3.531 \(\int \frac{\sqrt{a+c x^2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=103 \[ -\frac{\sqrt{a+c x^2} (a e-c d x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )}-\frac{a c \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{3/2}} \]

[Out]

-((a*e - c*d*x)*Sqrt[a + c*x^2])/(2*(c*d^2 + a*e^2)*(d + e*x)^2) - (a*c*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*
e^2]*Sqrt[a + c*x^2])])/(2*(c*d^2 + a*e^2)^(3/2))

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Rubi [A]  time = 0.0438794, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {721, 725, 206} \[ -\frac{\sqrt{a+c x^2} (a e-c d x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )}-\frac{a c \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^2]/(d + e*x)^3,x]

[Out]

-((a*e - c*d*x)*Sqrt[a + c*x^2])/(2*(c*d^2 + a*e^2)*(d + e*x)^2) - (a*c*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*
e^2]*Sqrt[a + c*x^2])])/(2*(c*d^2 + a*e^2)^(3/2))

Rule 721

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(-2*a*e + (2*c*
d)*x)*(a + c*x^2)^p)/(2*(m + 1)*(c*d^2 + a*e^2)), x] - Dist[(4*a*c*p)/(2*(m + 1)*(c*d^2 + a*e^2)), Int[(d + e*
x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2,
0] && GtQ[p, 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]

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 \frac{\sqrt{a+c x^2}}{(d+e x)^3} \, dx &=-\frac{(a e-c d x) \sqrt{a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}+\frac{(a c) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{2 \left (c d^2+a e^2\right )}\\ &=-\frac{(a e-c d x) \sqrt{a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac{(a c) \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 )}{2 \left (c d^2+a e^2\right )}\\ &=-\frac{(a e-c d x) \sqrt{a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac{a c \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{2 \left (c d^2+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.104056, size = 127, normalized size = 1.23 \[ \frac{\sqrt{a+c x^2} \sqrt{a e^2+c d^2} (c d x-a e)-a c (d+e x)^2 \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )+a c (d+e x)^2 \log (d+e x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^2]/(d + e*x)^3,x]

[Out]

(Sqrt[c*d^2 + a*e^2]*(-(a*e) + c*d*x)*Sqrt[a + c*x^2] + a*c*(d + e*x)^2*Log[d + e*x] - a*c*(d + e*x)^2*Log[a*e
- c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(2*(c*d^2 + a*e^2)^(3/2)*(d + e*x)^2)

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Maple [B]  time = 0.213, size = 1174, normalized size = 11.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(1/2)/(e*x+d)^3,x)

[Out]

-1/2/e/(a*e^2+c*d^2)/(d/e+x)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)-1/2*c*d/(a*e^2+c*d^2)^2/(
d/e+x)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(3/2)-1/2/e*c^2*d^2/(a*e^2+c*d^2)^2*(c*(d/e+x)^2-2*c*d/
e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)+1/2/e^2*c^(5/2)*d^3/(a*e^2+c*d^2)^2*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))+1/2/e*c^2*d^2/(a*e^2+c*d^2)^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*(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+1/2/e^3*c^3*d^4/(a*e^2+c*d^2)^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*(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))+1/2*c^2*d/(a*e
^2+c*d^2)^2*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x+1/2*c^(3/2)*d/(a*e^2+c*d^2)^2*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))*a+1/2/e/(a*e^2+c*d^2)*c*(c*(d/e+x)^2-
2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)-1/2/e^2/(a*e^2+c*d^2)*c^(3/2)*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))-1/2/e/(a*e^2+c*d^2)*c/((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-1/2/e^3/(a*e^2+c*d^2)*c^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*(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))*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.5001, size = 1064, normalized size = 10.33 \begin{align*} \left [\frac{{\left (a c e^{2} x^{2} + 2 \, a c d e x + a c d^{2}\right )} \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 \,{\left (a c d^{2} e + a^{2} e^{3} -{\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt{c x^{2} + a}}{4 \,{\left (c^{2} d^{6} + 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4} +{\left (c^{2} d^{4} e^{2} + 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} x^{2} + 2 \,{\left (c^{2} d^{5} e + 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x\right )}}, -\frac{{\left (a c e^{2} x^{2} + 2 \, a c d e x + a c d^{2}\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}}{a c d^{2} + a^{2} e^{2} +{\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) +{\left (a c d^{2} e + a^{2} e^{3} -{\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt{c x^{2} + a}}{2 \,{\left (c^{2} d^{6} + 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4} +{\left (c^{2} d^{4} e^{2} + 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} x^{2} + 2 \,{\left (c^{2} d^{5} e + 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

[1/4*((a*c*e^2*x^2 + 2*a*c*d*e*x + a*c*d^2)*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)) - 2*(a*
c*d^2*e + a^2*e^3 - (c^2*d^3 + a*c*d*e^2)*x)*sqrt(c*x^2 + a))/(c^2*d^6 + 2*a*c*d^4*e^2 + a^2*d^2*e^4 + (c^2*d^
4*e^2 + 2*a*c*d^2*e^4 + a^2*e^6)*x^2 + 2*(c^2*d^5*e + 2*a*c*d^3*e^3 + a^2*d*e^5)*x), -1/2*((a*c*e^2*x^2 + 2*a*
c*d*e*x + a*c*d^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)) + (a*c*d^2*e + a^2*e^3 - (c^2*d^3 + a*c*d*e^2)*x)*sqrt(c*x^2 + a))/(c^2*d^6
+ 2*a*c*d^4*e^2 + a^2*d^2*e^4 + (c^2*d^4*e^2 + 2*a*c*d^2*e^4 + a^2*e^6)*x^2 + 2*(c^2*d^5*e + 2*a*c*d^3*e^3 +
a^2*d*e^5)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + c x^{2}}}{\left (d + e x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**(1/2)/(e*x+d)**3,x)

[Out]

Integral(sqrt(a + c*x**2)/(d + e*x)**3, x)

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Giac [B]  time = 1.22874, size = 421, normalized size = 4.09 \begin{align*} -\frac{a c \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 )}{{\left (c d^{2} + a e^{2}\right )} \sqrt{-c d^{2} - a e^{2}}} + \frac{2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} c^{2} d^{2} e + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} c^{\frac{5}{2}} d^{3} - 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a c^{2} d^{2} e -{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a c^{\frac{3}{2}} d e^{2} +{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} a c e^{3} + a^{2} c^{\frac{3}{2}} d e^{2} +{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a^{2} c e^{3}}{{\left (c d^{2} e^{2} + a e^{4}\right )}{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} e + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} \sqrt{c} d - a e\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^(1/2)/(e*x+d)^3,x, algorithm="giac")

[Out]

-a*c*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))/((c*d^2 + a*e^2)*sqrt(-c*d^2 -
a*e^2)) + (2*(sqrt(c)*x - sqrt(c*x^2 + a))^3*c^2*d^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + a))^2*c^(5/2)*d^3 - 2*(s
qrt(c)*x - sqrt(c*x^2 + a))*a*c^2*d^2*e - (sqrt(c)*x - sqrt(c*x^2 + a))^2*a*c^(3/2)*d*e^2 + (sqrt(c)*x - sqrt(
c*x^2 + a))^3*a*c*e^3 + a^2*c^(3/2)*d*e^2 + (sqrt(c)*x - sqrt(c*x^2 + a))*a^2*c*e^3)/((c*d^2*e^2 + a*e^4)*((sq
rt(c)*x - sqrt(c*x^2 + a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + a))*sqrt(c)*d - a*e)^2)