3.24.19 \(\int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx\) [2319]

Optimal. Leaf size=113 \[ -\frac {2 \sqrt {1-2 x} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac {392 \sqrt {1-2 x} (2+3 x)^2}{825 \sqrt {3+5 x}}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (1243+1740 x)}{11000}+\frac {1071 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1000 \sqrt {10}} \]

[Out]

1071/10000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/15*(2+3*x)^3*(1-2*x)^(1/2)/(3+5*x)^(3/2)-392/825*(2+
3*x)^2*(1-2*x)^(1/2)/(3+5*x)^(1/2)+7/11000*(1243+1740*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {99, 155, 152, 56, 222} \begin {gather*} \frac {1071 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{1000 \sqrt {10}}-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}-\frac {392 \sqrt {1-2 x} (3 x+2)^2}{825 \sqrt {5 x+3}}+\frac {7 \sqrt {1-2 x} \sqrt {5 x+3} (1740 x+1243)}{11000} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(15*(3 + 5*x)^(3/2)) - (392*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(825*Sqrt[3 + 5*x]) + (7
*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(1243 + 1740*x))/11000 + (1071*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(1000*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 99

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/(b*(m + 1))), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 152

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

Rule 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

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 {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{15 (3+5 x)^{3/2}}+\frac {2}{15} \int \frac {(7-21 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac {392 \sqrt {1-2 x} (2+3 x)^2}{825 \sqrt {3+5 x}}+\frac {4}{825} \int \frac {\left (357-\frac {3045 x}{2}\right ) (2+3 x)}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac {392 \sqrt {1-2 x} (2+3 x)^2}{825 \sqrt {3+5 x}}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (1243+1740 x)}{11000}+\frac {1071 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{2000}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac {392 \sqrt {1-2 x} (2+3 x)^2}{825 \sqrt {3+5 x}}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (1243+1740 x)}{11000}+\frac {1071 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{1000 \sqrt {5}}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac {392 \sqrt {1-2 x} (2+3 x)^2}{825 \sqrt {3+5 x}}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (1243+1740 x)}{11000}+\frac {1071 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1000 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 69, normalized size = 0.61 \begin {gather*} \frac {\frac {10 \sqrt {1-2 x} \left (11567+75470 x+147015 x^2+89100 x^3\right )}{(3+5 x)^{3/2}}-35343 \sqrt {10} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{330000} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((10*Sqrt[1 - 2*x]*(11567 + 75470*x + 147015*x^2 + 89100*x^3))/(3 + 5*x)^(3/2) - 35343*Sqrt[10]*ArcTan[Sqrt[5/
2 - 5*x]/Sqrt[3 + 5*x]])/330000

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Maple [A]
time = 0.11, size = 130, normalized size = 1.15

method	result	size
default	\(\frac {\left (883575 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+1782000 x^{3} \sqrt {-10 x^{2}-x +3}+1060290 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +2940300 x^{2} \sqrt {-10 x^{2}-x +3}+318087 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1509400 x \sqrt {-10 x^{2}-x +3}+231340 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{660000 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/660000*(883575*10^(1/2)*arcsin(20/11*x+1/11)*x^2+1782000*x^3*(-10*x^2-x+3)^(1/2)+1060290*10^(1/2)*arcsin(20/
11*x+1/11)*x+2940300*x^2*(-10*x^2-x+3)^(1/2)+318087*10^(1/2)*arcsin(20/11*x+1/11)+1509400*x*(-10*x^2-x+3)^(1/2
)+231340*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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Maxima [F(-1)] Timed out
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)^(1/2)/(3+5*x)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 1.13, size = 96, normalized size = 0.85 \begin {gather*} -\frac {35343 \, \sqrt {10} {\left (25 \, x^{2} + 30 \, 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 (89100 \, x^{3} + 147015 \, x^{2} + 75470 \, x + 11567\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{660000 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/660000*(35343*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10
*x^2 + x - 3)) - 20*(89100*x^3 + 147015*x^2 + 75470*x + 11567)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x +
9)

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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 )^{3}}{\left (5 x + 3\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.72, size = 171, normalized size = 1.51 \begin {gather*} \frac {27}{25000} \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} - 3 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {1}{1650000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {2364 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} + \frac {1071}{10000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {591 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{103125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/25000*(4*sqrt(5)*(5*x + 3) - 3*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 1/1650000*sqrt(10)*((sqrt(2)*sqrt(-
10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 2364*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3)) + 1071/1000
0*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/103125*sqrt(10)*(5*x + 3)^(3/2)*(591*(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^3}{{\left (5\,x+3\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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