3.2116 \(\int \frac{(2+3 x)^5}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\)

Optimal. Leaf size=120 \[ \frac{7 (3 x+2)^4}{11 \sqrt{1-2 x} (5 x+3)}-\frac{36 \sqrt{1-2 x} (3 x+2)^3}{605 (5 x+3)}+\frac{10836 \sqrt{1-2 x} (3 x+2)^2}{15125}+\frac{504 \sqrt{1-2 x} (1500 x+4499)}{75625}-\frac{336 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{75625 \sqrt{55}} \]

[Out]

(10836*Sqrt[1 - 2*x]*(2 + 3*x)^2)/15125 - (36*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(605*(3 + 5*x)) + (7*(2 + 3*x)^4)/(11
*Sqrt[1 - 2*x]*(3 + 5*x)) + (504*Sqrt[1 - 2*x]*(4499 + 1500*x))/75625 - (336*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]
)/(75625*Sqrt[55])

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Rubi [A]  time = 0.0380979, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {98, 149, 153, 147, 63, 206} \[ \frac{7 (3 x+2)^4}{11 \sqrt{1-2 x} (5 x+3)}-\frac{36 \sqrt{1-2 x} (3 x+2)^3}{605 (5 x+3)}+\frac{10836 \sqrt{1-2 x} (3 x+2)^2}{15125}+\frac{504 \sqrt{1-2 x} (1500 x+4499)}{75625}-\frac{336 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{75625 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10836*Sqrt[1 - 2*x]*(2 + 3*x)^2)/15125 - (36*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(605*(3 + 5*x)) + (7*(2 + 3*x)^4)/(11
*Sqrt[1 - 2*x]*(3 + 5*x)) + (504*Sqrt[1 - 2*x]*(4499 + 1500*x))/75625 - (336*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]
)/(75625*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 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 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)^5}{(1-2 x)^{3/2} (3+5 x)^2} \, dx &=\frac{7 (2+3 x)^4}{11 \sqrt{1-2 x} (3+5 x)}-\frac{1}{11} \int \frac{(2+3 x)^3 (180+312 x)}{\sqrt{1-2 x} (3+5 x)^2} \, dx\\ &=-\frac{36 \sqrt{1-2 x} (2+3 x)^3}{605 (3+5 x)}+\frac{7 (2+3 x)^4}{11 \sqrt{1-2 x} (3+5 x)}-\frac{1}{605} \int \frac{(2+3 x)^2 (6468+10836 x)}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{10836 \sqrt{1-2 x} (2+3 x)^2}{15125}-\frac{36 \sqrt{1-2 x} (2+3 x)^3}{605 (3+5 x)}+\frac{7 (2+3 x)^4}{11 \sqrt{1-2 x} (3+5 x)}+\frac{\int \frac{(-453432-756000 x) (2+3 x)}{\sqrt{1-2 x} (3+5 x)} \, dx}{15125}\\ &=\frac{10836 \sqrt{1-2 x} (2+3 x)^2}{15125}-\frac{36 \sqrt{1-2 x} (2+3 x)^3}{605 (3+5 x)}+\frac{7 (2+3 x)^4}{11 \sqrt{1-2 x} (3+5 x)}+\frac{504 \sqrt{1-2 x} (4499+1500 x)}{75625}+\frac{168 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{75625}\\ &=\frac{10836 \sqrt{1-2 x} (2+3 x)^2}{15125}-\frac{36 \sqrt{1-2 x} (2+3 x)^3}{605 (3+5 x)}+\frac{7 (2+3 x)^4}{11 \sqrt{1-2 x} (3+5 x)}+\frac{504 \sqrt{1-2 x} (4499+1500 x)}{75625}-\frac{168 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{75625}\\ &=\frac{10836 \sqrt{1-2 x} (2+3 x)^2}{15125}-\frac{36 \sqrt{1-2 x} (2+3 x)^3}{605 (3+5 x)}+\frac{7 (2+3 x)^4}{11 \sqrt{1-2 x} (3+5 x)}+\frac{504 \sqrt{1-2 x} (4499+1500 x)}{75625}-\frac{336 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{75625 \sqrt{55}}\\ \end{align*}

Mathematica [C]  time = 0.0819921, size = 96, normalized size = 0.8 \[ \frac{\frac{1260 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{5}{11} (1-2 x)\right )}{\sqrt{1-2 x}}-\frac{55 \left (66825 x^4+344520 x^3+1300860 x^2-569364 x-744172\right )}{\sqrt{1-2 x} (5 x+3)}+84 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{378125} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*(-744172 - 569364*x + 1300860*x^2 + 344520*x^3 + 66825*x^4))/(Sqrt[1 - 2*x]*(3 + 5*x)) + 84*Sqrt[55]*Arc
Tanh[Sqrt[5/11]*Sqrt[1 - 2*x]] + (1260*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 - 2*x))/11])/Sqrt[1 - 2*x])/37812
5

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Maple [A]  time = 0.01, size = 72, normalized size = 0.6 \begin{align*}{\frac{243}{1000} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}-{\frac{2943}{1000} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{107109}{5000}\sqrt{1-2\,x}}+{\frac{16807}{968}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{2}{378125}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{336\,\sqrt{55}}{4159375}{\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)^5/(1-2*x)^(3/2)/(3+5*x)^2,x)

[Out]

243/1000*(1-2*x)^(5/2)-2943/1000*(1-2*x)^(3/2)+107109/5000*(1-2*x)^(1/2)+16807/968/(1-2*x)^(1/2)+2/378125*(1-2
*x)^(1/2)/(-2*x-6/5)-336/4159375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.56953, size = 124, normalized size = 1.03 \begin{align*} \frac{243}{1000} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - \frac{2943}{1000} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{168}{4159375} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{107109}{5000} \, \sqrt{-2 \, x + 1} - \frac{52521891 \, x + 31513117}{302500 \,{\left (5 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 11 \, \sqrt{-2 \, x + 1}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

243/1000*(-2*x + 1)^(5/2) - 2943/1000*(-2*x + 1)^(3/2) + 168/4159375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1
))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 107109/5000*sqrt(-2*x + 1) - 1/302500*(52521891*x + 31513117)/(5*(-2*x + 1
)^(3/2) - 11*sqrt(-2*x + 1))

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Fricas [A]  time = 1.64639, size = 263, normalized size = 2.19 \begin{align*} \frac{168 \, \sqrt{55}{\left (10 \, x^{2} + x - 3\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \,{\left (735075 \, x^{4} + 3789720 \, x^{3} + 14309460 \, x^{2} - 6264264 \, x - 8186648\right )} \sqrt{-2 \, x + 1}}{4159375 \,{\left (10 \, x^{2} + x - 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4159375*(168*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(735075*x^4 +
3789720*x^3 + 14309460*x^2 - 6264264*x - 8186648)*sqrt(-2*x + 1))/(10*x^2 + x - 3)

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 2.12869, size = 138, normalized size = 1.15 \begin{align*} \frac{243}{1000} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} - \frac{2943}{1000} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{168}{4159375} \, \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{107109}{5000} \, \sqrt{-2 \, x + 1} - \frac{52521891 \, x + 31513117}{302500 \,{\left (5 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 11 \, \sqrt{-2 \, x + 1}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

243/1000*(2*x - 1)^2*sqrt(-2*x + 1) - 2943/1000*(-2*x + 1)^(3/2) + 168/4159375*sqrt(55)*log(1/2*abs(-2*sqrt(55
) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 107109/5000*sqrt(-2*x + 1) - 1/302500*(52521891*x + 31
513117)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))