3.20.73 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx\) [1973]

Optimal. Leaf size=85 \[ -\frac {272}{225} \sqrt {1-2 x}-\frac {4}{45} (1-2 x)^{3/2}+\frac {98}{9} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-4/45*(1-2*x)^(3/2)-242/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+98/27*arctanh(1/7*21^(1/2)*(1-2*x)^(
1/2))*21^(1/2)-272/225*(1-2*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 85, 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 = {86, 159, 162, 65, 212} \begin {gather*} -\frac {4}{45} (1-2 x)^{3/2}-\frac {272}{225} \sqrt {1-2 x}+\frac {98}{9} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-272*Sqrt[1 - 2*x])/225 - (4*(1 - 2*x)^(3/2))/45 + (98*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/9 - (242*S
qrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/25

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 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[f*((e + f*x)^(p -
1)/(b*d*(p - 1))), x] + Dist[1/(b*d), Int[(b*d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p -
2)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]

Rule 159

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 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 {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx &=-\frac {4}{45} (1-2 x)^{3/2}+\frac {1}{15} \int \frac {(-9-136 x) \sqrt {1-2 x}}{(2+3 x) (3+5 x)} \, dx\\ &=-\frac {272}{225} \sqrt {1-2 x}-\frac {4}{45} (1-2 x)^{3/2}+\frac {2}{225} \int \frac {-\frac {1767}{2}-3469 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {272}{225} \sqrt {1-2 x}-\frac {4}{45} (1-2 x)^{3/2}-\frac {343}{9} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {1331}{25} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {272}{225} \sqrt {1-2 x}-\frac {4}{45} (1-2 x)^{3/2}+\frac {343}{9} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {1331}{25} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {272}{225} \sqrt {1-2 x}-\frac {4}{45} (1-2 x)^{3/2}+\frac {98}{9} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{25} \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 = 0.11, size = 71, normalized size = 0.84 \begin {gather*} \frac {60 \sqrt {1-2 x} (-73+10 x)+12250 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-6534 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3375} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(60*Sqrt[1 - 2*x]*(-73 + 10*x) + 12250*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 6534*Sqrt[55]*ArcTanh[Sqrt[
5/11]*Sqrt[1 - 2*x]])/3375

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Maple [A]
time = 0.17, size = 56, normalized size = 0.66

method	result	size
derivativedivides	\(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{45}-\frac {242 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{27}-\frac {272 \sqrt {1-2 x}}{225}\)	\(56\)
default	\(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{45}-\frac {242 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{27}-\frac {272 \sqrt {1-2 x}}{225}\)	\(56\)
risch	\(-\frac {4 \left (10 x -73\right ) \left (-1+2 x \right )}{225 \sqrt {1-2 x}}-\frac {242 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{27}\)	\(57\)
trager	\(\left (\frac {8 x}{45}-\frac {292}{225}\right ) \sqrt {1-2 x}+\frac {121 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{125}+\frac {49 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{27}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/45*(1-2*x)^(3/2)-242/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+98/27*arctanh(1/7*21^(1/2)*(1-2*x)^(
1/2))*21^(1/2)-272/225*(1-2*x)^(1/2)

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Maxima [A]
time = 0.54, size = 91, normalized size = 1.07 \begin {gather*} -\frac {4}{45} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {49}{27} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {272}{225} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/45*(-2*x + 1)^(3/2) + 121/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) -
49/27*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 272/225*sqrt(-2*x + 1)

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Fricas [A]
time = 0.73, size = 88, normalized size = 1.04 \begin {gather*} \frac {121}{125} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {49}{27} \, \sqrt {7} \sqrt {3} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + \frac {4}{225} \, {\left (10 \, x - 73\right )} \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/125*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 49/27*sqrt(7)*sqrt(3)*lo
g(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 4/225*(10*x - 73)*sqrt(-2*x + 1)

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Sympy [A]
time = 12.86, size = 144, normalized size = 1.69 \begin {gather*} - \frac {4 \left (1 - 2 x\right )^{\frac {3}{2}}}{45} - \frac {272 \sqrt {1 - 2 x}}{225} - \frac {686 \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 )}{9} + \frac {2662 \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 )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(1 - 2*x)**(3/2)/45 - 272*sqrt(1 - 2*x)/225 - 686*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))/9 + 2662*Piecewise((-sqrt(55)*acoth(sqrt(
55)*sqrt(1 - 2*x)/11)/55, x < -3/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, x > -3/5))/25

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Giac [A]
time = 1.23, size = 97, normalized size = 1.14 \begin {gather*} -\frac {4}{45} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{125} \, \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 {49}{27} \, \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 {272}{225} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/45*(-2*x + 1)^(3/2) + 121/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x
+ 1))) - 49/27*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 272/225*
sqrt(-2*x + 1)

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Mupad [B]
time = 0.10, size = 59, normalized size = 0.69 \begin {gather*} -\frac {272\,\sqrt {1-2\,x}}{225}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{45}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,98{}\mathrm {i}}{27}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*242i)/125 - (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*9
8i)/27 - (272*(1 - 2*x)^(1/2))/225 - (4*(1 - 2*x)^(3/2))/45

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