3.713 \(\int \frac{x^4}{(a+b x^2) (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=109 \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b (b c-a d)^{3/2}}-\frac{c x}{d \sqrt{c+d x^2} (b c-a d)}+\frac{\tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b d^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.100527, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {470, 523, 217, 206, 377, 205} \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b (b c-a d)^{3/2}}-\frac{c x}{d \sqrt{c+d x^2} (b c-a d)}+\frac{\tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b d^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

Rule 470

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

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

Mathematica [A]  time = 0.166266, size = 111, normalized size = 1.02 \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b (b c-a d)^{3/2}}+\frac{c x}{d \sqrt{c+d x^2} (a d-b c)}+\frac{\log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{b d^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.016, size = 720, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.00503, size = 2066, normalized size = 18.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(4*sqrt(d*x^2 + c)*b*c*d*x - 2*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2
+ c)*sqrt(d)*x - c) + (a*d^3*x^2 + a*c*d^2)*sqrt(-a/(b*c - a*d))*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^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)*s
qrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b
*d^4)*x^2), -1/4*(4*sqrt(d*x^2 + c)*b*c*d*x + 4*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)*sqrt(-d)*arctan(sqrt(-d)
*x/sqrt(d*x^2 + c)) + (a*d^3*x^2 + a*c*d^2)*sqrt(-a/(b*c - a*d))*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^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)*sq
rt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b*
d^4)*x^2), -1/2*(2*sqrt(d*x^2 + c)*b*c*d*x + (a*d^3*x^2 + a*c*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a
*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)) - (b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)*
sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b*d^4)*x^2)
, -1/2*(2*sqrt(d*x^2 + c)*b*c*d*x + 2*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*
x^2 + c)) + (a*d^3*x^2 + a*c*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sq
rt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b*d^4)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**2+a)/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**4/((a + b*x**2)*(c + d*x**2)**(3/2)), x)

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Giac [A]  time = 1.14158, size = 201, normalized size = 1.84 \begin{align*} -\frac{b^{2} c x}{{\left (b^{3} c d - a b^{2} d^{2}\right )} \sqrt{d x^{2} + c}} - \frac{a^{2} \sqrt{d} \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}}{\left (b^{2} c - a b d\right )}} - \frac{\log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{2 \, b d^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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