\(\int \frac {x^2 (c+d x^2)^{5/2}}{a+b x^2} \, dx\) [696]

Optimal result

Integrand size = 24, antiderivative size = 217 \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\sqrt {a} (b c-a d)^{5/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}+\frac {\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 \sqrt {d}} \]

[Out]

1/6*d*x^3*(d*x^2+c)^(3/2)/b-(-a*d+b*c)^(5/2)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))*a^(1/2)/b^4+1/
16*(-16*a^3*d^3+40*a^2*b*c*d^2-30*a*b^2*c^2*d+5*b^3*c^3)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/b^4/d^(1/2)+1/16*(
8*a^2*d^2-18*a*b*c*d+11*b^2*c^2)*x*(d*x^2+c)^(1/2)/b^3+1/8*d*(-2*a*d+3*b*c)*x^3*(d*x^2+c)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {488, 595, 596, 537, 223, 212, 385, 211} \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-18 a b c d+11 b^2 c^2\right )}{16 b^3}+\frac {\left (-16 a^3 d^3+40 a^2 b c d^2-30 a b^2 c^2 d+5 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 \sqrt {d}}-\frac {\sqrt {a} (b c-a d)^{5/2} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}+\frac {d x^3 \sqrt {c+d x^2} (3 b c-2 a d)}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b} \]

[In]

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

[Out]

((11*b^2*c^2 - 18*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^3) + (d*(3*b*c - 2*a*d)*x^3*Sqrt[c + d*x^2])/(
8*b^2) + (d*x^3*(c + d*x^2)^(3/2))/(6*b) - (Sqrt[a]*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt
[c + d*x^2])])/b^4 + ((5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(16*b^4*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 488

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*e*(m + n*(p + q) + 1))), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && 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 {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}+\frac {\int \frac {x^2 \sqrt {c+d x^2} \left (3 c (2 b c-a d)+3 d (3 b c-2 a d) x^2\right )}{a+b x^2} \, dx}{6 b} \\ & = \frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}+\frac {\int \frac {x^2 \left (3 c \left (8 b^2 c^2-13 a b c d+6 a^2 d^2\right )+3 d \left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x^2\right )}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{24 b^2} \\ & = \frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\int \frac {3 a c d \left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right )-3 d \left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{48 b^3 d} \\ & = \frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\left (a (b c-a d)^3\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{b^4}+\frac {\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 b^4} \\ & = \frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\left (a (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{b^4}+\frac {\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 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^4} \\ & = \frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\sqrt {a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}+\frac {\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 \sqrt {d}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(452\) vs. \(2(217)=434\).

Time = 2.36 (sec) , antiderivative size = 452, normalized size of antiderivative = 2.08 \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {b x \sqrt {c+d x^2} \left (24 a^2 d^2-6 a b d \left (9 c+2 d x^2\right )+b^2 \left (33 c^2+26 c d x^2+8 d^2 x^4\right )\right )+\frac {48 (b c-a d)^2 \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \left (b c-a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{\sqrt {a} d}-\frac {48 (b c-a d)^2 \left (-b c+a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{\sqrt {a} d}+\frac {6 \left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{\sqrt {d}}}{48 b^4} \]

[In]

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

[Out]

(b*x*Sqrt[c + d*x^2]*(24*a^2*d^2 - 6*a*b*d*(9*c + 2*d*x^2) + b^2*(33*c^2 + 26*c*d*x^2 + 8*d^2*x^4)) + (48*(b*c
- a*d)^2*Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*(b*c - a*d + Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d])*
ArcTan[(Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(-Sqrt[c] + Sqrt[c + d*x^2]))])/(Sqr
t[a]*d) - (48*(b*c - a*d)^2*(-(b*c) + a*d + Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d])*Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt
[c]*Sqrt[b*c - a*d]]*ArcTan[(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(-Sqrt[c] + Sqr
t[c + d*x^2]))])/(Sqrt[a]*d) + (6*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*ArcTanh[(Sqrt[d]*
x)/(-Sqrt[c] + Sqrt[c + d*x^2])])/Sqrt[d])/(48*b^4)

Maple [A] (verified)

Time = 3.05 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(-\frac {-\frac {b \sqrt {d \,x^{2}+c}\, \left (8 b^{2} d^{2} x^{4}-12 x^{2} a b \,d^{2}+26 x^{2} b^{2} c d +24 a^{2} d^{2}-54 a b c d +33 b^{2} c^{2}\right ) x}{24}+\frac {\left (16 a^{3} d^{3}-40 a^{2} b c \,d^{2}+30 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{8 \sqrt {d}}-\frac {2 \left (a d -b c \right )^{3} a \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}}{2 b^{4}}\)	\(185\)
risch	\(\frac {x \left (8 b^{2} d^{2} x^{4}-12 x^{2} a b \,d^{2}+26 x^{2} b^{2} c d +24 a^{2} d^{2}-54 a b c d +33 b^{2} c^{2}\right ) \sqrt {d \,x^{2}+c}}{48 b^{3}}-\frac {\frac {\left (16 a^{3} d^{3}-40 a^{2} b c \,d^{2}+30 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{b \sqrt {d}}-\frac {8 a \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{\sqrt {-a b}\, b \sqrt {-\frac {a d -b c}{b}}}+\frac {8 a \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{\sqrt {-a b}\, b \sqrt {-\frac {a d -b c}{b}}}}{16 b^{3}}\)	\(521\)
default	\(\text {Expression too large to display}\)	\(2153\)

[In]

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

[Out]

-1/2/b^4*(-1/24*b*(d*x^2+c)^(1/2)*(8*b^2*d^2*x^4-12*a*b*d^2*x^2+26*b^2*c*d*x^2+24*a^2*d^2-54*a*b*c*d+33*b^2*c^
2)*x+1/8*(16*a^3*d^3-40*a^2*b*c*d^2+30*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)^3*a/((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 = 1161, normalized size of antiderivative = 5.35 \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/96*(3*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*
sqrt(d)*x - c) - 24*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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)) - 2*(8*b^3*d^3*x^5 + 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c
^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d), -1/48*(3*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^
2*b*c*d^2 - 16*a^3*d^3)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 12*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*s
qrt(-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^3*d^3*x^5
+ 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b
^4*d), -1/96*(48*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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*b^3*c^3 - 30*a*b
^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(8*b^3*d^3
*x^5 + 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c)
)/(b^4*d), -1/48*(24*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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*b^3*c^3 - 30
*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b^3*d^3*x^5 + 2*(
13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^2 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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