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

Optimal. Leaf size=140 \[ \frac {5 b^{3/2} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 (b c-a d)^{7/2}}-\frac {5 b d}{2 \sqrt {c+d x^2} (b c-a d)^3}-\frac {1}{2 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {5 d}{6 \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {444, 51, 63, 208} \begin {gather*} \frac {5 b^{3/2} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 (b c-a d)^{7/2}}-\frac {5 b d}{2 \sqrt {c+d x^2} (b c-a d)^3}-\frac {1}{2 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {5 d}{6 \left (c+d x^2\right )^{3/2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d)/(6*(b*c - a*d)^2*(c + d*x^2)^(3/2)) - 1/(2*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) - (5*b*d)/(2*(b*c
- a*d)^3*Sqrt[c + d*x^2]) + (5*b^(3/2)*d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*(b*c - a*d)^(
7/2))

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]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {(5 d) \operatorname {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{5/2}} \, dx,x,x^2\right )}{4 (b c-a d)}\\ &=-\frac {5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {(5 b d) \operatorname {Subst}\left (\int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 (b c-a d)^2}\\ &=-\frac {5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {5 b d}{2 (b c-a d)^3 \sqrt {c+d x^2}}-\frac {\left (5 b^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 (b c-a d)^3}\\ &=-\frac {5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {5 b d}{2 (b c-a d)^3 \sqrt {c+d x^2}}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 (b c-a d)^3}\\ &=-\frac {5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {5 b d}{2 (b c-a d)^3 \sqrt {c+d x^2}}+\frac {5 b^{3/2} d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 (b c-a d)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.39 \begin {gather*} -\frac {d \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {b \left (d x^2+c\right )}{a d-b c}\right )}{3 \left (c+d x^2\right )^{3/2} (a d-b c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.36, size = 151, normalized size = 1.08 \begin {gather*} \frac {2 a^2 d^2-14 a b c d-10 a b d^2 x^2-3 b^2 c^2-20 b^2 c d x^2-15 b^2 d^2 x^4}{6 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)^3}-\frac {5 b^{3/2} d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{2 (a d-b c)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2*c^2 - 14*a*b*c*d + 2*a^2*d^2 - 20*b^2*c*d*x^2 - 10*a*b*d^2*x^2 - 15*b^2*d^2*x^4)/(6*(b*c - a*d)^3*(a +
b*x^2)*(c + d*x^2)^(3/2)) - (5*b^(3/2)*d*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x^2])/(b*c - a*d)])/(2
*(-(b*c) + a*d)^(7/2))

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fricas [B]  time = 1.14, size = 895, normalized size = 6.39 \begin {gather*} \left [-\frac {15 \, {\left (b^{2} d^{3} x^{6} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{4} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x^{2}\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 (15 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{6} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{4} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x^{2}\right )}}, -\frac {15 \, {\left (b^{2} d^{3} x^{6} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{4} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x^{2}\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 (15 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{6} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{4} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(15*(b^2*d^3*x^6 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^4 + (b^2*c^2*d + 2*a*b*c*d^2)*x^2)*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*(15*b^2*d^2*x^4 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a
*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4
- a^3*b*d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5
- a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x^2), -1/12*(15*(b^2*d^3*x^6 + a*b*c^2*d +
(2*b^2*c*d^2 + a*b*d^3)*x^4 + (b^2*c^2*d + 2*a*b*c*d^2)*x^2)*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*(15*b^2*d^2*x^4 + 3*b^2*c^2 + 14*a*b*c*d - 2*
a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*
c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 +
3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3
- 2*a^4*c*d^4)*x^2)]

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giac [A]  time = 0.56, size = 226, normalized size = 1.61 \begin {gather*} -\frac {5 \, b^{2} d \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {\sqrt {d x^{2} + c} b^{2} d}{2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )}} - \frac {6 \, {\left (d x^{2} + c\right )} b d + b c d - a d^{2}}{3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*b^2*d*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)
*sqrt(-b^2*c + a*b*d)) - 1/2*sqrt(d*x^2 + c)*b^2*d/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x^
2 + c)*b - b*c + a*d)) - 1/3*(6*(d*x^2 + c)*b*d + b*c*d - a*d^2)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a
^3*d^3)*(d*x^2 + c)^(3/2))

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maple [B]  time = 0.02, size = 1639, normalized size = 11.71

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-a*b)^(1/2)/a/b/(a*d-b*c)/(x-(-a*b)^(1/2)/b)/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b
*d-(a*d-b*c)/b)^(3/2)-5/12*d/(a*d-b*c)^2/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*
c)/b)^(3/2)+5/12*(-a*b)^(1/2)/b*d^2/(a*d-b*c)^2/c/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*
d-(a*d-b*c)/b)^(3/2)*x+5/6*(-a*b)^(1/2)/b*d^2/(a*d-b*c)^2/c^2/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)
^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x+5/4*b*d/(a*d-b*c)^3/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/
b)/b*d-(a*d-b*c)/b)^(1/2)-5/4*(-a*b)^(1/2)*d^2/(a*d-b*c)^3/c/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^
(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-5/4*b*d/(a*d-b*c)^3/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/
b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d
-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))-1/3*(-a*b)^(1/2)/a/b*d/(a*d-b*c)/c/((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*
(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*x-2/3*(-a*b)^(1/2)/a/b*d/(a*d-b*c)/c^2/((x-(-a*b)^(1/2)/b)^2*d+2*(-a
*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x+1/4*(-a*b)^(1/2)/a/b/(a*d-b*c)/(x+(-a*b)^(1/2)/b)/((x+(-
a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)-5/12*d/(a*d-b*c)^2/((x+(-a*b)^(1/2)
/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)-5/12*(-a*b)^(1/2)/b*d^2/(a*d-b*c)^2/c/((x+(-a
*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*x-5/6*(-a*b)^(1/2)/b*d^2/(a*d-b*c)^2
/c^2/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x+5/4*b*d/(a*d-b*c)^3/((
x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)+5/4*(-a*b)^(1/2)*d^2/(a*d-b*c)^
3/c/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-5/4*b*d/(a*d-b*c)^3/(-(
a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(
1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))+1/3*(-a*b)^(1/2)/a/b
*d/(a*d-b*c)/c/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*x+2/3*(-a*b)^(
1/2)/a/b*d/(a*d-b*c)/c^2/((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x

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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/(b*x^2+a)^2/(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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 1.24, size = 171, normalized size = 1.22 \begin {gather*} \frac {\frac {5\,b^2\,d\,{\left (d\,x^2+c\right )}^2}{2\,{\left (a\,d-b\,c\right )}^3}-\frac {d}{3\,\left (a\,d-b\,c\right )}+\frac {5\,b\,d\,\left (d\,x^2+c\right )}{3\,{\left (a\,d-b\,c\right )}^2}}{b\,{\left (d\,x^2+c\right )}^{5/2}+{\left (d\,x^2+c\right )}^{3/2}\,\left (a\,d-b\,c\right )}+\frac {5\,b^{3/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{2\,{\left (a\,d-b\,c\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*b^2*d*(c + d*x^2)^2)/(2*(a*d - b*c)^3) - d/(3*(a*d - b*c)) + (5*b*d*(c + d*x^2))/(3*(a*d - b*c)^2))/(b*(c
+ d*x^2)^(5/2) + (c + d*x^2)^(3/2)*(a*d - b*c)) + (5*b^(3/2)*d*atan((b^(1/2)*(c + d*x^2)^(1/2)*(a^3*d^3 - b^3*
c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(a*d - b*c)^(7/2)))/(2*(a*d - b*c)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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