3.26.33 \(\int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx\) [2533]

Optimal. Leaf size=71 \[ \frac {15}{4} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}}-\frac {33}{4} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]

[Out]

-33/8*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+(3+5*x)^(3/2)/(1-2*x)^(1/2)+15/4*(1-2*x)^(1/2)*(3+5*x)^(1/2
)

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Rubi [A]
time = 0.01, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {49, 52, 56, 222} \begin {gather*} -\frac {33}{4} \sqrt {\frac {5}{2}} \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}+\frac {15}{4} \sqrt {1-2 x} \sqrt {5 x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/4 + (3 + 5*x)^(3/2)/Sqrt[1 - 2*x] - (33*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 +
5*x]])/4

Rule 49

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 52

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 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 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 {(3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx &=\frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}}-\frac {15}{2} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\\ &=\frac {15}{4} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}}-\frac {165}{8} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {15}{4} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}}-\frac {1}{4} \left (33 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=\frac {15}{4} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}}-\frac {33}{4} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 75, normalized size = 1.06 \begin {gather*} \frac {\sqrt {1-2 x} \sqrt {3+5 x} (-27+10 x)+33 \sqrt {10} (-1+2 x) \tan ^{-1}\left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{-4+8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(-27 + 10*x) + 33*Sqrt[10]*(-1 + 2*x)*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 -
10*x])])/(-4 + 8*x)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (3+5 x \right )^{\frac {3}{2}}}{\left (1-2 x \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 62, normalized size = 0.87 \begin {gather*} -\frac {33}{16} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{2 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {33 \, \sqrt {-10 \, x^{2} - x + 3}}{4 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-33/16*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 1/2*(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) - 33/4*sqrt(-10*
x^2 - x + 3)/(2*x - 1)

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Fricas [A]
time = 0.62, size = 82, normalized size = 1.15 \begin {gather*} \frac {33 \, \sqrt {5} \sqrt {2} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 4 \, {\left (10 \, x - 27\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{16 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(33*sqrt(5)*sqrt(2)*(2*x - 1)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) + 4*(10*x - 27)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)

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Sympy [C] Result contains complex when optimal does not.
time = 1.70, size = 143, normalized size = 2.01 \begin {gather*} \begin {cases} \frac {25 i \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{2 \sqrt {10 x - 5}} - \frac {165 i \sqrt {x + \frac {3}{5}}}{4 \sqrt {10 x - 5}} + \frac {33 \sqrt {10} i \operatorname {acosh}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{8} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {11}{10} \\- \frac {33 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{8} - \frac {25 \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{2 \sqrt {5 - 10 x}} + \frac {165 \sqrt {x + \frac {3}{5}}}{4 \sqrt {5 - 10 x}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((25*I*(x + 3/5)**(3/2)/(2*sqrt(10*x - 5)) - 165*I*sqrt(x + 3/5)/(4*sqrt(10*x - 5)) + 33*sqrt(10)*I*a
cosh(sqrt(110)*sqrt(x + 3/5)/11)/8, Abs(x + 3/5) > 11/10), (-33*sqrt(10)*asin(sqrt(110)*sqrt(x + 3/5)/11)/8 -
25*(x + 3/5)**(3/2)/(2*sqrt(5 - 10*x)) + 165*sqrt(x + 3/5)/(4*sqrt(5 - 10*x)), True))

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Giac [A]
time = 0.67, size = 58, normalized size = 0.82 \begin {gather*} -\frac {33}{8} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, \sqrt {5} {\left (5 \, x + 3\right )} - 33 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{20 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-33/8*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/20*(2*sqrt(5)*(5*x + 3) - 33*sqrt(5))*sqrt(5*x + 3)*sqr
t(-10*x + 5)/(2*x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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