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3.2573 \(\int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=84 \[ \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {\sqrt {1-2 x} (38770 x+24439)}{99825 (5 x+3)^{3/2}}-\frac {27 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{25 \sqrt {10}} \]

[Out]

-27/250*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(2+3*x)^2/(3+5*x)^(3/2)/(1-2*x)^(1/2)-1/99825*(24439
+38770*x)*(1-2*x)^(1/2)/(3+5*x)^(3/2)

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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, 145, 54, 216} \[ \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {\sqrt {1-2 x} (38770 x+24439)}{99825 (5 x+3)^{3/2}}-\frac {27 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{25 \sqrt {10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (Sqrt[1 - 2*x]*(24439 + 38770*x))/(99825*(3 + 5*x)^(3/2))
 - (27*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(25*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 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
 e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
 + 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

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)^{3/2} (3+5 x)^{5/2}} \, dx &=\frac {7 (2+3 x)^2}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {1}{11} \int \frac {(2+3 x) \left (19+\frac {99 x}{2}\right )}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx\\ &=\frac {7 (2+3 x)^2}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {\sqrt {1-2 x} (24439+38770 x)}{99825 (3+5 x)^{3/2}}-\frac {27}{50} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {7 (2+3 x)^2}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {\sqrt {1-2 x} (24439+38770 x)}{99825 (3+5 x)^{3/2}}-\frac {27 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{25 \sqrt {5}}\\ &=\frac {7 (2+3 x)^2}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {\sqrt {1-2 x} (24439+38770 x)}{99825 (3+5 x)^{3/2}}-\frac {27 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{25 \sqrt {10}}\\ \end {align*}

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Mathematica [C]  time = 0.91, size = 189, normalized size = 2.25 \[ \frac {343 \left (\frac {160 (2 x-1) (3 x+2)^3 \, _4F_3\left (\frac {1}{2},2,2,\frac {7}{2};1,1,\frac {9}{2};-\frac {5}{11} (2 x-1)\right )}{79233}-\frac {200 (x+3) \left (6 x^2+x-2\right )^2 \, _2F_1\left (\frac {3}{2},\frac {9}{2};\frac {11}{2};-\frac {5}{11} (2 x-1)\right )}{124509}+\frac {\sqrt {10-20 x} \sqrt {5 x+3} \left (43200 x^5+28080 x^4-400032 x^3+1229303 x^2+2053496 x+1669914\right )-27951 \left (108 x^3+513 x^2+1296 x+374\right ) \sin ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{154350 \sqrt {55} (1-2 x)^{5/2}}\right )}{121 \sqrt {22-44 x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(343*((Sqrt[10 - 20*x]*Sqrt[3 + 5*x]*(1669914 + 2053496*x + 1229303*x^2 - 400032*x^3 + 28080*x^4 + 43200*x^5)
- 27951*(374 + 1296*x + 513*x^2 + 108*x^3)*ArcSin[Sqrt[5/11]*Sqrt[1 - 2*x]])/(154350*Sqrt[55]*(1 - 2*x)^(5/2))
 - (200*(3 + x)*(-2 + x + 6*x^2)^2*Hypergeometric2F1[3/2, 9/2, 11/2, (-5*(-1 + 2*x))/11])/124509 + (160*(-1 +
2*x)*(2 + 3*x)^3*HypergeometricPFQ[{1/2, 2, 2, 7/2}, {1, 1, 9/2}, (-5*(-1 + 2*x))/11])/79233))/(121*Sqrt[22 -
44*x])

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fricas [A]  time = 1.52, size = 101, normalized size = 1.20 \[ \frac {107811 \, \sqrt {10} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (649265 \, x^{2} + 772408 \, x + 229661\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1996500 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1996500*(107811*sqrt(10)*(50*x^3 + 35*x^2 - 12*x - 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*
x + 1)/(10*x^2 + x - 3)) - 20*(649265*x^2 + 772408*x + 229661)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2
- 12*x - 9)

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giac [B]  time = 1.40, size = 165, normalized size = 1.96 \[ -\frac {1}{7986000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {2460 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} - \frac {27}{250} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {343 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{6655 \, {\left (2 \, x - 1\right )}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {615 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{499125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7986000*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 2460*(sqrt(2)*sqrt(-10*x + 5) -
sqrt(22))/sqrt(5*x + 3)) - 27/250*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 343/6655*sqrt(5)*sqrt(5*x + 3
)*sqrt(-10*x + 5)/(2*x - 1) + 1/499125*sqrt(10)*(5*x + 3)^(3/2)*(615*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5
*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3

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maple [B]  time = 0.02, size = 134, normalized size = 1.60 \[ -\frac {\sqrt {-2 x +1}\, \left (5390550 \sqrt {10}\, x^{3} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+3773385 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+12985300 \sqrt {-10 x^{2}-x +3}\, x^{2}-1293732 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+15448160 \sqrt {-10 x^{2}-x +3}\, x -970299 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+4593220 \sqrt {-10 x^{2}-x +3}\right )}{1996500 \left (2 x -1\right ) \sqrt {-10 x^{2}-x +3}\, \left (5 x +3\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1996500*(-2*x+1)^(1/2)*(5390550*10^(1/2)*x^3*arcsin(20/11*x+1/11)+3773385*10^(1/2)*x^2*arcsin(20/11*x+1/11)
-1293732*10^(1/2)*x*arcsin(20/11*x+1/11)+12985300*(-10*x^2-x+3)^(1/2)*x^2-970299*10^(1/2)*arcsin(20/11*x+1/11)
+15448160*(-10*x^2-x+3)^(1/2)*x+4593220*(-10*x^2-x+3)^(1/2))/(2*x-1)/(-10*x^2-x+3)^(1/2)/(5*x+3)^(3/2)

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maxima [A]  time = 1.15, size = 78, normalized size = 0.93 \[ -\frac {27}{500} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {129853 \, x}{99825 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {382849}{499125 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {2}{4125 \, {\left (5 \, \sqrt {-10 \, x^{2} - x + 3} x + 3 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/500*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 129853/99825*x/sqrt(-10*x^2 - x + 3) + 382849/499125/sqrt(-10
*x^2 - x + 3) - 2/4125/(5*sqrt(-10*x^2 - x + 3)*x + 3*sqrt(-10*x^2 - x + 3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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