3.3.4 \(\int \frac {a+b x^2+c x^4}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {x \left (4 c d^2-e (2 a e+b d)\right )}{3 d^2 e^2 \sqrt {d+e x^2}}+\frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{5/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1157, 385, 217, 206} \begin {gather*} -\frac {x \left (4 c d^2-e (2 a e+b d)\right )}{3 d^2 e^2 \sqrt {d+e x^2}}+\frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(d + e*x^2)^(5/2),x]

[Out]

((a + (d*(c*d - b*e))/e^2)*x)/(3*d*(d + e*x^2)^(3/2)) - ((4*c*d^2 - e*(b*d + 2*a*e))*x)/(3*d^2*e^2*Sqrt[d + e*
x^2]) + (c*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/e^(5/2)

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 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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {\left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \left (d+e x^2\right )^{3/2}}-\frac {\int \frac {-2 a+\frac {d (c d-b e)}{e^2}-\frac {3 c d x^2}{e}}{\left (d+e x^2\right )^{3/2}} \, dx}{3 d}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \left (d+e x^2\right )^{3/2}}-\frac {\left (4 c d^2-e (b d+2 a e)\right ) x}{3 d^2 e^2 \sqrt {d+e x^2}}+\frac {c \int \frac {1}{\sqrt {d+e x^2}} \, dx}{e^2}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \left (d+e x^2\right )^{3/2}}-\frac {\left (4 c d^2-e (b d+2 a e)\right ) x}{3 d^2 e^2 \sqrt {d+e x^2}}+\frac {c \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{e^2}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \left (d+e x^2\right )^{3/2}}-\frac {\left (4 c d^2-e (b d+2 a e)\right ) x}{3 d^2 e^2 \sqrt {d+e x^2}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 112, normalized size = 1.11 \begin {gather*} \frac {\sqrt {e} x \left (e^2 \left (3 a d+2 a e x^2+b d x^2\right )-c d^2 \left (3 d+4 e x^2\right )\right )+3 c d^{5/2} \left (d+e x^2\right ) \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 d^2 e^{5/2} \left (d+e x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(d + e*x^2)^(5/2),x]

[Out]

(Sqrt[e]*x*(-(c*d^2*(3*d + 4*e*x^2)) + e^2*(3*a*d + b*d*x^2 + 2*a*e*x^2)) + 3*c*d^(5/2)*(d + e*x^2)*Sqrt[1 + (
e*x^2)/d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(3*d^2*e^(5/2)*(d + e*x^2)^(3/2))

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IntegrateAlgebraic [A]  time = 0.20, size = 95, normalized size = 0.94 \begin {gather*} \frac {3 a d e^2 x+2 a e^3 x^3+b d e^2 x^3-3 c d^3 x-4 c d^2 e x^3}{3 d^2 e^2 \left (d+e x^2\right )^{3/2}}-\frac {c \log \left (\sqrt {d+e x^2}-\sqrt {e} x\right )}{e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2 + c*x^4)/(d + e*x^2)^(5/2),x]

[Out]

(-3*c*d^3*x + 3*a*d*e^2*x - 4*c*d^2*e*x^3 + b*d*e^2*x^3 + 2*a*e^3*x^3)/(3*d^2*e^2*(d + e*x^2)^(3/2)) - (c*Log[
-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/e^(5/2)

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fricas [A]  time = 0.85, size = 289, normalized size = 2.86 \begin {gather*} \left [\frac {3 \, {\left (c d^{2} e^{2} x^{4} + 2 \, c d^{3} e x^{2} + c d^{4}\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - 2 \, {\left ({\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} x^{3} + 3 \, {\left (c d^{3} e - a d e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{6 \, {\left (d^{2} e^{5} x^{4} + 2 \, d^{3} e^{4} x^{2} + d^{4} e^{3}\right )}}, -\frac {3 \, {\left (c d^{2} e^{2} x^{4} + 2 \, c d^{3} e x^{2} + c d^{4}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left ({\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} x^{3} + 3 \, {\left (c d^{3} e - a d e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{2} e^{5} x^{4} + 2 \, d^{3} e^{4} x^{2} + d^{4} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(c*d^2*e^2*x^4 + 2*c*d^3*e*x^2 + c*d^4)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - 2*((
4*c*d^2*e^2 - b*d*e^3 - 2*a*e^4)*x^3 + 3*(c*d^3*e - a*d*e^3)*x)*sqrt(e*x^2 + d))/(d^2*e^5*x^4 + 2*d^3*e^4*x^2
+ d^4*e^3), -1/3*(3*(c*d^2*e^2*x^4 + 2*c*d^3*e*x^2 + c*d^4)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + ((4*
c*d^2*e^2 - b*d*e^3 - 2*a*e^4)*x^3 + 3*(c*d^3*e - a*d*e^3)*x)*sqrt(e*x^2 + d))/(d^2*e^5*x^4 + 2*d^3*e^4*x^2 +
d^4*e^3)]

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giac [A]  time = 0.23, size = 88, normalized size = 0.87 \begin {gather*} -c e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -x e^{\frac {1}{2}} + \sqrt {x^{2} e + d} \right |}\right ) - \frac {{\left (\frac {{\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} x^{2} e^{\left (-3\right )}}{d^{2}} + \frac {3 \, {\left (c d^{3} e - a d e^{3}\right )} e^{\left (-3\right )}}{d^{2}}\right )} x}{3 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

-c*e^(-5/2)*log(abs(-x*e^(1/2) + sqrt(x^2*e + d))) - 1/3*((4*c*d^2*e^2 - b*d*e^3 - 2*a*e^4)*x^2*e^(-3)/d^2 + 3
*(c*d^3*e - a*d*e^3)*e^(-3)/d^2)*x/(x^2*e + d)^(3/2)

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maple [A]  time = 0.01, size = 124, normalized size = 1.23 \begin {gather*} -\frac {c \,x^{3}}{3 \left (e \,x^{2}+d \right )^{\frac {3}{2}} e}+\frac {a x}{3 \left (e \,x^{2}+d \right )^{\frac {3}{2}} d}-\frac {b x}{3 \left (e \,x^{2}+d \right )^{\frac {3}{2}} e}+\frac {2 a x}{3 \sqrt {e \,x^{2}+d}\, d^{2}}+\frac {b x}{3 \sqrt {e \,x^{2}+d}\, d e}-\frac {c x}{\sqrt {e \,x^{2}+d}\, e^{2}}+\frac {c \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{e^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)/(e*x^2+d)^(5/2),x)

[Out]

-1/3*c*x^3/e/(e*x^2+d)^(3/2)-c/e^2*x/(e*x^2+d)^(1/2)+c/e^(5/2)*ln(e^(1/2)*x+(e*x^2+d)^(1/2))-1/3*b/e*x/(e*x^2+
d)^(3/2)+1/3*b/d/e*x/(e*x^2+d)^(1/2)+1/3*a*x/d/(e*x^2+d)^(3/2)+2/3*a/d^2*x/(e*x^2+d)^(1/2)

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maxima [A]  time = 1.01, size = 135, normalized size = 1.34 \begin {gather*} -\frac {1}{3} \, c x {\left (\frac {3 \, x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} + \frac {2 \, d}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}}\right )} + \frac {2 \, a x}{3 \, \sqrt {e x^{2} + d} d^{2}} + \frac {a x}{3 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d} - \frac {c x}{3 \, \sqrt {e x^{2} + d} e^{2}} - \frac {b x}{3 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} e} + \frac {b x}{3 \, \sqrt {e x^{2} + d} d e} + \frac {c \operatorname {arsinh}\left (\frac {e x}{\sqrt {d e}}\right )}{e^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*c*x*(3*x^2/((e*x^2 + d)^(3/2)*e) + 2*d/((e*x^2 + d)^(3/2)*e^2)) + 2/3*a*x/(sqrt(e*x^2 + d)*d^2) + 1/3*a*x
/((e*x^2 + d)^(3/2)*d) - 1/3*c*x/(sqrt(e*x^2 + d)*e^2) - 1/3*b*x/((e*x^2 + d)^(3/2)*e) + 1/3*b*x/(sqrt(e*x^2 +
d)*d*e) + c*arcsinh(e*x/sqrt(d*e))/e^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c\,x^4+b\,x^2+a}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)/(d + e*x^2)^(5/2),x)

[Out]

int((a + b*x^2 + c*x^4)/(d + e*x^2)^(5/2), x)

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sympy [B]  time = 18.95, size = 450, normalized size = 4.46 \begin {gather*} a \left (\frac {3 d x}{3 d^{\frac {7}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 3 d^{\frac {5}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {2 e x^{3}}{3 d^{\frac {7}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 3 d^{\frac {5}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + \frac {b x^{3}}{3 d^{\frac {5}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 3 d^{\frac {3}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}}} + c \left (\frac {3 d^{\frac {39}{2}} e^{11} \sqrt {1 + \frac {e x^{2}}{d}} \operatorname {asinh}{\left (\frac {\sqrt {e} x}{\sqrt {d}} \right )}}{3 d^{\frac {39}{2}} e^{\frac {27}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 3 d^{\frac {37}{2}} e^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {3 d^{\frac {37}{2}} e^{12} x^{2} \sqrt {1 + \frac {e x^{2}}{d}} \operatorname {asinh}{\left (\frac {\sqrt {e} x}{\sqrt {d}} \right )}}{3 d^{\frac {39}{2}} e^{\frac {27}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 3 d^{\frac {37}{2}} e^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {e x^{2}}{d}}} - \frac {3 d^{19} e^{\frac {23}{2}} x}{3 d^{\frac {39}{2}} e^{\frac {27}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 3 d^{\frac {37}{2}} e^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {e x^{2}}{d}}} - \frac {4 d^{18} e^{\frac {25}{2}} x^{3}}{3 d^{\frac {39}{2}} e^{\frac {27}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 3 d^{\frac {37}{2}} e^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {e x^{2}}{d}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/(e*x**2+d)**(5/2),x)

[Out]

a*(3*d*x/(3*d**(7/2)*sqrt(1 + e*x**2/d) + 3*d**(5/2)*e*x**2*sqrt(1 + e*x**2/d)) + 2*e*x**3/(3*d**(7/2)*sqrt(1
+ e*x**2/d) + 3*d**(5/2)*e*x**2*sqrt(1 + e*x**2/d))) + b*x**3/(3*d**(5/2)*sqrt(1 + e*x**2/d) + 3*d**(3/2)*e*x*
*2*sqrt(1 + e*x**2/d)) + c*(3*d**(39/2)*e**11*sqrt(1 + e*x**2/d)*asinh(sqrt(e)*x/sqrt(d))/(3*d**(39/2)*e**(27/
2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*sqrt(1 + e*x**2/d)) + 3*d**(37/2)*e**12*x**2*sqrt(1 + e*x**
2/d)*asinh(sqrt(e)*x/sqrt(d))/(3*d**(39/2)*e**(27/2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*sqrt(1 +
e*x**2/d)) - 3*d**19*e**(23/2)*x/(3*d**(39/2)*e**(27/2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*sqrt(1
+ e*x**2/d)) - 4*d**18*e**(25/2)*x**3/(3*d**(39/2)*e**(27/2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*
sqrt(1 + e*x**2/d)))

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