3.22.29 \(\int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\) [2129]

Optimal. Leaf size=88 \[ \frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}-\frac {2589 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}} \]

[Out]

-2589/366025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/22/(3+5*x)^2/(1-2*x)^(1/2)-613/605*(1-2*x)^(1/2)
/(3+5*x)^2-2589/13310*(1-2*x)^(1/2)/(3+5*x)

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Rubi [A]
time = 0.02, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {91, 79, 44, 65, 212} \begin {gather*} -\frac {2589 \sqrt {1-2 x}}{13310 (5 x+3)}-\frac {613 \sqrt {1-2 x}}{605 (5 x+3)^2}+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^2}-\frac {2589 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(22*Sqrt[1 - 2*x]*(3 + 5*x)^2) - (613*Sqrt[1 - 2*x])/(605*(3 + 5*x)^2) - (2589*Sqrt[1 - 2*x])/(13310*(3 + 5
*x)) - (2589*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6655*Sqrt[55])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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)^3} \, dx &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {1}{22} \int \frac {-431+99 x}{\sqrt {1-2 x} (3+5 x)^3} \, dx\\ &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}+\frac {2589 \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx}{1210}\\ &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}+\frac {2589 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{13310}\\ &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}-\frac {2589 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{13310}\\ &=\frac {49}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {613 \sqrt {1-2 x}}{605 (3+5 x)^2}-\frac {2589 \sqrt {1-2 x}}{13310 (3+5 x)}-\frac {2589 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 58, normalized size = 0.66 \begin {gather*} \frac {\frac {55 \left (8392+29561 x+25890 x^2\right )}{\sqrt {1-2 x} (3+5 x)^2}-5178 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{732050} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((55*(8392 + 29561*x + 25890*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)^2) - 5178*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/732050

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Maple [A]
time = 0.12, size = 57, normalized size = 0.65

method	result	size
risch	\(\frac {25890 x^{2}+29561 x +8392}{13310 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {2589 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}\)	\(46\)
derivativedivides	\(\frac {\frac {139 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {141 \sqrt {1-2 x}}{605}}{\left (-6-10 x \right )^{2}}-\frac {2589 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}+\frac {98}{1331 \sqrt {1-2 x}}\)	\(57\)
default	\(\frac {\frac {139 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {141 \sqrt {1-2 x}}{605}}{\left (-6-10 x \right )^{2}}-\frac {2589 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}+\frac {98}{1331 \sqrt {1-2 x}}\)	\(57\)
trager	\(-\frac {\left (25890 x^{2}+29561 x +8392\right ) \sqrt {1-2 x}}{13310 \left (3+5 x \right )^{2} \left (-1+2 x \right )}-\frac {2589 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{732050}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/1331*(139/50*(1-2*x)^(3/2)-1551/250*(1-2*x)^(1/2))/(-6-10*x)^2-2589/366025*arctanh(1/11*55^(1/2)*(1-2*x)^(1
/2))*55^(1/2)+98/1331/(1-2*x)^(1/2)

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Maxima [A]
time = 0.53, size = 83, normalized size = 0.94 \begin {gather*} \frac {2589}{732050} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {12945 \, {\left (2 \, x - 1\right )}^{2} + 110902 \, x + 3839}{6655 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \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)^3,x, algorithm="maxima")

[Out]

2589/732050*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/6655*(12945*(2*x -
1)^2 + 110902*x + 3839)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))

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Fricas [A]
time = 1.50, size = 84, normalized size = 0.95 \begin {gather*} \frac {2589 \, \sqrt {55} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (25890 \, x^{2} + 29561 \, x + 8392\right )} \sqrt {-2 \, x + 1}}{732050 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\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)^3,x, algorithm="fricas")

[Out]

1/732050*(2589*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(
25890*x^2 + 29561*x + 8392)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)

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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)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.67, size = 77, normalized size = 0.88 \begin {gather*} \frac {2589}{732050} \, \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 {98}{1331 \, \sqrt {-2 \, x + 1}} + \frac {695 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1551 \, \sqrt {-2 \, x + 1}}{26620 \, {\left (5 \, x + 3\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

2589/732050*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 98/1331/sqr
t(-2*x + 1) + 1/26620*(695*(-2*x + 1)^(3/2) - 1551*sqrt(-2*x + 1))/(5*x + 3)^2

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Mupad [B]
time = 0.07, size = 62, normalized size = 0.70 \begin {gather*} \frac {\frac {10082\,x}{15125}+\frac {2589\,{\left (2\,x-1\right )}^2}{33275}+\frac {349}{15125}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}}-\frac {2589\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{366025} \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)^3),x)

[Out]

((10082*x)/15125 + (2589*(2*x - 1)^2)/33275 + 349/15125)/((121*(1 - 2*x)^(1/2))/25 - (22*(1 - 2*x)^(3/2))/5 +
(1 - 2*x)^(5/2)) - (2589*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/366025

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