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

Optimal. Leaf size=140 \[ -\frac {129 \sqrt {1-2 x} (3 x+2)^4}{50 (5 x+3)}-\frac {(1-2 x)^{3/2} (3 x+2)^4}{10 (5 x+3)^2}+\frac {2643 \sqrt {1-2 x} (3 x+2)^3}{1750}+\frac {1404 \sqrt {1-2 x} (3 x+2)^2}{3125}+\frac {9 \sqrt {1-2 x} (1375 x+32)}{31250}-\frac {12279 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}} \]

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Rubi [A]  time = 0.05, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {97, 149, 153, 147, 63, 206} \begin {gather*} -\frac {129 \sqrt {1-2 x} (3 x+2)^4}{50 (5 x+3)}-\frac {(1-2 x)^{3/2} (3 x+2)^4}{10 (5 x+3)^2}+\frac {2643 \sqrt {1-2 x} (3 x+2)^3}{1750}+\frac {1404 \sqrt {1-2 x} (3 x+2)^2}{3125}+\frac {9 \sqrt {1-2 x} (1375 x+32)}{31250}-\frac {12279 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1404*Sqrt[1 - 2*x]*(2 + 3*x)^2)/3125 + (2643*Sqrt[1 - 2*x]*(2 + 3*x)^3)/1750 - ((1 - 2*x)^(3/2)*(2 + 3*x)^4)/
(10*(3 + 5*x)^2) - (129*Sqrt[1 - 2*x]*(2 + 3*x)^4)/(50*(3 + 5*x)) + (9*Sqrt[1 - 2*x]*(32 + 1375*x))/31250 - (1
2279*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15625*Sqrt[55])

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)^{3/2} (2+3 x)^4}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{3/2} (2+3 x)^4}{10 (3+5 x)^2}+\frac {1}{10} \int \frac {(6-33 x) \sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{3/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {129 \sqrt {1-2 x} (2+3 x)^4}{50 (3+5 x)}+\frac {1}{50} \int \frac {(870-2643 x) (2+3 x)^3}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {2643 \sqrt {1-2 x} (2+3 x)^3}{1750}-\frac {(1-2 x)^{3/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {129 \sqrt {1-2 x} (2+3 x)^4}{50 (3+5 x)}-\frac {\int \frac {(2+3 x)^2 (-5397+19656 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{1750}\\ &=\frac {1404 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {2643 \sqrt {1-2 x} (2+3 x)^3}{1750}-\frac {(1-2 x)^{3/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {129 \sqrt {1-2 x} (2+3 x)^4}{50 (3+5 x)}+\frac {\int \frac {(33978-86625 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{43750}\\ &=\frac {1404 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {2643 \sqrt {1-2 x} (2+3 x)^3}{1750}-\frac {(1-2 x)^{3/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {129 \sqrt {1-2 x} (2+3 x)^4}{50 (3+5 x)}+\frac {9 \sqrt {1-2 x} (32+1375 x)}{31250}+\frac {12279 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{31250}\\ &=\frac {1404 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {2643 \sqrt {1-2 x} (2+3 x)^3}{1750}-\frac {(1-2 x)^{3/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {129 \sqrt {1-2 x} (2+3 x)^4}{50 (3+5 x)}+\frac {9 \sqrt {1-2 x} (32+1375 x)}{31250}-\frac {12279 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{31250}\\ &=\frac {1404 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {2643 \sqrt {1-2 x} (2+3 x)^3}{1750}-\frac {(1-2 x)^{3/2} (2+3 x)^4}{10 (3+5 x)^2}-\frac {129 \sqrt {1-2 x} (2+3 x)^4}{50 (3+5 x)}+\frac {9 \sqrt {1-2 x} (32+1375 x)}{31250}-\frac {12279 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 73, normalized size = 0.52 \begin {gather*} \frac {-\frac {55 \sqrt {1-2 x} \left (2025000 x^5+3267000 x^4-496350 x^3-2120880 x^2-489445 x+96776\right )}{(5 x+3)^2}-171906 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{12031250} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*Sqrt[1 - 2*x]*(96776 - 489445*x - 2120880*x^2 - 496350*x^3 + 3267000*x^4 + 2025000*x^5))/(3 + 5*x)^2 - 1
71906*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/12031250

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IntegrateAlgebraic [A]  time = 0.19, size = 97, normalized size = 0.69 \begin {gather*} \frac {\left (253125 (1-2 x)^5-2082375 (1-2 x)^4+5550075 (1-2 x)^3-4566345 (1-2 x)^2-1432550 (1-2 x)+1890966\right ) \sqrt {1-2 x}}{218750 (5 (1-2 x)-11)^2}-\frac {12279 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1890966 - 1432550*(1 - 2*x) - 4566345*(1 - 2*x)^2 + 5550075*(1 - 2*x)^3 - 2082375*(1 - 2*x)^4 + 253125*(1 -
2*x)^5)*Sqrt[1 - 2*x])/(218750*(-11 + 5*(1 - 2*x))^2) - (12279*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15625*Sqrt[
55])

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fricas [A]  time = 1.58, size = 89, normalized size = 0.64 \begin {gather*} \frac {85953 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (2025000 \, x^{5} + 3267000 \, x^{4} - 496350 \, x^{3} - 2120880 \, x^{2} - 489445 \, x + 96776\right )} \sqrt {-2 \, x + 1}}{12031250 \, {\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)^(3/2)*(2+3*x)^4/(3+5*x)^3,x, algorithm="fricas")

[Out]

1/12031250*(85953*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(202500
0*x^5 + 3267000*x^4 - 496350*x^3 - 2120880*x^2 - 489445*x + 96776)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.08, size = 118, normalized size = 0.84 \begin {gather*} -\frac {81}{1750} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {1107}{6250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {36}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {12279}{1718750} \, \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 {228}{3125} \, \sqrt {-2 \, x + 1} + \frac {1295 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2871 \, \sqrt {-2 \, x + 1}}{62500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/1750*(2*x - 1)^3*sqrt(-2*x + 1) - 1107/6250*(2*x - 1)^2*sqrt(-2*x + 1) + 36/3125*(-2*x + 1)^(3/2) + 12279/
1718750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 228/3125*sqrt(-
2*x + 1) + 1/62500*(1295*(-2*x + 1)^(3/2) - 2871*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 84, normalized size = 0.60 \begin {gather*} -\frac {12279 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{859375}+\frac {81 \left (-2 x +1\right )^{\frac {7}{2}}}{1750}-\frac {1107 \left (-2 x +1\right )^{\frac {5}{2}}}{6250}+\frac {36 \left (-2 x +1\right )^{\frac {3}{2}}}{3125}+\frac {228 \sqrt {-2 x +1}}{3125}+\frac {\frac {259 \left (-2 x +1\right )^{\frac {3}{2}}}{3125}-\frac {2871 \sqrt {-2 x +1}}{15625}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/1750*(-2*x+1)^(7/2)-1107/6250*(-2*x+1)^(5/2)+36/3125*(-2*x+1)^(3/2)+228/3125*(-2*x+1)^(1/2)+4/125*(259/100*
(-2*x+1)^(3/2)-2871/500*(-2*x+1)^(1/2))/(-10*x-6)^2-12279/859375*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2
)

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maxima [A]  time = 1.20, size = 110, normalized size = 0.79 \begin {gather*} \frac {81}{1750} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {1107}{6250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {36}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {12279}{1718750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {228}{3125} \, \sqrt {-2 \, x + 1} + \frac {1295 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2871 \, \sqrt {-2 \, x + 1}}{15625 \, {\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)^(3/2)*(2+3*x)^4/(3+5*x)^3,x, algorithm="maxima")

[Out]

81/1750*(-2*x + 1)^(7/2) - 1107/6250*(-2*x + 1)^(5/2) + 36/3125*(-2*x + 1)^(3/2) + 12279/1718750*sqrt(55)*log(
-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 228/3125*sqrt(-2*x + 1) + 1/15625*(1295*(-2*x
+ 1)^(3/2) - 2871*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.06, size = 92, normalized size = 0.66 \begin {gather*} \frac {228\,\sqrt {1-2\,x}}{3125}+\frac {36\,{\left (1-2\,x\right )}^{3/2}}{3125}-\frac {1107\,{\left (1-2\,x\right )}^{5/2}}{6250}+\frac {81\,{\left (1-2\,x\right )}^{7/2}}{1750}-\frac {\frac {2871\,\sqrt {1-2\,x}}{390625}-\frac {259\,{\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 )\,12279{}\mathrm {i}}{859375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*12279i)/859375 + (228*(1 - 2*x)^(1/2))/3125 + (36*(1 - 2*x)^(
3/2))/3125 - (1107*(1 - 2*x)^(5/2))/6250 + (81*(1 - 2*x)^(7/2))/1750 - ((2871*(1 - 2*x)^(1/2))/390625 - (259*(
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)**(3/2)*(2+3*x)**4/(3+5*x)**3,x)

[Out]

Timed out

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