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

Optimal. Leaf size=84 \[ \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}-\frac {111311 x+66967}{39930 \sqrt {1-2 x} \sqrt {5 x+3}}+\frac {27 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{10 \sqrt {10}} \]

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Rubi [A]  time = 0.02, antiderivative size = 84, 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 = {98, 144, 54, 216} \begin {gather*} \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}-\frac {111311 x+66967}{39930 \sqrt {1-2 x} \sqrt {5 x+3}}+\frac {27 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{10 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2)/(33*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) - (66967 + 111311*x)/(39930*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) +
(27*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(10*Sqrt[10])

Rule 54

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 98

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

Rule 144

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :>
Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(
m + n + 2)) + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m +
1) + d^2*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b*d*(b*c - a*d)^2*(m + 1)*(n + 1)), x] -
Dist[(a^2*d^2*f*h*(2 + 3*n + n^2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h
*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b
*c - a*d)^2*(m + 1)*(n + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, h
}, x] && LtQ[m, -1] && LtQ[n, -1]

Rule 216

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)^3}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {1}{33} \int \frac {(2+3 x) \left (85+\frac {297 x}{2}\right )}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {66967+111311 x}{39930 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {27}{20} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {66967+111311 x}{39930 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {27 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{10 \sqrt {5}}\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {66967+111311 x}{39930 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {27 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{10 \sqrt {10}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 143, normalized size = 1.70 \begin {gather*} \frac {\sqrt {10-20 x} \sqrt {5 x+3} \left (21600 x^5-43740 x^4+79209 x^3+272474 x^2+678368 x+129582\right )-3993 \left (513 x^3+2538 x^2+936 x+334\right ) \sin ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{79860 \sqrt {10} (1-2 x)^3}-\frac {250 \sqrt {\frac {2}{11}} (1-2 x)^{3/2} (3 x+2)^3 \, _2F_1\left (\frac {3}{2},\frac {9}{2};\frac {11}{2};-\frac {5}{11} (2 x-1)\right )}{131769} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[10 - 20*x]*Sqrt[3 + 5*x]*(129582 + 678368*x + 272474*x^2 + 79209*x^3 - 43740*x^4 + 21600*x^5) - 3993*(33
4 + 936*x + 2538*x^2 + 513*x^3)*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(79860*Sqrt[10]*(1 - 2*x)^3) - (250*Sqrt[2/1
1]*(1 - 2*x)^(3/2)*(2 + 3*x)^3*Hypergeometric2F1[3/2, 9/2, 11/2, (-5*(-1 + 2*x))/11])/131769

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IntegrateAlgebraic [A]  time = 0.12, size = 91, normalized size = 1.08 \begin {gather*} \frac {(5 x+3)^{3/2} \left (-\frac {12 (1-2 x)^2}{(5 x+3)^2}-\frac {21315 (1-2 x)}{5 x+3}+3430\right )}{39930 (1-2 x)^{3/2}}-\frac {27 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{10 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3 + 5*x)^(3/2)*(3430 - (12*(1 - 2*x)^2)/(3 + 5*x)^2 - (21315*(1 - 2*x))/(3 + 5*x)))/(39930*(1 - 2*x)^(3/2))
- (27*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/(10*Sqrt[10])

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fricas [A]  time = 1.52, size = 101, normalized size = 1.20 \begin {gather*} -\frac {107811 \, \sqrt {10} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, 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 (298852 \, x^{2} + 124263 \, x - 33087\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{798600 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/798600*(107811*sqrt(10)*(20*x^3 - 8*x^2 - 7*x + 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x
+ 1)/(10*x^2 + x - 3)) - 20*(298852*x^2 + 124263*x - 33087)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*
x + 3)

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giac [A]  time = 1.36, size = 118, normalized size = 1.40 \begin {gather*} \frac {27}{100} \, \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 )}}{66550 \, \sqrt {5 \, x + 3}} + \frac {49 \, {\left (244 \, \sqrt {5} {\left (5 \, x + 3\right )} - 957 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{199650 \, {\left (2 \, x - 1\right )}^{2}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{33275 \, {\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)^3/(1-2*x)^(5/2)/(3+5*x)^(3/2),x, algorithm="giac")

[Out]

27/100*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/66550*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sq
rt(5*x + 3) + 49/199650*(244*sqrt(5)*(5*x + 3) - 957*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2 + 2/33
275*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))

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maple [B]  time = 0.02, size = 134, normalized size = 1.60 \begin {gather*} \frac {\sqrt {-2 x +1}\, \left (2156220 \sqrt {10}\, x^{3} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-862488 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+5977040 \sqrt {-10 x^{2}-x +3}\, x^{2}-754677 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+2485260 \sqrt {-10 x^{2}-x +3}\, x +323433 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-661740 \sqrt {-10 x^{2}-x +3}\right )}{798600 \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}\, \sqrt {5 x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/798600*(-2*x+1)^(1/2)*(2156220*10^(1/2)*x^3*arcsin(20/11*x+1/11)-862488*10^(1/2)*x^2*arcsin(20/11*x+1/11)-75
4677*10^(1/2)*x*arcsin(20/11*x+1/11)+5977040*(-10*x^2-x+3)^(1/2)*x^2+323433*10^(1/2)*arcsin(20/11*x+1/11)+2485
260*(-10*x^2-x+3)^(1/2)*x-661740*(-10*x^2-x+3)^(1/2))/(2*x-1)^2/(-10*x^2-x+3)^(1/2)/(5*x+3)^(1/2)

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maxima [A]  time = 1.38, size = 78, normalized size = 0.93 \begin {gather*} \frac {27}{200} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {74713 \, x}{19965 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {273689}{79860 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {343}{132 \, {\left (2 \, \sqrt {-10 \, x^{2} - x + 3} x - \sqrt {-10 \, x^{2} - x + 3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/200*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 74713/19965*x/sqrt(-10*x^2 - x + 3) - 273689/79860/sqrt(-10*x^
2 - x + 3) - 343/132/(2*sqrt(-10*x^2 - x + 3)*x - sqrt(-10*x^2 - x + 3))

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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