3.2536 \(\int \frac{(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\)

Optimal. Leaf size=122 \[ \frac{4 (5 x+3)^{5/2}}{77 \sqrt{1-2 x} (3 x+2)^2}-\frac{25 \sqrt{1-2 x} (5 x+3)^{3/2}}{1078 (3 x+2)^2}-\frac{75 \sqrt{1-2 x} \sqrt{5 x+3}}{1372 (3 x+2)}-\frac{825 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{1372 \sqrt{7}} \]

[Out]

(-75*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(1372*(2 + 3*x)) - (25*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(1078*(2 + 3*x)^2) + (
4*(3 + 5*x)^(5/2))/(77*Sqrt[1 - 2*x]*(2 + 3*x)^2) - (825*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1372*
Sqrt[7])

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Rubi [A]  time = 0.0290412, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {96, 94, 93, 204} \[ \frac{4 (5 x+3)^{5/2}}{77 \sqrt{1-2 x} (3 x+2)^2}-\frac{25 \sqrt{1-2 x} (5 x+3)^{3/2}}{1078 (3 x+2)^2}-\frac{75 \sqrt{1-2 x} \sqrt{5 x+3}}{1372 (3 x+2)}-\frac{825 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{1372 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-75*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(1372*(2 + 3*x)) - (25*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(1078*(2 + 3*x)^2) + (
4*(3 + 5*x)^(5/2))/(77*Sqrt[1 - 2*x]*(2 + 3*x)^2) - (825*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1372*
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 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[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{(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx &=\frac{4 (3+5 x)^{5/2}}{77 \sqrt{1-2 x} (2+3 x)^2}+\frac{25}{77} \int \frac{(3+5 x)^{3/2}}{\sqrt{1-2 x} (2+3 x)^3} \, dx\\ &=-\frac{25 \sqrt{1-2 x} (3+5 x)^{3/2}}{1078 (2+3 x)^2}+\frac{4 (3+5 x)^{5/2}}{77 \sqrt{1-2 x} (2+3 x)^2}+\frac{75}{196} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)^2} \, dx\\ &=-\frac{75 \sqrt{1-2 x} \sqrt{3+5 x}}{1372 (2+3 x)}-\frac{25 \sqrt{1-2 x} (3+5 x)^{3/2}}{1078 (2+3 x)^2}+\frac{4 (3+5 x)^{5/2}}{77 \sqrt{1-2 x} (2+3 x)^2}+\frac{825 \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{2744}\\ &=-\frac{75 \sqrt{1-2 x} \sqrt{3+5 x}}{1372 (2+3 x)}-\frac{25 \sqrt{1-2 x} (3+5 x)^{3/2}}{1078 (2+3 x)^2}+\frac{4 (3+5 x)^{5/2}}{77 \sqrt{1-2 x} (2+3 x)^2}+\frac{825 \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )}{1372}\\ &=-\frac{75 \sqrt{1-2 x} \sqrt{3+5 x}}{1372 (2+3 x)}-\frac{25 \sqrt{1-2 x} (3+5 x)^{3/2}}{1078 (2+3 x)^2}+\frac{4 (3+5 x)^{5/2}}{77 \sqrt{1-2 x} (2+3 x)^2}-\frac{825 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{1372 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0447779, size = 85, normalized size = 0.7 \[ \frac{7 \sqrt{5 x+3} \left (2550 x^2+2245 x+396\right )-825 \sqrt{7-14 x} (3 x+2)^2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{9604 \sqrt{1-2 x} (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[3 + 5*x]*(396 + 2245*x + 2550*x^2) - 825*Sqrt[7 - 14*x]*(2 + 3*x)^2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt
[3 + 5*x])])/(9604*Sqrt[1 - 2*x]*(2 + 3*x)^2)

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Maple [B]  time = 0.013, size = 209, normalized size = 1.7 \begin{align*}{\frac{1}{19208\, \left ( 2+3\,x \right ) ^{2} \left ( 2\,x-1 \right ) } \left ( 14850\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{3}+12375\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-3300\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x-35700\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-3300\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) -31430\,x\sqrt{-10\,{x}^{2}-x+3}-5544\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}\sqrt{3+5\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/19208*(14850*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3+12375*7^(1/2)*arctan(1/14*(37*x+
20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-3300*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x-35700*x
^2*(-10*x^2-x+3)^(1/2)-3300*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-31430*x*(-10*x^2-x+3)^(
1/2)-5544*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)^2/(2*x-1)/(-10*x^2-x+3)^(1/2)

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Maxima [A]  time = 2.75142, size = 193, normalized size = 1.58 \begin{align*} \frac{825}{19208} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) + \frac{2125 \, x}{2058 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{625}{4116 \, \sqrt{-10 \, x^{2} - x + 3}} - \frac{1}{126 \,{\left (9 \, \sqrt{-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt{-10 \, x^{2} - x + 3} x + 4 \, \sqrt{-10 \, x^{2} - x + 3}\right )}} + \frac{235}{1764 \,{\left (3 \, \sqrt{-10 \, x^{2} - x + 3} x + 2 \, \sqrt{-10 \, x^{2} - x + 3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

825/19208*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 2125/2058*x/sqrt(-10*x^2 - x + 3) + 625/
4116/sqrt(-10*x^2 - x + 3) - 1/126/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sqrt(-10*x^2
- x + 3)) + 235/1764/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-10*x^2 - x + 3))

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Fricas [A]  time = 1.78298, size = 296, normalized size = 2.43 \begin{align*} -\frac{825 \, \sqrt{7}{\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\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 (2550 \, x^{2} + 2245 \, x + 396\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{19208 \,{\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/19208*(825*sqrt(7)*(18*x^3 + 15*x^2 - 4*x - 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)
/(10*x^2 + x - 3)) + 14*(2550*x^2 + 2245*x + 396)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(18*x^3 + 15*x^2 - 4*x - 4)

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

[Out]

Exception raised: ValueError

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Giac [B]  time = 2.53175, size = 382, normalized size = 3.13 \begin{align*} \frac{165}{38416} \, \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{44 \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{1715 \,{\left (2 \, x - 1\right )}} - \frac{11 \,{\left (13 \, \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} + 6280 \, \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 )}\right )}}{98 \,{\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 )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

165/38416*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)))) - 44/1715*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x
- 1) - 11/98*(13*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 + 6280*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)^2