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]

2/21*(3+5*x)^(5/2)/(1-2*x)^(3/2)/(2+3*x)-55/2401*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-10/14
7*(3+5*x)^(3/2)/(2+3*x)/(1-2*x)^(1/2)-5/343*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.03, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [A]  time = 0.05, size = 86, normalized size = 0.70 \[ -\frac {-7 \sqrt {5 x+3} \left (3090 x^2+3070 x+657\right )-165 \sqrt {7-14 x} \left (6 x^2+x-2\right ) \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7203 (1-2 x)^{3/2} (3 x+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7203*(-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])])/((1 - 2*x)^(3/2)*(2 + 3*x))

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fricas [A]  time = 1.23, size = 101, normalized size = 0.83 \[ -\frac {165 \, \sqrt {7} {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (3090 \, x^{2} + 3070 \, x + 657\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14406 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \]

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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giac [B]  time = 1.91, size = 232, normalized size = 1.90 \[ \frac {11}{9604} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {22 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{343 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} + \frac {22 \, {\left (47 \, \sqrt {5} {\left (5 \, x + 3\right )} - 66 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{25725 \, {\left (2 \, x - 1\right )}^{2}} \]

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

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maple [B]  time = 0.02, size = 209, normalized size = 1.71 \[ \frac {\left (1980 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-660 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+43260 \sqrt {-10 x^{2}-x +3}\, x^{2}-825 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+42980 \sqrt {-10 x^{2}-x +3}\, x +330 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+9198 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{14406 \left (3 x +2\right ) \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.14, size = 138, normalized size = 1.13 \[ \frac {55}{4802} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {2575 \, x}{1029 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {625 \, x^{2}}{18 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {135}{1372 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {138125 \, x}{5292 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {1}{567 \, {\left (3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 2 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} + \frac {50315}{15876 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (5\,x+3\right )}^{5/2}}{{\left (1-2\,x\right )}^{5/2}\,{\left (3\,x+2\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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