\(\int \frac {x^4 (c+d x^2)^{5/2}}{(a+b x^2)^2} \, dx\) [749]

Optimal result

Integrand size = 24, antiderivative size = 258 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^4}+\frac {d (7 b c-8 a d) x^3 \sqrt {c+d x^2}}{8 b^3}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\sqrt {a} (3 b c-8 a d) (b c-a d)^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 b^5}+\frac {\left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^5 \sqrt {d}} \]

[Out]

2/3*d*x^3*(d*x^2+c)^(3/2)/b^2-1/2*x^3*(d*x^2+c)^(5/2)/b/(b*x^2+a)-1/2*(-8*a*d+3*b*c)*(-a*d+b*c)^(3/2)*arctan(x
*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))*a^(1/2)/b^5+1/16*(-64*a^3*d^3+120*a^2*b*c*d^2-60*a*b^2*c^2*d+5*b^3*
c^3)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/b^5/d^(1/2)+1/16*(32*a^2*d^2-52*a*b*c*d+19*b^2*c^2)*x*(d*x^2+c)^(1/2)/
b^4+1/8*d*(-8*a*d+7*b*c)*x^3*(d*x^2+c)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {478, 595, 596, 537, 223, 212, 385, 211} \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{16 b^4}+\frac {\left (-64 a^3 d^3+120 a^2 b c d^2-60 a b^2 c^2 d+5 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^5 \sqrt {d}}-\frac {\sqrt {a} (3 b c-8 a d) (b c-a d)^{3/2} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 b^5}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{8 b^3}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2} \]

[In]

Int[(x^4*(c + d*x^2)^(5/2))/(a + b*x^2)^2,x]

[Out]

((19*b^2*c^2 - 52*a*b*c*d + 32*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^4) + (d*(7*b*c - 8*a*d)*x^3*Sqrt[c + d*x^2])/
(8*b^3) + (2*d*x^3*(c + d*x^2)^(3/2))/(3*b^2) - (x^3*(c + d*x^2)^(5/2))/(2*b*(a + b*x^2)) - (Sqrt[a]*(3*b*c -
8*a*d)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*b^5) + ((5*b^3*c^3 - 60*a*b
^2*c^2*d + 120*a^2*b*c*d^2 - 64*a^3*d^3)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*b^5*Sqrt[d])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

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

Rule 223

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 595

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[f*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*g*(m + n*(p + q + 1) + 1))), x] + Dis
t[1/(b*(m + n*(p + q + 1) + 1)), Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*((b*e - a*f)*(m + 1) + b
*e*n*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*n*q*(b*c - a*d) + b*e*d*n*(p + q + 1))*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 596

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q +
1) + 1))), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {x^2 \left (c+d x^2\right )^{3/2} \left (3 c+8 d x^2\right )}{a+b x^2} \, dx}{2 b} \\ & = \frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {x^2 \sqrt {c+d x^2} \left (6 c (3 b c-4 a d)+6 d (7 b c-8 a d) x^2\right )}{a+b x^2} \, dx}{12 b^2} \\ & = \frac {d (7 b c-8 a d) x^3 \sqrt {c+d x^2}}{8 b^3}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}+\frac {\int \frac {x^2 \left (6 c \left (12 b^2 c^2-37 a b c d+24 a^2 d^2\right )+6 d \left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right ) x^2\right )}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{48 b^3} \\ & = \frac {\left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^4}+\frac {d (7 b c-8 a d) x^3 \sqrt {c+d x^2}}{8 b^3}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\int \frac {6 a c d \left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right )-6 d \left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{96 b^4 d} \\ & = \frac {\left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^4}+\frac {d (7 b c-8 a d) x^3 \sqrt {c+d x^2}}{8 b^3}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\left (a (3 b c-8 a d) (b c-a d)^2\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 b^5}+\frac {\left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 b^5} \\ & = \frac {\left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^4}+\frac {d (7 b c-8 a d) x^3 \sqrt {c+d x^2}}{8 b^3}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\left (a (3 b c-8 a d) (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 b^5}+\frac {\left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 b^5} \\ & = \frac {\left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^4}+\frac {d (7 b c-8 a d) x^3 \sqrt {c+d x^2}}{8 b^3}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\sqrt {a} (3 b c-8 a d) (b c-a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 b^5}+\frac {\left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^5 \sqrt {d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.20 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {b x \sqrt {c+d x^2} \left (33 b^2 c^2-108 a b c d+72 a^2 d^2+2 b d (13 b c-12 a d) x^2+8 b^2 d^2 x^4+\frac {24 a (b c-a d)^2}{a+b x^2}\right )+24 \sqrt {a} (b c-a d)^{3/2} (-3 b c+8 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )+\frac {3 \left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) \log \left (d x+\sqrt {d} \sqrt {c+d x^2}\right )}{\sqrt {d}}}{48 b^5} \]

[In]

Integrate[(x^4*(c + d*x^2)^(5/2))/(a + b*x^2)^2,x]

[Out]

(b*x*Sqrt[c + d*x^2]*(33*b^2*c^2 - 108*a*b*c*d + 72*a^2*d^2 + 2*b*d*(13*b*c - 12*a*d)*x^2 + 8*b^2*d^2*x^4 + (2
4*a*(b*c - a*d)^2)/(a + b*x^2)) + 24*Sqrt[a]*(b*c - a*d)^(3/2)*(-3*b*c + 8*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sq
rt[a]*Sqrt[c + d*x^2])] + (3*(5*b^3*c^3 - 60*a*b^2*c^2*d + 120*a^2*b*c*d^2 - 64*a^3*d^3)*Log[d*x + Sqrt[d]*Sqr
t[c + d*x^2]])/Sqrt[d])/(48*b^5)

Maple [A] (verified)

Time = 3.21 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(-\frac {-\frac {b \sqrt {d \,x^{2}+c}\, \left (8 b^{2} d^{2} x^{4}-24 x^{2} a b \,d^{2}+26 x^{2} b^{2} c d +72 a^{2} d^{2}-108 a b c d +33 b^{2} c^{2}\right ) x}{12}+\frac {\left (64 a^{3} d^{3}-120 a^{2} b c \,d^{2}+60 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{4 \sqrt {d}}+2 \left (a d -b c \right )^{2} a \left (-\frac {b \sqrt {d \,x^{2}+c}\, x}{b \,x^{2}+a}-\frac {\left (8 a d -3 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{4 b^{5}}\)	\(219\)
risch	\(\text {Expression too large to display}\)	\(1101\)
default	\(\text {Expression too large to display}\)	\(5359\)

[In]

int(x^4*(d*x^2+c)^(5/2)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4/b^5*(-1/12*b*(d*x^2+c)^(1/2)*(8*b^2*d^2*x^4-24*a*b*d^2*x^2+26*b^2*c*d*x^2+72*a^2*d^2-108*a*b*c*d+33*b^2*c
^2)*x+1/4*(64*a^3*d^3-120*a^2*b*c*d^2+60*a*b^2*c^2*d-5*b^3*c^3)/d^(1/2)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))+2*(
a*d-b*c)^2*a*(-b*(d*x^2+c)^(1/2)*x/(b*x^2+a)-(8*a*d-3*b*c)/((a*d-b*c)*a)^(1/2)*arctanh((d*x^2+c)^(1/2)/x*a/((a
*d-b*c)*a)^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 1.69 (sec) , antiderivative size = 1697, normalized size of antiderivative = 6.58 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate(x^4*(d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[-1/96*(3*(5*a*b^3*c^3 - 60*a^2*b^2*c^2*d + 120*a^3*b*c*d^2 - 64*a^4*d^3 + (5*b^4*c^3 - 60*a*b^3*c^2*d + 120*a
^2*b^2*c*d^2 - 64*a^3*b*d^3)*x^2)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 12*(3*a*b^2*c^2*d
- 11*a^2*b*c*d^2 + 8*a^3*d^3 + (3*b^3*c^2*d - 11*a*b^2*c*d^2 + 8*a^2*b*d^3)*x^2)*sqrt(-a*b*c + a^2*d)*log(((b^
2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c*x)*s
qrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 2*(8*b^4*d^3*x^7 + 2*(13*b^4*c*d^2 - 8*a*b
^3*d^3)*x^5 + (33*b^4*c^2*d - 82*a*b^3*c*d^2 + 48*a^2*b^2*d^3)*x^3 + 3*(19*a*b^3*c^2*d - 52*a^2*b^2*c*d^2 + 32
*a^3*b*d^3)*x)*sqrt(d*x^2 + c))/(b^6*d*x^2 + a*b^5*d), -1/48*(3*(5*a*b^3*c^3 - 60*a^2*b^2*c^2*d + 120*a^3*b*c*
d^2 - 64*a^4*d^3 + (5*b^4*c^3 - 60*a*b^3*c^2*d + 120*a^2*b^2*c*d^2 - 64*a^3*b*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-
d)*x/sqrt(d*x^2 + c)) - 6*(3*a*b^2*c^2*d - 11*a^2*b*c*d^2 + 8*a^3*d^3 + (3*b^3*c^2*d - 11*a*b^2*c*d^2 + 8*a^2*
b*d^3)*x^2)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c
*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - (
8*b^4*d^3*x^7 + 2*(13*b^4*c*d^2 - 8*a*b^3*d^3)*x^5 + (33*b^4*c^2*d - 82*a*b^3*c*d^2 + 48*a^2*b^2*d^3)*x^3 + 3*
(19*a*b^3*c^2*d - 52*a^2*b^2*c*d^2 + 32*a^3*b*d^3)*x)*sqrt(d*x^2 + c))/(b^6*d*x^2 + a*b^5*d), -1/96*(24*(3*a*b
^2*c^2*d - 11*a^2*b*c*d^2 + 8*a^3*d^3 + (3*b^3*c^2*d - 11*a*b^2*c*d^2 + 8*a^2*b*d^3)*x^2)*sqrt(a*b*c - a^2*d)*
arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 -
a^2*c*d)*x)) + 3*(5*a*b^3*c^3 - 60*a^2*b^2*c^2*d + 120*a^3*b*c*d^2 - 64*a^4*d^3 + (5*b^4*c^3 - 60*a*b^3*c^2*d
+ 120*a^2*b^2*c*d^2 - 64*a^3*b*d^3)*x^2)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(8*b^4*d
^3*x^7 + 2*(13*b^4*c*d^2 - 8*a*b^3*d^3)*x^5 + (33*b^4*c^2*d - 82*a*b^3*c*d^2 + 48*a^2*b^2*d^3)*x^3 + 3*(19*a*b
^3*c^2*d - 52*a^2*b^2*c*d^2 + 32*a^3*b*d^3)*x)*sqrt(d*x^2 + c))/(b^6*d*x^2 + a*b^5*d), -1/48*(12*(3*a*b^2*c^2*
d - 11*a^2*b*c*d^2 + 8*a^3*d^3 + (3*b^3*c^2*d - 11*a*b^2*c*d^2 + 8*a^2*b*d^3)*x^2)*sqrt(a*b*c - a^2*d)*arctan(
1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*
d)*x)) + 3*(5*a*b^3*c^3 - 60*a^2*b^2*c^2*d + 120*a^3*b*c*d^2 - 64*a^4*d^3 + (5*b^4*c^3 - 60*a*b^3*c^2*d + 120*
a^2*b^2*c*d^2 - 64*a^3*b*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b^4*d^3*x^7 + 2*(13*b^4*c*
d^2 - 8*a*b^3*d^3)*x^5 + (33*b^4*c^2*d - 82*a*b^3*c*d^2 + 48*a^2*b^2*d^3)*x^3 + 3*(19*a*b^3*c^2*d - 52*a^2*b^2
*c*d^2 + 32*a^3*b*d^3)*x)*sqrt(d*x^2 + c))/(b^6*d*x^2 + a*b^5*d)]

Sympy [F]

\[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^{4} \left (c + d x^{2}\right )^{\frac {5}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \]

[In]

integrate(x**4*(d*x**2+c)**(5/2)/(b*x**2+a)**2,x)

[Out]

Integral(x**4*(c + d*x**2)**(5/2)/(a + b*x**2)**2, x)

Maxima [F]

\[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} x^{4}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \]

[In]

integrate(x^4*(d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(5/2)*x^4/(b*x^2 + a)^2, x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (222) = 444\).

Time = 0.35 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.00 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, d^{2} x^{2}}{b^{2}} + \frac {13 \, b^{12} c d^{5} - 12 \, a b^{11} d^{6}}{b^{14} d^{4}}\right )} x^{2} + \frac {3 \, {\left (11 \, b^{12} c^{2} d^{4} - 36 \, a b^{11} c d^{5} + 24 \, a^{2} b^{10} d^{6}\right )}}{b^{14} d^{4}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (3 \, a b^{3} c^{3} \sqrt {d} - 14 \, a^{2} b^{2} c^{2} d^{\frac {3}{2}} + 19 \, a^{3} b c d^{\frac {5}{2}} - 8 \, a^{4} d^{\frac {7}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} b^{5}} - \frac {{\left (5 \, b^{3} c^{3} - 60 \, a b^{2} c^{2} d + 120 \, a^{2} b c d^{2} - 64 \, a^{3} d^{3}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{32 \, b^{5} \sqrt {d}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{3} c^{3} \sqrt {d} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b^{2} c^{2} d^{\frac {3}{2}} + 5 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{3} b c d^{\frac {5}{2}} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{4} d^{\frac {7}{2}} - a b^{3} c^{4} \sqrt {d} + 2 \, a^{2} b^{2} c^{3} d^{\frac {3}{2}} - a^{3} b c^{2} d^{\frac {5}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} b^{5}} \]

[In]

integrate(x^4*(d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

1/48*(2*(4*d^2*x^2/b^2 + (13*b^12*c*d^5 - 12*a*b^11*d^6)/(b^14*d^4))*x^2 + 3*(11*b^12*c^2*d^4 - 36*a*b^11*c*d^
5 + 24*a^2*b^10*d^6)/(b^14*d^4))*sqrt(d*x^2 + c)*x + 1/2*(3*a*b^3*c^3*sqrt(d) - 14*a^2*b^2*c^2*d^(3/2) + 19*a^
3*b*c*d^(5/2) - 8*a^4*d^(7/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2
*d^2))/(sqrt(a*b*c*d - a^2*d^2)*b^5) - 1/32*(5*b^3*c^3 - 60*a*b^2*c^2*d + 120*a^2*b*c*d^2 - 64*a^3*d^3)*log((s
qrt(d)*x - sqrt(d*x^2 + c))^2)/(b^5*sqrt(d)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b^3*c^3*sqrt(d) - 4*(sqrt(d)
*x - sqrt(d*x^2 + c))^2*a^2*b^2*c^2*d^(3/2) + 5*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^3*b*c*d^(5/2) - 2*(sqrt(d)*x
- sqrt(d*x^2 + c))^2*a^4*d^(7/2) - a*b^3*c^4*sqrt(d) + 2*a^2*b^2*c^3*d^(3/2) - a^3*b*c^2*d^(5/2))/(((sqrt(d)*
x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c
^2)*b^5)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^4\,{\left (d\,x^2+c\right )}^{5/2}}{{\left (b\,x^2+a\right )}^2} \,d x \]

[In]

int((x^4*(c + d*x^2)^(5/2))/(a + b*x^2)^2,x)

[Out]

int((x^4*(c + d*x^2)^(5/2))/(a + b*x^2)^2, x)