3.4.47 \(\int \frac {1}{(d+e x)^2 \sqrt {a+c x^2}} \, dx\) [347]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 = {745, 739, 212} \begin {gather*} -\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}} \end {gather*}

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 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 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 745

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]

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) \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 )}{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*}

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Mathematica [A]
time = 0.04, size = 101, normalized size = 1.11 \begin {gather*} -\frac {e \sqrt {a+c x^2}}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {2 c d \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\left (-c d^2-a e^2\right )^{3/2}} \end {gather*}

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(214\) vs. \(2(83)=166\).
time = 0.09, size = 215, normalized size = 2.36

method	result	size
default	\(\frac {-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {c d e \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}}{e^{2}}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^2*(-1/(a*e^2+c*d^2)*e^2/(x+d/e)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)-c*d*e/(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*(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 = 0.31, size = 93, normalized size = 1.02 \begin {gather*} \frac {c d \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 (-3\right )}}{{\left (c d^{2} e^{\left (-2\right )} + a\right )}^{\frac {3}{2}}} - \frac {\sqrt {c x^{2} + a}}{c d^{3} e^{\left (-1\right )} + c d^{2} x + a x e^{2} + a d e} \end {gather*}

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]

c*d*arcsinh(c*d*x/(sqrt(a*c)*abs(x*e + d)) - a*e/(sqrt(a*c)*abs(x*e + d)))*e^(-3)/(c*d^2*e^(-2) + a)^(3/2) - s
qrt(c*x^2 + a)/(c*d^3*e^(-1) + c*d^2*x + a*x*e^2 + a*d*e)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (84) = 168\).
time = 1.69, size = 373, normalized size = 4.10 \begin {gather*} \left [\frac {{\left (c d x e + c d^{2}\right )} \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 ) - 2 \, {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c^{2} d^{4} x e + c^{2} d^{5} + 2 \, a c d^{2} x e^{3} + 2 \, a c d^{3} e^{2} + a^{2} x e^{5} + a^{2} d e^{4}\right )}}, \frac {{\left (c d x e + 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}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) - {\left (c d^{2} e + a e^{3}\right )} \sqrt {c x^{2} + a}}{c^{2} d^{4} x e + c^{2} d^{5} + 2 \, a c d^{2} x e^{3} + 2 \, a c d^{3} e^{2} + a^{2} x e^{5} + a^{2} d e^{4}}\right ] \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a + c x^{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}

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 [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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