3.2164 \(\int \frac{(3+5 x)^3}{(1-2 x)^{5/2} (2+3 x)^2} \, dx\)

Optimal. Leaf size=80 \[ \frac{11 (5 x+3)^2}{21 (1-2 x)^{3/2} (3 x+2)}-\frac{10 (1450 x+969)}{1029 \sqrt{1-2 x} (3 x+2)}-\frac{200 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{1029 \sqrt{21}} \]

[Out]

(11*(3 + 5*x)^2)/(21*(1 - 2*x)^(3/2)*(2 + 3*x)) - (10*(969 + 1450*x))/(1029*Sqrt[1 - 2*x]*(2 + 3*x)) - (200*Ar
cTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(1029*Sqrt[21])

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Rubi [A]  time = 0.0193615, 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, 144, 63, 206} \[ \frac{11 (5 x+3)^2}{21 (1-2 x)^{3/2} (3 x+2)}-\frac{10 (1450 x+969)}{1029 \sqrt{1-2 x} (3 x+2)}-\frac{200 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{1029 \sqrt{21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(3 + 5*x)^2)/(21*(1 - 2*x)^(3/2)*(2 + 3*x)) - (10*(969 + 1450*x))/(1029*Sqrt[1 - 2*x]*(2 + 3*x)) - (200*Ar
cTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(1029*Sqrt[21])

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 144

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :>
Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(
m + n + 2)) + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m +
1) + d^2*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b*d*(b*c - a*d)^2*(m + 1)*(n + 1)), x] -
Dist[(a^2*d^2*f*h*(2 + 3*n + 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
*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b
*c - a*d)^2*(m + 1)*(n + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, h
}, x] && LtQ[m, -1] && LtQ[n, -1]

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{(3+5 x)^3}{(1-2 x)^{5/2} (2+3 x)^2} \, dx &=\frac{11 (3+5 x)^2}{21 (1-2 x)^{3/2} (2+3 x)}-\frac{1}{21} \int \frac{(3+5 x) (130+180 x)}{(1-2 x)^{3/2} (2+3 x)^2} \, dx\\ &=\frac{11 (3+5 x)^2}{21 (1-2 x)^{3/2} (2+3 x)}-\frac{10 (969+1450 x)}{1029 \sqrt{1-2 x} (2+3 x)}+\frac{100 \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx}{1029}\\ &=\frac{11 (3+5 x)^2}{21 (1-2 x)^{3/2} (2+3 x)}-\frac{10 (969+1450 x)}{1029 \sqrt{1-2 x} (2+3 x)}-\frac{100 \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{1029}\\ &=\frac{11 (3+5 x)^2}{21 (1-2 x)^{3/2} (2+3 x)}-\frac{10 (969+1450 x)}{1029 \sqrt{1-2 x} (2+3 x)}-\frac{200 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{1029 \sqrt{21}}\\ \end{align*}

Mathematica [A]  time = 0.0439729, size = 70, normalized size = 0.88 \[ -\frac{-21 \left (42475 x^2+21050 x-4839\right )-200 \sqrt{21-42 x} \left (6 x^2+x-2\right ) \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{21609 (1-2 x)^{3/2} (3 x+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-21*(-4839 + 21050*x + 42475*x^2) - 200*Sqrt[21 - 42*x]*(-2 + x + 6*x^2)*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(
21609*(1 - 2*x)^(3/2)*(2 + 3*x))

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Maple [A]  time = 0.013, size = 54, normalized size = 0.7 \begin{align*} -{\frac{2}{3087}\sqrt{1-2\,x} \left ( -2\,x-{\frac{4}{3}} \right ) ^{-1}}-{\frac{200\,\sqrt{21}}{21609}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+{\frac{1331}{294} \left ( 1-2\,x \right ) ^{-{\frac{3}{2}}}}-{\frac{4719}{686}{\frac{1}{\sqrt{1-2\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3087*(1-2*x)^(1/2)/(-2*x-4/3)-200/21609*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+1331/294/(1-2*x)^(3/2)
-4719/686/(1-2*x)^(1/2)

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Maxima [A]  time = 2.5391, size = 100, normalized size = 1.25 \begin{align*} \frac{100}{21609} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) - \frac{42475 \,{\left (2 \, x - 1\right )}^{2} + 254100 \, x - 61831}{2058 \,{\left (3 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - 7 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

100/21609*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/2058*(42475*(2*x - 1)
^2 + 254100*x - 61831)/(3*(-2*x + 1)^(5/2) - 7*(-2*x + 1)^(3/2))

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Fricas [A]  time = 1.3478, size = 240, normalized size = 3. \begin{align*} \frac{100 \, \sqrt{21}{\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (\frac{3 \, x + \sqrt{21} \sqrt{-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \,{\left (42475 \, x^{2} + 21050 \, x - 4839\right )} \sqrt{-2 \, x + 1}}{21609 \,{\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21609*(100*sqrt(21)*(12*x^3 - 4*x^2 - 5*x + 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 21*(4247
5*x^2 + 21050*x - 4839)*sqrt(-2*x + 1))/(12*x^3 - 4*x^2 - 5*x + 2)

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.8135, size = 104, normalized size = 1.3 \begin{align*} \frac{100}{21609} \, \sqrt{21} \log \left (\frac{{\left | -2 \, \sqrt{21} + 6 \, \sqrt{-2 \, x + 1} \right |}}{2 \,{\left (\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}\right )}}\right ) - \frac{121 \,{\left (117 \, x - 20\right )}}{1029 \,{\left (2 \, x - 1\right )} \sqrt{-2 \, x + 1}} + \frac{\sqrt{-2 \, x + 1}}{1029 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

100/21609*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 121/1029*(117*
x - 20)/((2*x - 1)*sqrt(-2*x + 1)) + 1/1029*sqrt(-2*x + 1)/(3*x + 2)