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

Optimal. Leaf size=115 \[ -\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 \sqrt {b c-a d}}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 c^{3/2}}-\frac {\sqrt {c+d x^2}}{2 a c x^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 103, 156, 63, 208} \[ -\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 \sqrt {b c-a d}}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 c^{3/2}}-\frac {\sqrt {c+d x^2}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[c + d*x^2]/(2*a*c*x^2) + ((2*b*c + a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*a^2*c^(3/2)) - (b^(3/2)*Arc
Tanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a^2*Sqrt[b*c - a*d])

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

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Mathematica [A]  time = 0.38, size = 109, normalized size = 0.95 \[ \frac {-\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d}}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {a \sqrt {c+d x^2}}{c x^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.30, size = 734, normalized size = 6.38 \[ \left [\frac {b c^{2} x^{2} \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 ) + {\left (2 \, b c + a d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} c^{2} x^{2}}, \frac {b c^{2} x^{2} \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 ) - 2 \, {\left (2 \, b c + a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} c^{2} x^{2}}, \frac {2 \, b c^{2} x^{2} \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 ) + {\left (2 \, b c + a d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} c^{2} x^{2}}, \frac {b c^{2} x^{2} \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 ) - {\left (2 \, b c + a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - \sqrt {d x^{2} + c} a c}{2 \, a^{2} c^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 103, normalized size = 0.90 \[ \frac {b^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c + a d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{2} \sqrt {-c} c} - \frac {\sqrt {d x^{2} + c}}{2 \, a c x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 385, normalized size = 3.35 \[ -\frac {b \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, a^{2}}-\frac {b \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, a^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 a \,c^{\frac {3}{2}}}+\frac {b \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{a^{2} \sqrt {c}}-\frac {\sqrt {d \,x^{2}+c}}{2 a c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/a^2/(-(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*(d
*x^2+c)^(1/2)/a/c/x^2+1/2/a*d/c^(3/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)-1/2*b/a^2/(-(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^(1/2)*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 \[ \int \frac {1}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.11, size = 396, normalized size = 3.44 \[ \frac {\ln \left (\sqrt {d\,x^2+c}\,{\left (b^4\,c-a\,b^3\,d\right )}^{3/2}+b^6\,c^2+a^2\,b^4\,d^2-2\,a\,b^5\,c\,d\right )\,\sqrt {b^4\,c-a\,b^3\,d}}{2\,a^3\,d-2\,a^2\,b\,c}-\frac {\ln \left (\sqrt {d\,x^2+c}\,{\left (b^4\,c-a\,b^3\,d\right )}^{3/2}-b^6\,c^2-a^2\,b^4\,d^2+2\,a\,b^5\,c\,d\right )\,\sqrt {b^4\,c-a\,b^3\,d}}{2\,\left (a^3\,d-a^2\,b\,c\right )}-\frac {\sqrt {d\,x^2+c}}{2\,a\,c\,x^2}-\frac {\mathrm {atan}\left (\frac {b^4\,d^4\,\sqrt {d\,x^2+c}\,3{}\mathrm {i}}{2\,\sqrt {c^3}\,\left (\frac {3\,b^4\,d^4}{2\,c}+\frac {5\,a\,b^3\,d^5}{4\,c^2}+\frac {a^2\,b^2\,d^6}{4\,c^3}\right )}+\frac {b^2\,d^6\,\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{4\,\sqrt {c^3}\,\left (\frac {5\,b^3\,d^5}{4\,a}+\frac {b^2\,d^6}{4\,c}+\frac {3\,b^4\,c\,d^4}{2\,a^2}\right )}+\frac {b^3\,d^5\,\sqrt {d\,x^2+c}\,5{}\mathrm {i}}{4\,\sqrt {c^3}\,\left (\frac {3\,b^4\,d^4}{2\,a}+\frac {5\,b^3\,d^5}{4\,c}+\frac {a\,b^2\,d^6}{4\,c^2}\right )}\right )\,\left (a\,d+2\,b\,c\right )\,1{}\mathrm {i}}{2\,a^2\,\sqrt {c^3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((c + d*x^2)^(1/2)*(b^4*c - a*b^3*d)^(3/2) + b^6*c^2 + a^2*b^4*d^2 - 2*a*b^5*c*d)*(b^4*c - a*b^3*d)^(1/2))
/(2*a^3*d - 2*a^2*b*c) - (log((c + d*x^2)^(1/2)*(b^4*c - a*b^3*d)^(3/2) - b^6*c^2 - a^2*b^4*d^2 + 2*a*b^5*c*d)
*(b^4*c - a*b^3*d)^(1/2))/(2*(a^3*d - a^2*b*c)) - (c + d*x^2)^(1/2)/(2*a*c*x^2) - (atan((b^4*d^4*(c + d*x^2)^(
1/2)*3i)/(2*(c^3)^(1/2)*((3*b^4*d^4)/(2*c) + (5*a*b^3*d^5)/(4*c^2) + (a^2*b^2*d^6)/(4*c^3))) + (b^2*d^6*(c + d
*x^2)^(1/2)*1i)/(4*(c^3)^(1/2)*((5*b^3*d^5)/(4*a) + (b^2*d^6)/(4*c) + (3*b^4*c*d^4)/(2*a^2))) + (b^3*d^5*(c +
d*x^2)^(1/2)*5i)/(4*(c^3)^(1/2)*((3*b^4*d^4)/(2*a) + (5*b^3*d^5)/(4*c) + (a*b^2*d^6)/(4*c^2))))*(a*d + 2*b*c)*
1i)/(2*a^2*(c^3)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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