3.8.3 \(\int \frac {1}{x^4 (a+b x^2) (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=176 \[ \frac {b^3 \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} (b c-a d)^{3/2}}+\frac {\sqrt {c+d x^2} (3 b c-4 a d) (2 a d+b c)}{3 a^2 c^3 x (b c-a d)}-\frac {\sqrt {c+d x^2} (b c-4 a d)}{3 a c^2 x^3 (b c-a d)}-\frac {d}{c x^3 \sqrt {c+d x^2} (b c-a d)} \]

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Rubi [A]  time = 0.22, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {472, 583, 12, 377, 205} \begin {gather*} \frac {b^3 \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} (b c-a d)^{3/2}}+\frac {\sqrt {c+d x^2} (3 b c-4 a d) (2 a d+b c)}{3 a^2 c^3 x (b c-a d)}-\frac {\sqrt {c+d x^2} (b c-4 a d)}{3 a c^2 x^3 (b c-a d)}-\frac {d}{c x^3 \sqrt {c+d x^2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

Rule 12

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

Rule 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]

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 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

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

Rubi steps

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

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Mathematica [A]  time = 5.25, size = 124, normalized size = 0.70 \begin {gather*} \frac {b^3 \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} (b c-a d)^{3/2}}+\frac {\sqrt {c+d x^2} \left (\frac {x^2 (5 a d+3 b c)}{a^2}+\frac {3 d^3 x^4}{\left (c+d x^2\right ) (a d-b c)}-\frac {c}{a}\right )}{3 c^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.55, size = 225, normalized size = 1.28 \begin {gather*} \frac {-a^2 c^2 d+4 a^2 c d^2 x^2+8 a^2 d^3 x^4+a b c^3-a b c^2 d x^2-2 a b c d^2 x^4-3 b^2 c^3 x^2-3 b^2 c^2 d x^4}{3 a^2 c^3 x^3 \sqrt {c+d x^2} (a d-b c)}-\frac {b^3 \tan ^{-1}\left (\frac {b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}-\frac {b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{a^{5/2} (b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^4*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

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fricas [B]  time = 1.75, size = 706, normalized size = 4.01 \begin {gather*} \left [\frac {3 \, {\left (b^{3} c^{3} d x^{5} + b^{3} c^{4} x^{3}\right )} \sqrt {-a b c + a^{2} 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 c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2} - {\left (3 \, a b^{3} c^{3} d - a^{2} b^{2} c^{2} d^{2} - 10 \, a^{3} b c d^{3} + 8 \, a^{4} d^{4}\right )} x^{4} - {\left (3 \, a b^{3} c^{4} - 2 \, a^{2} b^{2} c^{3} d - 5 \, a^{3} b c^{2} d^{2} + 4 \, a^{4} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left ({\left (a^{3} b^{2} c^{5} d - 2 \, a^{4} b c^{4} d^{2} + a^{5} c^{3} d^{3}\right )} x^{5} + {\left (a^{3} b^{2} c^{6} - 2 \, a^{4} b c^{5} d + a^{5} c^{4} d^{2}\right )} x^{3}\right )}}, \frac {3 \, {\left (b^{3} c^{3} d x^{5} + b^{3} c^{4} x^{3}\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2} - {\left (3 \, a b^{3} c^{3} d - a^{2} b^{2} c^{2} d^{2} - 10 \, a^{3} b c d^{3} + 8 \, a^{4} d^{4}\right )} x^{4} - {\left (3 \, a b^{3} c^{4} - 2 \, a^{2} b^{2} c^{3} d - 5 \, a^{3} b c^{2} d^{2} + 4 \, a^{4} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left ({\left (a^{3} b^{2} c^{5} d - 2 \, a^{4} b c^{4} d^{2} + a^{5} c^{3} d^{3}\right )} x^{5} + {\left (a^{3} b^{2} c^{6} - 2 \, a^{4} b c^{5} d + a^{5} c^{4} d^{2}\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(-a*b*c + a^2*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*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*
x^4 + 2*a*b*x^2 + a^2)) - 4*(a^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 - (3*a*b^3*c^3*d - a^2*b^2*c^2*d^2 - 10
*a^3*b*c*d^3 + 8*a^4*d^4)*x^4 - (3*a*b^3*c^4 - 2*a^2*b^2*c^3*d - 5*a^3*b*c^2*d^2 + 4*a^4*c*d^3)*x^2)*sqrt(d*x^
2 + c))/((a^3*b^2*c^5*d - 2*a^4*b*c^4*d^2 + a^5*c^3*d^3)*x^5 + (a^3*b^2*c^6 - 2*a^4*b*c^5*d + a^5*c^4*d^2)*x^3
), 1/6*(3*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2
- a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - 2*(a^2*b^2*c^4 - 2*a^3*b*c^3*d + a
^4*c^2*d^2 - (3*a*b^3*c^3*d - a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 8*a^4*d^4)*x^4 - (3*a*b^3*c^4 - 2*a^2*b^2*c^3
*d - 5*a^3*b*c^2*d^2 + 4*a^4*c*d^3)*x^2)*sqrt(d*x^2 + c))/((a^3*b^2*c^5*d - 2*a^4*b*c^4*d^2 + a^5*c^3*d^3)*x^5
+ (a^3*b^2*c^6 - 2*a^4*b*c^5*d + a^5*c^4*d^2)*x^3)]

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giac [A]  time = 3.62, size = 275, normalized size = 1.56 \begin {gather*} \frac {b^{3} \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 )}{{\left (a^{2} b c - a^{3} d\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {d^{3} x}{{\left (b c^{4} - a c^{3} d\right )} \sqrt {d x^{2} + c}} - \frac {2 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} \sqrt {d} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c d^{\frac {3}{2}} + 3 \, b c^{3} \sqrt {d} + 5 \, a c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

b^3*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))/((a^2*b*c -
a^3*d)*sqrt(a*b*c*d - a^2*d^2)) - d^3*x/((b*c^4 - a*c^3*d)*sqrt(d*x^2 + c)) - 2/3*(3*(sqrt(d)*x - sqrt(d*x^2
+ c))^4*b*c*sqrt(d) + 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^2*sq
rt(d) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*d^(3/2) + 3*b*c^3*sqrt(d) + 5*a*c^2*d^(3/2))/(((sqrt(d)*x - sqr
t(d*x^2 + c))^2 - c)^3*a^2*c^2)

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maple [B]  time = 0.02, size = 762, normalized size = 4.33 \begin {gather*} \frac {b^{3} \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, a^{2}}-\frac {b^{3} \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, a^{2}}+\frac {b^{3}}{2 \sqrt {-a b}\, \left (a d -b c \right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a^{2}}-\frac {b^{3}}{2 \sqrt {-a b}\, \left (a d -b c \right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a^{2}}+\frac {b^{2} d x}{2 \left (a d -b c \right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a^{2} c}+\frac {b^{2} d x}{2 \left (a d -b c \right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a^{2} c}+\frac {8 d^{2} x}{3 \sqrt {d \,x^{2}+c}\, a \,c^{3}}+\frac {2 b d x}{\sqrt {d \,x^{2}+c}\, a^{2} c^{2}}+\frac {4 d}{3 \sqrt {d \,x^{2}+c}\, a \,c^{2} x}+\frac {b}{\sqrt {d \,x^{2}+c}\, a^{2} c x}-\frac {1}{3 \sqrt {d \,x^{2}+c}\, a c \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x)

[Out]

1/2*b^3/a^2/(-a*b)^(1/2)/(a*d-b*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)+1/2*b^2/a^2/(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)^(1/2)
*d*x-1/2*b^3/a^2/(-a*b)^(1/2)/(a*d-b*c)/(-(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))-1/2*b^3/a^2/(-a*b)^(1/2)/(a*d-b*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)+1/2*b^2/a^2/(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)^(1/2)*d*x+1/2*b^3/a^2/(-a*b)^(1/2)/(a*d-b*c)/(-(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/a^2*b/c/x/(d*x^2+c)^(1/2)+2/a^2*b*d/c^2*x/(d*x^2+c)^(1/2)-1/3/a
/c/x^3/(d*x^2+c)^(1/2)+4/3/a*d/c^2/x/(d*x^2+c)^(1/2)+8/3/a*d^2/c^3*x/(d*x^2+c)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)*(d*x^2 + c)^(3/2)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^4\,\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^(3/2)),x)

[Out]

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**2+a)/(d*x**2+c)**(3/2),x)

[Out]

Integral(1/(x**4*(a + b*x**2)*(c + d*x**2)**(3/2)), x)

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