3.2503 \(\int \frac{1}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=55 \[ \frac{6 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}}-\frac{10 \sqrt{1-2 x}}{11 \sqrt{5 x+3}} \]

[Out]

(-10*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (6*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

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Rubi [A]  time = 0.0131668, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {96, 93, 204} \[ \frac{6 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}}-\frac{10 \sqrt{1-2 x}}{11 \sqrt{5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (6*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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) (3+5 x)^{3/2}} \, dx &=-\frac{10 \sqrt{1-2 x}}{11 \sqrt{3+5 x}}-3 \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{10 \sqrt{1-2 x}}{11 \sqrt{3+5 x}}-6 \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )\\ &=-\frac{10 \sqrt{1-2 x}}{11 \sqrt{3+5 x}}+\frac{6 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{\sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0183409, size = 55, normalized size = 1. \[ \frac{6 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}}-\frac{10 \sqrt{1-2 x}}{11 \sqrt{5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (6*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

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Maple [B]  time = 0.014, size = 101, normalized size = 1.8 \begin{align*} -{\frac{1}{77} \left ( 165\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+99\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +70\,\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)/(3+5*x)^(3/2)/(1-2*x)^(1/2),x)

[Out]

-1/77*(165*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+99*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/
2)/(-10*x^2-x+3)^(1/2))+70*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-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 )} \sqrt{-2 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.53002, size = 205, normalized size = 3.73 \begin{align*} \frac{33 \, \sqrt{7}{\left (5 \, x + 3\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 ) - 70 \, \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{77 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/77*(33*sqrt(7)*(5*x + 3)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 70
*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.90377, size = 180, normalized size = 3.27 \begin{align*} -\frac{3}{70} \, \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{1}{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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/70*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)))) - 1/22*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqr
t(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))