### 3.530 $$\int \frac{\sqrt{a+c x^2}}{(d+e x)^2} \, dx$$

Optimal. Leaf size=110 $\frac{c d \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^2 \sqrt{a e^2+c d^2}}-\frac{\sqrt{a+c x^2}}{e (d+e x)}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{e^2}$

[Out]

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

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Rubi [A]  time = 0.0681209, antiderivative size = 110, 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 = {733, 844, 217, 206, 725} $\frac{c d \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^2 \sqrt{a e^2+c d^2}}-\frac{\sqrt{a+c x^2}}{e (d+e x)}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{e^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[a + c*x^2]/(d + e*x)^2,x]

[Out]

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

Rule 733

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 + 1)), x] - Dist[(2*c*p)/(e*(m + 1)), Int[x*(d + e*x)^(m + 1)*(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] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 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)^2} \, dx &=-\frac{\sqrt{a+c x^2}}{e (d+e x)}+\frac{c \int \frac{x}{(d+e x) \sqrt{a+c x^2}} \, dx}{e}\\ &=-\frac{\sqrt{a+c x^2}}{e (d+e x)}+\frac{c \int \frac{1}{\sqrt{a+c x^2}} \, dx}{e^2}-\frac{(c d) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{e^2}\\ &=-\frac{\sqrt{a+c x^2}}{e (d+e x)}+\frac{c \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{e^2}+\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 )}{e^2}\\ &=-\frac{\sqrt{a+c x^2}}{e (d+e x)}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{e^2}+\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 )}{e^2 \sqrt{c d^2+a e^2}}\\ \end{align*}

Mathematica [A]  time = 0.13688, size = 134, normalized size = 1.22 $\frac{\frac{c d \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )}{\sqrt{a e^2+c d^2}}-\frac{c d \log (d+e x)}{\sqrt{a e^2+c d^2}}-\frac{e \sqrt{a+c x^2}}{d+e x}+\sqrt{c} \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.203, size = 649, normalized size = 5.9 \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)^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)^(3/2)-1/e*c*d/(a*e^2+c*d^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^2*c^(3/2)*d^2/(a*e^2+c*d^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/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))*a+1/e^3*c^2*d^3/(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))+1/(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)*x+1/(a*e^2+c*d^2)*c^(1/2)*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

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.25039, size = 1891, normalized size = 17.19 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((c*d^3 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + (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*d^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*e^5)*x), 1/2*(2*(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^3 + a*d*e^2 + (
c*d^2*e + a*e^3)*x)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(c*d^2*e + a*e^3)*sqrt(c*x^2 +
a))/(c*d^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*e^5)*x), -1/2*(2*(c*d^3 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(-c)*a
rctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (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*d^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*e^5)*x), ((c*d*e*x + c*d^2)*sq
rt(-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^3 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (c*d^2*e
+ a*e^3)*sqrt(c*x^2 + a))/(c*d^3*e^2 + a*d*e^4 + (c*d^2*e^3 + a*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 )^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError