3.723 \(\int \frac{x^3}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=103 \[ -\frac{a}{\sqrt{c+d x^2} (b c-a d)^2}-\frac{c}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac{a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}} \]

[Out]

-c/(3*d*(b*c - a*d)*(c + d*x^2)^(3/2)) - a/((b*c - a*d)^2*Sqrt[c + d*x^2]) + (a*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[
c + d*x^2])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

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Rubi [A]  time = 0.0920235, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 78, 51, 63, 208} \[ -\frac{a}{\sqrt{c+d x^2} (b c-a d)^2}-\frac{c}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac{a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

-c/(3*d*(b*c - a*d)*(c + d*x^2)^(3/2)) - a/((b*c - a*d)^2*Sqrt[c + d*x^2]) + (a*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[
c + d*x^2])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

Rule 446

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 0.0287797, size = 77, normalized size = 0.75 \[ \frac{c (a d-b c)-3 a d \left (c+d x^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

(c*(-(b*c) + a*d) - 3*a*d*(c + d*x^2)*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^2))/(b*c - a*d)])/(3*d*(b*c
- a*d)^2*(c + d*x^2)^(3/2))

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Maple [B]  time = 0.013, size = 1123, normalized size = 10.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

-1/3/b/d/(d*x^2+c)^(3/2)+1/6/b*a/(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)^(3/2)+1/6/b^2*a*d*(-a*b)^(1/2)/(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)^(3/2)*x+1/3/b^2*a*d*(-a*b)^(1/2)/(a*d-b*c)/c^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-1/2*a/(a*d-b*c)^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*a/(a*d-b*c)^2*(-a*b)^(1/2)/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*a/(a*d-b*c)^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)))+1/6/b*a/(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)^(3/2)-1/6/b^2*a*d*(-a*b)^(1/2)/(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)^(3/2)*x-1/3/b^2*a*d*(-a*b)^(1/2)/(a*d-b*c)/c^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-1/2*a/(a*d-b*c)^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*a/(a*d-b*c)^2*(-a*b)^(1/
2)/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*a/(a*d-b*c)^
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^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.86032, size = 1106, normalized size = 10.74 \begin{align*} \left [\frac{3 \,{\left (a d^{3} x^{4} + 2 \, a c d^{2} x^{2} + a c^{2} d\right )} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \,{\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \,{\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} +{\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \,{\left (3 \, a d^{2} x^{2} + b c^{2} + 2 \, a c d\right )} \sqrt{d x^{2} + c}}{12 \,{\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} +{\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} x^{4} + 2 \,{\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}, -\frac{3 \,{\left (a d^{3} x^{4} + 2 \, a c d^{2} x^{2} + a c^{2} d\right )} \sqrt{-\frac{b}{b c - a d}} \arctan \left (\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b}{b c - a d}}}{2 \,{\left (b d x^{2} + b c\right )}}\right ) + 2 \,{\left (3 \, a d^{2} x^{2} + b c^{2} + 2 \, a c d\right )} \sqrt{d x^{2} + c}}{6 \,{\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} +{\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} x^{4} + 2 \,{\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a*d^3*x^4 + 2*a*c*d^2*x^2 + a*c^2*d)*sqrt(b/(b*c - a*d))*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d +
a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d
*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(3*a*d^2*x^2 + b*c^2 + 2*a*c*d)*sqrt(d*x^2 + c
))/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3 + (b^2*c^2*d^3 - 2*a*b*c*d^4 + a^2*d^5)*x^4 + 2*(b^2*c^3*d^2 - 2*a
*b*c^2*d^3 + a^2*c*d^4)*x^2), -1/6*(3*(a*d^3*x^4 + 2*a*c*d^2*x^2 + a*c^2*d)*sqrt(-b/(b*c - a*d))*arctan(1/2*(b
*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) + 2*(3*a*d^2*x^2 + b*c^2 + 2*a*c*d
)*sqrt(d*x^2 + c))/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3 + (b^2*c^2*d^3 - 2*a*b*c*d^4 + a^2*d^5)*x^4 + 2*(b
^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Integral(x**3/((a + b*x**2)*(c + d*x**2)**(5/2)), x)

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Giac [A]  time = 1.14421, size = 171, normalized size = 1.66 \begin{align*} -\frac{\frac{3 \, a b d \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} + \frac{b c^{2} + 3 \,{\left (d x^{2} + c\right )} a d - a c d}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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