3.22.11 \(\int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)} \, dx\) [2111]

Optimal. Leaf size=72 \[ \frac {4}{77 \sqrt {1-2 x}}+\frac {6}{7} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {10}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

6/49*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-10/121*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+4/77/(1
-2*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 162, 65, 212} \begin {gather*} \frac {4}{77 \sqrt {1-2 x}}+\frac {6}{7} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {10}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4/(77*Sqrt[1 - 2*x]) + (6*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7 - (10*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sq
rt[1 - 2*x]])/11

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 87

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

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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 {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)} \, dx &=\frac {4}{77 \sqrt {1-2 x}}+\frac {1}{77} \int \frac {53+30 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac {4}{77 \sqrt {1-2 x}}-\frac {9}{7} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {25}{11} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {4}{77 \sqrt {1-2 x}}+\frac {9}{7} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {25}{11} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {4}{77 \sqrt {1-2 x}}+\frac {6}{7} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {10}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 72, normalized size = 1.00 \begin {gather*} \frac {4}{77 \sqrt {1-2 x}}+\frac {6}{7} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {10}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4/(77*Sqrt[1 - 2*x]) + (6*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7 - (10*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sq
rt[1 - 2*x]])/11

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Maple [A]
time = 0.15, size = 47, normalized size = 0.65

method	result	size
derivativedivides	\(\frac {6 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}-\frac {10 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}+\frac {4}{77 \sqrt {1-2 x}}\)	\(47\)
default	\(\frac {6 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}-\frac {10 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}+\frac {4}{77 \sqrt {1-2 x}}\)	\(47\)
risch	\(\frac {6 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}-\frac {10 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}+\frac {4}{77 \sqrt {1-2 x}}\)	\(47\)
trager	\(-\frac {4 \sqrt {1-2 x}}{77 \left (-1+2 x \right )}+\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{121}-\frac {3 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x -5 \RootOf \left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{49}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/49*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-10/121*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+4/77/(1
-2*x)^(1/2)

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Maxima [A]
time = 0.57, size = 82, normalized size = 1.14 \begin {gather*} \frac {5}{121} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {3}{49} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {4}{77 \, \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/121*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 3/49*sqrt(21)*log(-(sqrt(21
) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4/77/sqrt(-2*x + 1)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (46) = 92\).
time = 1.52, size = 102, normalized size = 1.42 \begin {gather*} \frac {245 \, \sqrt {11} \sqrt {5} {\left (2 \, x - 1\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 363 \, \sqrt {7} \sqrt {3} {\left (2 \, x - 1\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) - 308 \, \sqrt {-2 \, x + 1}}{5929 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5929*(245*sqrt(11)*sqrt(5)*(2*x - 1)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 363*sqrt(7
)*sqrt(3)*(2*x - 1)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) - 308*sqrt(-2*x + 1))/(2*x - 1)

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Sympy [A]
time = 7.94, size = 133, normalized size = 1.85 \begin {gather*} - \frac {18 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: x < - \frac {2}{3} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: x > - \frac {2}{3} \end {cases}\right )}{7} + \frac {50 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x < - \frac {3}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x > - \frac {3}{5} \end {cases}\right )}{11} + \frac {4}{77 \sqrt {1 - 2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-18*Piecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, x < -2/3), (-sqrt(21)*atanh(sqrt(21)*sqrt(1 - 2*x
)/7)/21, x > -2/3))/7 + 50*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, x < -3/5), (-sqrt(55)*ata
nh(sqrt(55)*sqrt(1 - 2*x)/11)/55, x > -3/5))/11 + 4/(77*sqrt(1 - 2*x))

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Giac [A]
time = 1.35, size = 88, normalized size = 1.22 \begin {gather*} \frac {5}{121} \, \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 {3}{49} \, \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 {4}{77 \, \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/121*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 3/49*sqrt(21)*log
(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4/77/sqrt(-2*x + 1)

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Mupad [B]
time = 0.10, size = 46, normalized size = 0.64 \begin {gather*} \frac {6\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{49}-\frac {10\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{121}+\frac {4}{77\,\sqrt {1-2\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/49 - (10*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/121 +
4/(77*(1 - 2*x)^(1/2))

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