3.2152 \(\int \frac{(3+5 x)^2}{(1-2 x)^{5/2} (2+3 x)} \, dx\)

Optimal. Leaf size=54 \[ -\frac{407}{98 \sqrt{1-2 x}}+\frac{121}{42 (1-2 x)^{3/2}}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{49 \sqrt{21}} \]

[Out]

121/(42*(1 - 2*x)^(3/2)) - 407/(98*Sqrt[1 - 2*x]) - (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])

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Rubi [A]  time = 0.02376, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {87, 63, 206} \[ -\frac{407}{98 \sqrt{1-2 x}}+\frac{121}{42 (1-2 x)^{3/2}}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{49 \sqrt{21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

121/(42*(1 - 2*x)^(3/2)) - 407/(98*Sqrt[1 - 2*x]) - (2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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 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)^2}{(1-2 x)^{5/2} (2+3 x)} \, dx &=\int \left (\frac{121}{14 (1-2 x)^{5/2}}-\frac{407}{98 (1-2 x)^{3/2}}+\frac{1}{49 \sqrt{1-2 x} (2+3 x)}\right ) \, dx\\ &=\frac{121}{42 (1-2 x)^{3/2}}-\frac{407}{98 \sqrt{1-2 x}}+\frac{1}{49} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=\frac{121}{42 (1-2 x)^{3/2}}-\frac{407}{98 \sqrt{1-2 x}}-\frac{1}{49} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{121}{42 (1-2 x)^{3/2}}-\frac{407}{98 \sqrt{1-2 x}}-\frac{2 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{49 \sqrt{21}}\\ \end{align*}

Mathematica [C]  time = 0.0205002, size = 40, normalized size = 0.74 \[ \frac{2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{3}{7}-\frac{6 x}{7}\right )+35 (45 x-7)}{189 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*(-7 + 45*x) + 2*Hypergeometric2F1[-3/2, 1, -1/2, 3/7 - (6*x)/7])/(189*(1 - 2*x)^(3/2))

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Maple [A]  time = 0.009, size = 38, normalized size = 0.7 \begin{align*}{\frac{121}{42} \left ( 1-2\,x \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,\sqrt{21}}{1029}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }-{\frac{407}{98}{\frac{1}{\sqrt{1-2\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/42/(1-2*x)^(3/2)-2/1029*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-407/98/(1-2*x)^(1/2)

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Maxima [A]  time = 1.86067, size = 69, normalized size = 1.28 \begin{align*} \frac{1}{1029} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) + \frac{11 \,{\left (111 \, x - 17\right )}}{147 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1029*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 11/147*(111*x - 17)/(-2*x
+ 1)^(3/2)

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Fricas [A]  time = 1.59275, size = 188, normalized size = 3.48 \begin{align*} \frac{\sqrt{21}{\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (\frac{3 \, x + \sqrt{21} \sqrt{-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 77 \,{\left (111 \, x - 17\right )} \sqrt{-2 \, x + 1}}{1029 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1029*(sqrt(21)*(4*x^2 - 4*x + 1)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 77*(111*x - 17)*sqrt(-
2*x + 1))/(4*x^2 - 4*x + 1)

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Sympy [A]  time = 23.8529, size = 90, normalized size = 1.67 \begin{align*} \frac{2 \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{407}{98 \sqrt{1 - 2 x}} + \frac{121}{42 \left (1 - 2 x\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*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 - 407/(98*sqrt(1 - 2*x)) + 121/(42*(1 - 2*x)**(3/2))

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Giac [A]  time = 2.33053, size = 82, normalized size = 1.52 \begin{align*} \frac{1}{1029} \, \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{11 \,{\left (111 \, x - 17\right )}}{147 \,{\left (2 \, x - 1\right )} \sqrt{-2 \, x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1029*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 11/147*(111*x - 1
7)/((2*x - 1)*sqrt(-2*x + 1))