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]

(-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])

________________________________________________________________________________________

Rubi [A]  time = 0.0289082, antiderivative size = 107, normalized size of antiderivative = 1., 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 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 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 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 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{(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*}

Mathematica [C]  time = 0.0923498, 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

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 66, normalized size = 0.6 \begin{align*}{\frac{81}{250}\sqrt{1-2\,x}}+{\frac{2401}{2662}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{4}{6655\, \left ( -10\,x-6 \right ) ^{2}} \left ({\frac{271}{20} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{3003}{100}\sqrt{1-2\,x}} \right ) }-{\frac{111\,\sqrt{55}}{73205}{\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((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.59487, size = 124, normalized size = 1.16 \begin{align*} \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 )}} \end{align*}

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))

________________________________________________________________________________________

Fricas [A]  time = 1.62296, size = 270, normalized size = 2.52 \begin{align*} \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 )}} \end{align*}

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)

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Giac [A]  time = 2.49091, size = 116, normalized size = 1.08 \begin{align*} \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}} \end{align*}

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