3.24.6 \(\int \frac {\sqrt {1-2 x}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx\) [2306]

Optimal. Leaf size=59 \[ \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x}-\frac {11 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{\sqrt {7}} \]

[Out]

-11/7*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {96, 95, 210} \begin {gather*} \frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 x+2}-\frac {11 \text {ArcTan}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x) - (11*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

Rule 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx &=\frac {\sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x}+\frac {11}{2} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {\sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x}+11 \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=\frac {\sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x}-\frac {11 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{\sqrt {7}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.28, size = 79, normalized size = 1.34 \begin {gather*} \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x}-\frac {11 i \tanh ^{-1}\left (\frac {1}{7} \left (2 \sqrt {70}+3 \sqrt {70} x+3 i \sqrt {7-14 x} \sqrt {3+5 x}\right )\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x) - ((11*I)*ArcTanh[(2*Sqrt[70] + 3*Sqrt[70]*x + (3*I)*Sqrt[7 - 14*x]*Sq
rt[3 + 5*x])/7])/Sqrt[7]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(107\) vs. \(2(48)=96\).
time = 0.11, size = 108, normalized size = 1.83

method	result	size
default	\(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (33 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +22 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+14 \sqrt {-10 x^{2}-x +3}\right )}{14 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )}\)	\(108\)
risch	\(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\left (2+3 x \right ) \sqrt {-\left (3+5 x \right ) \left (-1+2 x \right )}\, \sqrt {1-2 x}}+\frac {11 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{14 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(33*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+22*7^(1/2)*a
rctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+14*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(2+3*x)

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Maxima [A]
time = 0.55, size = 49, normalized size = 0.83 \begin {gather*} \frac {11}{14} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {\sqrt {-10 \, x^{2} - x + 3}}{3 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/14*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + sqrt(-10*x^2 - x + 3)/(3*x + 2)

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Fricas [A]
time = 0.68, size = 71, normalized size = 1.20 \begin {gather*} -\frac {11 \, \sqrt {7} {\left (3 \, 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 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/14*(11*sqrt(7)*(3*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 1
4*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (48) = 96\).
time = 0.91, size = 197, normalized size = 3.34 \begin {gather*} \frac {11}{140} \, \sqrt {5} {\left (\sqrt {70} \sqrt {2} {\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 {280 \, \sqrt {2} {\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 )}}{{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/140*sqrt(5)*(sqrt(70)*sqrt(2)*(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)))) + 280*sqrt(2)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(2
2))/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))

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Mupad [B]
time = 6.80, size = 716, normalized size = 12.14 \begin {gather*} \frac {2\,{\left (\sqrt {1-2\,x}-1\right )}^3}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )}-\frac {11\,\sqrt {7}\,\mathrm {atan}\left (\frac {2904\,\sqrt {3}\,\sqrt {7}}{875\,\left (\frac {2904\,{\left (\sqrt {1-2\,x}-1\right )}^2}{35\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {53724\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {5808}{175}\right )}+\frac {1452\,\sqrt {7}\,\left (\sqrt {1-2\,x}-1\right )}{875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )\,\left (\frac {2904\,{\left (\sqrt {1-2\,x}-1\right )}^2}{35\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {53724\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {5808}{175}\right )}-\frac {1452\,\sqrt {3}\,\sqrt {7}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{175\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2\,\left (\frac {2904\,{\left (\sqrt {1-2\,x}-1\right )}^2}{35\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {53724\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {5808}{175}\right )}\right )}{7}-\frac {4\,\left (\sqrt {1-2\,x}-1\right )}{5\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )}+\frac {37\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*((1 - 2*x)^(1/2) - 1)^3)/((3^(1/2) - (5*x + 3)^(1/2))^3*((14*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x +
3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + (6*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(5
*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + 4/25))
- (11*7^(1/2)*atan((2904*3^(1/2)*7^(1/2))/(875*((2904*((1 - 2*x)^(1/2) - 1)^2)/(35*(3^(1/2) - (5*x + 3)^(1/2)
)^2) + (53724*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(875*(3^(1/2) - (5*x + 3)^(1/2))) - 5808/175)) + (1452*7^(1/2)*((
1 - 2*x)^(1/2) - 1))/(875*(3^(1/2) - (5*x + 3)^(1/2))*((2904*((1 - 2*x)^(1/2) - 1)^2)/(35*(3^(1/2) - (5*x + 3)
^(1/2))^2) + (53724*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(875*(3^(1/2) - (5*x + 3)^(1/2))) - 5808/175)) - (1452*3^(1
/2)*7^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(175*(3^(1/2) - (5*x + 3)^(1/2))^2*((2904*((1 - 2*x)^(1/2) - 1)^2)/(35*(3
^(1/2) - (5*x + 3)^(1/2))^2) + (53724*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(875*(3^(1/2) - (5*x + 3)^(1/2))) - 5808/
175))))/7 - (4*((1 - 2*x)^(1/2) - 1))/(5*(3^(1/2) - (5*x + 3)^(1/2))*((14*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2
) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + (6*3^(1/2)*((1 - 2*x)^(1/2)
- 1)^3)/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))
) + 4/25)) + (37*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2*((14*((1 - 2*x)^(1/2) - 1)
^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + (6*3^(1/2)*((
1 - 2*x)^(1/2) - 1)^3)/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (
5*x + 3)^(1/2))) + 4/25))

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