3.1974 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx\)

Optimal. Leaf size=92 \[ \frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x}+\frac {140}{3} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

7/3*(1-2*x)^(3/2)/(2+3*x)-242/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+140/9*arctanh(1/7*21^(1/2)*(1-2
*x)^(1/2))*21^(1/2)+26/15*(1-2*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {98, 154, 156, 63, 206} \[ \frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x}+\frac {140}{3} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(26*Sqrt[1 - 2*x])/15 + (7*(1 - 2*x)^(3/2))/(3*(2 + 3*x)) + (140*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3
- (242*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5

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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 156

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 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 {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx &=\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {1}{3} \int \frac {\sqrt {1-2 x} (96+39 x)}{(2+3 x) (3+5 x)} \, dx\\ &=\frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {2}{45} \int \frac {954-\frac {813 x}{2}}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}-\frac {490}{3} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {1331}{5} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {490}{3} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {1331}{5} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {140}{3} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 78, normalized size = 0.85 \[ \frac {1}{225} \left (\frac {15 \sqrt {1-2 x} (8 x+87)}{3 x+2}+3500 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2178 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*Sqrt[1 - 2*x]*(87 + 8*x))/(2 + 3*x) + 3500*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 2178*Sqrt[55]*ArcT
anh[Sqrt[5/11]*Sqrt[1 - 2*x]])/225

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fricas [A]  time = 1.16, size = 107, normalized size = 1.16 \[ \frac {1089 \, \sqrt {11} \sqrt {5} {\left (3 \, x + 2\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 1750 \, \sqrt {7} \sqrt {3} {\left (3 \, x + 2\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 15 \, {\left (8 \, x + 87\right )} \sqrt {-2 \, x + 1}}{225 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(1089*sqrt(11)*sqrt(5)*(3*x + 2)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 1750*sqrt(
7)*sqrt(3)*(3*x + 2)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 15*(8*x + 87)*sqrt(-2*x + 1)
)/(3*x + 2)

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giac [A]  time = 0.96, size = 104, normalized size = 1.13 \[ \frac {121}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {70}{9} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {8}{45} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/25*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 70/9*sqrt(21)*lo
g(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 8/45*sqrt(-2*x + 1) + 49/9*sqrt(-2*
x + 1)/(3*x + 2)

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maple [A]  time = 0.01, size = 63, normalized size = 0.68 \[ \frac {140 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{9}-\frac {242 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{25}+\frac {8 \sqrt {-2 x +1}}{45}-\frac {98 \sqrt {-2 x +1}}{27 \left (-2 x -\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/45*(-2*x+1)^(1/2)-242/25*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-98/27*(-2*x+1)^(1/2)/(-2*x-4/3)+140/
9*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.28, size = 98, normalized size = 1.07 \[ \frac {121}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {70}{9} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {8}{45} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 70/9*sqrt(21)*log(-(sqrt(2
1) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 8/45*sqrt(-2*x + 1) + 49/9*sqrt(-2*x + 1)/(3*x + 2)

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mupad [B]  time = 0.11, size = 66, normalized size = 0.72 \[ \frac {98\,\sqrt {1-2\,x}}{27\,\left (2\,x+\frac {4}{3}\right )}+\frac {8\,\sqrt {1-2\,x}}{45}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,140{}\mathrm {i}}{9}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*242i)/25 - (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*14
0i)/9 + (98*(1 - 2*x)^(1/2))/(27*(2*x + 4/3)) + (8*(1 - 2*x)^(1/2))/45

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

[Out]

Timed out

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