3.2118 \(\int \frac{(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\)

Optimal. Leaf size=80 \[ \frac{7 (3 x+2)^2}{11 \sqrt{1-2 x} (5 x+3)}+\frac{18 \sqrt{1-2 x} (935 x+559)}{3025 (5 x+3)}-\frac{204 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3025 \sqrt{55}} \]

[Out]

(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*(3 + 5*x)) + (18*Sqrt[1 - 2*x]*(559 + 935*x))/(3025*(3 + 5*x)) - (204*ArcTan
h[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*Sqrt[55])

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Rubi [A]  time = 0.0199295, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {98, 146, 63, 206} \[ \frac{7 (3 x+2)^2}{11 \sqrt{1-2 x} (5 x+3)}+\frac{18 \sqrt{1-2 x} (935 x+559)}{3025 (5 x+3)}-\frac{204 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3025 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*(3 + 5*x)) + (18*Sqrt[1 - 2*x]*(559 + 935*x))/(3025*(3 + 5*x)) - (204*ArcTan
h[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*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 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)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)}-\frac{1}{11} \int \frac{(2+3 x) (54+102 x)}{\sqrt{1-2 x} (3+5 x)^2} \, dx\\ &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)}+\frac{18 \sqrt{1-2 x} (559+935 x)}{3025 (3+5 x)}+\frac{102 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{3025}\\ &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)}+\frac{18 \sqrt{1-2 x} (559+935 x)}{3025 (3+5 x)}-\frac{102 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{3025}\\ &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)}+\frac{18 \sqrt{1-2 x} (559+935 x)}{3025 (3+5 x)}-\frac{204 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3025 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0353781, size = 58, normalized size = 0.72 \[ \frac{\frac{55 \left (-16335 x^2+19806 x+17762\right )}{\sqrt{1-2 x} (5 x+3)}-204 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{166375} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((55*(17762 + 19806*x - 16335*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)) - 204*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]
)/166375

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Maple [A]  time = 0.01, size = 54, normalized size = 0.7 \begin{align*}{\frac{27}{50}\sqrt{1-2\,x}}+{\frac{343}{242}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{2}{15125}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{204\,\sqrt{55}}{166375}{\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)^3/(1-2*x)^(3/2)/(3+5*x)^2,x)

[Out]

27/50*(1-2*x)^(1/2)+343/242/(1-2*x)^(1/2)+2/15125*(1-2*x)^(1/2)/(-2*x-6/5)-204/166375*arctanh(1/11*55^(1/2)*(1
-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 2.05567, size = 100, normalized size = 1.25 \begin{align*} \frac{102}{166375} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{27}{50} \, \sqrt{-2 \, x + 1} - \frac{42879 \, x + 25723}{3025 \,{\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)^3/(1-2*x)^(3/2)/(3+5*x)^2,x, algorithm="maxima")

[Out]

102/166375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 27/50*sqrt(-2*x + 1) -
1/3025*(42879*x + 25723)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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Fricas [A]  time = 1.62327, size = 216, normalized size = 2.7 \begin{align*} \frac{102 \, \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 (16335 \, x^{2} - 19806 \, x - 17762\right )} \sqrt{-2 \, x + 1}}{166375 \,{\left (10 \, x^{2} + x - 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/166375*(102*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(16335*x^2 - 1
9806*x - 17762)*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)**3/(1-2*x)**(3/2)/(3+5*x)**2,x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 2.3301, size = 104, normalized size = 1.3 \begin{align*} \frac{102}{166375} \, \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{27}{50} \, \sqrt{-2 \, x + 1} - \frac{42879 \, x + 25723}{3025 \,{\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)^3/(1-2*x)^(3/2)/(3+5*x)^2,x, algorithm="giac")

[Out]

102/166375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 27/50*sqrt(-
2*x + 1) - 1/3025*(42879*x + 25723)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))