\(\int (2+3 x) \sqrt {-5+7 x^2} \, dx\) [560]

Optimal result

Integrand size = 17, antiderivative size = 55 \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=x \sqrt {-5+7 x^2}+\frac {1}{7} \left (-5+7 x^2\right )^{3/2}-\frac {5 \text {arctanh}\left (\frac {\sqrt {7} x}{\sqrt {-5+7 x^2}}\right )}{\sqrt {7}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {655, 201, 223, 212} \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=-\frac {5 \text {arctanh}\left (\frac {\sqrt {7} x}{\sqrt {7 x^2-5}}\right )}{\sqrt {7}}+\frac {1}{7} \left (7 x^2-5\right )^{3/2}+x \sqrt {7 x^2-5} \]

[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 201

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 212

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

Rule 223

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 655

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]

Rubi steps \begin{align*} \text {integral}& = \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 \text {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] (verified)

Time = 0.41 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.11 \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=\frac {1}{7} \sqrt {-5+7 x^2} \left (-5+7 x+7 x^2\right )-\frac {10 \text {arctanh}\left (\frac {\sqrt {-5+7 x^2}}{\sqrt {5}+\sqrt {7} x}\right )}{\sqrt {7}} \]

[In]

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

[Out]

(Sqrt[-5 + 7*x^2]*(-5 + 7*x + 7*x^2))/7 - (10*ArcTanh[Sqrt[-5 + 7*x^2]/(Sqrt[5] + Sqrt[7]*x)])/Sqrt[7]

Maple [A] (verified)

Time = 1.93 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80

method	result	size
risch	\(\frac {\left (7 x^{2}+7 x -5\right ) \sqrt {7 x^{2}-5}}{7}-\frac {5 \ln \left (x \sqrt {7}+\sqrt {7 x^{2}-5}\right ) \sqrt {7}}{7}\)	\(44\)
default	\(x \sqrt {7 x^{2}-5}-\frac {5 \ln \left (x \sqrt {7}+\sqrt {7 x^{2}-5}\right ) \sqrt {7}}{7}+\frac {\left (7 x^{2}-5\right )^{\frac {3}{2}}}{7}\)	\(45\)
trager	\(\left (x^{2}+x -\frac {5}{7}\right ) \sqrt {7 x^{2}-5}+\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-7\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}-7\right ) \sqrt {7 x^{2}-5}+7 x \right )}{7}\)	\(48\)
meijerg	\(\frac {5 i \sqrt {\operatorname {signum}\left (-1+\frac {7 x^{2}}{5}\right )}\, \sqrt {7}\, \left (-\frac {2 i \sqrt {35}\, \sqrt {\pi }\, x \sqrt {-\frac {7 x^{2}}{5}+1}}{5}-2 i \sqrt {\pi }\, \arcsin \left (\frac {x \sqrt {7}\, \sqrt {5}}{5}\right )\right )}{14 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (-1+\frac {7 x^{2}}{5}\right )}}+\frac {15 \sqrt {5}\, \sqrt {\operatorname {signum}\left (-1+\frac {7 x^{2}}{5}\right )}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {2 \sqrt {\pi }\, \left (2-\frac {14 x^{2}}{5}\right ) \sqrt {-\frac {7 x^{2}}{5}+1}}{3}\right )}{28 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (-1+\frac {7 x^{2}}{5}\right )}}\)	\(126\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=\frac {1}{7} \, {\left (7 \, x^{2} + 7 \, x - 5\right )} \sqrt {7 \, x^{2} - 5} + \frac {5}{14} \, \sqrt {7} \log \left (-2 \, \sqrt {7} \sqrt {7 \, x^{2} - 5} x + 14 \, x^{2} - 5\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=x^{2} \sqrt {7 x^{2} - 5} + x \sqrt {7 x^{2} - 5} - \frac {5 \sqrt {7 x^{2} - 5}}{7} - \frac {5 \sqrt {7} \log {\left (7 x + \sqrt {7} \sqrt {7 x^{2} - 5} \right )}}{7} \]

[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)*log(7*x + sqrt(7)*sqrt(7*x**2 -
5))/7

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=\frac {1}{7} \, {\left (7 \, x^{2} - 5\right )}^{\frac {3}{2}} + \sqrt {7 \, x^{2} - 5} x - \frac {5}{7} \, \sqrt {7} \log \left (2 \, \sqrt {7} \sqrt {7 \, x^{2} - 5} + 14 \, x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=\frac {1}{7} \, {\left (7 \, {\left (x + 1\right )} x - 5\right )} \sqrt {7 \, x^{2} - 5} + \frac {5}{7} \, \sqrt {7} \log \left ({\left | -\sqrt {7} x + \sqrt {7 \, x^{2} - 5} \right |}\right ) \]

[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)))

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80 \[ \int (2+3 x) \sqrt {-5+7 x^2} \, dx=x\,\sqrt {7\,x^2-5}-\frac {5\,\sqrt {7}\,\ln \left (\sqrt {7}\,x+\sqrt {7\,x^2-5}\right )}{7}+\frac {{\left (7\,x^2-5\right )}^{3/2}}{7} \]

[In]

int((3*x + 2)*(7*x^2 - 5)^(1/2),x)

[Out]

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