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

Optimal. Leaf size=72 \[ \frac {7 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}+\frac {103}{44} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {103 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{4 \sqrt {10}} \]

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Rubi [A]  time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {78, 50, 54, 216} \begin {gather*} \frac {7 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}+\frac {103}{44} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {103 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{4 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(103*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/44 + (7*(3 + 5*x)^(3/2))/(11*Sqrt[1 - 2*x]) - (103*ArcSin[Sqrt[2/11]*Sqrt[3
+ 5*x]])/(4*Sqrt[10])

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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) \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx &=\frac {7 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}-\frac {103}{22} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\\ &=\frac {103}{44} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {7 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}-\frac {103}{8} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {103}{44} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {7 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}-\frac {103 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{4 \sqrt {5}}\\ &=\frac {103}{44} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {7 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}-\frac {103 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{4 \sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 73, normalized size = 1.01 \begin {gather*} \frac {10 (17-6 x) \sqrt {2 x-1} \sqrt {5 x+3}-103 \sqrt {10} (2 x-1) \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{40 \sqrt {-(1-2 x)^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10*(17 - 6*x)*Sqrt[-1 + 2*x]*Sqrt[3 + 5*x] - 103*Sqrt[10]*(-1 + 2*x)*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/(40*
Sqrt[-(1 - 2*x)^2])

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IntegrateAlgebraic [A]  time = 0.12, size = 93, normalized size = 1.29 \begin {gather*} \frac {\sqrt {5 x+3} \left (\frac {103 (1-2 x)}{5 x+3}+28\right )}{4 \sqrt {1-2 x} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )}+\frac {103 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{4 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 5*x]*(28 + (103*(1 - 2*x))/(3 + 5*x)))/(4*Sqrt[1 - 2*x]*(2 + (5*(1 - 2*x))/(3 + 5*x))) + (103*ArcTan
[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/(4*Sqrt[10])

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fricas [A]  time = 1.34, size = 76, normalized size = 1.06 \begin {gather*} \frac {103 \, \sqrt {10} {\left (2 \, x - 1\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 (6 \, x - 17\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{80 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(103*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) +
20*(6*x - 17)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)

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giac [A]  time = 1.19, size = 58, normalized size = 0.81 \begin {gather*} -\frac {103}{40} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (6 \, \sqrt {5} {\left (5 \, x + 3\right )} - 103 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{100 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-103/40*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/100*(6*sqrt(5)*(5*x + 3) - 103*sqrt(5))*sqrt(5*x + 3)
*sqrt(-10*x + 5)/(2*x - 1)

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maple [A]  time = 0.01, size = 89, normalized size = 1.24 \begin {gather*} -\frac {\left (206 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-120 \sqrt {-10 x^{2}-x +3}\, x -103 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+340 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{80 \left (2 x -1\right ) \sqrt {-10 x^{2}-x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*(206*10^(1/2)*x*arcsin(20/11*x+1/11)-103*10^(1/2)*arcsin(20/11*x+1/11)-120*(-10*x^2-x+3)^(1/2)*x+340*(-1
0*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(2*x-1)/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.18, size = 50, normalized size = 0.69 \begin {gather*} -\frac {103}{80} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {3}{4} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {7 \, \sqrt {-10 \, x^{2} - x + 3}}{2 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-103/80*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 3/4*sqrt(-10*x^2 - x + 3) - 7/2*sqrt(-10*x^2 - x + 3)/(2*x -
1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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