3.2367 \(\int \frac {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx\)

Optimal. Leaf size=93 \[ \frac {\sqrt {5 x+3} (1-2 x)^{3/2}}{2 (3 x+2)^2}+\frac {33 \sqrt {5 x+3} \sqrt {1-2 x}}{4 (3 x+2)}-\frac {363 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{4 \sqrt {7}} \]

[Out]

-363/28*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+1/2*(1-2*x)^(3/2)*(3+5*x)^(1/2)/(2+3*x)^2+33/4
*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.02, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {94, 93, 204} \[ \frac {\sqrt {5 x+3} (1-2 x)^{3/2}}{2 (3 x+2)^2}+\frac {33 \sqrt {5 x+3} \sqrt {1-2 x}}{4 (3 x+2)}-\frac {363 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{4 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/(2*(2 + 3*x)^2) + (33*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(4*(2 + 3*x)) - (363*ArcTan
[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(4*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 {(1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx &=\frac {(1-2 x)^{3/2} \sqrt {3+5 x}}{2 (2+3 x)^2}+\frac {33}{4} \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx\\ &=\frac {(1-2 x)^{3/2} \sqrt {3+5 x}}{2 (2+3 x)^2}+\frac {33 \sqrt {1-2 x} \sqrt {3+5 x}}{4 (2+3 x)}+\frac {363}{8} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {(1-2 x)^{3/2} \sqrt {3+5 x}}{2 (2+3 x)^2}+\frac {33 \sqrt {1-2 x} \sqrt {3+5 x}}{4 (2+3 x)}+\frac {363}{4} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=\frac {(1-2 x)^{3/2} \sqrt {3+5 x}}{2 (2+3 x)^2}+\frac {33 \sqrt {1-2 x} \sqrt {3+5 x}}{4 (2+3 x)}-\frac {363 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{4 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.74 \[ \frac {\sqrt {1-2 x} \sqrt {5 x+3} (95 x+68)}{4 (3 x+2)^2}-\frac {363 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{4 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(68 + 95*x))/(4*(2 + 3*x)^2) - (363*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])]
)/(4*Sqrt[7])

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fricas [A]  time = 0.81, size = 86, normalized size = 0.92 \[ -\frac {363 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (95 \, x + 68\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{56 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/56*(363*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x
- 3)) - 14*(95*x + 68)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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giac [B]  time = 1.54, size = 250, normalized size = 2.69 \[ \frac {363}{560} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {605 \, \sqrt {10} {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {168 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {672 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{2 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

363/560*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)))) + 605/2*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 + 168*(sqrt(2)*sqrt(-10*x + 5) - sqrt
(22))/sqrt(5*x + 3) - 672*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqr
t(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2

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maple [B]  time = 0.01, size = 154, normalized size = 1.66 \[ \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (3267 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+4356 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+1330 \sqrt {-10 x^{2}-x +3}\, x +1452 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+952 \sqrt {-10 x^{2}-x +3}\right )}{56 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(3267*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+4356*7^
(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+1452*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2
-x+3)^(1/2))+1330*(-10*x^2-x+3)^(1/2)*x+952*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)^2

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maxima [A]  time = 1.32, size = 76, normalized size = 0.82 \[ \frac {363}{56} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {7 \, \sqrt {-10 \, x^{2} - x + 3}}{6 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {95 \, \sqrt {-10 \, x^{2} - x + 3}}{12 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

363/56*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 7/6*sqrt(-10*x^2 - x + 3)/(9*x^2 + 12*x + 4
) + 95/12*sqrt(-10*x^2 - x + 3)/(3*x + 2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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