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

Optimal. Leaf size=156 \[ -\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 (b c-a d)^{3/2}}+\frac {(3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 c^{5/2}}-\frac {d (b c-3 a d)}{2 a c^2 \sqrt {c+d x^2} (b c-a d)}-\frac {1}{2 a c x^2 \sqrt {c+d x^2}} \]

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Rubi [A]  time = 0.22, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 103, 152, 156, 63, 208} \begin {gather*} -\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 (b c-a d)^{3/2}}+\frac {(3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 c^{5/2}}-\frac {d (b c-3 a d)}{2 a c^2 \sqrt {c+d x^2} (b c-a d)}-\frac {1}{2 a c x^2 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 152

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 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^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {1}{2 a c x^2 \sqrt {c+d x^2}}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (2 b c+3 a d)+\frac {3 b d x}{2}}{x (a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{2 a c}\\ &=-\frac {d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt {c+d x^2}}-\frac {1}{2 a c x^2 \sqrt {c+d x^2}}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{4} (b c-a d) (2 b c+3 a d)-\frac {1}{4} b d (b c-3 a d) x}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{a c^2 (b c-a d)}\\ &=-\frac {d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt {c+d x^2}}-\frac {1}{2 a c x^2 \sqrt {c+d x^2}}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a^2 (b c-a d)}-\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{4 a^2 c^2}\\ &=-\frac {d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt {c+d x^2}}-\frac {1}{2 a c x^2 \sqrt {c+d x^2}}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{a^2 d (b c-a d)}-\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 a^2 c^2 d}\\ &=-\frac {d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt {c+d x^2}}-\frac {1}{2 a c x^2 \sqrt {c+d x^2}}+\frac {(2 b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 c^{5/2}}-\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 117, normalized size = 0.75 \begin {gather*} \frac {2 b^2 c^2 x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^2+c\right )}{b c-a d}\right )+(a d-b c) \left (x^2 (3 a d+2 b c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^2}{c}+1\right )+a c\right )}{2 a^2 c^2 x^2 \sqrt {c+d x^2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.31, size = 162, normalized size = 1.04 \begin {gather*} \frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{a^2 (a d-b c)^{3/2}}+\frac {(3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 c^{5/2}}+\frac {-a c d-3 a d^2 x^2+b c^2+b c d x^2}{2 a c^2 x^2 \sqrt {c+d x^2} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 2.66, size = 1291, normalized size = 8.28

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.45, size = 172, normalized size = 1.10 \begin {gather*} \frac {b^{3} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (a^{2} b c - a^{3} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {{\left (d x^{2} + c\right )} b c d - 3 \, {\left (d x^{2} + c\right )} a d^{2} + 2 \, a c d^{2}}{2 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} {\left ({\left (d x^{2} + c\right )}^{\frac {3}{2}} - \sqrt {d x^{2} + c} c\right )}} - \frac {{\left (2 \, b c + 3 \, a d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{2} \sqrt {-c} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 763, normalized size = 4.89 \begin {gather*} \frac {b^{2} \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 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, a^{2}}+\frac {b^{2} \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 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, a^{2}}-\frac {b^{2}}{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}}-\frac {b^{2}}{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}}-\frac {\sqrt {-a b}\, b 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 {\sqrt {-a b}\, b 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 {3 d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 a \,c^{\frac {5}{2}}}+\frac {b \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{a^{2} c^{\frac {3}{2}}}-\frac {3 d}{2 \sqrt {d \,x^{2}+c}\, a \,c^{2}}-\frac {b}{\sqrt {d \,x^{2}+c}\, a^{2} c}-\frac {1}{2 \sqrt {d \,x^{2}+c}\, a c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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mupad [B]  time = 1.98, size = 3025, normalized size = 19.39

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2/(b*c^2 - a*c*d) + (d*(c + d*x^2)*(3*a*d - b*c))/(2*a*c^2*(a*d - b*c)))/(c*(c + d*x^2)^(1/2) - (c + d*x^2)
^(3/2)) + (atan((((-b^5*(a*d - b*c)^3)^(1/2)*(((c + d*x^2)^(1/2)*(128*a^3*b^10*c^13*d^2 - 320*a^4*b^9*c^12*d^3
+ 16*a^5*b^8*c^11*d^4 + 496*a^6*b^7*c^10*d^5 - 160*a^7*b^6*c^9*d^6 - 544*a^8*b^5*c^8*d^7 + 528*a^9*b^4*c^7*d^
8 - 144*a^10*b^3*c^6*d^9))/2 - ((-b^5*(a*d - b*c)^3)^(1/2)*(416*a^8*b^6*c^12*d^5 - 32*a^6*b^8*c^14*d^3 - 1024*
a^9*b^5*c^11*d^6 + 1056*a^10*b^4*c^10*d^7 - 512*a^11*b^3*c^9*d^8 + 96*a^12*b^2*c^8*d^9 + ((-b^5*(a*d - b*c)^3)
^(1/2)*(c + d*x^2)^(1/2)*(512*a^7*b^8*c^16*d^2 - 2816*a^8*b^7*c^15*d^3 + 6400*a^9*b^6*c^14*d^4 - 7680*a^10*b^5
*c^13*d^5 + 5120*a^11*b^4*c^12*d^6 - 1792*a^12*b^3*c^11*d^7 + 256*a^13*b^2*c^10*d^8))/(4*a^2*(a*d - b*c)^3)))/
(2*a^2*(a*d - b*c)^3))*1i)/(a^2*(a*d - b*c)^3) + ((-b^5*(a*d - b*c)^3)^(1/2)*(((c + d*x^2)^(1/2)*(128*a^3*b^10
*c^13*d^2 - 320*a^4*b^9*c^12*d^3 + 16*a^5*b^8*c^11*d^4 + 496*a^6*b^7*c^10*d^5 - 160*a^7*b^6*c^9*d^6 - 544*a^8*
b^5*c^8*d^7 + 528*a^9*b^4*c^7*d^8 - 144*a^10*b^3*c^6*d^9))/2 - ((-b^5*(a*d - b*c)^3)^(1/2)*(32*a^6*b^8*c^14*d^
3 - 416*a^8*b^6*c^12*d^5 + 1024*a^9*b^5*c^11*d^6 - 1056*a^10*b^4*c^10*d^7 + 512*a^11*b^3*c^9*d^8 - 96*a^12*b^2
*c^8*d^9 + ((-b^5*(a*d - b*c)^3)^(1/2)*(c + d*x^2)^(1/2)*(512*a^7*b^8*c^16*d^2 - 2816*a^8*b^7*c^15*d^3 + 6400*
a^9*b^6*c^14*d^4 - 7680*a^10*b^5*c^13*d^5 + 5120*a^11*b^4*c^12*d^6 - 1792*a^12*b^3*c^11*d^7 + 256*a^13*b^2*c^1
0*d^8))/(4*a^2*(a*d - b*c)^3)))/(2*a^2*(a*d - b*c)^3))*1i)/(a^2*(a*d - b*c)^3))/(32*a^2*b^10*c^11*d^3 - 144*a^
3*b^9*c^10*d^4 + 96*a^4*b^8*c^9*d^5 + 256*a^5*b^7*c^8*d^6 - 384*a^6*b^6*c^7*d^7 + 144*a^7*b^5*c^6*d^8 - ((-b^5
*(a*d - b*c)^3)^(1/2)*(((c + d*x^2)^(1/2)*(128*a^3*b^10*c^13*d^2 - 320*a^4*b^9*c^12*d^3 + 16*a^5*b^8*c^11*d^4
+ 496*a^6*b^7*c^10*d^5 - 160*a^7*b^6*c^9*d^6 - 544*a^8*b^5*c^8*d^7 + 528*a^9*b^4*c^7*d^8 - 144*a^10*b^3*c^6*d^
9))/2 - ((-b^5*(a*d - b*c)^3)^(1/2)*(416*a^8*b^6*c^12*d^5 - 32*a^6*b^8*c^14*d^3 - 1024*a^9*b^5*c^11*d^6 + 1056
*a^10*b^4*c^10*d^7 - 512*a^11*b^3*c^9*d^8 + 96*a^12*b^2*c^8*d^9 + ((-b^5*(a*d - b*c)^3)^(1/2)*(c + d*x^2)^(1/2
)*(512*a^7*b^8*c^16*d^2 - 2816*a^8*b^7*c^15*d^3 + 6400*a^9*b^6*c^14*d^4 - 7680*a^10*b^5*c^13*d^5 + 5120*a^11*b
^4*c^12*d^6 - 1792*a^12*b^3*c^11*d^7 + 256*a^13*b^2*c^10*d^8))/(4*a^2*(a*d - b*c)^3)))/(2*a^2*(a*d - b*c)^3)))
/(a^2*(a*d - b*c)^3) + ((-b^5*(a*d - b*c)^3)^(1/2)*(((c + d*x^2)^(1/2)*(128*a^3*b^10*c^13*d^2 - 320*a^4*b^9*c^
12*d^3 + 16*a^5*b^8*c^11*d^4 + 496*a^6*b^7*c^10*d^5 - 160*a^7*b^6*c^9*d^6 - 544*a^8*b^5*c^8*d^7 + 528*a^9*b^4*
c^7*d^8 - 144*a^10*b^3*c^6*d^9))/2 - ((-b^5*(a*d - b*c)^3)^(1/2)*(32*a^6*b^8*c^14*d^3 - 416*a^8*b^6*c^12*d^5 +
1024*a^9*b^5*c^11*d^6 - 1056*a^10*b^4*c^10*d^7 + 512*a^11*b^3*c^9*d^8 - 96*a^12*b^2*c^8*d^9 + ((-b^5*(a*d - b
*c)^3)^(1/2)*(c + d*x^2)^(1/2)*(512*a^7*b^8*c^16*d^2 - 2816*a^8*b^7*c^15*d^3 + 6400*a^9*b^6*c^14*d^4 - 7680*a^
10*b^5*c^13*d^5 + 5120*a^11*b^4*c^12*d^6 - 1792*a^12*b^3*c^11*d^7 + 256*a^13*b^2*c^10*d^8))/(4*a^2*(a*d - b*c)
^3)))/(2*a^2*(a*d - b*c)^3)))/(a^2*(a*d - b*c)^3)))*(-b^5*(a*d - b*c)^3)^(1/2)*1i)/(a^2*(a*d - b*c)^3) + (atan
h((440*a^4*b^8*c^11*d^5*(c + d*x^2)^(1/2))/((c^5)^(1/2)*(440*a^4*b^8*c^9*d^5 - 240*a^3*b^9*c^10*d^4 + 480*a^5*
b^7*c^8*d^6 - 1464*a^6*b^6*c^7*d^7 + 496*a^7*b^5*c^6*d^8 + 936*a^8*b^4*c^5*d^9 - 864*a^9*b^3*c^4*d^10 + 216*a^
10*b^2*c^3*d^11)) - (240*a^3*b^9*c^12*d^4*(c + d*x^2)^(1/2))/((c^5)^(1/2)*(440*a^4*b^8*c^9*d^5 - 240*a^3*b^9*c
^10*d^4 + 480*a^5*b^7*c^8*d^6 - 1464*a^6*b^6*c^7*d^7 + 496*a^7*b^5*c^6*d^8 + 936*a^8*b^4*c^5*d^9 - 864*a^9*b^3
*c^4*d^10 + 216*a^10*b^2*c^3*d^11)) + (480*a^5*b^7*c^10*d^6*(c + d*x^2)^(1/2))/((c^5)^(1/2)*(440*a^4*b^8*c^9*d
^5 - 240*a^3*b^9*c^10*d^4 + 480*a^5*b^7*c^8*d^6 - 1464*a^6*b^6*c^7*d^7 + 496*a^7*b^5*c^6*d^8 + 936*a^8*b^4*c^5
*d^9 - 864*a^9*b^3*c^4*d^10 + 216*a^10*b^2*c^3*d^11)) - (1464*a^6*b^6*c^9*d^7*(c + d*x^2)^(1/2))/((c^5)^(1/2)*
(440*a^4*b^8*c^9*d^5 - 240*a^3*b^9*c^10*d^4 + 480*a^5*b^7*c^8*d^6 - 1464*a^6*b^6*c^7*d^7 + 496*a^7*b^5*c^6*d^8
+ 936*a^8*b^4*c^5*d^9 - 864*a^9*b^3*c^4*d^10 + 216*a^10*b^2*c^3*d^11)) + (496*a^7*b^5*c^8*d^8*(c + d*x^2)^(1/
2))/((c^5)^(1/2)*(440*a^4*b^8*c^9*d^5 - 240*a^3*b^9*c^10*d^4 + 480*a^5*b^7*c^8*d^6 - 1464*a^6*b^6*c^7*d^7 + 49
6*a^7*b^5*c^6*d^8 + 936*a^8*b^4*c^5*d^9 - 864*a^9*b^3*c^4*d^10 + 216*a^10*b^2*c^3*d^11)) + (936*a^8*b^4*c^7*d^
9*(c + d*x^2)^(1/2))/((c^5)^(1/2)*(440*a^4*b^8*c^9*d^5 - 240*a^3*b^9*c^10*d^4 + 480*a^5*b^7*c^8*d^6 - 1464*a^6
*b^6*c^7*d^7 + 496*a^7*b^5*c^6*d^8 + 936*a^8*b^4*c^5*d^9 - 864*a^9*b^3*c^4*d^10 + 216*a^10*b^2*c^3*d^11)) - (8
64*a^9*b^3*c^6*d^10*(c + d*x^2)^(1/2))/((c^5)^(1/2)*(440*a^4*b^8*c^9*d^5 - 240*a^3*b^9*c^10*d^4 + 480*a^5*b^7*
c^8*d^6 - 1464*a^6*b^6*c^7*d^7 + 496*a^7*b^5*c^6*d^8 + 936*a^8*b^4*c^5*d^9 - 864*a^9*b^3*c^4*d^10 + 216*a^10*b
^2*c^3*d^11)) + (216*a^10*b^2*c^5*d^11*(c + d*x^2)^(1/2))/((c^5)^(1/2)*(440*a^4*b^8*c^9*d^5 - 240*a^3*b^9*c^10
*d^4 + 480*a^5*b^7*c^8*d^6 - 1464*a^6*b^6*c^7*d^7 + 496*a^7*b^5*c^6*d^8 + 936*a^8*b^4*c^5*d^9 - 864*a^9*b^3*c^
4*d^10 + 216*a^10*b^2*c^3*d^11)))*(3*a*d + 2*b*c))/(2*a^2*(c^5)^(1/2))

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

[Out]

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

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