3.4.28 \(\int \frac {x^2}{(d+e x) \sqrt {a+c x^2}} \, dx\) [328]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, 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 = {1668, 12, 858, 223, 212, 739} \begin {gather*} -\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} \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 223

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 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 858

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 1668

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]))

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 \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{e^2}-\frac {d^2 \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 )}{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*}

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Mathematica [A]
time = 0.34, size = 116, normalized size = 1.06 \begin {gather*} \frac {\frac {e \sqrt {a+c x^2}}{c}-\frac {2 d^2 \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\sqrt {-c d^2-a e^2}}+\frac {d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{\sqrt {c}}}{e^2} \end {gather*}

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

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Maple [A]
time = 0.07, size = 172, normalized size = 1.58

method	result	size
default	\(\frac {\sqrt {c \,x^{2}+a}}{c e}-\frac {d \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{e^{2} \sqrt {c}}-\frac {d^{2} \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{3} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}\)	\(172\)
risch	\(\frac {\sqrt {c \,x^{2}+a}}{c e}-\frac {d \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{e^{2} \sqrt {c}}-\frac {d^{2} \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{3} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^2+a)^(1/2)/c/e-d/e^2*ln(x*c^(1/2)+(c*x^2+a)^(1/2))/c^(1/2)-1/e^3*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 = 0.29, size = 89, normalized size = 0.82 \begin {gather*} \frac {d^{2} \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 )}}{\sqrt {c d^{2} e^{\left (-2\right )} + a}} - \frac {d \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) e^{\left (-2\right )}}{\sqrt {c}} + \frac {\sqrt {c x^{2} + a} e^{\left (-1\right )}}{c} \end {gather*}

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]

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

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Fricas [A]
time = 2.62, size = 721, normalized size = 6.61 \begin {gather*} \left [\frac {\sqrt {c d^{2} + a e^{2}} c d^{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 ) + {\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}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{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 \, 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^{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}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{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 {gather*}

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

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 = 1.54, size = 105, normalized size = 0.96 \begin {gather*} \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 {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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