3.153 \(\int \frac{c+d x^2+e x^4+f x^6}{\sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \left (6 a^2 b e-5 a^3 f-8 a b^2 d+16 b^3 c\right )}{16 b^{7/2}}+\frac{x \sqrt{a+b x^2} \left (5 a^2 f-6 a b e+8 b^2 d\right )}{16 b^3}+\frac{x^3 \sqrt{a+b x^2} (6 b e-5 a f)}{24 b^2}+\frac{f x^5 \sqrt{a+b x^2}}{6 b} \]

[Out]

((8*b^2*d - 6*a*b*e + 5*a^2*f)*x*Sqrt[a + b*x^2])/(16*b^3) + ((6*b*e - 5*a*f)*x^3*Sqrt[a + b*x^2])/(24*b^2) +
(f*x^5*Sqrt[a + b*x^2])/(6*b) + ((16*b^3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x
^2]])/(16*b^(7/2))

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Rubi [A]  time = 0.119289, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {1815, 1159, 388, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \left (6 a^2 b e-5 a^3 f-8 a b^2 d+16 b^3 c\right )}{16 b^{7/2}}+\frac{x \sqrt{a+b x^2} \left (5 a^2 f-6 a b e+8 b^2 d\right )}{16 b^3}+\frac{x^3 \sqrt{a+b x^2} (6 b e-5 a f)}{24 b^2}+\frac{f x^5 \sqrt{a+b x^2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/Sqrt[a + b*x^2],x]

[Out]

((8*b^2*d - 6*a*b*e + 5*a^2*f)*x*Sqrt[a + b*x^2])/(16*b^3) + ((6*b*e - 5*a*f)*x^3*Sqrt[a + b*x^2])/(24*b^2) +
(f*x^5*Sqrt[a + b*x^2])/(6*b) + ((16*b^3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x
^2]])/(16*b^(7/2))

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 1159

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(c^p*x^(4*p - 1)*
(d + e*x^2)^(q + 1))/(e*(4*p + 2*q + 1)), x] + Dist[1/(e*(4*p + 2*q + 1)), Int[(d + e*x^2)^q*ExpandToSum[e*(4*
p + 2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 2*q + 1)*x^(4*p), x], x], x] /
; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[
q, -1]

Rule 388

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

Rubi steps

\begin{align*} \int \frac{c+d x^2+e x^4+f x^6}{\sqrt{a+b x^2}} \, dx &=\frac{f x^5 \sqrt{a+b x^2}}{6 b}+\frac{\int \frac{6 b c+6 b d x^2+(6 b e-5 a f) x^4}{\sqrt{a+b x^2}} \, dx}{6 b}\\ &=\frac{(6 b e-5 a f) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{f x^5 \sqrt{a+b x^2}}{6 b}+\frac{\int \frac{24 b^2 c+3 \left (8 b^2 d-6 a b e+5 a^2 f\right ) x^2}{\sqrt{a+b x^2}} \, dx}{24 b^2}\\ &=\frac{\left (8 b^2 d-6 a b e+5 a^2 f\right ) x \sqrt{a+b x^2}}{16 b^3}+\frac{(6 b e-5 a f) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{f x^5 \sqrt{a+b x^2}}{6 b}-\frac{1}{16} \left (-16 c+\frac{a \left (8 b^2 d-6 a b e+5 a^2 f\right )}{b^3}\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=\frac{\left (8 b^2 d-6 a b e+5 a^2 f\right ) x \sqrt{a+b x^2}}{16 b^3}+\frac{(6 b e-5 a f) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{f x^5 \sqrt{a+b x^2}}{6 b}-\frac{1}{16} \left (-16 c+\frac{a \left (8 b^2 d-6 a b e+5 a^2 f\right )}{b^3}\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=\frac{\left (8 b^2 d-6 a b e+5 a^2 f\right ) x \sqrt{a+b x^2}}{16 b^3}+\frac{(6 b e-5 a f) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{f x^5 \sqrt{a+b x^2}}{6 b}+\frac{\left (16 c-\frac{a \left (8 b^2 d-6 a b e+5 a^2 f\right )}{b^3}\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.108642, size = 118, normalized size = 0.81 \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \left (6 a^2 b e-5 a^3 f-8 a b^2 d+16 b^3 c\right )+\sqrt{b} x \sqrt{a+b x^2} \left (15 a^2 f-2 a b \left (9 e+5 f x^2\right )+4 b^2 \left (6 d+3 e x^2+2 f x^4\right )\right )}{48 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/Sqrt[a + b*x^2],x]

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(15*a^2*f - 2*a*b*(9*e + 5*f*x^2) + 4*b^2*(6*d + 3*e*x^2 + 2*f*x^4)) + 3*(16*b^3*c
- 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(48*b^(7/2))

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Maple [A]  time = 0.006, size = 203, normalized size = 1.4 \begin{align*}{\frac{f{x}^{5}}{6\,b}\sqrt{b{x}^{2}+a}}-{\frac{5\,af{x}^{3}}{24\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{5\,{a}^{2}fx}{16\,{b}^{3}}\sqrt{b{x}^{2}+a}}-{\frac{5\,{a}^{3}f}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}+{\frac{e{x}^{3}}{4\,b}\sqrt{b{x}^{2}+a}}-{\frac{3\,aex}{8\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,{a}^{2}e}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}+{\frac{dx}{2\,b}\sqrt{b{x}^{2}+a}}-{\frac{ad}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{c\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x)

[Out]

1/6*f*x^5*(b*x^2+a)^(1/2)/b-5/24*f/b^2*a*x^3*(b*x^2+a)^(1/2)+5/16*f/b^3*a^2*x*(b*x^2+a)^(1/2)-5/16*f/b^(7/2)*a
^3*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/4*e*x^3/b*(b*x^2+a)^(1/2)-3/8*e/b^2*a*x*(b*x^2+a)^(1/2)+3/8*e/b^(5/2)*a^2*l
n(x*b^(1/2)+(b*x^2+a)^(1/2))+1/2*d*x/b*(b*x^2+a)^(1/2)-1/2*d*a/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+c*ln(x*b^
(1/2)+(b*x^2+a)^(1/2))/b^(1/2)

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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((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52251, size = 581, normalized size = 4.01 \begin{align*} \left [-\frac{3 \,{\left (16 \, b^{3} c - 8 \, a b^{2} d + 6 \, a^{2} b e - 5 \, a^{3} f\right )} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (8 \, b^{3} f x^{5} + 2 \,{\left (6 \, b^{3} e - 5 \, a b^{2} f\right )} x^{3} + 3 \,{\left (8 \, b^{3} d - 6 \, a b^{2} e + 5 \, a^{2} b f\right )} x\right )} \sqrt{b x^{2} + a}}{96 \, b^{4}}, -\frac{3 \,{\left (16 \, b^{3} c - 8 \, a b^{2} d + 6 \, a^{2} b e - 5 \, a^{3} f\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (8 \, b^{3} f x^{5} + 2 \,{\left (6 \, b^{3} e - 5 \, a b^{2} f\right )} x^{3} + 3 \,{\left (8 \, b^{3} d - 6 \, a b^{2} e + 5 \, a^{2} b f\right )} x\right )} \sqrt{b x^{2} + a}}{48 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(16*b^3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a)
- 2*(8*b^3*f*x^5 + 2*(6*b^3*e - 5*a*b^2*f)*x^3 + 3*(8*b^3*d - 6*a*b^2*e + 5*a^2*b*f)*x)*sqrt(b*x^2 + a))/b^4,
-1/48*(3*(16*b^3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*f*
x^5 + 2*(6*b^3*e - 5*a*b^2*f)*x^3 + 3*(8*b^3*d - 6*a*b^2*e + 5*a^2*b*f)*x)*sqrt(b*x^2 + a))/b^4]

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Sympy [A]  time = 12.1099, size = 362, normalized size = 2.5 \begin{align*} \frac{5 a^{\frac{5}{2}} f x}{16 b^{3} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a^{\frac{3}{2}} e x}{8 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 a^{\frac{3}{2}} f x^{3}}{48 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{\sqrt{a} d x \sqrt{1 + \frac{b x^{2}}{a}}}{2 b} - \frac{\sqrt{a} e x^{3}}{8 b \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{\sqrt{a} f x^{5}}{24 b \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{5 a^{3} f \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16 b^{\frac{7}{2}}} + \frac{3 a^{2} e \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 b^{\frac{5}{2}}} - \frac{a d \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{3}{2}}} + c \left (\begin{cases} \frac{\sqrt{- \frac{a}{b}} \operatorname{asin}{\left (x \sqrt{- \frac{b}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge b < 0 \\\frac{\sqrt{\frac{a}{b}} \operatorname{asinh}{\left (x \sqrt{\frac{b}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge b > 0 \\\frac{\sqrt{- \frac{a}{b}} \operatorname{acosh}{\left (x \sqrt{- \frac{b}{a}} \right )}}{\sqrt{- a}} & \text{for}\: b > 0 \wedge a < 0 \end{cases}\right ) + \frac{e x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{f x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**6+e*x**4+d*x**2+c)/(b*x**2+a)**(1/2),x)

[Out]

5*a**(5/2)*f*x/(16*b**3*sqrt(1 + b*x**2/a)) - 3*a**(3/2)*e*x/(8*b**2*sqrt(1 + b*x**2/a)) + 5*a**(3/2)*f*x**3/(
48*b**2*sqrt(1 + b*x**2/a)) + sqrt(a)*d*x*sqrt(1 + b*x**2/a)/(2*b) - sqrt(a)*e*x**3/(8*b*sqrt(1 + b*x**2/a)) -
sqrt(a)*f*x**5/(24*b*sqrt(1 + b*x**2/a)) - 5*a**3*f*asinh(sqrt(b)*x/sqrt(a))/(16*b**(7/2)) + 3*a**2*e*asinh(s
qrt(b)*x/sqrt(a))/(8*b**(5/2)) - a*d*asinh(sqrt(b)*x/sqrt(a))/(2*b**(3/2)) + c*Piecewise((sqrt(-a/b)*asin(x*sq
rt(-b/a))/sqrt(a), (a > 0) & (b < 0)), (sqrt(a/b)*asinh(x*sqrt(b/a))/sqrt(a), (a > 0) & (b > 0)), (sqrt(-a/b)*
acosh(x*sqrt(-b/a))/sqrt(-a), (b > 0) & (a < 0))) + e*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + f*x**7/(6*sqrt(a)*
sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.19111, size = 174, normalized size = 1.2 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (\frac{4 \, f x^{2}}{b} - \frac{5 \, a b^{3} f - 6 \, b^{4} e}{b^{5}}\right )} x^{2} + \frac{3 \,{\left (8 \, b^{4} d + 5 \, a^{2} b^{2} f - 6 \, a b^{3} e\right )}}{b^{5}}\right )} \sqrt{b x^{2} + a} x - \frac{{\left (16 \, b^{3} c - 8 \, a b^{2} d - 5 \, a^{3} f + 6 \, a^{2} b e\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{16 \, b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/48*(2*(4*f*x^2/b - (5*a*b^3*f - 6*b^4*e)/b^5)*x^2 + 3*(8*b^4*d + 5*a^2*b^2*f - 6*a*b^3*e)/b^5)*sqrt(b*x^2 +
a)*x - 1/16*(16*b^3*c - 8*a*b^2*d - 5*a^3*f + 6*a^2*b*e)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)