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

Optimal. Leaf size=163 \[ -\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}}+\frac {\sqrt {c+d x^2} (2 b c-5 a d)}{2 b^3}+\frac {\left (c+d x^2\right )^{3/2} (2 b c-5 a d)}{6 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right ) (b c-a d)} \]

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Rubi [A]  time = 0.14, antiderivative size = 163, 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 = {446, 78, 50, 63, 208} \begin {gather*} \frac {\left (c+d x^2\right )^{3/2} (2 b c-5 a d)}{6 b^2 (b c-a d)}+\frac {\sqrt {c+d x^2} (2 b c-5 a d)}{2 b^3}-\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}}+\frac {a \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c - 5*a*d)*Sqrt[c + d*x^2])/(2*b^3) + ((2*b*c - 5*a*d)*(c + d*x^2)^(3/2))/(6*b^2*(b*c - a*d)) + (a*(c +
d*x^2)^(5/2))/(2*b*(b*c - a*d)*(a + b*x^2)) - ((2*b*c - 5*a*d)*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2
])/Sqrt[b*c - a*d]])/(2*b^(7/2))

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 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 {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x (c+d x)^{3/2}}{(a+b x)^2} \, dx,x,x^2\right )\\ &=\frac {a \left (c+d x^2\right )^{5/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-5 a d) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right )}{4 b (b c-a d)}\\ &=\frac {(2 b c-5 a d) \left (c+d x^2\right )^{3/2}}{6 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {(2 b c-5 a d) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{4 b^2}\\ &=\frac {(2 b c-5 a d) \sqrt {c+d x^2}}{2 b^3}+\frac {(2 b c-5 a d) \left (c+d x^2\right )^{3/2}}{6 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {((2 b c-5 a d) (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b^3}\\ &=\frac {(2 b c-5 a d) \sqrt {c+d x^2}}{2 b^3}+\frac {(2 b c-5 a d) \left (c+d x^2\right )^{3/2}}{6 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{2 b (b c-a d) \left (a+b x^2\right )}+\frac {((2 b c-5 a d) (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 b^3 d}\\ &=\frac {(2 b c-5 a d) \sqrt {c+d x^2}}{2 b^3}+\frac {(2 b c-5 a d) \left (c+d x^2\right )^{3/2}}{6 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 125, normalized size = 0.77 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-15 a^2 d+a b \left (11 c-10 d x^2\right )+2 b^2 x^2 \left (4 c+d x^2\right )\right )}{6 b^3 \left (a+b x^2\right )}-\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-15*a^2*d + a*b*(11*c - 10*d*x^2) + 2*b^2*x^2*(4*c + d*x^2)))/(6*b^3*(a + b*x^2)) - ((2*b*c
- 5*a*d)*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*b^(7/2))

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IntegrateAlgebraic [A]  time = 0.32, size = 150, normalized size = 0.92 \begin {gather*} \frac {\left (-5 a^2 d^2+7 a b c d-2 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{2 b^{7/2} \sqrt {a d-b c}}+\frac {\sqrt {c+d x^2} \left (-15 a^2 d+11 a b c-10 a b d x^2+8 b^2 c x^2+2 b^2 d x^4\right )}{6 b^3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(11*a*b*c - 15*a^2*d + 8*b^2*c*x^2 - 10*a*b*d*x^2 + 2*b^2*d*x^4))/(6*b^3*(a + b*x^2)) + ((-2*
b^2*c^2 + 7*a*b*c*d - 5*a^2*d^2)*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x^2])/(b*c - a*d)])/(2*b^(7/2)*
Sqrt[-(b*c) + a*d])

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fricas [A]  time = 0.94, size = 413, normalized size = 2.53 \begin {gather*} \left [-\frac {3 \, {\left (2 \, a b c - 5 \, a^{2} d + {\left (2 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \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 ) - 4 \, {\left (2 \, b^{2} d x^{4} + 11 \, a b c - 15 \, a^{2} d + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, -\frac {3 \, {\left (2 \, a b c - 5 \, a^{2} d + {\left (2 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \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 ) - 2 \, {\left (2 \, b^{2} d x^{4} + 11 \, a b c - 15 \, a^{2} d + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*(2*a*b*c - 5*a^2*d + (2*b^2*c - 5*a*b*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)) - 4*(2*b^2*d*x^4 + 11*a*b*c - 15*a^2*d + 2*(4*b^2*c - 5*a*b*d)*x^2)*sqr
t(d*x^2 + c))/(b^4*x^2 + a*b^3), -1/12*(3*(2*a*b*c - 5*a^2*d + (2*b^2*c - 5*a*b*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*(2*b^2*d*x^4 + 11*a*b*c - 15*a^2*d + 2*(4*b^2*c - 5*a*b*d)*x^2)*sqrt(d*x^2 + c))/(b^4*x^2 + a*b^3)]

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giac [A]  time = 0.36, size = 173, normalized size = 1.06 \begin {gather*} \frac {{\left (2 \, b^{2} c^{2} - 7 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b^{3}} + \frac {\sqrt {d x^{2} + c} a b c d - \sqrt {d x^{2} + c} a^{2} d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{3}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{4} + 3 \, \sqrt {d x^{2} + c} b^{4} c - 6 \, \sqrt {d x^{2} + c} a b^{3} d}{3 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*c^2 - 7*a*b*c*d + 5*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)*b
^3) + 1/2*(sqrt(d*x^2 + c)*a*b*c*d - sqrt(d*x^2 + c)*a^2*d^2)/(((d*x^2 + c)*b - b*c + a*d)*b^3) + 1/3*((d*x^2
+ c)^(3/2)*b^4 + 3*sqrt(d*x^2 + c)*b^4*c - 6*sqrt(d*x^2 + c)*a*b^3*d)/b^6

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maple [B]  time = 0.02, size = 4673, normalized size = 28.67 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/b^3*(-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/
b^3*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+1/4*(-a*b)^(1/2)/b^2*d/(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)^(3/2)*x+1/b^3/(-(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))*a*d*c-3/4*a^3/b^4*d^3/(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))-1/4*(-a*b)^(1/2)/b^2*d/(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)^(3/2)*x-3/8*(-a*b)^(1/2)/b^2*d^(1/2)/(a*d-b*c)*c^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))-3/4*(-a*b)^(1/2)/b^4*a^2*d^(5/2)/(a*d-b*c)*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))+3/4*a/b^2*d/(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)*c+3/8*(-a*b)^(1/2)/b^2*d^(1/2)
/(a*d-b*c)*c^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/b^3/(-(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))*a*d*c+3/4*a/b^2*d/(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)*c+3/4*(-a*b)^(1/2)/b^4*a^2*d^(5/2)/(a*d-b*c)*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))-3/4*a^3/b^4*d^3/(
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^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))*c^2-1/2/b^3*((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)*a*d-1/2/b^2/(-(a*d-b*c)/b)^(1/2)*l
n((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^2-1/2/b^3*((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)*a*d+1/6/b^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)^(3/2)+1/6/b^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)^(3/2)-1/2/b^4*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))*a-1/2/b^4/(-(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))*a^2*d^2+1/2/b^4*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))*a-1/2/b^4/(-(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)
)*a^2*d^2-1/4/b^3*(-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/b^3*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-1/4*(-a*b)^(1/2)/b^2/(a*d-b*c)/(x+(-a*b)^(1/2)/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)^(5/2)+1/4*a/b^2*d/(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)^(3/2)-3/4*a^2/b^3*d^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)+1/4*(-a*b)^(1/2)/b^2/(a*d-b*c)/(x-(-a*b
)^(1/2)/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)^(5/2)+1/4*a/b^2*d/(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)^(3/2)-3/4*a^2/b^3*d^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)+3/2*a^2/b^3*d^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))*c-3/4*a/b^2*d/(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))*c^2+3/8*(-a*
b)^(1/2)/b^2*d/(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)*x-
9/8*(-a*b)^(1/2)/b^3*a*d^(3/2)/(a*d-b*c)*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-3/8*(-a*b)^(1/2)/b^2*d/(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)*x+9/8*(-a*b)^(1/2)/b^3*a*d^(3/2)/(a*d-
b*c)*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+3/8*(-a*b)^(1/2)/b^3*a*d^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)*x-3/8*(-a*b)^(1/2)/b^3*a*d^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)*x+3/2*a^2/b^3*d^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))*c-3/4*a/b^2*d/(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))*c^2+1/2/b^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/b^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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.99, size = 183, normalized size = 1.12 \begin {gather*} \frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b^2}-\sqrt {d\,x^2+c}\,\left (\frac {c}{b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{b^4}\right )-\frac {\left (\frac {a^2\,d^2}{2}-\frac {a\,b\,c\,d}{2}\right )\,\sqrt {d\,x^2+c}}{b^4\,\left (d\,x^2+c\right )-b^4\,c+a\,b^3\,d}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\sqrt {a\,d-b\,c}\,\left (5\,a\,d-2\,b\,c\right )}{5\,a^2\,d^2-7\,a\,b\,c\,d+2\,b^2\,c^2}\right )\,\sqrt {a\,d-b\,c}\,\left (5\,a\,d-2\,b\,c\right )}{2\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x^2)^(3/2)/(3*b^2) - (c + d*x^2)^(1/2)*(c/b^2 - (2*b^2*c - 2*a*b*d)/b^4) - (((a^2*d^2)/2 - (a*b*c*d)/2)
*(c + d*x^2)^(1/2))/(b^4*(c + d*x^2) - b^4*c + a*b^3*d) + (atan((b^(1/2)*(c + d*x^2)^(1/2)*(a*d - b*c)^(1/2)*(
5*a*d - 2*b*c))/(5*a^2*d^2 + 2*b^2*c^2 - 7*a*b*c*d))*(a*d - b*c)^(1/2)*(5*a*d - 2*b*c))/(2*b^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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