3.2513 \(\int \frac{1}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=77 \[ \frac{950 \sqrt{1-2 x}}{363 \sqrt{5 x+3}}-\frac{10 \sqrt{1-2 x}}{33 (5 x+3)^{3/2}}-\frac{18 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}} \]

[Out]

(-10*Sqrt[1 - 2*x])/(33*(3 + 5*x)^(3/2)) + (950*Sqrt[1 - 2*x])/(363*Sqrt[3 + 5*x]) - (18*ArcTan[Sqrt[1 - 2*x]/
(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

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Rubi [A]  time = 0.026939, antiderivative size = 77, 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 = {104, 152, 12, 93, 204} \[ \frac{950 \sqrt{1-2 x}}{363 \sqrt{5 x+3}}-\frac{10 \sqrt{1-2 x}}{33 (5 x+3)^{3/2}}-\frac{18 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x])/(33*(3 + 5*x)^(3/2)) + (950*Sqrt[1 - 2*x])/(363*Sqrt[3 + 5*x]) - (18*ArcTan[Sqrt[1 - 2*x]/
(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

Rule 104

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] &&
IntegersQ[2*m, 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) (3+5 x)^{5/2}} \, dx &=-\frac{10 \sqrt{1-2 x}}{33 (3+5 x)^{3/2}}-\frac{2}{33} \int \frac{\frac{59}{2}-30 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac{10 \sqrt{1-2 x}}{33 (3+5 x)^{3/2}}+\frac{950 \sqrt{1-2 x}}{363 \sqrt{3+5 x}}+\frac{4}{363} \int \frac{3267}{4 \sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{10 \sqrt{1-2 x}}{33 (3+5 x)^{3/2}}+\frac{950 \sqrt{1-2 x}}{363 \sqrt{3+5 x}}+9 \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{10 \sqrt{1-2 x}}{33 (3+5 x)^{3/2}}+\frac{950 \sqrt{1-2 x}}{363 \sqrt{3+5 x}}+18 \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}}{33 (3+5 x)^{3/2}}+\frac{950 \sqrt{1-2 x}}{363 \sqrt{3+5 x}}-\frac{18 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{\sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0511082, size = 60, normalized size = 0.78 \[ \frac{10 \sqrt{1-2 x} (475 x+274)}{363 (5 x+3)^{3/2}}-\frac{18 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[1 - 2*x]*(274 + 475*x))/(363*(3 + 5*x)^(3/2)) - (18*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sq
rt[7]

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Maple [B]  time = 0.013, size = 147, normalized size = 1.9 \begin{align*}{\frac{1}{2541} \left ( 81675\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}+98010\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+29403\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +33250\,x\sqrt{-10\,{x}^{2}-x+3}+19180\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2541*(81675*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+98010*7^(1/2)*arctan(1/14*(37*x+2
0)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+29403*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+33250*x*(-1
0*x^2-x+3)^(1/2)+19180*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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

[Out]

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

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Fricas [A]  time = 1.87648, size = 258, normalized size = 3.35 \begin{align*} -\frac{3267 \, \sqrt{7}{\left (25 \, x^{2} + 30 \, x + 9\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 \,{\left (475 \, x + 274\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{2541 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2541*(3267*sqrt(7)*(25*x^2 + 30*x + 9)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) - 70*(475*x + 274)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.61855, size = 262, normalized size = 3.4 \begin{align*} -\frac{1}{5808} \, \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 )}^{3} + \frac{9}{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{31}{242} \, \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)^(5/2)/(1-2*x)^(1/2),x, algorithm="giac")

[Out]

-1/5808*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)))^3 + 9/70*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)))) + 31/242*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)))