3.6.29 \(\int \frac {\sqrt {a+c x^2}}{d+e x} \, dx\) [529]

Optimal. Leaf size=103 \[ \frac {\sqrt {a+c x^2}}{e}-\frac {\sqrt {c} d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{e^2}-\frac {\sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^2} \]

[Out]

-d*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))*c^(1/2)/e^2-arctanh((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))*(a
*e^2+c*d^2)^(1/2)/e^2+(c*x^2+a)^(1/2)/e

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Rubi [A]
time = 0.06, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {749, 858, 223, 212, 739} \begin {gather*} -\frac {\sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^2}-\frac {\sqrt {c} d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{e^2}+\frac {\sqrt {a+c x^2}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^2]/(d + e*x),x]

[Out]

Sqrt[a + c*x^2]/e - (Sqrt[c]*d*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/e^2 - (Sqrt[c*d^2 + a*e^2]*ArcTanh[(a*e -
c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/e^2

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 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 749

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+c x^2}}{d+e x} \, dx &=\frac {\sqrt {a+c x^2}}{e}+\frac {\int \frac {a e-c d x}{(d+e x) \sqrt {a+c x^2}} \, dx}{e}\\ &=\frac {\sqrt {a+c x^2}}{e}+\left (a+\frac {c d^2}{e^2}\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx-\frac {(c d) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{e^2}\\ &=\frac {\sqrt {a+c x^2}}{e}+\left (-a-\frac {c d^2}{e^2}\right ) \text {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )-\frac {(c d) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{e^2}\\ &=\frac {\sqrt {a+c x^2}}{e}-\frac {\sqrt {c} d \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{e^2}-\frac {\sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^2}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 110, normalized size = 1.07 \begin {gather*} \frac {e \sqrt {a+c x^2}+2 \sqrt {-c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )+\sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^2]/(d + e*x),x]

[Out]

(e*Sqrt[a + c*x^2] + 2*Sqrt[-(c*d^2) - a*e^2]*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a
*e^2]] + Sqrt[c]*d*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/e^2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(260\) vs. \(2(89)=178\).
time = 0.46, size = 261, normalized size = 2.53

method	result	size
default	\(\frac {\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}-\frac {\sqrt {c}\, d \ln \left (\frac {-\frac {c d}{e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}\right )}{e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{e}\)	\(261\)
risch	\(\frac {\sqrt {c \,x^{2}+a}}{e}-\frac {\sqrt {c}\, d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{e^{2}}-\frac {\ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) a}{e \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) c \,d^{2}}{e^{3} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\)	\(297\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(1/2)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

1/e*((c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)-c^(1/2)*d/e*ln((-c*d/e+c*(x+d/e))/c^(1/2)+(c*(x+d/e
)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))-(a*e^2+c*d^2)/e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/
e^2-2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)
))

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Maxima [A]
time = 0.30, size = 83, normalized size = 0.81 \begin {gather*} -\sqrt {c} d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{\left (-2\right )} + \sqrt {c d^{2} e^{\left (-2\right )} + a} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | x e + d \right |}} - \frac {a e}{\sqrt {a c} {\left | x e + d \right |}}\right ) e^{\left (-1\right )} + \sqrt {c x^{2} + a} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d),x, algorithm="maxima")

[Out]

-sqrt(c)*d*arcsinh(c*x/sqrt(a*c))*e^(-2) + sqrt(c*d^2*e^(-2) + a)*arcsinh(c*d*x/(sqrt(a*c)*abs(x*e + d)) - a*e
/(sqrt(a*c)*abs(x*e + d)))*e^(-1) + sqrt(c*x^2 + a)*e^(-1)

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Fricas [A]
time = 1.94, size = 565, normalized size = 5.49 \begin {gather*} \left [\frac {1}{2} \, {\left (\sqrt {c} d \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, \sqrt {c x^{2} + a} e + \sqrt {c d^{2} + a e^{2}} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right )\right )} e^{\left (-2\right )}, \frac {1}{2} \, {\left (2 \, \sqrt {-c} d \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + 2 \, \sqrt {c x^{2} + a} e + \sqrt {c d^{2} + a e^{2}} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right )\right )} e^{\left (-2\right )}, \frac {1}{2} \, {\left (\sqrt {c} d \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) + 2 \, \sqrt {c x^{2} + a} e\right )} e^{\left (-2\right )}, {\left (\sqrt {-c} d \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) + \sqrt {c x^{2} + a} e\right )} e^{\left (-2\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d),x, algorithm="fricas")

[Out]

[1/2*(sqrt(c)*d*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*sqrt(c*x^2 + a)*e + sqrt(c*d^2 + a*e^2)*lo
g(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2
*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)))*e^(-2), 1/2*(2*sqrt(-c)*d*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + 2*sqrt(c
*x^2 + a)*e + sqrt(c*d^2 + a*e^2)*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x -
a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)))*e^(-2), 1/2*(sqrt(c)*d*log(-2*c*x^2
+ 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*sqrt(-c*d^2 - a*e^2)*arctan(-sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c
*x^2 + a)/(c^2*d^2*x^2 + a*c*d^2 + (a*c*x^2 + a^2)*e^2)) + 2*sqrt(c*x^2 + a)*e)*e^(-2), (sqrt(-c)*d*arctan(sqr
t(-c)*x/sqrt(c*x^2 + a)) + sqrt(-c*d^2 - a*e^2)*arctan(-sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^
2*d^2*x^2 + a*c*d^2 + (a*c*x^2 + a^2)*e^2)) + sqrt(c*x^2 + a)*e)*e^(-2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + c x^{2}}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**(1/2)/(e*x+d),x)

[Out]

Integral(sqrt(a + c*x**2)/(d + e*x), x)

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Giac [A]
time = 1.83, size = 109, normalized size = 1.06 \begin {gather*} \sqrt {c} d e^{\left (-2\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \frac {2 \, {\left (c d^{2} + a e^{2}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{\left (-2\right )}}{\sqrt {-c d^{2} - a e^{2}}} + \sqrt {c x^{2} + a} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d),x, algorithm="giac")

[Out]

sqrt(c)*d*e^(-2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a))) + 2*(c*d^2 + a*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 +
a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-2)/sqrt(-c*d^2 - a*e^2) + sqrt(c*x^2 + a)*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+a}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^(1/2)/(d + e*x),x)

[Out]

int((a + c*x^2)^(1/2)/(d + e*x), x)

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