3.2604 \(\int \frac{(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx\)

Optimal. Leaf size=122 \[ \frac{2 (5 x+3)^{5/2}}{21 (1-2 x)^{3/2} (3 x+2)}-\frac{10 (5 x+3)^{3/2}}{147 \sqrt{1-2 x} (3 x+2)}-\frac{5 \sqrt{1-2 x} \sqrt{5 x+3}}{343 (3 x+2)}-\frac{55 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{343 \sqrt{7}} \]

[Out]

(-5*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(343*(2 + 3*x)) - (10*(3 + 5*x)^(3/2))/(147*Sqrt[1 - 2*x]*(2 + 3*x)) + (2*(3
+ 5*x)^(5/2))/(21*(1 - 2*x)^(3/2)*(2 + 3*x)) - (55*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(343*Sqrt[7]
)

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Rubi [A]  time = 0.0285955, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {94, 93, 204} \[ \frac{2 (5 x+3)^{5/2}}{21 (1-2 x)^{3/2} (3 x+2)}-\frac{10 (5 x+3)^{3/2}}{147 \sqrt{1-2 x} (3 x+2)}-\frac{5 \sqrt{1-2 x} \sqrt{5 x+3}}{343 (3 x+2)}-\frac{55 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{343 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(343*(2 + 3*x)) - (10*(3 + 5*x)^(3/2))/(147*Sqrt[1 - 2*x]*(2 + 3*x)) + (2*(3
+ 5*x)^(5/2))/(21*(1 - 2*x)^(3/2)*(2 + 3*x)) - (55*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(343*Sqrt[7]
)

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)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx &=\frac{2 (3+5 x)^{5/2}}{21 (1-2 x)^{3/2} (2+3 x)}-\frac{5}{21} \int \frac{(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)^2} \, dx\\ &=-\frac{10 (3+5 x)^{3/2}}{147 \sqrt{1-2 x} (2+3 x)}+\frac{2 (3+5 x)^{5/2}}{21 (1-2 x)^{3/2} (2+3 x)}+\frac{5}{49} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)^2} \, dx\\ &=-\frac{5 \sqrt{1-2 x} \sqrt{3+5 x}}{343 (2+3 x)}-\frac{10 (3+5 x)^{3/2}}{147 \sqrt{1-2 x} (2+3 x)}+\frac{2 (3+5 x)^{5/2}}{21 (1-2 x)^{3/2} (2+3 x)}+\frac{55}{686} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=-\frac{5 \sqrt{1-2 x} \sqrt{3+5 x}}{343 (2+3 x)}-\frac{10 (3+5 x)^{3/2}}{147 \sqrt{1-2 x} (2+3 x)}+\frac{2 (3+5 x)^{5/2}}{21 (1-2 x)^{3/2} (2+3 x)}+\frac{55}{343} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )\\ &=-\frac{5 \sqrt{1-2 x} \sqrt{3+5 x}}{343 (2+3 x)}-\frac{10 (3+5 x)^{3/2}}{147 \sqrt{1-2 x} (2+3 x)}+\frac{2 (3+5 x)^{5/2}}{21 (1-2 x)^{3/2} (2+3 x)}-\frac{55 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{343 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0461044, size = 86, normalized size = 0.7 \[ -\frac{-7 \sqrt{5 x+3} \left (3090 x^2+3070 x+657\right )-165 \sqrt{7-14 x} \left (6 x^2+x-2\right ) \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{7203 (1-2 x)^{3/2} (3 x+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-7*Sqrt[3 + 5*x]*(657 + 3070*x + 3090*x^2) - 165*Sqrt[7 - 14*x]*(-2 + x + 6*x^2)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[
7]*Sqrt[3 + 5*x])])/(7203*(1 - 2*x)^(3/2)*(2 + 3*x))

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Maple [B]  time = 0.013, size = 209, normalized size = 1.7 \begin{align*}{\frac{1}{ \left ( 28812+43218\,x \right ) \left ( 2\,x-1 \right ) ^{2}} \left ( 1980\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{3}-660\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-825\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+43260\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+330\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +42980\,x\sqrt{-10\,{x}^{2}-x+3}+9198\,\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)^(5/2)/(1-2*x)^(5/2)/(2+3*x)^2,x)

[Out]

1/14406*(1980*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3-660*7^(1/2)*arctan(1/14*(37*x+20)
*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-825*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+43260*x^2*(
-10*x^2-x+3)^(1/2)+330*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+42980*x*(-10*x^2-x+3)^(1/2)+
9198*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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Maxima [A]  time = 4.10085, size = 186, normalized size = 1.52 \begin{align*} \frac{55}{4802} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) - \frac{2575 \, x}{1029 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{625 \, x^{2}}{18 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} - \frac{135}{1372 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{138125 \, x}{5292 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} - \frac{1}{567 \,{\left (3 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}} x + 2 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}\right )}} + \frac{50315}{15876 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

55/4802*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 2575/1029*x/sqrt(-10*x^2 - x + 3) + 625/18
*x^2/(-10*x^2 - x + 3)^(3/2) - 135/1372/sqrt(-10*x^2 - x + 3) + 138125/5292*x/(-10*x^2 - x + 3)^(3/2) - 1/567/
(3*(-10*x^2 - x + 3)^(3/2)*x + 2*(-10*x^2 - x + 3)^(3/2)) + 50315/15876/(-10*x^2 - x + 3)^(3/2)

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Fricas [A]  time = 1.65283, size = 293, normalized size = 2.4 \begin{align*} -\frac{165 \, \sqrt{7}{\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\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 (3090 \, x^{2} + 3070 \, x + 657\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{14406 \,{\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/14406*(165*sqrt(7)*(12*x^3 - 4*x^2 - 5*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/
(10*x^2 + x - 3)) - 14*(3090*x^2 + 3070*x + 657)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(12*x^3 - 4*x^2 - 5*x + 2)

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

[Out]

Exception raised: ValueError

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Giac [B]  time = 2.92707, size = 313, normalized size = 2.57 \begin{align*} \frac{11}{9604} \, \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{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 )}}{343 \,{\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 )}} + \frac{22 \,{\left (47 \, \sqrt{5}{\left (5 \, x + 3\right )} - 66 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{25725 \,{\left (2 \, x - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/9604*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)))) - 22/343*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) + 22/25725*(47*sqrt(5)*(5*x + 3
) - 66*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2