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

Optimal. Leaf size=172 \[ \frac {a \left (c+d x^2\right )^2 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 d^2}+\frac {(5 b-4 a c) \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{8 d^2}+\frac {b c \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{d^2}+\frac {3 b (b-4 a c) \tanh ^{-1}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{8 \sqrt {a} d^2} \]

[Out]

3/8*b*(-4*a*c+b)*arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^(1/2))/d^2/a^(1/2)+b*c*((a*d*x^2+a*c+b)/(d*x^2+c)
)^(1/2)/d^2+1/8*(-4*a*c+5*b)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/d^2+1/4*a*(d*x^2+c)^2*((a*d*x^2+a*c+b
)/(d*x^2+c))^(1/2)/d^2

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Rubi [A]  time = 0.54, antiderivative size = 222, normalized size of antiderivative = 1.29, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6722, 1975, 446, 78, 50, 63, 217, 206} \[ \frac {c \sqrt {a+\frac {b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )^2}{b d^2}+\frac {(b-4 a c) \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{4 b d^2}+\frac {3 (b-4 a c) \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{8 d^2}+\frac {3 b (b-4 a c) \sqrt {c+d x^2} \sqrt {a+\frac {b}{c+d x^2}} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {a \left (c+d x^2\right )+b}}\right )}{8 \sqrt {a} d^2 \sqrt {a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b/(c + d*x^2))^(3/2),x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 78

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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]]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 104, normalized size = 0.60 \[ \frac {\sqrt {a} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}} \left (-2 a c^2+2 a d^2 x^4+13 b c+5 b d x^2\right )+3 b (b-4 a c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{8 \sqrt {a} d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(13*b*c - 2*a*c^2 + 5*b*d*x^2 + 2*a*d^2*x^4) + 3*b*(b - 4*a*c)*
ArcTanh[Sqrt[a + b/(c + d*x^2)]/Sqrt[a]])/(8*Sqrt[a]*d^2)

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fricas [A]  time = 0.93, size = 335, normalized size = 1.95 \[ \left [\frac {3 \, {\left (4 \, a b c - b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, a^{2} d^{2} x^{4} + 5 \, a b d x^{2} - 2 \, a^{2} c^{2} + 13 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, a d^{2}}, \frac {3 \, {\left (4 \, a b c - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x^{4} + 5 \, a b d x^{2} - 2 \, a^{2} c^{2} + 13 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, a d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(4*a*b*c - b^2)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 - 4*(
2*a*d^2*x^4 + (4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + 4*(2*a^2*d^2
*x^4 + 5*a*b*d*x^2 - 2*a^2*c^2 + 13*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a*d^2), 1/16*(3*(4*a*b*c -
b^2)*sqrt(-a)*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a
^2*c + a*b)) + 2*(2*a^2*d^2*x^4 + 5*a*b*d*x^2 - 2*a^2*c^2 + 13*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(
a*d^2)]

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giac [B]  time = 1.99, size = 438, normalized size = 2.55 \[ \frac {1}{8} \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (\frac {2 \, a x^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{d} - \frac {2 \, a^{2} c d^{2} \mathrm {sgn}\left (d x^{2} + c\right ) - 5 \, a b d^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{a d^{4}}\right )} + \frac {{\left (4 \, a^{\frac {3}{2}} b c \mathrm {sgn}\left (d x^{2} + c\right ) - \sqrt {a} b^{2} \mathrm {sgn}\left (d x^{2} + c\right )\right )} \log \left ({\left | -2 \, a^{\frac {5}{2}} c^{3} d - 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{2} c^{2} {\left | d \right |} - 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a^{\frac {3}{2}} c d - a^{\frac {3}{2}} b c^{2} d - 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{3} a {\left | d \right |} - 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a b c {\left | d \right |} - {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} \sqrt {a} b d \right |}\right )}{16 \, a d {\left | d \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*a*x^2*sgn(d*x^2 + c)/d - (2*a^2*c*d^2*sgn(d*x^2 +
c) - 5*a*b*d^2*sgn(d*x^2 + c))/(a*d^4)) + 1/16*(4*a^(3/2)*b*c*sgn(d*x^2 + c) - sqrt(a)*b^2*sgn(d*x^2 + c))*lo
g(abs(-2*a^(5/2)*c^3*d - 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a^2*c^2*a
bs(d) - 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^2*a^(3/2)*c*d - a^(3/2)*b*
c^2*d - 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^3*a*abs(d) - 2*(sqrt(a*d^2
)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a*b*c*abs(d) - (sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^
4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^2*sqrt(a)*b*d))/(a*d*abs(d))

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maple [B]  time = 0.06, size = 593, normalized size = 3.45 \[ \frac {\left (-12 a b c \,d^{2} x^{2} \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )+4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a \,d^{2} x^{4}+3 b^{2} d^{2} x^{2} \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )-12 a b \,c^{2} d \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )+3 b^{2} c d \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )+10 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b d \,x^{2}-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a \,c^{2}+16 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,d^{2}}\, b c +10 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{16 \sqrt {a \,d^{2}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)*x^4*a*d^2-12*ln(1/2*(2*a*d^2*x^2+2*a*c*d
+b*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2))/(a*d^2)^(1/2))*x^2*a*b*c*d^2+3*ln(1/2*(2
*a*d^2*x^2+2*a*c*d+b*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2))/(a*d^2)^(1/2))*x^2*b^2
*d^2+10*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)*x^2*b*d-12*ln(1/2*(2*a*d^2*x^2+2*a*c*d+b
*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2))/(a*d^2)^(1/2))*a*b*c^2*d-4*(a*d^2*x^4+2*a*
c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)*a*c^2+3*ln(1/2*(2*a*d^2*x^2+2*a*c*d+b*d+2*(a*d^2*x^4+2*a*c*d*x^
2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2))/(a*d^2)^(1/2))*b^2*c*d+16*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(a*d^2)^
(1/2)*b*c+10*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)*b*c)/d^2*((a*d*x^2+a*c+b)/(d*x^2+c)
)^(1/2)/(a*d^2)^(1/2)/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)

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maxima [A]  time = 1.71, size = 247, normalized size = 1.44 \[ \frac {b c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{d^{2}} + \frac {3 \, {\left (4 \, a c - b\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{16 \, \sqrt {a} d^{2}} - \frac {{\left (4 \, a b c - 5 \, b^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c - 3 \, a b^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{2} d^{2} - \frac {2 \, {\left (a d x^{2} + a c + b\right )} a d^{2}}{d x^{2} + c} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} d^{2}}{{\left (d x^{2} + c\right )}^{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/d^2 + 3/16*(4*a*c - b)*b*log(-(sqrt(a) - sqrt((a*d*x^2 + a*c + b)/(d
*x^2 + c)))/(sqrt(a) + sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))))/(sqrt(a)*d^2) - 1/8*((4*a*b*c - 5*b^2)*((a*d*x^
2 + a*c + b)/(d*x^2 + c))^(3/2) - (4*a^2*b*c - 3*a*b^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^2*d^2 - 2*(a
*d*x^2 + a*c + b)*a*d^2/(d*x^2 + c) + (a*d*x^2 + a*c + b)^2*d^2/(d*x^2 + c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/(c + d*x^2))^(3/2),x)

[Out]

int(x^3*(a + b/(c + d*x^2))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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