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

Optimal. Leaf size=93 \[ \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}+\frac {25}{8} \sqrt {1-2 x} (5 x+3)^{3/2}+\frac {825}{32} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {1815}{32} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 93, 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 {(5 x+3)^{5/2}}{\sqrt {1-2 x}}+\frac {25}{8} \sqrt {1-2 x} (5 x+3)^{3/2}+\frac {825}{32} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {1815}{32} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(825*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/32 + (25*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/8 + (3 + 5*x)^(5/2)/Sqrt[1 - 2*x] -
(1815*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/32

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 {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx &=\frac {(3+5 x)^{5/2}}{\sqrt {1-2 x}}-\frac {25}{2} \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx\\ &=\frac {25}{8} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {(3+5 x)^{5/2}}{\sqrt {1-2 x}}-\frac {825}{16} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\\ &=\frac {825}{32} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {25}{8} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {(3+5 x)^{5/2}}{\sqrt {1-2 x}}-\frac {9075}{64} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {825}{32} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {25}{8} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {(3+5 x)^{5/2}}{\sqrt {1-2 x}}-\frac {1}{32} \left (1815 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=\frac {825}{32} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {25}{8} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {(3+5 x)^{5/2}}{\sqrt {1-2 x}}-\frac {1815}{32} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.42 \begin {gather*} \frac {121 \sqrt {\frac {11}{2}} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};-\frac {5}{11} (2 x-1)\right )}{4 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(121*Sqrt[11/2]*Hypergeometric2F1[-5/2, -1/2, 1/2, (-5*(-1 + 2*x))/11])/(4*Sqrt[1 - 2*x])

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IntegrateAlgebraic [A]  time = 0.15, size = 111, normalized size = 1.19 \begin {gather*} \frac {121 \sqrt {5 x+3} \left (\frac {375 (1-2 x)^2}{(5 x+3)^2}+\frac {250 (1-2 x)}{5 x+3}+32\right )}{32 \sqrt {1-2 x} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^2}+\frac {1815}{32} \sqrt {\frac {5}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(121*Sqrt[3 + 5*x]*(32 + (375*(1 - 2*x)^2)/(3 + 5*x)^2 + (250*(1 - 2*x))/(3 + 5*x)))/(32*Sqrt[1 - 2*x]*(2 + (5
*(1 - 2*x))/(3 + 5*x))^2) + (1815*Sqrt[5/2]*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/32

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fricas [A]  time = 1.32, size = 87, normalized size = 0.94 \begin {gather*} \frac {1815 \, \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 (200 \, x^{2} + 790 \, x - 1413\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{128 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(1815*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*(200*x^2 + 790*x - 1413)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)

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giac [A]  time = 1.05, size = 71, normalized size = 0.76 \begin {gather*} -\frac {1815}{64} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} + 55 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 1815 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{160 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1815/64*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/160*(2*(4*sqrt(5)*(5*x + 3) + 55*sqrt(5))*(5*x + 3)
- 1815*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.23, size = 75, normalized size = 0.81 \begin {gather*} -\frac {125 \, x^{3}}{4 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {2275 \, x^{2}}{16 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1815}{128} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {4695 \, x}{32 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {4239}{32 \, \sqrt {-10 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/4*x^3/sqrt(-10*x^2 - x + 3) - 2275/16*x^2/sqrt(-10*x^2 - x + 3) + 1815/128*sqrt(10)*arcsin(-20/11*x - 1/1
1) + 4695/32*x/sqrt(-10*x^2 - x + 3) + 4239/32/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 (5\,x+3\right )}^{5/2}}{{\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)^(5/2)/(1 - 2*x)^(3/2),x)

[Out]

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

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sympy [A]  time = 7.75, size = 187, normalized size = 2.01 \begin {gather*} \begin {cases} \frac {125 i \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{4 \sqrt {10 x - 5}} + \frac {1375 i \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{16 \sqrt {10 x - 5}} - \frac {9075 i \sqrt {x + \frac {3}{5}}}{32 \sqrt {10 x - 5}} + \frac {1815 \sqrt {10} i \operatorname {acosh}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{64} & \text {for}\: \frac {10 \left |{x + \frac {3}{5}}\right |}{11} > 1 \\- \frac {1815 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{64} - \frac {125 \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{4 \sqrt {5 - 10 x}} - \frac {1375 \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{16 \sqrt {5 - 10 x}} + \frac {9075 \sqrt {x + \frac {3}{5}}}{32 \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)**(5/2)/(1-2*x)**(3/2),x)

[Out]

Piecewise((125*I*(x + 3/5)**(5/2)/(4*sqrt(10*x - 5)) + 1375*I*(x + 3/5)**(3/2)/(16*sqrt(10*x - 5)) - 9075*I*sq
rt(x + 3/5)/(32*sqrt(10*x - 5)) + 1815*sqrt(10)*I*acosh(sqrt(110)*sqrt(x + 3/5)/11)/64, 10*Abs(x + 3/5)/11 > 1
), (-1815*sqrt(10)*asin(sqrt(110)*sqrt(x + 3/5)/11)/64 - 125*(x + 3/5)**(5/2)/(4*sqrt(5 - 10*x)) - 1375*(x + 3
/5)**(3/2)/(16*sqrt(5 - 10*x)) + 9075*sqrt(x + 3/5)/(32*sqrt(5 - 10*x)), True))

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