3.2546 \(\int \frac{(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\)

Optimal. Leaf size=122 \[ \frac{2 (5 x+3)^{5/2}}{7 \sqrt{1-2 x} (3 x+2)^2}+\frac{5 \sqrt{1-2 x} (5 x+3)^{3/2}}{98 (3 x+2)^2}+\frac{165 \sqrt{1-2 x} \sqrt{5 x+3}}{1372 (3 x+2)}+\frac{1815 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{1372 \sqrt{7}} \]

[Out]

(165*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(1372*(2 + 3*x)) + (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(98*(2 + 3*x)^2) + (2*(
3 + 5*x)^(5/2))/(7*Sqrt[1 - 2*x]*(2 + 3*x)^2) + (1815*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1372*Sqr
t[7])

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Rubi [A]  time = 0.0294422, 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} \[ \frac{2 (5 x+3)^{5/2}}{7 \sqrt{1-2 x} (3 x+2)^2}+\frac{5 \sqrt{1-2 x} (5 x+3)^{3/2}}{98 (3 x+2)^2}+\frac{165 \sqrt{1-2 x} \sqrt{5 x+3}}{1372 (3 x+2)}+\frac{1815 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{1372 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(165*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(1372*(2 + 3*x)) + (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(98*(2 + 3*x)^2) + (2*(
3 + 5*x)^(5/2))/(7*Sqrt[1 - 2*x]*(2 + 3*x)^2) + (1815*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1372*Sqr
t[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)^{3/2} (2+3 x)^3} \, dx &=\frac{2 (3+5 x)^{5/2}}{7 \sqrt{1-2 x} (2+3 x)^2}-\frac{5}{7} \int \frac{(3+5 x)^{3/2}}{\sqrt{1-2 x} (2+3 x)^3} \, dx\\ &=\frac{5 \sqrt{1-2 x} (3+5 x)^{3/2}}{98 (2+3 x)^2}+\frac{2 (3+5 x)^{5/2}}{7 \sqrt{1-2 x} (2+3 x)^2}-\frac{165}{196} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)^2} \, dx\\ &=\frac{165 \sqrt{1-2 x} \sqrt{3+5 x}}{1372 (2+3 x)}+\frac{5 \sqrt{1-2 x} (3+5 x)^{3/2}}{98 (2+3 x)^2}+\frac{2 (3+5 x)^{5/2}}{7 \sqrt{1-2 x} (2+3 x)^2}-\frac{1815 \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{2744}\\ &=\frac{165 \sqrt{1-2 x} \sqrt{3+5 x}}{1372 (2+3 x)}+\frac{5 \sqrt{1-2 x} (3+5 x)^{3/2}}{98 (2+3 x)^2}+\frac{2 (3+5 x)^{5/2}}{7 \sqrt{1-2 x} (2+3 x)^2}-\frac{1815 \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )}{1372}\\ &=\frac{165 \sqrt{1-2 x} \sqrt{3+5 x}}{1372 (2+3 x)}+\frac{5 \sqrt{1-2 x} (3+5 x)^{3/2}}{98 (2+3 x)^2}+\frac{2 (3+5 x)^{5/2}}{7 \sqrt{1-2 x} (2+3 x)^2}+\frac{1815 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{1372 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0418913, size = 85, normalized size = 0.7 \[ \frac{7 \sqrt{5 x+3} \left (8110 x^2+11525 x+4068\right )+1815 \sqrt{7-14 x} (3 x+2)^2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{9604 \sqrt{1-2 x} (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[3 + 5*x]*(4068 + 11525*x + 8110*x^2) + 1815*Sqrt[7 - 14*x]*(2 + 3*x)^2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*S
qrt[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 ( 32670\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{3}+27225\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-7260\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+113540\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-7260\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +161350\,x\sqrt{-10\,{x}^{2}-x+3}+56952\,\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)^(3/2)/(2+3*x)^3,x)

[Out]

-1/19208*(32670*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3+27225*7^(1/2)*arctan(1/14*(37*x
+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-7260*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+113540
*x^2*(-10*x^2-x+3)^(1/2)-7260*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+161350*x*(-10*x^2-x+3
)^(1/2)+56952*(-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 = 3.77205, size = 193, normalized size = 1.58 \begin{align*} -\frac{1815}{19208} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) + \frac{20275 \, x}{6174 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{83665}{37044 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{1}{378 \,{\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{125}{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)^(5/2)/(1-2*x)^(3/2)/(2+3*x)^3,x, algorithm="maxima")

[Out]

-1815/19208*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 20275/6174*x/sqrt(-10*x^2 - x + 3) + 8
3665/37044/sqrt(-10*x^2 - x + 3) + 1/378/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sqrt(-1
0*x^2 - x + 3)) - 125/1764/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-10*x^2 - x + 3))

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Fricas [A]  time = 1.74149, size = 298, normalized size = 2.44 \begin{align*} \frac{1815 \, \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 (8110 \, x^{2} + 11525 \, x + 4068\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)^(5/2)/(1-2*x)^(3/2)/(2+3*x)^3,x, algorithm="fricas")

[Out]

1/19208*(1815*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*(8110*x^2 + 11525*x + 4068)*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)**(5/2)/(1-2*x)**(3/2)/(2+3*x)**3,x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 3.39227, size = 381, normalized size = 3.12 \begin{align*} -\frac{363}{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{242 \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{1715 \,{\left (2 \, x - 1\right )}} + \frac{121 \,{\left (\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} + 360 \, \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)^(5/2)/(1-2*x)^(3/2)/(2+3*x)^3,x, algorithm="giac")

[Out]

-363/38416*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)))) - 242/1715*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*
x - 1) + 121/98*(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 + 360*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