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

Optimal. Leaf size=73 \[ -\frac {\sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}-\frac {6}{55} \sqrt {1-2 x} (3 x+11)-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \]

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Rubi [A]  time = 0.02, antiderivative size = 73, 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, 147, 63, 206} \begin {gather*} -\frac {\sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}-\frac {6}{55} \sqrt {1-2 x} (3 x+11)-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Sqrt[1 - 2*x]*(11 + 3*x))/55 - (Sqrt[1 - 2*x]*(2 + 3*x)^2)/(55*(3 + 5*x)) - (8*ArcTanh[Sqrt[5/11]*Sqrt[1 -
2*x]])/(55*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 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 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)^2} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^2}{55 (3+5 x)}-\frac {1}{55} \int \frac {(-74-90 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {6}{55} \sqrt {1-2 x} (11+3 x)-\frac {\sqrt {1-2 x} (2+3 x)^2}{55 (3+5 x)}+\frac {4}{55} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {6}{55} \sqrt {1-2 x} (11+3 x)-\frac {\sqrt {1-2 x} (2+3 x)^2}{55 (3+5 x)}-\frac {4}{55} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {6}{55} \sqrt {1-2 x} (11+3 x)-\frac {\sqrt {1-2 x} (2+3 x)^2}{55 (3+5 x)}-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 58, normalized size = 0.79 \begin {gather*} -\frac {\sqrt {1-2 x} \left (99 x^2+396 x+202\right )}{55 (5 x+3)}-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/55*(Sqrt[1 - 2*x]*(202 + 396*x + 99*x^2))/(3 + 5*x) - (8*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

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IntegrateAlgebraic [A]  time = 0.11, size = 70, normalized size = 0.96 \begin {gather*} \frac {\left (99 (1-2 x)^2-990 (1-2 x)+1699\right ) \sqrt {1-2 x}}{110 (5 (1-2 x)-11)}-\frac {8 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1699 - 990*(1 - 2*x) + 99*(1 - 2*x)^2)*Sqrt[1 - 2*x])/(110*(-11 + 5*(1 - 2*x))) - (8*ArcTanh[Sqrt[5/11]*Sqrt
[1 - 2*x]])/(55*Sqrt[55])

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fricas [A]  time = 1.38, size = 64, normalized size = 0.88 \begin {gather*} \frac {4 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (99 \, x^{2} + 396 \, x + 202\right )} \sqrt {-2 \, x + 1}}{3025 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3025*(4*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(99*x^2 + 396*x + 202)*sq
rt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.24, size = 74, normalized size = 1.01 \begin {gather*} \frac {9}{50} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {4}{3025} \, \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 {351}{250} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{1375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/50*(-2*x + 1)^(3/2) + 4/3025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x +
1))) - 351/250*sqrt(-2*x + 1) - 1/1375*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 54, normalized size = 0.74 \begin {gather*} -\frac {8 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{3025}+\frac {9 \left (-2 x +1\right )^{\frac {3}{2}}}{50}-\frac {351 \sqrt {-2 x +1}}{250}+\frac {2 \sqrt {-2 x +1}}{6875 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/50*(-2*x+1)^(3/2)-351/250*(-2*x+1)^(1/2)+2/6875*(-2*x+1)^(1/2)/(-2*x-6/5)-8/3025*arctanh(1/11*55^(1/2)*(-2*x
+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.10, size = 71, normalized size = 0.97 \begin {gather*} \frac {9}{50} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {4}{3025} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {351}{250} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{1375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/50*(-2*x + 1)^(3/2) + 4/3025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 35
1/250*sqrt(-2*x + 1) - 1/1375*sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 0.07, size = 55, normalized size = 0.75 \begin {gather*} \frac {9\,{\left (1-2\,x\right )}^{3/2}}{50}-\frac {351\,\sqrt {1-2\,x}}{250}-\frac {2\,\sqrt {1-2\,x}}{6875\,\left (2\,x+\frac {6}{5}\right )}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,8{}\mathrm {i}}{3025} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*8i)/3025 - (2*(1 - 2*x)^(1/2))/(6875*(2*x + 6/5)) - (351*(1 -
2*x)^(1/2))/250 + (9*(1 - 2*x)^(3/2))/50

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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