3.22.43 \(\int \frac {3+5 x}{(1-2 x)^{5/2} (2+3 x)^3} \, dx\) [2143]

Optimal. Leaf size=96 \[ \frac {5}{49 (1-2 x)^{3/2}}+\frac {45}{343 \sqrt {1-2 x}}+\frac {1}{42 (1-2 x)^{3/2} (2+3 x)^2}-\frac {3}{14 (1-2 x)^{3/2} (2+3 x)}-\frac {45}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

[Out]

5/49/(1-2*x)^(3/2)+1/42/(1-2*x)^(3/2)/(2+3*x)^2-3/14/(1-2*x)^(3/2)/(2+3*x)-45/2401*arctanh(1/7*21^(1/2)*(1-2*x
)^(1/2))*21^(1/2)+45/343/(1-2*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 44, 53, 65, 212} \begin {gather*} \frac {45}{343 \sqrt {1-2 x}}-\frac {3}{14 (1-2 x)^{3/2} (3 x+2)}+\frac {5}{49 (1-2 x)^{3/2}}+\frac {1}{42 (1-2 x)^{3/2} (3 x+2)^2}-\frac {45}{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)^3),x]

[Out]

5/(49*(1 - 2*x)^(3/2)) + 45/(343*Sqrt[1 - 2*x]) + 1/(42*(1 - 2*x)^(3/2)*(2 + 3*x)^2) - 3/(14*(1 - 2*x)^(3/2)*(
2 + 3*x)) - (45*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/343

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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)^3} \, dx &=\frac {1}{42 (1-2 x)^{3/2} (2+3 x)^2}+\frac {3}{2} \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^2} \, dx\\ &=\frac {1}{42 (1-2 x)^{3/2} (2+3 x)^2}-\frac {3}{14 (1-2 x)^{3/2} (2+3 x)}+\frac {15}{14} \int \frac {1}{(1-2 x)^{5/2} (2+3 x)} \, dx\\ &=\frac {5}{49 (1-2 x)^{3/2}}+\frac {1}{42 (1-2 x)^{3/2} (2+3 x)^2}-\frac {3}{14 (1-2 x)^{3/2} (2+3 x)}+\frac {45}{98} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)} \, dx\\ &=\frac {5}{49 (1-2 x)^{3/2}}+\frac {45}{343 \sqrt {1-2 x}}+\frac {1}{42 (1-2 x)^{3/2} (2+3 x)^2}-\frac {3}{14 (1-2 x)^{3/2} (2+3 x)}+\frac {135}{686} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {5}{49 (1-2 x)^{3/2}}+\frac {45}{343 \sqrt {1-2 x}}+\frac {1}{42 (1-2 x)^{3/2} (2+3 x)^2}-\frac {3}{14 (1-2 x)^{3/2} (2+3 x)}-\frac {135}{686} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {5}{49 (1-2 x)^{3/2}}+\frac {45}{343 \sqrt {1-2 x}}+\frac {1}{42 (1-2 x)^{3/2} (2+3 x)^2}-\frac {3}{14 (1-2 x)^{3/2} (2+3 x)}-\frac {45}{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 = 0.18, size = 65, normalized size = 0.68 \begin {gather*} \frac {-\frac {7 \left (-1087-2277 x+2160 x^2+4860 x^3\right )}{2 (1-2 x)^{3/2} (2+3 x)^2}-135 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{7203} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-7*(-1087 - 2277*x + 2160*x^2 + 4860*x^3))/(2*(1 - 2*x)^(3/2)*(2 + 3*x)^2) - 135*Sqrt[21]*ArcTanh[Sqrt[3/7]*
Sqrt[1 - 2*x]])/7203

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Maple [A]
time = 0.11, size = 66, normalized size = 0.69

method	result	size
risch	\(\frac {4860 x^{3}+2160 x^{2}-2277 x -1087}{2058 \left (2+3 x \right )^{2} \sqrt {1-2 x}\, \left (-1+2 x \right )}-\frac {45 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(58\)
derivativedivides	\(\frac {44}{1029 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {256}{2401 \sqrt {1-2 x}}+\frac {\frac {531 \left (1-2 x \right )^{\frac {3}{2}}}{2401}-\frac {171 \sqrt {1-2 x}}{343}}{\left (-4-6 x \right )^{2}}-\frac {45 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(66\)
default	\(\frac {44}{1029 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {256}{2401 \sqrt {1-2 x}}+\frac {\frac {531 \left (1-2 x \right )^{\frac {3}{2}}}{2401}-\frac {171 \sqrt {1-2 x}}{343}}{\left (-4-6 x \right )^{2}}-\frac {45 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(66\)
trager	\(-\frac {\left (4860 x^{3}+2160 x^{2}-2277 x -1087\right ) \sqrt {1-2 x}}{2058 \left (6 x^{2}+x -2\right )^{2}}-\frac {45 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{4802}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

44/1029/(1-2*x)^(3/2)+256/2401/(1-2*x)^(1/2)+324/2401*(59/36*(1-2*x)^(3/2)-133/36*(1-2*x)^(1/2))/(-4-6*x)^2-45
/2401*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)

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Maxima [A]
time = 0.87, size = 92, normalized size = 0.96 \begin {gather*} \frac {45}{4802} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1215 \, {\left (2 \, x - 1\right )}^{3} + 4725 \, {\left (2 \, x - 1\right )}^{2} + 7056 \, x - 5684}{1029 \, {\left (9 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 42 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 49 \, {\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)^3,x, algorithm="maxima")

[Out]

45/4802*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/1029*(1215*(2*x - 1)^3
+ 4725*(2*x - 1)^2 + 7056*x - 5684)/(9*(-2*x + 1)^(7/2) - 42*(-2*x + 1)^(5/2) + 49*(-2*x + 1)^(3/2))

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Fricas [A]
time = 1.14, size = 105, normalized size = 1.09 \begin {gather*} \frac {135 \, \sqrt {7} \sqrt {3} {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 7 \, {\left (4860 \, x^{3} + 2160 \, x^{2} - 2277 \, x - 1087\right )} \sqrt {-2 \, x + 1}}{14406 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\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)^3,x, algorithm="fricas")

[Out]

1/14406*(135*sqrt(7)*sqrt(3)*(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x -
5)/(3*x + 2)) - 7*(4860*x^3 + 2160*x^2 - 2277*x - 1087)*sqrt(-2*x + 1))/(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.65, size = 89, normalized size = 0.93 \begin {gather*} \frac {45}{4802} \, \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 (384 \, x - 269\right )}}{7203 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {9 \, {\left (59 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 133 \, \sqrt {-2 \, x + 1}\right )}}{9604 \, {\left (3 \, x + 2\right )}^{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)^3,x, algorithm="giac")

[Out]

45/4802*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4/7203*(384*x -
269)/((2*x - 1)*sqrt(-2*x + 1)) + 9/9604*(59*(-2*x + 1)^(3/2) - 133*sqrt(-2*x + 1))/(3*x + 2)^2

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Mupad [B]
time = 0.07, size = 72, normalized size = 0.75 \begin {gather*} -\frac {45\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{2401}-\frac {\frac {16\,x}{21}+\frac {25\,{\left (2\,x-1\right )}^2}{49}+\frac {45\,{\left (2\,x-1\right )}^3}{343}-\frac {116}{189}}{\frac {49\,{\left (1-2\,x\right )}^{3/2}}{9}-\frac {14\,{\left (1-2\,x\right )}^{5/2}}{3}+{\left (1-2\,x\right )}^{7/2}} \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)^3),x)

[Out]

- (45*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/2401 - ((16*x)/21 + (25*(2*x - 1)^2)/49 + (45*(2*x - 1)^3)
/343 - 116/189)/((49*(1 - 2*x)^(3/2))/9 - (14*(1 - 2*x)^(5/2))/3 + (1 - 2*x)^(7/2))

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