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

Optimal. Leaf size=94 \[ -\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}-\frac {1}{5} \sqrt {5 x+3} (1-2 x)^{3/2}-\frac {33}{50} \sqrt {5 x+3} \sqrt {1-2 x}-\frac {363 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{50 \sqrt {10}} \]

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Rubi [A]  time = 0.02, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {47, 50, 54, 216} \begin {gather*} -\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}-\frac {1}{5} \sqrt {5 x+3} (1-2 x)^{3/2}-\frac {33}{50} \sqrt {5 x+3} \sqrt {1-2 x}-\frac {363 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{50 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 - 2*x)^(5/2))/(5*Sqrt[3 + 5*x]) - (33*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/50 - ((1 - 2*x)^(3/2)*Sqrt[3 + 5*x])
/5 - (363*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(50*Sqrt[10])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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 {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx &=-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {3+5 x}}-2 \int \frac {(1-2 x)^{3/2}}{\sqrt {3+5 x}} \, dx\\ &=-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {3+5 x}}-\frac {1}{5} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {33}{10} \int \frac {\sqrt {1-2 x}}{\sqrt {3+5 x}} \, dx\\ &=-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {3+5 x}}-\frac {33}{50} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{5} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {363}{100} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {3+5 x}}-\frac {33}{50} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{5} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {363 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{50 \sqrt {5}}\\ &=-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {3+5 x}}-\frac {33}{50} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{5} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {363 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{50 \sqrt {10}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.41 \begin {gather*} -\frac {2}{77} \sqrt {\frac {2}{11}} (1-2 x)^{7/2} \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};-\frac {5}{11} (2 x-1)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2/11]*(1 - 2*x)^(7/2)*Hypergeometric2F1[3/2, 7/2, 9/2, (-5*(-1 + 2*x))/11])/77

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IntegrateAlgebraic [A]  time = 0.17, size = 109, normalized size = 1.16 \begin {gather*} \frac {363 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{50 \sqrt {10}}-\frac {121 \sqrt {1-2 x} \left (\frac {20 (1-2 x)^2}{(5 x+3)^2}+\frac {25 (1-2 x)}{5 x+3}+6\right )}{50 \sqrt {5 x+3} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-121*Sqrt[1 - 2*x]*(6 + (20*(1 - 2*x)^2)/(3 + 5*x)^2 + (25*(1 - 2*x))/(3 + 5*x)))/(50*Sqrt[3 + 5*x]*(2 + (5*(
1 - 2*x))/(3 + 5*x))^2) + (363*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/(50*Sqrt[10])

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fricas [A]  time = 1.21, size = 81, normalized size = 0.86 \begin {gather*} \frac {363 \, \sqrt {10} {\left (5 \, 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 (20 \, x^{2} - 75 \, x - 149\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1000 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1000*(363*sqrt(10)*(5*x + 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
+ 20*(20*x^2 - 75*x - 149)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.63, size = 111, normalized size = 1.18 \begin {gather*} \frac {1}{1250} \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} - 99 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {363}{500} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {121 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{1250 \, \sqrt {5 \, x + 3}} + \frac {242 \, \sqrt {10} \sqrt {5 \, x + 3}}{625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1250*(4*sqrt(5)*(5*x + 3) - 99*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 363/500*sqrt(10)*arcsin(1/11*sqrt(22
)*sqrt(5*x + 3)) - 121/1250*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 242/625*sqrt(10)*sqr
t(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-2 x +1\right )^{\frac {5}{2}}}{\left (5 x +3\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.26, size = 75, normalized size = 0.80 \begin {gather*} -\frac {4 \, x^{3}}{5 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {17 \, x^{2}}{5 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {363}{1000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {223 \, x}{50 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {149}{50 \, \sqrt {-10 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*x^3/sqrt(-10*x^2 - x + 3) + 17/5*x^2/sqrt(-10*x^2 - x + 3) + 363/1000*sqrt(10)*arcsin(-20/11*x - 1/11) +
223/50*x/sqrt(-10*x^2 - x + 3) - 149/50/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 (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((1 - 2*x)^(5/2)/(5*x + 3)^(3/2),x)

[Out]

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

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sympy [A]  time = 6.55, size = 230, normalized size = 2.45 \begin {gather*} \begin {cases} \frac {4 i \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{5 \sqrt {10 x - 5}} - \frac {121 i \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{25 \sqrt {10 x - 5}} + \frac {121 i \sqrt {x + \frac {3}{5}}}{250 \sqrt {10 x - 5}} + \frac {363 \sqrt {10} i \operatorname {acosh}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{500} + \frac {2662 i}{625 \sqrt {x + \frac {3}{5}} \sqrt {10 x - 5}} & \text {for}\: \frac {10 \left |{x + \frac {3}{5}}\right |}{11} > 1 \\- \frac {363 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{500} - \frac {4 \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{5 \sqrt {5 - 10 x}} + \frac {121 \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{25 \sqrt {5 - 10 x}} - \frac {121 \sqrt {x + \frac {3}{5}}}{250 \sqrt {5 - 10 x}} - \frac {2662}{625 \sqrt {5 - 10 x} \sqrt {x + \frac {3}{5}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((4*I*(x + 3/5)**(5/2)/(5*sqrt(10*x - 5)) - 121*I*(x + 3/5)**(3/2)/(25*sqrt(10*x - 5)) + 121*I*sqrt(x
+ 3/5)/(250*sqrt(10*x - 5)) + 363*sqrt(10)*I*acosh(sqrt(110)*sqrt(x + 3/5)/11)/500 + 2662*I/(625*sqrt(x + 3/5
)*sqrt(10*x - 5)), 10*Abs(x + 3/5)/11 > 1), (-363*sqrt(10)*asin(sqrt(110)*sqrt(x + 3/5)/11)/500 - 4*(x + 3/5)*
*(5/2)/(5*sqrt(5 - 10*x)) + 121*(x + 3/5)**(3/2)/(25*sqrt(5 - 10*x)) - 121*sqrt(x + 3/5)/(250*sqrt(5 - 10*x))
- 2662/(625*sqrt(5 - 10*x)*sqrt(x + 3/5)), True))

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