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

Optimal. Leaf size=32 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}} \]

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Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {93, 204} \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 32, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.47, size = 40, normalized size = 1.25 \begin {gather*} -\frac {1}{7} \, \sqrt {7} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*sqrt(7)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))

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giac [B]  time = 1.05, size = 73, normalized size = 2.28 \begin {gather*} \frac {1}{70} \, \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/70*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))))

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maple [B]  time = 0.01, size = 55, normalized size = 1.72 \begin {gather*} \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )}{7 \sqrt {-10 x^{2}-x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(-10*x^2-x+3)^(1/2)*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2)
)

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maxima [A]  time = 1.11, size = 28, normalized size = 0.88 \begin {gather*} \frac {1}{7} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2))

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mupad [B]  time = 6.27, size = 145, normalized size = 4.53 \begin {gather*} \frac {2\,\sqrt {7}\,\mathrm {atan}\left (\frac {20\,\sqrt {21}\,x-\sqrt {3}\,\sqrt {7}+\sqrt {7}\,\sqrt {5\,x+3}+10\,\sqrt {21}\,\sqrt {1-2\,x}+2\,\sqrt {21}+\sqrt {3}\,\sqrt {7}\,\sqrt {1-2\,x}-4\,\sqrt {3}\,\sqrt {21}\,\sqrt {5\,x+3}-\sqrt {7}\,\sqrt {1-2\,x}\,\sqrt {5\,x+3}}{200\,x-77\,\sqrt {3}\,\sqrt {5\,x+3}-11\,\sqrt {1-2\,x}+37\,\sqrt {3}\,\sqrt {1-2\,x}\,\sqrt {5\,x+3}+131}\right )}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*7^(1/2)*atan((20*21^(1/2)*x - 3^(1/2)*7^(1/2) + 7^(1/2)*(5*x + 3)^(1/2) + 10*21^(1/2)*(1 - 2*x)^(1/2) + 2*2
1^(1/2) + 3^(1/2)*7^(1/2)*(1 - 2*x)^(1/2) - 4*3^(1/2)*21^(1/2)*(5*x + 3)^(1/2) - 7^(1/2)*(1 - 2*x)^(1/2)*(5*x
+ 3)^(1/2))/(200*x - 77*3^(1/2)*(5*x + 3)^(1/2) - 11*(1 - 2*x)^(1/2) + 37*3^(1/2)*(1 - 2*x)^(1/2)*(5*x + 3)^(1
/2) + 131)))/7

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {1 - 2 x} \left (3 x + 2\right ) \sqrt {5 x + 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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