3.22.42 \(\int \frac {3+5 x}{(1-2 x)^{5/2} (2+3 x)^2} \, dx\) [2142]

Optimal. Leaf size=76 \[ \frac {20}{147 (1-2 x)^{3/2}}+\frac {60}{343 \sqrt {1-2 x}}+\frac {1}{21 (1-2 x)^{3/2} (2+3 x)}-\frac {60}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

[Out]

20/147/(1-2*x)^(3/2)+1/21/(1-2*x)^(3/2)/(2+3*x)-60/2401*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+60/343/(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 {60}{343 \sqrt {1-2 x}}+\frac {1}{21 (1-2 x)^{3/2} (3 x+2)}+\frac {20}{147 (1-2 x)^{3/2}}-\frac {60}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

20/(147*(1 - 2*x)^(3/2)) + 60/(343*Sqrt[1 - 2*x]) + 1/(21*(1 - 2*x)^(3/2)*(2 + 3*x)) - (60*Sqrt[3/7]*ArcTanh[S
qrt[3/7]*Sqrt[1 - 2*x]])/343

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 {3+5 x}{(1-2 x)^{5/2} (2+3 x)^2} \, dx &=\frac {1}{21 (1-2 x)^{3/2} (2+3 x)}+\frac {10}{7} \int \frac {1}{(1-2 x)^{5/2} (2+3 x)} \, dx\\ &=\frac {20}{147 (1-2 x)^{3/2}}+\frac {1}{21 (1-2 x)^{3/2} (2+3 x)}+\frac {30}{49} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)} \, dx\\ &=\frac {20}{147 (1-2 x)^{3/2}}+\frac {60}{343 \sqrt {1-2 x}}+\frac {1}{21 (1-2 x)^{3/2} (2+3 x)}+\frac {90}{343} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {20}{147 (1-2 x)^{3/2}}+\frac {60}{343 \sqrt {1-2 x}}+\frac {1}{21 (1-2 x)^{3/2} (2+3 x)}-\frac {90}{343} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {20}{147 (1-2 x)^{3/2}}+\frac {60}{343 \sqrt {1-2 x}}+\frac {1}{21 (1-2 x)^{3/2} (2+3 x)}-\frac {60}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 72, normalized size = 0.95 \begin {gather*} \frac {2 \left (-539-420 (1-2 x)+270 (1-2 x)^2\right )}{1029 (-7+3 (1-2 x)) (1-2 x)^{3/2}}-\frac {60}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-539 - 420*(1 - 2*x) + 270*(1 - 2*x)^2))/(1029*(-7 + 3*(1 - 2*x))*(1 - 2*x)^(3/2)) - (60*Sqrt[3/7]*ArcTanh
[Sqrt[3/7]*Sqrt[1 - 2*x]])/343

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Maple [A]
time = 0.11, size = 54, normalized size = 0.71

method	result	size
risch	\(\frac {1080 x^{2}-240 x -689}{1029 \sqrt {1-2 x}\, \left (-1+2 x \right ) \left (2+3 x \right )}-\frac {60 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(53\)
derivativedivides	\(\frac {22}{147 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {62}{343 \sqrt {1-2 x}}-\frac {2 \sqrt {1-2 x}}{343 \left (-\frac {4}{3}-2 x \right )}-\frac {60 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(54\)
default	\(\frac {22}{147 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {62}{343 \sqrt {1-2 x}}-\frac {2 \sqrt {1-2 x}}{343 \left (-\frac {4}{3}-2 x \right )}-\frac {60 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(54\)
trager	\(-\frac {\left (1080 x^{2}-240 x -689\right ) \sqrt {1-2 x}}{1029 \left (-1+2 x \right )^{2} \left (2+3 x \right )}-\frac {30 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{2401}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

22/147/(1-2*x)^(3/2)+62/343/(1-2*x)^(1/2)-2/343*(1-2*x)^(1/2)/(-4/3-2*x)-60/2401*arctanh(1/7*21^(1/2)*(1-2*x)^
(1/2))*21^(1/2)

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Maxima [A]
time = 0.53, size = 74, normalized size = 0.97 \begin {gather*} \frac {30}{2401} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2 \, {\left (270 \, {\left (2 \, x - 1\right )}^{2} + 840 \, x - 959\right )}}{1029 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 7 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

30/2401*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 2/1029*(270*(2*x - 1)^2 +
840*x - 959)/(3*(-2*x + 1)^(5/2) - 7*(-2*x + 1)^(3/2))

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Fricas [A]
time = 1.22, size = 90, normalized size = 1.18 \begin {gather*} \frac {90 \, \sqrt {7} \sqrt {3} {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 7 \, {\left (1080 \, x^{2} - 240 \, x - 689\right )} \sqrt {-2 \, x + 1}}{7203 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7203*(90*sqrt(7)*sqrt(3)*(12*x^3 - 4*x^2 - 5*x + 2)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)
) - 7*(1080*x^2 - 240*x - 689)*sqrt(-2*x + 1))/(12*x^3 - 4*x^2 - 5*x + 2)

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Sympy [A]
time = 211.42, size = 204, normalized size = 2.68 \begin {gather*} \frac {12 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{49} + \frac {186 \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 )}{343} + \frac {62}{343 \sqrt {1 - 2 x}} + \frac {22}{147 \left (1 - 2 x\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/4 + log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/4 - 1/(4*(sqr
t(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (sqrt(1 - 2*x) > -sqrt(21)/3) & (sqrt
(1 - 2*x) < sqrt(21)/3)))/49 + 186*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))/343 + 62/(343*sqrt(1 - 2*x)) + 22/(147*(1 - 2*x)**(3/2))

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Giac [A]
time = 2.72, size = 77, normalized size = 1.01 \begin {gather*} \frac {30}{2401} \, \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 \, {\left (93 \, x - 85\right )}}{1029 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {3 \, \sqrt {-2 \, x + 1}}{343 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

30/2401*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4/1029*(93*x - 8
5)/((2*x - 1)*sqrt(-2*x + 1)) + 3/343*sqrt(-2*x + 1)/(3*x + 2)

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Mupad [B]
time = 1.21, size = 56, normalized size = 0.74 \begin {gather*} -\frac {\frac {80\,x}{147}+\frac {60\,{\left (2\,x-1\right )}^2}{343}-\frac {274}{441}}{\frac {7\,{\left (1-2\,x\right )}^{3/2}}{3}-{\left (1-2\,x\right )}^{5/2}}-\frac {60\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{2401} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((80*x)/147 + (60*(2*x - 1)^2)/343 - 274/441)/((7*(1 - 2*x)^(3/2))/3 - (1 - 2*x)^(5/2)) - (60*21^(1/2)*atanh
((21^(1/2)*(1 - 2*x)^(1/2))/7))/2401

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