3.676 \(\int \frac{x^4 \sqrt{c+d x^2}}{a+b x^2} \, dx\)

Optimal. Leaf size=157 \[ -\frac{\left (-8 a^2 d^2+4 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 b^3 d^{3/2}}+\frac{a^{3/2} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^3}+\frac{x \sqrt{c+d x^2} (b c-4 a d)}{8 b^2 d}+\frac{x^3 \sqrt{c+d x^2}}{4 b} \]

[Out]

((b*c - 4*a*d)*x*Sqrt[c + d*x^2])/(8*b^2*d) + (x^3*Sqrt[c + d*x^2])/(4*b) + (a^(3/2)*Sqrt[b*c - a*d]*ArcTan[(S
qrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/b^3 - ((b^2*c^2 + 4*a*b*c*d - 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqr
t[c + d*x^2]])/(8*b^3*d^(3/2))

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Rubi [A]  time = 0.231156, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {478, 582, 523, 217, 206, 377, 205} \[ -\frac{\left (-8 a^2 d^2+4 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 b^3 d^{3/2}}+\frac{a^{3/2} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^3}+\frac{x \sqrt{c+d x^2} (b c-4 a d)}{8 b^2 d}+\frac{x^3 \sqrt{c+d x^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - 4*a*d)*x*Sqrt[c + d*x^2])/(8*b^2*d) + (x^3*Sqrt[c + d*x^2])/(4*b) + (a^(3/2)*Sqrt[b*c - a*d]*ArcTan[(S
qrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/b^3 - ((b^2*c^2 + 4*a*b*c*d - 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqr
t[c + d*x^2]])/(8*b^3*d^(3/2))

Rule 478

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

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

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 377

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.211981, size = 148, normalized size = 0.94 \[ \frac{-\left (-8 a^2 d^2+4 a b c d+b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )+8 a^{3/2} d^{3/2} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )+b \sqrt{d} x \sqrt{c+d x^2} \left (-4 a d+b c+2 b d x^2\right )}{8 b^3 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*x*Sqrt[c + d*x^2]*(b*c - 4*a*d + 2*b*d*x^2) + 8*a^(3/2)*d^(3/2)*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c -
a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])] - (b^2*c^2 + 4*a*b*c*d - 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(8*
b^3*d^(3/2))

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Maple [B]  time = 0.019, size = 1088, normalized size = 6.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/b*x*(d*x^2+c)^(3/2)/d-1/8/b*c/d*x*(d*x^2+c)^(1/2)-1/8/b*c^2/d^(3/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))-1/2/b^2*
a*x*(d*x^2+c)^(1/2)-1/2/b^2*a*c/d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))-1/2/b^2*a^2/(-a*b)^(1/2)*((x+1/b*(-a*b)^
(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/2/b^3*a^2*d^(1/2)*ln((-d*(-a*b)^(1/2)/
b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/
b)^(1/2))-1/2/b^3*a^3/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1
/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/
2))/(x+1/b*(-a*b)^(1/2)))*d+1/2/b^2*a^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/
b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2)
)-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*c+1/2/b^2*a^2/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1
/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/2/b^3*a^2*d^(1/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)
/d^(1/2)+((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+1/2/b^3*a^3/(-a
*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/
2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))
*d-1/2/b^2*a^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*
(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-
1/b*(-a*b)^(1/2)))*c

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.39072, size = 1885, normalized size = 12.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \sqrt{c + d x^{2}}}{a + b x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4*sqrt(c + d*x**2)/(a + b*x**2), x)

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Giac [A]  time = 1.15251, size = 254, normalized size = 1.62 \begin{align*} \frac{1}{8} \, \sqrt{d x^{2} + c} x{\left (\frac{2 \, x^{2}}{b} + \frac{b^{5} c d - 4 \, a b^{4} d^{2}}{b^{6} d^{2}}\right )} - \frac{{\left (a^{2} b c \sqrt{d} - a^{3} d^{\frac{3}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{\sqrt{a b c d - a^{2} d^{2}} b^{3}} + \frac{{\left (b^{2} c^{2} \sqrt{d} + 4 \, a b c d^{\frac{3}{2}} - 8 \, a^{2} d^{\frac{5}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{16 \, b^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(d*x^2 + c)*x*(2*x^2/b + (b^5*c*d - 4*a*b^4*d^2)/(b^6*d^2)) - (a^2*b*c*sqrt(d) - a^3*d^(3/2))*arctan(1
/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*b^3) +
1/16*(b^2*c^2*sqrt(d) + 4*a*b*c*d^(3/2) - 8*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/(b^3*d^2)