frac {127 (1-2 x)^{3/2} (3 x+2)^4}{50 (5 x+3)}-\frac {(1-2 x)^{5/2} (3 x+2)^4}{10 (5 x+3)^2}+\frac {1117}{750} (1-2 x)^{3/2} (3 x+2)^3+\frac {1903 (1-2 x)^{3/2} (3 x+2)^2}{4375}+\frac {(1-2 x)^{3/2} (24939 x+734)}{93750}+\frac {11763 \sqrt {1-2 x}}{78125}-\frac {11763 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \]

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Rubi [A]  time = 0.06, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {97, 149, 153, 147, 50, 63, 206} \begin {gather*} -\frac {127 (1-2 x)^{3/2} (3 x+2)^4}{50 (5 x+3)}-\frac {(1-2 x)^{5/2} (3 x+2)^4}{10 (5 x+3)^2}+\frac {1117}{750} (1-2 x)^{3/2} (3 x+2)^3+\frac {1903 (1-2 x)^{3/2} (3 x+2)^2}{4375}+\frac {(1-2 x)^{3/2} (24939 x+734)}{93750}+\frac {11763 \sqrt {1-2 x}}{78125}-\frac {11763 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11763*Sqrt[1 - 2*x])/78125 + (1903*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/4375 + (1117*(1 - 2*x)^(3/2)*(2 + 3*x)^3)/750
- ((1 - 2*x)^(5/2)*(2 + 3*x)^4)/(10*(3 + 5*x)^2) - (127*(1 - 2*x)^(3/2)*(2 + 3*x)^4)/(50*(3 + 5*x)) + ((1 - 2
*x)^(3/2)*(734 + 24939*x))/93750 - (11763*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/78125

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 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 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 149

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

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 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)^4}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}+\frac {1}{10} \int \frac {(2-39 x) (1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {127 (1-2 x)^{3/2} (2+3 x)^4}{50 (3+5 x)}+\frac {1}{50} \int \frac {(342-3351 x) \sqrt {1-2 x} (2+3 x)^3}{3+5 x} \, dx\\ &=\frac {1117}{750} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {127 (1-2 x)^{3/2} (2+3 x)^4}{50 (3+5 x)}-\frac {\int \frac {\sqrt {1-2 x} (2+3 x)^2 (-621+34254 x)}{3+5 x} \, dx}{2250}\\ &=\frac {1903 (1-2 x)^{3/2} (2+3 x)^2}{4375}+\frac {1117}{750} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {127 (1-2 x)^{3/2} (2+3 x)^4}{50 (3+5 x)}+\frac {\int \frac {(43470-174573 x) \sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx}{78750}\\ &=\frac {1903 (1-2 x)^{3/2} (2+3 x)^2}{4375}+\frac {1117}{750} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {127 (1-2 x)^{3/2} (2+3 x)^4}{50 (3+5 x)}+\frac {(1-2 x)^{3/2} (734+24939 x)}{93750}+\frac {11763 \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx}{31250}\\ &=\frac {11763 \sqrt {1-2 x}}{78125}+\frac {1903 (1-2 x)^{3/2} (2+3 x)^2}{4375}+\frac {1117}{750} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {127 (1-2 x)^{3/2} (2+3 x)^4}{50 (3+5 x)}+\frac {(1-2 x)^{3/2} (734+24939 x)}{93750}+\frac {129393 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{156250}\\ &=\frac {11763 \sqrt {1-2 x}}{78125}+\frac {1903 (1-2 x)^{3/2} (2+3 x)^2}{4375}+\frac {1117}{750} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {127 (1-2 x)^{3/2} (2+3 x)^4}{50 (3+5 x)}+\frac {(1-2 x)^{3/2} (734+24939 x)}{93750}-\frac {129393 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{156250}\\ &=\frac {11763 \sqrt {1-2 x}}{78125}+\frac {1903 (1-2 x)^{3/2} (2+3 x)^2}{4375}+\frac {1117}{750} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {127 (1-2 x)^{3/2} (2+3 x)^4}{50 (3+5 x)}+\frac {(1-2 x)^{3/2} (734+24939 x)}{93750}-\frac {11763 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 78, normalized size = 0.50 \begin {gather*} \frac {\frac {5 \sqrt {1-2 x} \left (15750000 x^6+15075000 x^5-16051500 x^4-11139550 x^3+9372960 x^2+6891315 x+871208\right )}{(5 x+3)^2}-164682 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5468750} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(871208 + 6891315*x + 9372960*x^2 - 11139550*x^3 - 16051500*x^4 + 15075000*x^5 + 15750000*x^
6))/(3 + 5*x)^2 - 164682*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5468750

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IntegrateAlgebraic [A]  time = 0.19, size = 108, normalized size = 0.70 \begin {gather*} \frac {\left (984375 (1-2 x)^6-7790625 (1-2 x)^5+20174625 (1-2 x)^4-16909975 (1-2 x)^3+2195760 (1-2 x)^2-15095850 (1-2 x)[phone]\right ) \sqrt {1-2 x}}{1093750 (5 (1-2 x)-11)^2}-\frac {11763 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((19926522 - 15095850*(1 - 2*x) + 2195760*(1 - 2*x)^2 - 16909975*(1 - 2*x)^3 + 20174625*(1 - 2*x)^4 - 7790625*
(1 - 2*x)^5 + 984375*(1 - 2*x)^6)*Sqrt[1 - 2*x])/(1093750*(-11 + 5*(1 - 2*x))^2) - (11763*Sqrt[11/5]*ArcTanh[S
qrt[5/11]*Sqrt[1 - 2*x]])/78125

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fricas [A]  time = 1.76, size = 100, normalized size = 0.65 \begin {gather*} \frac {82341 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (15750000 \, x^{6} + 15075000 \, x^{5} - 16051500 \, x^{4} - 11139550 \, x^{3} + 9372960 \, x^{2} + 6891315 \, x + 871208\right )} \sqrt {-2 \, x + 1}}{5468750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5468750*(82341*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3
)) + 5*(15750000*x^6 + 15075000*x^5 - 16051500*x^4 - 11139550*x^3 + 9372960*x^2 + 6891315*x + 871208)*sqrt(-2*
x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.16, size = 134, normalized size = 0.86 \begin {gather*} \frac {9}{250} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} + \frac {1107}{8750} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {108}{15625} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {76}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11763}{781250} \, \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 {2404}{15625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (1275 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2827 \, \sqrt {-2 \, x + 1}\right )}}{312500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/250*(2*x - 1)^4*sqrt(-2*x + 1) + 1107/8750*(2*x - 1)^3*sqrt(-2*x + 1) + 108/15625*(2*x - 1)^2*sqrt(-2*x + 1)
+ 76/3125*(-2*x + 1)^(3/2) + 11763/781250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5
*sqrt(-2*x + 1))) + 2404/15625*sqrt(-2*x + 1) + 11/312500*(1275*(-2*x + 1)^(3/2) - 2827*sqrt(-2*x + 1))/(5*x +
3)^2

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maple [A]  time = 0.01, size = 93, normalized size = 0.60 \begin {gather*} -\frac {11763 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{390625}+\frac {9 \left (-2 x +1\right )^{\frac {9}{2}}}{250}-\frac {1107 \left (-2 x +1\right )^{\frac {7}{2}}}{8750}+\frac {108 \left (-2 x +1\right )^{\frac {5}{2}}}{15625}+\frac {76 \left (-2 x +1\right )^{\frac {3}{2}}}{3125}+\frac {2404 \sqrt {-2 x +1}}{15625}+\frac {\frac {561 \left (-2 x +1\right )^{\frac {3}{2}}}{3125}-\frac {31097 \sqrt {-2 x +1}}{78125}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/250*(-2*x+1)^(9/2)-1107/8750*(-2*x+1)^(7/2)+108/15625*(-2*x+1)^(5/2)+76/3125*(-2*x+1)^(3/2)+2404/15625*(-2*x
+1)^(1/2)+44/625*(51/20*(-2*x+1)^(3/2)-2827/500*(-2*x+1)^(1/2))/(-10*x-6)^2-11763/390625*arctanh(1/11*55^(1/2)
*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.33, size = 119, normalized size = 0.77 \begin {gather*} \frac {9}{250} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - \frac {1107}{8750} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {108}{15625} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {76}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11763}{781250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2404}{15625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (1275 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2827 \, \sqrt {-2 \, x + 1}\right )}}{78125 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/250*(-2*x + 1)^(9/2) - 1107/8750*(-2*x + 1)^(7/2) + 108/15625*(-2*x + 1)^(5/2) + 76/3125*(-2*x + 1)^(3/2) +
11763/781250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2404/15625*sqrt(-2*x
+ 1) + 11/78125*(1275*(-2*x + 1)^(3/2) - 2827*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.07, size = 101, normalized size = 0.65 \begin {gather*} \frac {2404\,\sqrt {1-2\,x}}{15625}+\frac {76\,{\left (1-2\,x\right )}^{3/2}}{3125}+\frac {108\,{\left (1-2\,x\right )}^{5/2}}{15625}-\frac {1107\,{\left (1-2\,x\right )}^{7/2}}{8750}+\frac {9\,{\left (1-2\,x\right )}^{9/2}}{250}-\frac {\frac {31097\,\sqrt {1-2\,x}}{1953125}-\frac {561\,{\left (1-2\,x\right )}^{3/2}}{78125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,11763{}\mathrm {i}}{390625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*11763i)/390625 + (2404*(1 - 2*x)^(1/2))/15625 + (76*(1 - 2*x)
^(3/2))/3125 + (108*(1 - 2*x)^(5/2))/15625 - (1107*(1 - 2*x)^(7/2))/8750 + (9*(1 - 2*x)^(9/2))/250 - ((31097*(
1 - 2*x)^(1/2))/1953125 - (561*(1 - 2*x)^(3/2))/78125)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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