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

Optimal. Leaf size=56 \[ -\frac {2}{49 \sqrt {1-2 x}}+\frac {11}{21 (1-2 x)^{3/2}}+\frac {2}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 206} \begin {gather*} -\frac {2}{49 \sqrt {1-2 x}}+\frac {11}{21 (1-2 x)^{3/2}}+\frac {2}{49} \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)),x]

[Out]

11/(21*(1 - 2*x)^(3/2)) - 2/(49*Sqrt[1 - 2*x]) + (2*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/49

Rule 51

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 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 78

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] || LtQ
[p, n]))))

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 {3+5 x}{(1-2 x)^{5/2} (2+3 x)} \, dx &=\frac {11}{21 (1-2 x)^{3/2}}-\frac {1}{7} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)} \, dx\\ &=\frac {11}{21 (1-2 x)^{3/2}}-\frac {2}{49 \sqrt {1-2 x}}-\frac {3}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {11}{21 (1-2 x)^{3/2}}-\frac {2}{49 \sqrt {1-2 x}}+\frac {3}{49} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {11}{21 (1-2 x)^{3/2}}-\frac {2}{49 \sqrt {1-2 x}}+\frac {2}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 38, normalized size = 0.68 \begin {gather*} -\frac {(6-12 x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {3}{7}-\frac {6 x}{7}\right )-77}{147 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/147*(-77 + (6 - 12*x)*Hypergeometric2F1[-1/2, 1, 1/2, 3/7 - (6*x)/7])/(1 - 2*x)^(3/2)

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IntegrateAlgebraic [A]  time = 0.07, size = 52, normalized size = 0.93 \begin {gather*} \frac {77-6 (1-2 x)}{147 (1-2 x)^{3/2}}+\frac {2}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(77 - 6*(1 - 2*x))/(147*(1 - 2*x)^(3/2)) + (2*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/49

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fricas [B]  time = 1.23, size = 76, normalized size = 1.36 \begin {gather*} \frac {3 \, \sqrt {7} \sqrt {3} {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 7 \, {\left (12 \, x + 71\right )} \sqrt {-2 \, x + 1}}{1029 \, {\left (4 \, x^{2} - 4 \, x + 1\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),x, algorithm="fricas")

[Out]

1/1029*(3*sqrt(7)*sqrt(3)*(4*x^2 - 4*x + 1)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 7*(12
*x + 71)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.29, size = 61, normalized size = 1.09 \begin {gather*} -\frac {1}{343} \, \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 {12 \, x + 71}{147 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \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),x, algorithm="giac")

[Out]

-1/343*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/147*(12*x + 71)
/((2*x - 1)*sqrt(-2*x + 1))

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maple [A]  time = 0.01, size = 38, normalized size = 0.68 \begin {gather*} \frac {2 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{343}+\frac {11}{21 \left (-2 x +1\right )^{\frac {3}{2}}}-\frac {2}{49 \sqrt {-2 x +1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/21/(-2*x+1)^(3/2)+2/343*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)-2/49/(-2*x+1)^(1/2)

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maxima [A]  time = 1.20, size = 51, normalized size = 0.91 \begin {gather*} -\frac {1}{343} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {12 \, x + 71}{147 \, {\left (-2 \, x + 1\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),x, algorithm="maxima")

[Out]

-1/343*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/147*(12*x + 71)/(-2*x +
1)^(3/2)

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mupad [B]  time = 0.08, size = 32, normalized size = 0.57 \begin {gather*} \frac {\frac {4\,x}{49}+\frac {71}{147}}{{\left (1-2\,x\right )}^{3/2}}+\frac {2\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{343} \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)),x)

[Out]

((4*x)/49 + 71/147)/(1 - 2*x)^(3/2) + (2*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/343

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sympy [A]  time = 28.08, size = 90, normalized size = 1.61 \begin {gather*} - \frac {6 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 < - \frac {7}{3} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 > - \frac {7}{3} \end {cases}\right )}{49} - \frac {2}{49 \sqrt {1 - 2 x}} + \frac {11}{21 \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),x)

[Out]

-6*Piecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 < -7/3), (-sqrt(21)*atanh(sqrt(21)*sqrt(1
- 2*x)/7)/21, 2*x - 1 > -7/3))/49 - 2/(49*sqrt(1 - 2*x)) + 11/(21*(1 - 2*x)**(3/2))

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