3.26.64 \(\int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx\) [2564]

Optimal. Leaf size=72 \[ \frac {49}{22 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {1229 \sqrt {1-2 x}}{1210 \sqrt {3+5 x}}-\frac {9 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{5 \sqrt {10}} \]

[Out]

-9/50*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+49/22/(1-2*x)^(1/2)/(3+5*x)^(1/2)-1229/1210*(1-2*x)^(1/2)/(
3+5*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {91, 79, 56, 222} \begin {gather*} -\frac {9 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}-\frac {1229 \sqrt {1-2 x}}{1210 \sqrt {5 x+3}}+\frac {49}{22 \sqrt {1-2 x} \sqrt {5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(22*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) - (1229*Sqrt[1 - 2*x])/(1210*Sqrt[3 + 5*x]) - (9*ArcSin[Sqrt[2/11]*Sqrt[3
+ 5*x]])/(5*Sqrt[10])

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin {align*} \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx &=\frac {49}{22 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {1}{22} \int \frac {-\frac {127}{2}+99 x}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx\\ &=\frac {49}{22 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {1229 \sqrt {1-2 x}}{1210 \sqrt {3+5 x}}-\frac {9}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {49}{22 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {1229 \sqrt {1-2 x}}{1210 \sqrt {3+5 x}}-\frac {9 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{5 \sqrt {5}}\\ &=\frac {49}{22 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {1229 \sqrt {1-2 x}}{1210 \sqrt {3+5 x}}-\frac {9 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{5 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 59, normalized size = 0.82 \begin {gather*} \frac {733+1229 x}{605 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {9 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{5 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(733 + 1229*x)/(605*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (9*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(5*Sqrt[10])

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Maple [A]
time = 0.09, size = 103, normalized size = 1.43

method	result	size
default	\(-\frac {\sqrt {1-2 x}\, \left (10890 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+1089 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -3267 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+24580 x \sqrt {-10 x^{2}-x +3}+14660 \sqrt {-10 x^{2}-x +3}\right )}{12100 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12100*(1-2*x)^(1/2)*(10890*10^(1/2)*arcsin(20/11*x+1/11)*x^2+1089*10^(1/2)*arcsin(20/11*x+1/11)*x-3267*10^(
1/2)*arcsin(20/11*x+1/11)+24580*x*(-10*x^2-x+3)^(1/2)+14660*(-10*x^2-x+3)^(1/2))/(-1+2*x)/(-10*x^2-x+3)^(1/2)/
(3+5*x)^(1/2)

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Maxima [A]
time = 0.69, size = 41, normalized size = 0.57 \begin {gather*} \frac {9}{100} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {1229 \, x}{605 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {733}{605 \, \sqrt {-10 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/100*sqrt(10)*arcsin(-20/11*x - 1/11) + 1229/605*x/sqrt(-10*x^2 - x + 3) + 733/605/sqrt(-10*x^2 - x + 3)

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Fricas [A]
time = 0.49, size = 82, normalized size = 1.14 \begin {gather*} \frac {1089 \, \sqrt {10} {\left (10 \, x^{2} + x - 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (1229 \, x + 733\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{12100 \, {\left (10 \, x^{2} + x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12100*(1089*sqrt(10)*(10*x^2 + x - 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 +
x - 3)) - 20*(1229*x + 733)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(10*x^2 + x - 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (51) = 102\).
time = 0.69, size = 105, normalized size = 1.46 \begin {gather*} -\frac {9}{50} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{6050 \, \sqrt {5 \, x + 3}} - \frac {49 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{605 \, {\left (2 \, x - 1\right )}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{3025 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/50*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/6050*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt
(5*x + 3) - 49/605*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) + 2/3025*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sq
rt(-10*x + 5) - sqrt(22))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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