3.2443 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=115 \[ \frac {(1-2 x)^{5/2}}{2 (3 x+2)^2 \sqrt {5 x+3}}+\frac {55 (1-2 x)^{3/2}}{4 (3 x+2) \sqrt {5 x+3}}-\frac {1815 \sqrt {1-2 x}}{4 \sqrt {5 x+3}}+\frac {1815}{4} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

[Out]

1815/4*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+1/2*(1-2*x)^(5/2)/(2+3*x)^2/(3+5*x)^(1/2)+55/4*
(1-2*x)^(3/2)/(2+3*x)/(3+5*x)^(1/2)-1815/4*(1-2*x)^(1/2)/(3+5*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 115, 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} \[ \frac {(1-2 x)^{5/2}}{2 (3 x+2)^2 \sqrt {5 x+3}}+\frac {55 (1-2 x)^{3/2}}{4 (3 x+2) \sqrt {5 x+3}}-\frac {1815 \sqrt {1-2 x}}{4 \sqrt {5 x+3}}+\frac {1815}{4} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1815*Sqrt[1 - 2*x])/(4*Sqrt[3 + 5*x]) + (1 - 2*x)^(5/2)/(2*(2 + 3*x)^2*Sqrt[3 + 5*x]) + (55*(1 - 2*x)^(3/2))
/(4*(2 + 3*x)*Sqrt[3 + 5*x]) + (1815*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/4

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)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx &=\frac {(1-2 x)^{5/2}}{2 (2+3 x)^2 \sqrt {3+5 x}}+\frac {55}{4} \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{3/2}} \, dx\\ &=\frac {(1-2 x)^{5/2}}{2 (2+3 x)^2 \sqrt {3+5 x}}+\frac {55 (1-2 x)^{3/2}}{4 (2+3 x) \sqrt {3+5 x}}+\frac {1815}{8} \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac {1815 \sqrt {1-2 x}}{4 \sqrt {3+5 x}}+\frac {(1-2 x)^{5/2}}{2 (2+3 x)^2 \sqrt {3+5 x}}+\frac {55 (1-2 x)^{3/2}}{4 (2+3 x) \sqrt {3+5 x}}-\frac {12705}{8} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {1815 \sqrt {1-2 x}}{4 \sqrt {3+5 x}}+\frac {(1-2 x)^{5/2}}{2 (2+3 x)^2 \sqrt {3+5 x}}+\frac {55 (1-2 x)^{3/2}}{4 (2+3 x) \sqrt {3+5 x}}-\frac {12705}{4} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {1815 \sqrt {1-2 x}}{4 \sqrt {3+5 x}}+\frac {(1-2 x)^{5/2}}{2 (2+3 x)^2 \sqrt {3+5 x}}+\frac {55 (1-2 x)^{3/2}}{4 (2+3 x) \sqrt {3+5 x}}+\frac {1815}{4} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 74, normalized size = 0.64 \[ \frac {1815}{4} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-\frac {\sqrt {1-2 x} \left (16657 x^2+21843 x+7148\right )}{4 (3 x+2)^2 \sqrt {5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[1 - 2*x]*(7148 + 21843*x + 16657*x^2))/((2 + 3*x)^2*Sqrt[3 + 5*x]) + (1815*Sqrt[7]*ArcTan[Sqrt[1 -
2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/4

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fricas [A]  time = 0.89, size = 101, normalized size = 0.88 \[ \frac {1815 \, \sqrt {7} {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 2 \, {\left (16657 \, x^{2} + 21843 \, x + 7148\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{8 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(1815*sqrt(7)*(45*x^3 + 87*x^2 + 56*x + 12)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(
10*x^2 + x - 3)) - 2*(16657*x^2 + 21843*x + 7148)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(45*x^3 + 87*x^2 + 56*x + 12)

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giac [B]  time = 2.45, size = 311, normalized size = 2.70 \[ -\frac {363}{16} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {121}{10} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {847 \, \sqrt {10} {\left (9 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {1960 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {7840 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{2 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-363/16*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)))) - 121/10*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))) - 847/2*sqrt(10)*(9*((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 + 1960*(sqrt(2)*sqr
t(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 7840*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*s
qrt(-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

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maple [B]  time = 0.02, size = 202, normalized size = 1.76 \[ -\frac {\left (81675 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+157905 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+33314 \sqrt {-10 x^{2}-x +3}\, x^{2}+101640 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+43686 \sqrt {-10 x^{2}-x +3}\, x +21780 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+14296 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{8 \left (3 x +2\right )^{2} \sqrt {-10 x^{2}-x +3}\, \sqrt {5 x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(81675*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+157905*7^(1/2)*x^2*arctan(1/14*(37*
x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+101640*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+33314*(
-10*x^2-x+3)^(1/2)*x^2+21780*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+43686*(-10*x^2-x+3)^(1
/2)*x+14296*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(3*x+2)^2/(-10*x^2-x+3)^(1/2)/(5*x+3)^(1/2)

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maxima [A]  time = 1.43, size = 143, normalized size = 1.24 \[ -\frac {1815}{8} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {16657 \, x}{18 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {52169}{108 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {343}{54 \, {\left (9 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt {-10 \, x^{2} - x + 3} x + 4 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} + \frac {833}{12 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1815/8*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 16657/18*x/sqrt(-10*x^2 - x + 3) - 52169/1
08/sqrt(-10*x^2 - x + 3) + 343/54/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sqrt(-10*x^2 -
x + 3)) + 833/12/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-10*x^2 - x + 3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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