3.22.19 \(\int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\) [2119]

Optimal. Leaf size=68 \[ \frac {49}{22 \sqrt {1-2 x} (3+5 x)}-\frac {1227 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {138 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \]

[Out]

-138/33275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/22/(3+5*x)/(1-2*x)^(1/2)-1227/1210*(1-2*x)^(1/2)/(
3+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 68, 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 = {91, 79, 65, 212} \begin {gather*} -\frac {1227 \sqrt {1-2 x}}{1210 (5 x+3)}+\frac {49}{22 \sqrt {1-2 x} (5 x+3)}-\frac {138 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(22*Sqrt[1 - 2*x]*(3 + 5*x)) - (1227*Sqrt[1 - 2*x])/(1210*(3 + 5*x)) - (138*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/(605*Sqrt[55])

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)}-\frac {1}{22} \int \frac {-186+99 x}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)}-\frac {1227 \sqrt {1-2 x}}{1210 (3+5 x)}+\frac {69}{605} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)}-\frac {1227 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {69}{605} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)}-\frac {1227 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {138 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 53, normalized size = 0.78 \begin {gather*} \frac {734+1227 x}{605 \sqrt {1-2 x} (3+5 x)}-\frac {138 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(734 + 1227*x)/(605*Sqrt[1 - 2*x]*(3 + 5*x)) - (138*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(605*Sqrt[55])

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Maple [A]
time = 0.17, size = 45, normalized size = 0.66

method	result	size
risch	\(\frac {734+1227 x}{605 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {138 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}\)	\(41\)
derivativedivides	\(\frac {2 \sqrt {1-2 x}}{3025 \left (-\frac {6}{5}-2 x \right )}-\frac {138 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}+\frac {49}{121 \sqrt {1-2 x}}\)	\(45\)
default	\(\frac {2 \sqrt {1-2 x}}{3025 \left (-\frac {6}{5}-2 x \right )}-\frac {138 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}+\frac {49}{121 \sqrt {1-2 x}}\)	\(45\)
trager	\(-\frac {\left (734+1227 x \right ) \sqrt {1-2 x}}{605 \left (10 x^{2}+x -3\right )}+\frac {69 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{33275}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3025*(1-2*x)^(1/2)/(-6/5-2*x)-138/33275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/121/(1-2*x)^(1/2)

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Maxima [A]
time = 0.53, size = 65, normalized size = 0.96 \begin {gather*} \frac {69}{33275} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (1227 \, x + 734\right )}}{605 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

69/33275*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2/605*(1227*x + 734)/(5*
(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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Fricas [A]
time = 0.52, size = 65, normalized size = 0.96 \begin {gather*} \frac {69 \, \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 (1227 \, x + 734\right )} \sqrt {-2 \, x + 1}}{33275 \, {\left (10 \, x^{2} + x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/33275*(69*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(1227*x + 734)*s
qrt(-2*x + 1))/(10*x^2 + x - 3)

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Sympy [F(-1)] Timed out
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)**2/(1-2*x)**(3/2)/(3+5*x)**2,x)

[Out]

Timed out

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Giac [A]
time = 2.04, size = 68, normalized size = 1.00 \begin {gather*} \frac {69}{33275} \, \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 {2 \, {\left (1227 \, x + 734\right )}}{605 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

69/33275*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2/605*(1227*x
+ 734)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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Mupad [B]
time = 0.07, size = 46, normalized size = 0.68 \begin {gather*} \frac {\frac {2454\,x}{3025}+\frac {1468}{3025}}{\frac {11\,\sqrt {1-2\,x}}{5}-{\left (1-2\,x\right )}^{3/2}}-\frac {138\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{33275} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2454*x)/3025 + 1468/3025)/((11*(1 - 2*x)^(1/2))/5 - (1 - 2*x)^(3/2)) - (138*55^(1/2)*atanh((55^(1/2)*(1 - 2*
x)^(1/2))/11))/33275

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