3.2614 \(\int \frac{1}{(1-2 x)^{5/2} (2+3 x) \sqrt{3+5 x}} \, dx\)

Optimal. Leaf size=79 \[ \frac{676 \sqrt{5 x+3}}{17787 \sqrt{1-2 x}}+\frac{4 \sqrt{5 x+3}}{231 (1-2 x)^{3/2}}-\frac{18 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{49 \sqrt{7}} \]

[Out]

(4*Sqrt[3 + 5*x])/(231*(1 - 2*x)^(3/2)) + (676*Sqrt[3 + 5*x])/(17787*Sqrt[1 - 2*x]) - (18*ArcTan[Sqrt[1 - 2*x]
/(Sqrt[7]*Sqrt[3 + 5*x])])/(49*Sqrt[7])

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Rubi [A]  time = 0.0258861, antiderivative size = 79, 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} \[ \frac{676 \sqrt{5 x+3}}{17787 \sqrt{1-2 x}}+\frac{4 \sqrt{5 x+3}}{231 (1-2 x)^{3/2}}-\frac{18 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{49 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 + 5*x])/(231*(1 - 2*x)^(3/2)) + (676*Sqrt[3 + 5*x])/(17787*Sqrt[1 - 2*x]) - (18*ArcTan[Sqrt[1 - 2*x]
/(Sqrt[7]*Sqrt[3 + 5*x])])/(49*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}{(1-2 x)^{5/2} (2+3 x) \sqrt{3+5 x}} \, dx &=\frac{4 \sqrt{3+5 x}}{231 (1-2 x)^{3/2}}-\frac{2}{231} \int \frac{-\frac{139}{2}-30 x}{(1-2 x)^{3/2} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=\frac{4 \sqrt{3+5 x}}{231 (1-2 x)^{3/2}}+\frac{676 \sqrt{3+5 x}}{17787 \sqrt{1-2 x}}+\frac{4 \int \frac{3267}{4 \sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{17787}\\ &=\frac{4 \sqrt{3+5 x}}{231 (1-2 x)^{3/2}}+\frac{676 \sqrt{3+5 x}}{17787 \sqrt{1-2 x}}+\frac{9}{49} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=\frac{4 \sqrt{3+5 x}}{231 (1-2 x)^{3/2}}+\frac{676 \sqrt{3+5 x}}{17787 \sqrt{1-2 x}}+\frac{18}{49} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )\\ &=\frac{4 \sqrt{3+5 x}}{231 (1-2 x)^{3/2}}+\frac{676 \sqrt{3+5 x}}{17787 \sqrt{1-2 x}}-\frac{18 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{49 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0549434, size = 71, normalized size = 0.9 \[ \frac{56 \sqrt{5 x+3} (123-169 x)+6534 \sqrt{7-14 x} (2 x-1) \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{124509 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(56*(123 - 169*x)*Sqrt[3 + 5*x] + 6534*Sqrt[7 - 14*x]*(-1 + 2*x)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])]
)/(124509*(1 - 2*x)^(3/2))

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Maple [B]  time = 0.015, size = 154, normalized size = 2. \begin{align*}{\frac{1}{124509\, \left ( 2\,x-1 \right ) ^{2}} \left ( 13068\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-13068\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+3267\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) -9464\,x\sqrt{-10\,{x}^{2}-x+3}+6888\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{3+5\,x}\sqrt{1-2\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/124509*(13068*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-13068*7^(1/2)*arctan(1/14*(37*x
+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+3267*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-9464*x*(-1
0*x^2-x+3)^(1/2)+6888*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)*(1-2*x)^(1/2)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58109, size = 255, normalized size = 3.23 \begin{align*} -\frac{3267 \, \sqrt{7}{\left (4 \, x^{2} - 4 \, x + 1\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 ) + 56 \,{\left (169 \, x - 123\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{124509 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/124509*(3267*sqrt(7)*(4*x^2 - 4*x + 1)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) + 56*(169*x - 123)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.30546, size = 153, normalized size = 1.94 \begin{align*} \frac{9}{3430} \, \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{8 \,{\left (169 \, \sqrt{5}{\left (5 \, x + 3\right )} - 1122 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{444675 \,{\left (2 \, x - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/3430*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)))) - 8/444675*(169*sqrt(5)*(5*x + 3) - 1122*sqrt(5))*sqrt(5
*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2