3.529 \(\int \frac{\sqrt{a+c x^2}}{d+e x} \, dx\)

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

[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

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Rubi [A]  time = 0.0814617, antiderivative size = 103, normalized size of antiderivative = 1., 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 = {735, 844, 217, 206, 725} \[ -\frac{\sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^2}-\frac{\sqrt{c} d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{e^2}+\frac{\sqrt{a+c x^2}}{e} \]

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 735

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 844

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]

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])

Rule 725

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]

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 ) \operatorname{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) \operatorname{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*}

Mathematica [A]  time = 0.0478463, size = 99, normalized size = 0.96 \[ \frac{-\sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )-\sqrt{c} d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+e \sqrt{a+c x^2}}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.257, size = 381, normalized size = 3.7 \begin{align*}{\frac{1}{e}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}-{\frac{d}{{e}^{2}}\sqrt{c}\ln \left ({ \left ( -{\frac{cd}{e}}+ \left ({\frac{d}{e}}+x \right ) c \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) }-{\frac{a}{e}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-{\frac{c{d}^{2}}{{e}^{3}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)-1/e^2*c^(1/2)*d*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+(c*(d/
e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))-1/e/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e
*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))*a-1/e^3/(
(a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2*c*
d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))*c*d^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.98198, size = 1274, normalized size = 12.37 \begin{align*} \left [\frac{\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 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} -{\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt{c d^{2} + a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right )}{2 \, e^{2}}, \frac{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 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} -{\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt{c d^{2} + a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right )}{2 \, e^{2}}, \frac{\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 - 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}}{a c d^{2} + a^{2} e^{2} +{\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right )}{2 \, e^{2}}, \frac{\sqrt{-c} d \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) + \sqrt{c x^{2} + a} e - \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}}{a c d^{2} + a^{2} e^{2} +{\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right )}{e^{2}}\right ] \end{align*}

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

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

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 = 1.38916, size = 147, normalized size = 1.43 \begin{align*} \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{align*}

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)