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

Optimal. Leaf size=114 \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^2 \sqrt{b c-a d}}-\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^2 d^{3/2}}+\frac{x \sqrt{c+d x^2}}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.0939242, antiderivative size = 114, 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 = {479, 523, 217, 206, 377, 205} \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^2 \sqrt{b c-a d}}-\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^2 d^{3/2}}+\frac{x \sqrt{c+d x^2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && 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 ) \sqrt{c+d x^2}} \, dx &=\frac{x \sqrt{c+d x^2}}{2 b d}-\frac{\int \frac{a c+(b c+2 a d) x^2}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{2 b d}\\ &=\frac{x \sqrt{c+d x^2}}{2 b d}+\frac{a^2 \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{b^2}-\frac{(b c+2 a d) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{2 b^2 d}\\ &=\frac{x \sqrt{c+d x^2}}{2 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^2}-\frac{(b c+2 a d) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{2 b^2 d}\\ &=\frac{x \sqrt{c+d x^2}}{2 b d}+\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^2 \sqrt{b c-a d}}-\frac{(b c+2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^2 d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.115657, size = 112, normalized size = 0.98 \[ \frac{\frac{2 a^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{b c-a d}}-\frac{(2 a d+b c) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{d^{3/2}}+\frac{b x \sqrt{c+d x^2}}{d}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 386, normalized size = 3.4 \begin{align*}{\frac{x}{2\,bd}\sqrt{d{x}^{2}+c}}-{\frac{c}{2\,b}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{3}{2}}}}-{\frac{a}{{b}^{2}}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){\frac{1}{\sqrt{d}}}}+{\frac{{a}^{2}}{2\,{b}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}-{\frac{{a}^{2}}{2\,{b}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.06082, size = 1593, normalized size = 13.97 \begin{align*} \left [\frac{a d^{2} \sqrt{-\frac{a}{b c - a d}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} -{\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt{d x^{2} + c} b d x +{\left (b c + 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right )}{4 \, b^{2} d^{2}}, \frac{a d^{2} \sqrt{-\frac{a}{b c - a d}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} -{\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt{d x^{2} + c} b d x + 2 \,{\left (b c + 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right )}{4 \, b^{2} d^{2}}, -\frac{2 \, a d^{2} \sqrt{\frac{a}{b c - a d}} \arctan \left (-\frac{{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c} \sqrt{\frac{a}{b c - a d}}}{2 \,{\left (a d x^{3} + a c x\right )}}\right ) - 2 \, \sqrt{d x^{2} + c} b d x -{\left (b c + 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right )}{4 \, b^{2} d^{2}}, -\frac{a d^{2} \sqrt{\frac{a}{b c - a d}} \arctan \left (-\frac{{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c} \sqrt{\frac{a}{b c - a d}}}{2 \,{\left (a d x^{3} + a c x\right )}}\right ) - \sqrt{d x^{2} + c} b d x -{\left (b c + 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right )}{2 \, b^{2} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(a*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)*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a*d
)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*sqrt(d*x^2 + c)*b*d*x + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2
+ c)*sqrt(d)*x - c))/(b^2*d^2), 1/4*(a*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)) + 2*sqrt(d*x^2 + c)*b*d*x + 2*(b*c + 2*a*d)*s
qrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)))/(b^2*d^2), -1/4*(2*a*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)) - 2*sqrt(d*x^2 + c)*b*d*x - (b*c + 2*a
*d)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c))/(b^2*d^2), -1/2*(a*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)) - sqrt(d*x^2 + c)*b*d*x
- (b*c + 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)))/(b^2*d^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 ) \sqrt{c + d x^{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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.1516, size = 184, normalized size = 1.61 \begin{align*} -\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}} b^{2}} + \frac{\sqrt{d x^{2} + c} x}{2 \, b d} + \frac{{\left (b c \sqrt{d} + 2 \, a d^{\frac{3}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{4 \, b^{2} d^{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)^(1/2),x, algorithm="giac")

[Out]

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