3.2061 \(\int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^3} \, dx\)

Optimal. Leaf size=80 \[ -\frac {\sqrt {1-2 x} (3 x+2)^2}{110 (5 x+3)^2}-\frac {9 \sqrt {1-2 x} (715 x+432)}{6050 (5 x+3)}-\frac {1347 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

[Out]

-1347/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/110*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)^2-9/6050*(432
+715*x)*(1-2*x)^(1/2)/(3+5*x)

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Rubi [A]  time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.00, 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} \[ -\frac {\sqrt {1-2 x} (3 x+2)^2}{110 (5 x+3)^2}-\frac {9 \sqrt {1-2 x} (715 x+432)}{6050 (5 x+3)}-\frac {1347 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]*(2 + 3*x)^2)/(110*(3 + 5*x)^2) - (9*Sqrt[1 - 2*x]*(432 + 715*x))/(6050*(3 + 5*x)) - (1347*ArcT
anh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*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 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}{\sqrt {1-2 x} (3+5 x)^3} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^2}{110 (3+5 x)^2}-\frac {1}{110} \int \frac {(-144-195 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^2}{110 (3+5 x)^2}-\frac {9 \sqrt {1-2 x} (432+715 x)}{6050 (3+5 x)}+\frac {1347 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{6050}\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^2}{110 (3+5 x)^2}-\frac {9 \sqrt {1-2 x} (432+715 x)}{6050 (3+5 x)}-\frac {1347 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{6050}\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^2}{110 (3+5 x)^2}-\frac {9 \sqrt {1-2 x} (432+715 x)}{6050 (3+5 x)}-\frac {1347 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 58, normalized size = 0.72 \[ \frac {-\frac {55 \sqrt {1-2 x} \left (32670 x^2+39405 x+11884\right )}{(5 x+3)^2}-2694 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{332750} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*Sqrt[1 - 2*x]*(11884 + 39405*x + 32670*x^2))/(3 + 5*x)^2 - 2694*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/332750

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fricas [A]  time = 0.88, size = 74, normalized size = 0.92 \[ \frac {1347 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (32670 \, x^{2} + 39405 \, x + 11884\right )} \sqrt {-2 \, x + 1}}{332750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/332750*(1347*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(32670*x^2
+ 39405*x + 11884)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.25, size = 77, normalized size = 0.96 \[ \frac {1347}{332750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {27}{125} \, \sqrt {-2 \, x + 1} + \frac {1005 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2233 \, \sqrt {-2 \, x + 1}}{60500 \, {\left (5 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1347/332750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 27/125*sqrt
(-2*x + 1) + 1/60500*(1005*(-2*x + 1)^(3/2) - 2233*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 57, normalized size = 0.71 \[ -\frac {1347 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{166375}-\frac {27 \sqrt {-2 x +1}}{125}+\frac {\frac {201 \left (-2 x +1\right )^{\frac {3}{2}}}{3025}-\frac {203 \sqrt {-2 x +1}}{1375}}{\left (-10 x -6\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/125*(-2*x+1)^(1/2)+2/5*(201/1210*(-2*x+1)^(3/2)-203/550*(-2*x+1)^(1/2))/(-10*x-6)^2-1347/166375*arctanh(1/
11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.20, size = 83, normalized size = 1.04 \[ \frac {1347}{332750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {27}{125} \, \sqrt {-2 \, x + 1} + \frac {1005 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2233 \, \sqrt {-2 \, x + 1}}{15125 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1347/332750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 27/125*sqrt(-2*x + 1)
+ 1/15125*(1005*(-2*x + 1)^(3/2) - 2233*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 1.24, size = 63, normalized size = 0.79 \[ -\frac {1347\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{166375}-\frac {27\,\sqrt {1-2\,x}}{125}-\frac {\frac {203\,\sqrt {1-2\,x}}{34375}-\frac {201\,{\left (1-2\,x\right )}^{3/2}}{75625}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (1347*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/166375 - (27*(1 - 2*x)^(1/2))/125 - ((203*(1 - 2*x)^(1/
2))/34375 - (201*(1 - 2*x)^(3/2))/75625)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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

[Out]

Timed out

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