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

Optimal. Leaf size=109 \[ -\frac{d^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^2 \sqrt{a e^2+c d^2}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{\sqrt{c} e^2}+\frac{\sqrt{a+c x^2}}{c e} \]

[Out]

Sqrt[a + c*x^2]/(c*e) - (d*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(Sqrt[c]*e^2) - (d^2*ArcTanh[(a*e - c*d*x)/(S
qrt[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 = 0.127501, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1654, 12, 844, 217, 206, 725} \[ -\frac{d^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^2 \sqrt{a e^2+c d^2}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{\sqrt{c} e^2}+\frac{\sqrt{a+c x^2}}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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{x^2}{(d+e x) \sqrt{a+c x^2}} \, dx &=\frac{\sqrt{a+c x^2}}{c e}-\frac{\int \frac{c d e x}{(d+e x) \sqrt{a+c x^2}} \, dx}{c e^2}\\ &=\frac{\sqrt{a+c x^2}}{c e}-\frac{d \int \frac{x}{(d+e x) \sqrt{a+c x^2}} \, dx}{e}\\ &=\frac{\sqrt{a+c x^2}}{c e}-\frac{d \int \frac{1}{\sqrt{a+c x^2}} \, dx}{e^2}+\frac{d^2 \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{e^2}\\ &=\frac{\sqrt{a+c x^2}}{c e}-\frac{d \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{e^2}-\frac{d^2 \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}}{c e}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{\sqrt{c} e^2}-\frac{d^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 \sqrt{c d^2+a e^2}}\\ \end{align*}

Mathematica [A]  time = 0.0845403, size = 105, normalized size = 0.96 \[ \frac{-\frac{d^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\sqrt{a e^2+c d^2}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{\sqrt{c}}+\frac{e \sqrt{a+c x^2}}{c}}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Sqrt[a + c*x^2])/c - (d*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/Sqrt[c] - (d^2*ArcTanh[(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 [A]  time = 0.264, size = 172, normalized size = 1.6 \begin{align*}{\frac{1}{ce}\sqrt{c{x}^{2}+a}}-{\frac{d}{{e}^{2}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{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{ \left ({\frac{d}{e}}+x \right ) ^{2}c-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(x^2/(e*x+d)/(c*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.92065, size = 1590, normalized size = 14.59 \begin{align*} \left [\frac{\sqrt{c d^{2} + a e^{2}} c d^{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 ) +{\left (c d^{3} + a d e^{2}\right )} \sqrt{c} \log \left (-2 \, c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (c d^{2} e + a e^{3}\right )} \sqrt{c x^{2} + a}}{2 \,{\left (c^{2} d^{2} e^{2} + a c e^{4}\right )}}, -\frac{2 \, \sqrt{-c d^{2} - a e^{2}} c d^{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^{3} + a d e^{2}\right )} \sqrt{c} \log \left (-2 \, c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) - 2 \,{\left (c d^{2} e + a e^{3}\right )} \sqrt{c x^{2} + a}}{2 \,{\left (c^{2} d^{2} e^{2} + a c e^{4}\right )}}, \frac{\sqrt{c d^{2} + a e^{2}} c d^{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^{3} + a d e^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) + 2 \,{\left (c d^{2} e + a e^{3}\right )} \sqrt{c x^{2} + a}}{2 \,{\left (c^{2} d^{2} e^{2} + a c e^{4}\right )}}, -\frac{\sqrt{-c d^{2} - a e^{2}} c d^{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^{3} + a d e^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (c d^{2} e + a e^{3}\right )} \sqrt{c x^{2} + a}}{c^{2} d^{2} e^{2} + a c e^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15939, size = 142, normalized size = 1.3 \begin{align*} \frac{2 \, d^{2} \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}}} + \frac{d e^{\left (-2\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{\sqrt{c}} + \frac{\sqrt{c x^{2} + a} e^{\left (-1\right )}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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