3.566 \(\int \frac{1}{(d+e x)^2 \sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=91 \[ -\frac{e \sqrt{a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}-\frac{c d \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0339196, antiderivative size = 91, 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 = {731, 725, 206} \[ -\frac{e \sqrt{a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}-\frac{c d \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

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

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 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{1}{(d+e x)^2 \sqrt{a+c x^2}} \, dx &=-\frac{e \sqrt{a+c x^2}}{\left (c d^2+a e^2\right ) (d+e x)}+\frac{(c d) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{c d^2+a e^2}\\ &=-\frac{e \sqrt{a+c x^2}}{\left (c d^2+a e^2\right ) (d+e x)}-\frac{(c d) \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 )}{c d^2+a e^2}\\ &=-\frac{e \sqrt{a+c x^2}}{\left (c d^2+a e^2\right ) (d+e x)}-\frac{c d \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{\left (c d^2+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.101042, size = 115, normalized size = 1.26 \[ -\frac{e \sqrt{a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}-\frac{c d \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{3/2}}+\frac{c d \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

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

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Maple [B]  time = 0.193, size = 210, normalized size = 2.3 \begin{align*} -{\frac{1}{a{e}^{2}+c{d}^{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}}}} \left ({\frac{d}{e}}+x \right ) ^{-1}}-{\frac{cd}{e \left ( a{e}^{2}+c{d}^{2} \right ) }\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(1/(e*x+d)^2/(c*x^2+a)^(1/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.49099, size = 783, normalized size = 8.6 \begin{align*} \left [\frac{{\left (c d e x + 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 (c d^{2} e + a e^{3}\right )} \sqrt{c x^{2} + a}}{2 \,{\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} +{\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}, -\frac{{\left (c d e x + 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 (c d^{2} e + a e^{3}\right )} \sqrt{c x^{2} + a}}{c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} +{\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + c x^{2}} \left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 5.03177, size = 439, normalized size = 4.82 \begin{align*} -\frac{\sqrt{c d^{2} + a e^{2}} c d e \log \left ({\left | -\sqrt{c d^{2} + a e^{2}} c d +{\left (c d^{2} + a e^{2}\right )}{\left (\sqrt{c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}} + \frac{\sqrt{c d^{2} e^{2} + a e^{4}} e^{\left (-1\right )}}{x e + d}\right )} \right |}\right )}{{\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} \mathrm{sgn}\left (\frac{1}{x e + d}\right )} + \frac{{\left (c^{\frac{3}{2}} d^{2} + \sqrt{c d^{2} + a e^{2}} c d \log \left ({\left | c^{\frac{3}{2}} d^{2} - \sqrt{c d^{2} + a e^{2}} c d + a \sqrt{c} e^{2} \right |}\right ) + a \sqrt{c} e^{2}\right )} \mathrm{sgn}\left (\frac{1}{x e + d}\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} - \frac{\sqrt{c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}}}{c d^{2} \mathrm{sgn}\left (\frac{1}{x e + d}\right ) + a e^{2} \mathrm{sgn}\left (\frac{1}{x e + d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(c*d^2 + a*e^2)*c*d*e*log(abs(-sqrt(c*d^2 + a*e^2)*c*d + (c*d^2 + a*e^2)*(sqrt(c - 2*c*d/(x*e + d) + c*d^
2/(x*e + d)^2 + a*e^2/(x*e + d)^2) + sqrt(c*d^2*e^2 + a*e^4)*e^(-1)/(x*e + d))))/((c^2*d^4*e + 2*a*c*d^2*e^3 +
a^2*e^5)*sgn(1/(x*e + d))) + (c^(3/2)*d^2 + sqrt(c*d^2 + a*e^2)*c*d*log(abs(c^(3/2)*d^2 - sqrt(c*d^2 + a*e^2)
*c*d + a*sqrt(c)*e^2)) + a*sqrt(c)*e^2)*sgn(1/(x*e + d))/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) - sqrt(c - 2*c*d/
(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2)/(c*d^2*sgn(1/(x*e + d)) + a*e^2*sgn(1/(x*e + d)))