3.2127 \(\int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\)

Optimal. Leaf size=107 \[ \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}-\frac {71 \sqrt {1-2 x} (3 x+2)^2}{1210 (5 x+3)^2}+\frac {9 \sqrt {1-2 x} (5093 x+3044)}{13310 (5 x+3)}-\frac {111 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}} \]

[Out]

-111/73205*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+7/11*(2+3*x)^3/(3+5*x)^2/(1-2*x)^(1/2)-71/1210*(2+3*x
)^2*(1-2*x)^(1/2)/(3+5*x)^2+9/13310*(3044+5093*x)*(1-2*x)^(1/2)/(3+5*x)

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Rubi [A]  time = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {98, 149, 146, 63, 206} \[ \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}-\frac {71 \sqrt {1-2 x} (3 x+2)^2}{1210 (5 x+3)^2}+\frac {9 \sqrt {1-2 x} (5093 x+3044)}{13310 (5 x+3)}-\frac {111 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-71*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(1210*(3 + 5*x)^2) + (7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^2) + (9*Sqrt[
1 - 2*x]*(3044 + 5093*x))/(13310*(3 + 5*x)) - (111*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(1331*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 98

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

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(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*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && 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 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)^{3/2} (3+5 x)^3} \, dx &=\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {1}{11} \int \frac {(2+3 x)^2 (47+102 x)}{\sqrt {1-2 x} (3+5 x)^3} \, dx\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^2}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {\int \frac {(2+3 x) (3636+6945 x)}{\sqrt {1-2 x} (3+5 x)^2} \, dx}{1210}\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^2}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)^2}+\frac {9 \sqrt {1-2 x} (3044+5093 x)}{13310 (3+5 x)}+\frac {111 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{2662}\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^2}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)^2}+\frac {9 \sqrt {1-2 x} (3044+5093 x)}{13310 (3+5 x)}-\frac {111 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{2662}\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^2}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)^2}+\frac {9 \sqrt {1-2 x} (3044+5093 x)}{13310 (3+5 x)}-\frac {111 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 91, normalized size = 0.85 \[ \frac {\frac {2184 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {5}{11} (1-2 x)\right )}{\sqrt {1-2 x}}+\frac {11 \left (-490050 x^3+334350 x^2+930205 x+331904\right )}{\sqrt {1-2 x} (5 x+3)^2}-306 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{332750} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*(331904 + 930205*x + 334350*x^2 - 490050*x^3))/(Sqrt[1 - 2*x]*(3 + 5*x)^2) - 306*Sqrt[55]*ArcTanh[Sqrt[5/
11]*Sqrt[1 - 2*x]] + (2184*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 - 2*x))/11])/Sqrt[1 - 2*x])/332750

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fricas [A]  time = 0.97, size = 89, normalized size = 0.83 \[ \frac {111 \, \sqrt {55} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 11 \, {\left (215622 \, x^{3} - 149298 \, x^{2} - 411911 \, x - 146824\right )} \sqrt {-2 \, x + 1}}{146410 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/146410*(111*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 11*(2
15622*x^3 - 149298*x^2 - 411911*x - 146824)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)

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giac [A]  time = 1.31, size = 86, normalized size = 0.80 \[ \frac {111}{146410} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {81}{250} \, \sqrt {-2 \, x + 1} + \frac {2401}{2662 \, \sqrt {-2 \, x + 1}} + \frac {1355 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 3003 \, \sqrt {-2 \, x + 1}}{665500 \, {\left (5 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

111/146410*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 81/250*sqrt(
-2*x + 1) + 2401/2662/sqrt(-2*x + 1) + 1/665500*(1355*(-2*x + 1)^(3/2) - 3003*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.02, size = 66, normalized size = 0.62 \[ -\frac {111 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{73205}+\frac {81 \sqrt {-2 x +1}}{250}+\frac {2401}{2662 \sqrt {-2 x +1}}+\frac {\frac {271 \left (-2 x +1\right )^{\frac {3}{2}}}{33275}-\frac {273 \sqrt {-2 x +1}}{15125}}{\left (-10 x -6\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/250*(-2*x+1)^(1/2)+2401/2662/(-2*x+1)^(1/2)+4/6655*(271/20*(-2*x+1)^(3/2)-3003/100*(-2*x+1)^(1/2))/(-10*x-6
)^2-111/73205*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.41, size = 92, normalized size = 0.86 \[ \frac {111}{146410} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {81}{250} \, \sqrt {-2 \, x + 1} + \frac {7505835 \, {\left (2 \, x - 1\right )}^{2} + 66039512 \, x + 3295369}{332750 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

111/146410*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 81/250*sqrt(-2*x + 1)
+ 1/332750*(7505835*(2*x - 1)^2 + 66039512*x + 3295369)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt
(-2*x + 1))

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mupad [B]  time = 0.06, size = 71, normalized size = 0.66 \[ \frac {81\,\sqrt {1-2\,x}}{250}-\frac {111\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{73205}+\frac {\frac {3001796\,x}{378125}+\frac {1501167\,{\left (2\,x-1\right )}^2}{1663750}+\frac {299579}{756250}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(81*(1 - 2*x)^(1/2))/250 - (111*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/73205 + ((3001796*x)/378125 + (
1501167*(2*x - 1)^2)/1663750 + 299579/756250)/((121*(1 - 2*x)^(1/2))/25 - (22*(1 - 2*x)^(3/2))/5 + (1 - 2*x)^(
5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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