3.336 \(\int \frac {x^2}{(d+e x) (a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=95 \[ -\frac {a e+c d x}{c \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {d^2 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1647, 12, 725, 206} \[ -\frac {a e+c d x}{c \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {d^2 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((d + e*x)*(a + c*x^2)^(3/2)),x]

[Out]

-((a*e + c*d*x)/(c*(c*d^2 + a*e^2)*Sqrt[a + c*x^2])) - (d^2*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 12

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

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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^2}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx &=-\frac {a e+c d x}{c \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {\int \frac {a c d^2}{\left (c d^2+a e^2\right ) (d+e x) \sqrt {a+c x^2}} \, dx}{a c}\\ &=-\frac {a e+c d x}{c \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {d^2 \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{c d^2+a e^2}\\ &=-\frac {a e+c d x}{c \left (c d^2+a e^2\right ) \sqrt {a+c x^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 )}{c d^2+a e^2}\\ &=-\frac {a e+c d x}{c \left (c d^2+a e^2\right ) \sqrt {a+c x^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 )}{\left (c d^2+a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 95, normalized size = 1.00 \[ -\frac {a e+c d x}{c \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {d^2 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/((d + e*x)*(a + c*x^2)^(3/2)),x]

[Out]

-((a*e + c*d*x)/(c*(c*d^2 + a*e^2)*Sqrt[a + c*x^2])) - (d^2*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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fricas [B]  time = 1.11, size = 455, normalized size = 4.79 \[ \left [\frac {{\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (a c d^{2} e + a^{2} e^{3} + {\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4} + {\left (c^{4} d^{4} + 2 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{2}\right )}}, -\frac {{\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + {\left (a c d^{2} e + a^{2} e^{3} + {\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4} + {\left (c^{4} d^{4} + 2 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(e*x+d)/(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 174, normalized size = 1.83 \[ -\frac {2 \, d^{2} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{{\left (c d^{2} + a e^{2}\right )} \sqrt {-c d^{2} - a e^{2}}} - \frac {\frac {{\left (c^{2} d^{3} + a c d e^{2}\right )} x}{c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}} + \frac {a c d^{2} e + a^{2} e^{3}}{c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}}}{\sqrt {c x^{2} + a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.01, size = 311, normalized size = 3.27 \[ \frac {c \,d^{3} x}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a \,e^{2}}-\frac {d^{2} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e}+\frac {d^{2}}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e}-\frac {d x}{\sqrt {c \,x^{2}+a}\, a \,e^{2}}-\frac {1}{\sqrt {c \,x^{2}+a}\, c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(e*x+d)/(c*x^2+a)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.55, size = 171, normalized size = 1.80 \[ \frac {c d^{3} x}{\sqrt {c x^{2} + a} a c d^{2} e^{2} + \sqrt {c x^{2} + a} a^{2} e^{4}} + \frac {d^{2}}{\sqrt {c x^{2} + a} c d^{2} e + \sqrt {c x^{2} + a} a e^{3}} - \frac {d x}{\sqrt {c x^{2} + a} a e^{2}} + \frac {d^{2} \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} e^{3}} - \frac {1}{\sqrt {c x^{2} + a} c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(e*x+d)/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{{\left (c\,x^2+a\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + c*x^2)^(3/2)*(d + e*x)),x)

[Out]

int(x^2/((a + c*x^2)^(3/2)*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(e*x+d)/(c*x**2+a)**(3/2),x)

[Out]

Integral(x**2/((a + c*x**2)**(3/2)*(d + e*x)), x)

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