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

Optimal. Leaf size=114 \[ -\frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 \sqrt {b}}+\frac {\sqrt {c} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2}-\frac {c \sqrt {c+d x^2}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 114, 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, 98, 156, 63, 208} \[ -\frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 \sqrt {b}}+\frac {\sqrt {c} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2}-\frac {c \sqrt {c+d x^2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 98

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

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

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Mathematica [A]  time = 0.17, size = 108, normalized size = 0.95 \[ \frac {-\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {b}}+\sqrt {c} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\frac {a c \sqrt {c+d x^2}}{x^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.78, size = 732, normalized size = 6.42 \[ \left [-\frac {{\left (b c - a d\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \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 (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + {\left (2 \, b c - 3 \, 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} x^{2}}, -\frac {2 \, {\left (2 \, b c - 3 \, a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (b c - a d\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \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 (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} x^{2}}, -\frac {2 \, {\left (b c - a d\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c - 3 \, 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} x^{2}}, -\frac {{\left (b c - a d\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c - 3 \, 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} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((b*c - a*d)*x^2*sqrt((b*c - a*d)/b)*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*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 +
a^2)) + (2*b*c - 3*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*x^2), -1/4*(2*(2*b*c - 3*a*d)*sqrt(-c)*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (b*c - a*d)*x^2*sqrt((b*
c - a*d)/b)*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*(b^2*d*x^2
+ 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*sqrt(d*x^2 + c)*a*c)/
(a^2*x^2), -1/4*(2*(b*c - a*d)*x^2*sqrt(-(b*c - a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sq
rt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) + (2*b*c - 3*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*x^2), -1/2*((b*c - a*d)*x^2*sqrt(-(b*c - a*d)/b)*a
rctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2))
+ (2*b*c - 3*a*d)*sqrt(-c)*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + sqrt(d*x^2 + c)*a*c)/(a^2*x^2)]

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giac [A]  time = 0.41, size = 120, normalized size = 1.05 \[ \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{2} \sqrt {-c}} - \frac {\sqrt {d x^{2} + c} c}{2 \, a x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 2003, normalized size = 17.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.16, size = 560, normalized size = 4.91 \[ -\frac {c\,\sqrt {d\,x^2+c}}{2\,a\,x^2}-\frac {\sqrt {c}\,\mathrm {atanh}\left (\frac {29\,b^2\,c^{3/2}\,d^6\,\sqrt {d\,x^2+c}}{4\,\left (\frac {29\,b^2\,c^2\,d^6}{4}-3\,a\,b\,c\,d^7-\frac {23\,b^3\,c^3\,d^5}{4\,a}+\frac {3\,b^4\,c^4\,d^4}{2\,a^2}\right )}+\frac {23\,b^3\,c^{5/2}\,d^5\,\sqrt {d\,x^2+c}}{4\,\left (\frac {23\,b^3\,c^3\,d^5}{4}-\frac {29\,a\,b^2\,c^2\,d^6}{4}-\frac {3\,b^4\,c^4\,d^4}{2\,a}+3\,a^2\,b\,c\,d^7\right )}+\frac {3\,b^4\,c^{7/2}\,d^4\,\sqrt {d\,x^2+c}}{2\,\left (-3\,a^3\,b\,c\,d^7+\frac {29\,a^2\,b^2\,c^2\,d^6}{4}-\frac {23\,a\,b^3\,c^3\,d^5}{4}+\frac {3\,b^4\,c^4\,d^4}{2}\right )}-\frac {3\,a\,b\,\sqrt {c}\,d^7\,\sqrt {d\,x^2+c}}{\frac {29\,b^2\,c^2\,d^6}{4}-3\,a\,b\,c\,d^7-\frac {23\,b^3\,c^3\,d^5}{4\,a}+\frac {3\,b^4\,c^4\,d^4}{2\,a^2}}\right )\,\left (3\,a\,d-2\,b\,c\right )}{2\,a^2}-\frac {\mathrm {atanh}\left (\frac {3\,b^2\,c^2\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}}{2\,\left (-2\,a^3\,b\,c\,d^7+\frac {11\,a^2\,b^2\,c^2\,d^6}{2}-5\,a\,b^3\,c^3\,d^5+\frac {3\,b^4\,c^4\,d^4}{2}\right )}+\frac {2\,b\,c\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}}{5\,b^3\,c^3\,d^5-\frac {11\,a\,b^2\,c^2\,d^6}{2}-\frac {3\,b^4\,c^4\,d^4}{2\,a}+2\,a^2\,b\,c\,d^7}\right )\,\sqrt {-b\,{\left (a\,d-b\,c\right )}^3}}{a^2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (c*(c + d*x^2)^(1/2))/(2*a*x^2) - (c^(1/2)*atanh((29*b^2*c^(3/2)*d^6*(c + d*x^2)^(1/2))/(4*((29*b^2*c^2*d^6)
/4 - 3*a*b*c*d^7 - (23*b^3*c^3*d^5)/(4*a) + (3*b^4*c^4*d^4)/(2*a^2))) + (23*b^3*c^(5/2)*d^5*(c + d*x^2)^(1/2))
/(4*((23*b^3*c^3*d^5)/4 - (29*a*b^2*c^2*d^6)/4 - (3*b^4*c^4*d^4)/(2*a) + 3*a^2*b*c*d^7)) + (3*b^4*c^(7/2)*d^4*
(c + d*x^2)^(1/2))/(2*((3*b^4*c^4*d^4)/2 - (23*a*b^3*c^3*d^5)/4 + (29*a^2*b^2*c^2*d^6)/4 - 3*a^3*b*c*d^7)) - (
3*a*b*c^(1/2)*d^7*(c + d*x^2)^(1/2))/((29*b^2*c^2*d^6)/4 - 3*a*b*c*d^7 - (23*b^3*c^3*d^5)/(4*a) + (3*b^4*c^4*d
^4)/(2*a^2)))*(3*a*d - 2*b*c))/(2*a^2) - (atanh((3*b^2*c^2*d^4*(c + d*x^2)^(1/2)*(b^4*c^3 - a^3*b*d^3 + 3*a^2*
b^2*c*d^2 - 3*a*b^3*c^2*d)^(1/2))/(2*((3*b^4*c^4*d^4)/2 - 5*a*b^3*c^3*d^5 + (11*a^2*b^2*c^2*d^6)/2 - 2*a^3*b*c
*d^7)) + (2*b*c*d^5*(c + d*x^2)^(1/2)*(b^4*c^3 - a^3*b*d^3 + 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2*d)^(1/2))/(5*b^3*c^
3*d^5 - (11*a*b^2*c^2*d^6)/2 - (3*b^4*c^4*d^4)/(2*a) + 2*a^2*b*c*d^7))*(-b*(a*d - b*c)^3)^(1/2))/(a^2*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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