3.22.85 \(\int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^2} \, dx\) [2185]

Optimal. Leaf size=76 \[ \frac {76}{1815 (1-2 x)^{3/2}}+\frac {76}{1331 \sqrt {1-2 x}}-\frac {1}{55 (1-2 x)^{3/2} (3+5 x)}-\frac {76 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331} \]

[Out]

76/1815/(1-2*x)^(3/2)-1/55/(1-2*x)^(3/2)/(3+5*x)-76/14641*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+76/133
1/(1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {79, 53, 65, 212} \begin {gather*} \frac {76}{1331 \sqrt {1-2 x}}-\frac {1}{55 (1-2 x)^{3/2} (5 x+3)}+\frac {76}{1815 (1-2 x)^{3/2}}-\frac {76 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

76/(1815*(1 - 2*x)^(3/2)) + 76/(1331*Sqrt[1 - 2*x]) - 1/(55*(1 - 2*x)^(3/2)*(3 + 5*x)) - (76*Sqrt[5/11]*ArcTan
h[Sqrt[5/11]*Sqrt[1 - 2*x]])/1331

Rule 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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}{(1-2 x)^{5/2} (3+5 x)^2} \, dx &=-\frac {1}{55 (1-2 x)^{3/2} (3+5 x)}+\frac {38}{55} \int \frac {1}{(1-2 x)^{5/2} (3+5 x)} \, dx\\ &=\frac {76}{1815 (1-2 x)^{3/2}}-\frac {1}{55 (1-2 x)^{3/2} (3+5 x)}+\frac {38}{121} \int \frac {1}{(1-2 x)^{3/2} (3+5 x)} \, dx\\ &=\frac {76}{1815 (1-2 x)^{3/2}}+\frac {76}{1331 \sqrt {1-2 x}}-\frac {1}{55 (1-2 x)^{3/2} (3+5 x)}+\frac {190 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{1331}\\ &=\frac {76}{1815 (1-2 x)^{3/2}}+\frac {76}{1331 \sqrt {1-2 x}}-\frac {1}{55 (1-2 x)^{3/2} (3+5 x)}-\frac {190 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{1331}\\ &=\frac {76}{1815 (1-2 x)^{3/2}}+\frac {76}{1331 \sqrt {1-2 x}}-\frac {1}{55 (1-2 x)^{3/2} (3+5 x)}-\frac {76 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 60, normalized size = 0.79 \begin {gather*} \frac {2 \left (-\frac {11 \left (-1113-608 x+2280 x^2\right )}{2 (1-2 x)^{3/2} (3+5 x)}-114 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )}{43923} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-11*(-1113 - 608*x + 2280*x^2))/(2*(1 - 2*x)^(3/2)*(3 + 5*x)) - 114*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 -
2*x]]))/43923

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Maple [A]
time = 0.12, size = 54, normalized size = 0.71

method	result	size
risch	\(\frac {2280 x^{2}-608 x -1113}{3993 \left (3+5 x \right ) \sqrt {1-2 x}\, \left (-1+2 x \right )}-\frac {76 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}\)	\(53\)
derivativedivides	\(\frac {2 \sqrt {1-2 x}}{1331 \left (-\frac {6}{5}-2 x \right )}-\frac {76 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {14}{363 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {74}{1331 \sqrt {1-2 x}}\)	\(54\)
default	\(\frac {2 \sqrt {1-2 x}}{1331 \left (-\frac {6}{5}-2 x \right )}-\frac {76 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {14}{363 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {74}{1331 \sqrt {1-2 x}}\)	\(54\)
trager	\(-\frac {\left (2280 x^{2}-608 x -1113\right ) \sqrt {1-2 x}}{3993 \left (-1+2 x \right )^{2} \left (3+5 x \right )}+\frac {38 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{14641}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1331*(1-2*x)^(1/2)/(-6/5-2*x)-76/14641*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+14/363/(1-2*x)^(3/2)+74
/1331/(1-2*x)^(1/2)

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Maxima [A]
time = 0.49, size = 74, normalized size = 0.97 \begin {gather*} \frac {38}{14641} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2 \, {\left (570 \, {\left (2 \, x - 1\right )}^{2} + 1672 \, x - 1683\right )}}{3993 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

38/14641*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2/3993*(570*(2*x - 1)^2
+ 1672*x - 1683)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))

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Fricas [A]
time = 1.03, size = 90, normalized size = 1.18 \begin {gather*} \frac {114 \, \sqrt {11} \sqrt {5} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 11 \, {\left (2280 \, x^{2} - 608 \, x - 1113\right )} \sqrt {-2 \, x + 1}}{43923 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/43923*(114*sqrt(11)*sqrt(5)*(20*x^3 - 8*x^2 - 7*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x
+ 3)) - 11*(2280*x^2 - 608*x - 1113)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)

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Sympy [A]
time = 218.76, size = 204, normalized size = 2.68 \begin {gather*} - \frac {20 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{121} + \frac {370 \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 )}{1331} + \frac {74}{1331 \sqrt {1 - 2 x}} + \frac {14}{363 \left (1 - 2 x\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-20*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(
sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) &
(sqrt(1 - 2*x) < sqrt(55)/5)))/121 + 370*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))/1331 + 74/(1331*sqrt(1 - 2*x)) + 14/(363*(1 - 2*x)*
*(3/2))

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Giac [A]
time = 2.14, size = 77, normalized size = 1.01 \begin {gather*} \frac {38}{14641} \, \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 {4 \, {\left (111 \, x - 94\right )}}{3993 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {5 \, \sqrt {-2 \, x + 1}}{1331 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

38/14641*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 4/3993*(111*x
- 94)/((2*x - 1)*sqrt(-2*x + 1)) - 5/1331*sqrt(-2*x + 1)/(5*x + 3)

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Mupad [B]
time = 1.20, size = 56, normalized size = 0.74 \begin {gather*} -\frac {\frac {304\,x}{1815}+\frac {76\,{\left (2\,x-1\right )}^2}{1331}-\frac {102}{605}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {76\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{14641} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((304*x)/1815 + (76*(2*x - 1)^2)/1331 - 102/605)/((11*(1 - 2*x)^(3/2))/5 - (1 - 2*x)^(5/2)) - (76*55^(1/2)*a
tanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/14641

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