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

Optimal. Leaf size=132 \[ -\frac{\sqrt{a} (3 b c-2 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 b^2 (b c-a d)^{3/2}}+\frac{a x \sqrt{c+d x^2}}{2 b \left (a+b x^2\right ) (b c-a d)}+\frac{\tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b^2 \sqrt{d}} \]

[Out]

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

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Rubi [A]  time = 0.111655, antiderivative size = 132, 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} \[ -\frac{\sqrt{a} (3 b c-2 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 b^2 (b c-a d)^{3/2}}+\frac{a x \sqrt{c+d x^2}}{2 b \left (a+b x^2\right ) (b c-a d)}+\frac{\tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b^2 \sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.189716, size = 129, normalized size = 0.98 \[ \frac{\frac{a b x \sqrt{c+d x^2}}{\left (a+b x^2\right ) (b c-a d)}+\frac{\sqrt{a} (2 a d-3 b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{3/2}}+\frac{2 \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{\sqrt{d}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.019, size = 846, normalized size = 6.4 \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)^2/(d*x^2+c)^(1/2),x)

[Out]

1/b^2*ln(x*d^(1/2)+(d*x^2+c)^(1/2))/d^(1/2)-1/4/b^2*a/(a*d-b*c)/(x+1/b*(-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)-1/4/b^3*a*d*(-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)))-3/4/b^2*a/(-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)))+3/4/b^2*a/
(-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
)))-1/4/b^2*a/(a*d-b*c)/(x-1/b*(-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)+1/4/b^3*a*d*(-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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 4.88907, size = 2236, normalized size = 16.94 \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)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [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)**2/(d*x**2+c)**(1/2),x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 1.18719, size = 383, normalized size = 2.9 \begin{align*} -\frac{{\left (3 \, a b c \sqrt{d} - 2 \, a^{2} 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 )}{2 \,{\left (b^{3} c - a b^{2} d\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c \sqrt{d} - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} d^{\frac{3}{2}} - a b c^{2} \sqrt{d}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}{\left (b^{3} c - a b^{2} d\right )}} - \frac{\log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{2 \, b^{2} \sqrt{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*a*b*c*sqrt(d) - 2*a^2*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))/((b^3*c - a*b^2*d)*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c*sqrt(d) -
2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^(3/2) - a*b*c^2*sqrt(d))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqr
t(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*(b^3*c - a*b^2*d)) - 1/2*log(
(sqrt(d)*x - sqrt(d*x^2 + c))^2)/(b^2*sqrt(d))