3.8.23 \(\int \frac {x^3}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\) [723]

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

[Out]

-1/3*c/d/(-a*d+b*c)/(d*x^2+c)^(3/2)+a*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))*b^(1/2)/(-a*d+b*c)^(5/
2)-a/(-a*d+b*c)^2/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.00, 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 = {457, 79, 53, 65, 214} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*c/(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]*Sqr
t[c + d*x^2])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

Rule 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 214

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

Rule 457

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]]

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} \text {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 \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.23, size = 100, normalized size = 0.97 \begin {gather*} \frac {-b c^2-a d \left (2 c+3 d x^2\right )}{3 d (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {a \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1408\) vs. \(2(87)=174\).
time = 0.09, size = 1409, normalized size = 13.68

method	result	size
default	\(\text {Expression too large to display}\)	\(1409\)

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,method=_RETURNVERBOSE)

[Out]

-1/3/b/d/(d*x^2+c)^(3/2)-1/2*a/b^2*(-1/3/(a*d-b*c)*b/(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b
)^(1/2))-(a*d-b*c)/b)^(3/2)+d*(-a*b)^(1/2)/(a*d-b*c)*(2/3*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/(-4*d*
(a*d-b*c)/b+4*d^2*a/b)/(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+16
/3*d/(-4*d*(a*d-b*c)/b+4*d^2*a/b)^2*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/(d*(x-1/b*(-a*b)^(1/2))^2+2*
d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))-1/(a*d-b*c)*b*(-1/(a*d-b*c)*b/(d*(x-1/b*(-a*b)^(1/2)
)^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+2*d*(-a*b)^(1/2)/(a*d-b*c)*(2*d*(x-1/b*(-a*b)^(
1/2))+2*d*(-a*b)^(1/2)/b)/(-4*d*(a*d-b*c)/b+4*d^2*a/b)/(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a
*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/(a*d-b*c)*b/(-(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)*(d*(x-1/b*(-a*b)^(1/2))^2+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*a/b^2*(-1/3/(a*d-b*c)*b/(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*
(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)-d*(-a*b)^(1/2)/(a*d-b*c)*(2/3*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/
2)/b)/(-4*d*(a*d-b*c)/b+4*d^2*a/b)/(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)
/b)^(3/2)+16/3*d/(-4*d*(a*d-b*c)/b+4*d^2*a/b)^2*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/(d*(x+1/b*(-a*b)
^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))-1/(a*d-b*c)*b*(-1/(a*d-b*c)*b/(d*(x+1/b*
(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-2*d*(-a*b)^(1/2)/(a*d-b*c)*(2*d*(x+
1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/(-4*d*(a*d-b*c)/b+4*d^2*a/b)/(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/
b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/(a*d-b*c)*b/(-(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)*(d*(x+1/b*(-a*b)^(1/2))^2-2*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (87) = 174\).
time = 1.45, size = 535, normalized size = 5.19 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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 = 1.07, size = 127, normalized size = 1.23 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.54, size = 110, normalized size = 1.07 \begin {gather*} \frac {\frac {c}{3\,\left (a\,d-b\,c\right )}-\frac {a\,d\,\left (d\,x^2+c\right )}{{\left (a\,d-b\,c\right )}^2}}{d\,{\left (d\,x^2+c\right )}^{3/2}}-\frac {a\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )}{{\left (a\,d-b\,c\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c/(3*(a*d - b*c)) - (a*d*(c + d*x^2))/(a*d - b*c)^2)/(d*(c + d*x^2)^(3/2)) - (a*b^(1/2)*atan((b^(1/2)*(c + d*
x^2)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(a*d - b*c)^(5/2)))/(a*d - b*c)^(5/2)

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