3.2504 \(\int \frac{1}{\sqrt{1-2 x} (2+3 x)^2 (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=86 \[ -\frac{515 \sqrt{1-2 x}}{77 \sqrt{5 x+3}}+\frac{3 \sqrt{1-2 x}}{7 (3 x+2) \sqrt{5 x+3}}+\frac{321 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{7 \sqrt{7}} \]

[Out]

(-515*Sqrt[1 - 2*x])/(77*Sqrt[3 + 5*x]) + (3*Sqrt[1 - 2*x])/(7*(2 + 3*x)*Sqrt[3 + 5*x]) + (321*ArcTan[Sqrt[1 -
2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])

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Rubi [A]  time = 0.0269292, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {103, 152, 12, 93, 204} \[ -\frac{515 \sqrt{1-2 x}}{77 \sqrt{5 x+3}}+\frac{3 \sqrt{1-2 x}}{7 (3 x+2) \sqrt{5 x+3}}+\frac{321 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{7 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-515*Sqrt[1 - 2*x])/(77*Sqrt[3 + 5*x]) + (3*Sqrt[1 - 2*x])/(7*(2 + 3*x)*Sqrt[3 + 5*x]) + (321*ArcTan[Sqrt[1 -
2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{1-2 x} (2+3 x)^2 (3+5 x)^{3/2}} \, dx &=\frac{3 \sqrt{1-2 x}}{7 (2+3 x) \sqrt{3+5 x}}+\frac{1}{7} \int \frac{\frac{67}{2}-30 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac{515 \sqrt{1-2 x}}{77 \sqrt{3+5 x}}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) \sqrt{3+5 x}}-\frac{2}{77} \int \frac{3531}{4 \sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{515 \sqrt{1-2 x}}{77 \sqrt{3+5 x}}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) \sqrt{3+5 x}}-\frac{321}{14} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{515 \sqrt{1-2 x}}{77 \sqrt{3+5 x}}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) \sqrt{3+5 x}}-\frac{321}{7} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )\\ &=-\frac{515 \sqrt{1-2 x}}{77 \sqrt{3+5 x}}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) \sqrt{3+5 x}}+\frac{321 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{7 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0388287, size = 69, normalized size = 0.8 \[ \frac{321 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{7 \sqrt{7}}-\frac{\sqrt{1-2 x} (1545 x+997)}{77 (3 x+2) \sqrt{5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]*(997 + 1545*x))/(77*(2 + 3*x)*Sqrt[3 + 5*x]) + (321*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x
])])/(7*Sqrt[7])

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Maple [B]  time = 0.013, size = 154, normalized size = 1.8 \begin{align*} -{\frac{1}{2156+3234\,x} \left ( 52965\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}+67089\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+21186\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +21630\,x\sqrt{-10\,{x}^{2}-x+3}+13958\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}{\frac{1}{\sqrt{3+5\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1078*(52965*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+67089*7^(1/2)*arctan(1/14*(37*x+
20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+21186*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+21630*x*(-
10*x^2-x+3)^(1/2)+13958*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(2+3*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (5 \, x + 3\right )}^{\frac{3}{2}}{\left (3 \, x + 2\right )}^{2} \sqrt{-2 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.516, size = 258, normalized size = 3. \begin{align*} \frac{3531 \, \sqrt{7}{\left (15 \, x^{2} + 19 \, x + 6\right )} \arctan \left (\frac{\sqrt{7}{\left (37 \, x + 20\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{14 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \,{\left (1545 \, x + 997\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{1078 \,{\left (15 \, x^{2} + 19 \, x + 6\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1078*(3531*sqrt(7)*(15*x^2 + 19*x + 6)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) - 14*(1545*x + 997)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(15*x^2 + 19*x + 6)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [B]  time = 2.28657, size = 340, normalized size = 3.95 \begin{align*} -\frac{321}{980} \, \sqrt{70} \sqrt{10}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{70} \sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} - \frac{5}{22} \, \sqrt{10}{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )} - \frac{198 \, \sqrt{10}{\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}}{7 \,{\left ({\left (\frac{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}{\sqrt{5 \, x + 3}} - \frac{4 \, \sqrt{5 \, x + 3}}{\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}}\right )}^{2} + 280\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-321/980*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^
2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 5/22*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/
sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) - 198/7*sqrt(10)*((sqrt(2)*sqrt(-10*x +
5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5
) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)