3.8.92 \(\int \frac {1}{x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=149 \[ \frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{8 a^2 c^2 x^2}-\frac {\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{5/2} c^{5/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c x^4} \]

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Rubi [A]  time = 0.14, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {446, 103, 151, 12, 93, 208} \begin {gather*} -\frac {\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{5/2} c^{5/2}}+\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{8 a^2 c^2 x^2}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

-(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(4*a*c*x^4) + (3*(b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*a^2*c^2*x^
2) - ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*a^(
5/2)*c^(5/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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 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]]

Rubi steps

\begin {align*} \int \frac {1}{x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c x^4}-\frac {\operatorname {Subst}\left (\int \frac {\frac {3}{2} (b c+a d)+b d x}{x^2 \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 a c}\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c x^4}+\frac {3 (b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 a^2 c^2 x^2}+\frac {\operatorname {Subst}\left (\int \frac {3 b^2 c^2+2 a b c d+3 a^2 d^2}{4 x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 a^2 c^2}\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c x^4}+\frac {3 (b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 a^2 c^2 x^2}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{16 a^2 c^2}\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c x^4}+\frac {3 (b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 a^2 c^2 x^2}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{8 a^2 c^2}\\ &=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c x^4}+\frac {3 (b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 a^2 c^2 x^2}-\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{5/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 126, normalized size = 0.85 \begin {gather*} \frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (-2 a c+3 a d x^2+3 b c x^2\right )}{8 a^2 c^2 x^4}-\frac {\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{5/2} c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-2*a*c + 3*b*c*x^2 + 3*a*d*x^2))/(8*a^2*c^2*x^4) - ((3*b^2*c^2 + 2*a*b*c*d +
3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*a^(5/2)*c^(5/2))

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IntegrateAlgebraic [F]  time = 1.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

Defer[IntegrateAlgebraic][1/(x^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x]

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fricas [A]  time = 2.04, size = 360, normalized size = 2.42 \begin {gather*} \left [\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {a c} x^{4} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {a c}}{x^{4}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} - 3 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, a^{3} c^{3} x^{4}}, \frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-a c} x^{4} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-a c}}{2 \, {\left (a b c d x^{4} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - 3 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, a^{3} c^{3} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/32*((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*sqrt(a*c)*x^4*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2
+ 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a*c))/x^4) - 4*
(2*a^2*c^2 - 3*(a*b*c^2 + a^2*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a^3*c^3*x^4), 1/16*((3*b^2*c^2 + 2*a
*b*c*d + 3*a^2*d^2)*sqrt(-a*c)*x^4*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-
a*c)/(a*b*c*d*x^4 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x^2)) - 2*(2*a^2*c^2 - 3*(a*b*c^2 + a^2*c*d)*x^2)*sqrt(b*x^2
+ a)*sqrt(d*x^2 + c))/(a^3*c^3*x^4)]

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giac [B]  time = 1.97, size = 1015, normalized size = 6.81 \begin {gather*} -\frac {\sqrt {b d} b^{6} d^{2} {\left (\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} a^{2} b^{5} c^{2} d^{2}} - \frac {2 \, {\left (3 \, b^{8} c^{5} - 9 \, a b^{7} c^{4} d + 6 \, a^{2} b^{6} c^{3} d^{2} + 6 \, a^{3} b^{5} c^{2} d^{3} - 9 \, a^{4} b^{4} c d^{4} + 3 \, a^{5} b^{3} d^{5} - 9 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{6} c^{4} - 4 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b^{5} c^{3} d + 26 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{2} b^{4} c^{2} d^{2} - 4 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{3} b^{3} c d^{3} - 9 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{4} b^{2} d^{4} + 9 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} b^{4} c^{3} + 15 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a b^{3} c^{2} d + 15 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{2} b^{2} c d^{2} + 9 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{3} b d^{3} - 3 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} b^{2} c^{2} - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} a b c d - 3 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} a^{2} d^{2}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )}^{2} a^{2} b^{4} c^{2} d^{2}}\right )}}{8 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/8*sqrt(b*d)*b^6*d^2*((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*x^2 + a)*sqrt
(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a^2*b^5*c^2*d^2) - 2*(3*
b^8*c^5 - 9*a*b^7*c^4*d + 6*a^2*b^6*c^3*d^2 + 6*a^3*b^5*c^2*d^3 - 9*a^4*b^4*c*d^4 + 3*a^5*b^3*d^5 - 9*(sqrt(b*
x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^6*c^4 - 4*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^
2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b^5*c^3*d + 26*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d -
a*b*d))^2*a^2*b^4*c^2*d^2 - 4*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^3*b^3*c
*d^3 - 9*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^4*b^2*d^4 + 9*(sqrt(b*x^2 + a
)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*b^4*c^3 + 15*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c +
(b*x^2 + a)*b*d - a*b*d))^4*a*b^3*c^2*d + 15*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d
))^4*a^2*b^2*c*d^2 + 9*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a^3*b*d^3 - 3*(sq
rt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^6*b^2*c^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sq
rt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^6*a*b*c*d - 3*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d -
a*b*d))^6*a^2*d^2)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2
+ a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b*d +
(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4)^2*a^2*b^4*c^2*d^2))/abs(b)

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maple [B]  time = 0.03, size = 355, normalized size = 2.38 \begin {gather*} -\frac {\left (3 a^{2} d^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+2 a b c d \,x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+3 b^{2} c^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-6 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a d \,x^{2}-6 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b c \,x^{2}+4 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a c \right ) \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{16 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a^{2} c^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/16/a^2/c^2*(3*a^2*d^2*x^4*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)
+2*a*b*c*d*x^4*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)+3*b^2*c^2*x^4
*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)-6*(b*d*x^4+a*d*x^2+b*c*x^2+
a*c)^(1/2)*(a*c)^(1/2)*a*d*x^2-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*b*c*x^2+4*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a*c)*(d*x^2+c)^(1/2)*(b*x^2+a)^(1/2)/(a*c)^(1/2)/x^4/(b*d*x^4+a*d*x^2+b*c*x^2+a*c
)^(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(1/x^5/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),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 zero or nonzero?

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mupad [B]  time = 21.58, size = 962, normalized size = 6.46 \begin {gather*} \frac {\ln \left (\frac {\left (\sqrt {c}\,\sqrt {b\,x^2+a}-\sqrt {a}\,\sqrt {d\,x^2+c}\right )\,\left (b\,\sqrt {c}-\frac {\sqrt {a}\,d\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )\,\left (3\,\sqrt {a}\,b^2\,c^{5/2}+3\,a^{5/2}\,\sqrt {c}\,d^2+2\,a^{3/2}\,b\,c^{3/2}\,d\right )}{16\,a^3\,c^3}-\frac {\ln \left (\frac {\sqrt {b\,x^2+a}-\sqrt {a}}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )\,\left (3\,\sqrt {a}\,b^2\,c^{5/2}+3\,a^{5/2}\,\sqrt {c}\,d^2+2\,a^{3/2}\,b\,c^{3/2}\,d\right )}{16\,a^3\,c^3}-\frac {\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2\,\left (\frac {11\,a^2\,b^2\,d^2}{64}+\frac {5\,a\,b^3\,c\,d}{16}+\frac {11\,b^4\,c^2}{64}\right )}{a^{5/2}\,c^{5/2}\,d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}-\frac {b^4}{64\,a^{3/2}\,c^{3/2}\,d^2}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3\,\left (\frac {a^3\,b\,d^3}{32}-\frac {9\,a^2\,b^2\,c\,d^2}{16}-\frac {9\,a\,b^3\,c^2\,d}{16}+\frac {b^4\,c^3}{32}\right )}{a^3\,c^3\,d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}-\frac {\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (\frac {c\,b^4}{16}+\frac {a\,d\,b^3}{16}\right )}{a^2\,c^2\,d^2\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^5\,\left (\frac {a^3\,d^3}{8}-\frac {7\,a^2\,b\,c\,d^2}{32}-\frac {7\,a\,b^2\,c^2\,d}{32}+\frac {b^3\,c^3}{8}\right )}{a^3\,c^3\,d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^5}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4\,\left (-\frac {7\,a^4\,d^4}{64}+\frac {a^3\,b\,c\,d^3}{8}+\frac {45\,a^2\,b^2\,c^2\,d^2}{64}+\frac {a\,b^3\,c^3\,d}{8}-\frac {7\,b^4\,c^4}{64}\right )}{a^{7/2}\,c^{7/2}\,d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}}{\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^6}{{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^6}+\frac {b^2\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}-\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3\,\left (2\,c\,b^2+2\,a\,d\,b\right )}{\sqrt {a}\,\sqrt {c}\,d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}-\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^5\,\left (2\,a\,d+2\,b\,c\right )}{\sqrt {a}\,\sqrt {c}\,d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^5}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4\,\left (a^2\,d^2+4\,a\,b\,c\,d+b^2\,c^2\right )}{a\,c\,d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}}+\frac {d^2\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{64\,a^{3/2}\,c^{3/2}\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}+\frac {3\,d\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (a\,d+b\,c\right )}{32\,a^2\,c^2\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)),x)

[Out]

(log(((c^(1/2)*(a + b*x^2)^(1/2) - a^(1/2)*(c + d*x^2)^(1/2))*(b*c^(1/2) - (a^(1/2)*d*((a + b*x^2)^(1/2) - a^(
1/2)))/((c + d*x^2)^(1/2) - c^(1/2))))/((c + d*x^2)^(1/2) - c^(1/2)))*(3*a^(1/2)*b^2*c^(5/2) + 3*a^(5/2)*c^(1/
2)*d^2 + 2*a^(3/2)*b*c^(3/2)*d))/(16*a^3*c^3) - (log(((a + b*x^2)^(1/2) - a^(1/2))/((c + d*x^2)^(1/2) - c^(1/2
)))*(3*a^(1/2)*b^2*c^(5/2) + 3*a^(5/2)*c^(1/2)*d^2 + 2*a^(3/2)*b*c^(3/2)*d))/(16*a^3*c^3) - ((((a + b*x^2)^(1/
2) - a^(1/2))^2*((11*b^4*c^2)/64 + (11*a^2*b^2*d^2)/64 + (5*a*b^3*c*d)/16))/(a^(5/2)*c^(5/2)*d^2*((c + d*x^2)^
(1/2) - c^(1/2))^2) - b^4/(64*a^(3/2)*c^(3/2)*d^2) + (((a + b*x^2)^(1/2) - a^(1/2))^3*((b^4*c^3)/32 + (a^3*b*d
^3)/32 - (9*a^2*b^2*c*d^2)/16 - (9*a*b^3*c^2*d)/16))/(a^3*c^3*d^2*((c + d*x^2)^(1/2) - c^(1/2))^3) - (((a + b*
x^2)^(1/2) - a^(1/2))*((b^4*c)/16 + (a*b^3*d)/16))/(a^2*c^2*d^2*((c + d*x^2)^(1/2) - c^(1/2))) + (((a + b*x^2)
^(1/2) - a^(1/2))^5*((a^3*d^3)/8 + (b^3*c^3)/8 - (7*a*b^2*c^2*d)/32 - (7*a^2*b*c*d^2)/32))/(a^3*c^3*d*((c + d*
x^2)^(1/2) - c^(1/2))^5) + (((a + b*x^2)^(1/2) - a^(1/2))^4*((45*a^2*b^2*c^2*d^2)/64 - (7*b^4*c^4)/64 - (7*a^4
*d^4)/64 + (a*b^3*c^3*d)/8 + (a^3*b*c*d^3)/8))/(a^(7/2)*c^(7/2)*d^2*((c + d*x^2)^(1/2) - c^(1/2))^4))/(((a + b
*x^2)^(1/2) - a^(1/2))^6/((c + d*x^2)^(1/2) - c^(1/2))^6 + (b^2*((a + b*x^2)^(1/2) - a^(1/2))^2)/(d^2*((c + d*
x^2)^(1/2) - c^(1/2))^2) - (((a + b*x^2)^(1/2) - a^(1/2))^3*(2*b^2*c + 2*a*b*d))/(a^(1/2)*c^(1/2)*d^2*((c + d*
x^2)^(1/2) - c^(1/2))^3) - (((a + b*x^2)^(1/2) - a^(1/2))^5*(2*a*d + 2*b*c))/(a^(1/2)*c^(1/2)*d*((c + d*x^2)^(
1/2) - c^(1/2))^5) + (((a + b*x^2)^(1/2) - a^(1/2))^4*(a^2*d^2 + b^2*c^2 + 4*a*b*c*d))/(a*c*d^2*((c + d*x^2)^(
1/2) - c^(1/2))^4)) + (d^2*((a + b*x^2)^(1/2) - a^(1/2))^2)/(64*a^(3/2)*c^(3/2)*((c + d*x^2)^(1/2) - c^(1/2))^
2) + (3*d*((a + b*x^2)^(1/2) - a^(1/2))*(a*d + b*c))/(32*a^2*c^2*((c + d*x^2)^(1/2) - c^(1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \sqrt {a + b x^{2}} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(1/(x**5*sqrt(a + b*x**2)*sqrt(c + d*x**2)), x)

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