3.560 \(\int (2+3 x) \sqrt{-5+7 x^2} \, dx\)

Optimal. Leaf size=55 \[ \frac{1}{7} \left (7 x^2-5\right )^{3/2}+x \sqrt{7 x^2-5}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{7} x}{\sqrt{7 x^2-5}}\right )}{\sqrt{7}} \]

[Out]

x*Sqrt[-5 + 7*x^2] + (-5 + 7*x^2)^(3/2)/7 - (5*ArcTanh[(Sqrt[7]*x)/Sqrt[-5 + 7*x^2]])/Sqrt[7]

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Rubi [A]  time = 0.0152842, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {641, 195, 217, 206} \[ \frac{1}{7} \left (7 x^2-5\right )^{3/2}+x \sqrt{7 x^2-5}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{7} x}{\sqrt{7 x^2-5}}\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)*Sqrt[-5 + 7*x^2],x]

[Out]

x*Sqrt[-5 + 7*x^2] + (-5 + 7*x^2)^(3/2)/7 - (5*ArcTanh[(Sqrt[7]*x)/Sqrt[-5 + 7*x^2]])/Sqrt[7]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (2+3 x) \sqrt{-5+7 x^2} \, dx &=\frac{1}{7} \left (-5+7 x^2\right )^{3/2}+2 \int \sqrt{-5+7 x^2} \, dx\\ &=x \sqrt{-5+7 x^2}+\frac{1}{7} \left (-5+7 x^2\right )^{3/2}-5 \int \frac{1}{\sqrt{-5+7 x^2}} \, dx\\ &=x \sqrt{-5+7 x^2}+\frac{1}{7} \left (-5+7 x^2\right )^{3/2}-5 \operatorname{Subst}\left (\int \frac{1}{1-7 x^2} \, dx,x,\frac{x}{\sqrt{-5+7 x^2}}\right )\\ &=x \sqrt{-5+7 x^2}+\frac{1}{7} \left (-5+7 x^2\right )^{3/2}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{7} x}{\sqrt{-5+7 x^2}}\right )}{\sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0401515, size = 50, normalized size = 0.91 \[ \left (x^2+x-\frac{5}{7}\right ) \sqrt{7 x^2-5}-\frac{5 \log \left (\sqrt{7} \sqrt{7 x^2-5}+7 x\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)*Sqrt[-5 + 7*x^2],x]

[Out]

(-5/7 + x + x^2)*Sqrt[-5 + 7*x^2] - (5*Log[7*x + Sqrt[7]*Sqrt[-5 + 7*x^2]])/Sqrt[7]

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Maple [A]  time = 0.041, size = 45, normalized size = 0.8 \begin{align*} x\sqrt{7\,{x}^{2}-5}-{\frac{5\,\sqrt{7}}{7}\ln \left ( x\sqrt{7}+\sqrt{7\,{x}^{2}-5} \right ) }+{\frac{1}{7} \left ( 7\,{x}^{2}-5 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*x)*(7*x^2-5)^(1/2),x)

[Out]

x*(7*x^2-5)^(1/2)-5/7*ln(x*7^(1/2)+(7*x^2-5)^(1/2))*7^(1/2)+1/7*(7*x^2-5)^(3/2)

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Maxima [A]  time = 1.8913, size = 63, normalized size = 1.15 \begin{align*} \frac{1}{7} \,{\left (7 \, x^{2} - 5\right )}^{\frac{3}{2}} + \sqrt{7 \, x^{2} - 5} x - \frac{5}{7} \, \sqrt{7} \log \left (2 \, \sqrt{7} \sqrt{7 \, x^{2} - 5} + 14 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(7*x^2-5)^(1/2),x, algorithm="maxima")

[Out]

1/7*(7*x^2 - 5)^(3/2) + sqrt(7*x^2 - 5)*x - 5/7*sqrt(7)*log(2*sqrt(7)*sqrt(7*x^2 - 5) + 14*x)

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Fricas [A]  time = 2.41585, size = 136, normalized size = 2.47 \begin{align*} \frac{1}{7} \,{\left (7 \, x^{2} + 7 \, x - 5\right )} \sqrt{7 \, x^{2} - 5} + \frac{5}{14} \, \sqrt{7} \log \left (-2 \, \sqrt{7} \sqrt{7 \, x^{2} - 5} x + 14 \, x^{2} - 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(7*x^2-5)^(1/2),x, algorithm="fricas")

[Out]

1/7*(7*x^2 + 7*x - 5)*sqrt(7*x^2 - 5) + 5/14*sqrt(7)*log(-2*sqrt(7)*sqrt(7*x^2 - 5)*x + 14*x^2 - 5)

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Sympy [A]  time = 0.416093, size = 56, normalized size = 1.02 \begin{align*} x^{2} \sqrt{7 x^{2} - 5} + x \sqrt{7 x^{2} - 5} - \frac{5 \sqrt{7 x^{2} - 5}}{7} - \frac{5 \sqrt{7} \operatorname{acosh}{\left (\frac{\sqrt{35} x}{5} \right )}}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(7*x**2-5)**(1/2),x)

[Out]

x**2*sqrt(7*x**2 - 5) + x*sqrt(7*x**2 - 5) - 5*sqrt(7*x**2 - 5)/7 - 5*sqrt(7)*acosh(sqrt(35)*x/5)/7

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Giac [A]  time = 1.33174, size = 58, normalized size = 1.05 \begin{align*} \frac{1}{7} \,{\left (7 \,{\left (x + 1\right )} x - 5\right )} \sqrt{7 \, x^{2} - 5} + \frac{5}{7} \, \sqrt{7} \log \left ({\left | -\sqrt{7} x + \sqrt{7 \, x^{2} - 5} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(7*x^2-5)^(1/2),x, algorithm="giac")

[Out]

1/7*(7*(x + 1)*x - 5)*sqrt(7*x^2 - 5) + 5/7*sqrt(7)*log(abs(-sqrt(7)*x + sqrt(7*x^2 - 5)))