\(\int \frac {x^3 (c+d x^2)^{5/2}}{a+b x^2} \, dx\) [695]

Optimal result

Integrand size = 24, antiderivative size = 144 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=-\frac {a (b c-a d)^2 \sqrt {c+d x^2}}{b^4}-\frac {a (b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^3}-\frac {a \left (c+d x^2\right )^{5/2}}{5 b^2}+\frac {\left (c+d x^2\right )^{7/2}}{7 b d}+\frac {a (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{9/2}} \]

[Out]

-1/3*a*(-a*d+b*c)*(d*x^2+c)^(3/2)/b^3-1/5*a*(d*x^2+c)^(5/2)/b^2+1/7*(d*x^2+c)^(7/2)/b/d+a*(-a*d+b*c)^(5/2)*arc
tanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(9/2)-a*(-a*d+b*c)^2*(d*x^2+c)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 81, 52, 65, 214} \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {a (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{9/2}}-\frac {a \sqrt {c+d x^2} (b c-a d)^2}{b^4}-\frac {a \left (c+d x^2\right )^{3/2} (b c-a d)}{3 b^3}-\frac {a \left (c+d x^2\right )^{5/2}}{5 b^2}+\frac {\left (c+d x^2\right )^{7/2}}{7 b d} \]

[In]

Int[(x^3*(c + d*x^2)^(5/2))/(a + b*x^2),x]

[Out]

-((a*(b*c - a*d)^2*Sqrt[c + d*x^2])/b^4) - (a*(b*c - a*d)*(c + d*x^2)^(3/2))/(3*b^3) - (a*(c + d*x^2)^(5/2))/(
5*b^2) + (c + d*x^2)^(7/2)/(7*b*d) + (a*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/
b^(9/2)

Rule 52

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 65

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 81

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

Rule 214

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

Rule 457

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x (c+d x)^{5/2}}{a+b x} \, dx,x,x^2\right ) \\ & = \frac {\left (c+d x^2\right )^{7/2}}{7 b d}-\frac {a \text {Subst}\left (\int \frac {(c+d x)^{5/2}}{a+b x} \, dx,x,x^2\right )}{2 b} \\ & = -\frac {a \left (c+d x^2\right )^{5/2}}{5 b^2}+\frac {\left (c+d x^2\right )^{7/2}}{7 b d}-\frac {(a (b c-a d)) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right )}{2 b^2} \\ & = -\frac {a (b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^3}-\frac {a \left (c+d x^2\right )^{5/2}}{5 b^2}+\frac {\left (c+d x^2\right )^{7/2}}{7 b d}-\frac {\left (a (b c-a d)^2\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{2 b^3} \\ & = -\frac {a (b c-a d)^2 \sqrt {c+d x^2}}{b^4}-\frac {a (b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^3}-\frac {a \left (c+d x^2\right )^{5/2}}{5 b^2}+\frac {\left (c+d x^2\right )^{7/2}}{7 b d}-\frac {\left (a (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 b^4} \\ & = -\frac {a (b c-a d)^2 \sqrt {c+d x^2}}{b^4}-\frac {a (b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^3}-\frac {a \left (c+d x^2\right )^{5/2}}{5 b^2}+\frac {\left (c+d x^2\right )^{7/2}}{7 b d}-\frac {\left (a (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{b^4 d} \\ & = -\frac {a (b c-a d)^2 \sqrt {c+d x^2}}{b^4}-\frac {a (b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^3}-\frac {a \left (c+d x^2\right )^{5/2}}{5 b^2}+\frac {\left (c+d x^2\right )^{7/2}}{7 b d}+\frac {a (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2} \left (-105 a^3 d^3+15 b^3 \left (c+d x^2\right )^3+35 a^2 b d^2 \left (7 c+d x^2\right )-7 a b^2 d \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{105 b^4 d}+\frac {a (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{9/2}} \]

[In]

Integrate[(x^3*(c + d*x^2)^(5/2))/(a + b*x^2),x]

[Out]

(Sqrt[c + d*x^2]*(-105*a^3*d^3 + 15*b^3*(c + d*x^2)^3 + 35*a^2*b*d^2*(7*c + d*x^2) - 7*a*b^2*d*(23*c^2 + 11*c*
d*x^2 + 3*d^2*x^4)))/(105*b^4*d) + (a*(-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]
])/b^(9/2)

Maple [A] (verified)

Time = 3.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(-\frac {\sqrt {\left (a d -b c \right ) b}\, \left (-\frac {\left (d \,x^{2}+c \right )^{3} b^{3}}{7}+\frac {23 d \left (\frac {3}{23} d^{2} x^{4}+\frac {11}{23} c d \,x^{2}+c^{2}\right ) a \,b^{2}}{15}-\frac {7 \left (\frac {d \,x^{2}}{7}+c \right ) d^{2} a^{2} b}{3}+a^{3} d^{3}\right ) \sqrt {d \,x^{2}+c}-a d \left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, d \,b^{4}}\)	\(147\)
risch	\(-\frac {\left (-15 b^{3} d^{3} x^{6}+21 a \,b^{2} d^{3} x^{4}-45 b^{3} c \,d^{2} x^{4}-35 x^{2} a^{2} b \,d^{3}+77 x^{2} a \,b^{2} c \,d^{2}-45 x^{2} b^{3} c^{2} d +105 a^{3} d^{3}-245 a^{2} b c \,d^{2}+161 a \,b^{2} c^{2} d -15 b^{3} c^{3}\right ) \sqrt {d \,x^{2}+c}}{105 d \,b^{4}}+\frac {a \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}\right )}{b^{4}}\)	\(468\)
default	\(\text {Expression too large to display}\)	\(2087\)

[In]

int(x^3*(d*x^2+c)^(5/2)/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

-(((a*d-b*c)*b)^(1/2)*(-1/7*(d*x^2+c)^3*b^3+23/15*d*(3/23*d^2*x^4+11/23*c*d*x^2+c^2)*a*b^2-7/3*(1/7*d*x^2+c)*d
^2*a^2*b+a^3*d^3)*(d*x^2+c)^(1/2)-a*d*(a*d-b*c)^3*arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2)))/((a*d-b*c)*b)
^(1/2)/d/b^4

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 527, normalized size of antiderivative = 3.66 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\left [\frac {105 \, {\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (15 \, b^{3} d^{3} x^{6} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \, {\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{4} + {\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{420 \, b^{4} d}, \frac {105 \, {\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (15 \, b^{3} d^{3} x^{6} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \, {\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{4} + {\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{210 \, b^{4} d}\right ] \]

[In]

integrate(x^3*(d*x^2+c)^(5/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

[1/420*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*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*(15*b^3*d^3*x^6 + 15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 - 105*
a^3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^4 + (45*b^3*c^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^2)*sqrt(d*x^2
+ c))/(b^4*d), 1/210*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*sqrt(-(b*c - a*d)/b)*arctan(-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*(15*b^3*d^3*x^6 +
15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 - 105*a^3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^4 + (45*b^3*c
^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^2)*sqrt(d*x^2 + c))/(b^4*d)]

Sympy [A] (verification not implemented)

Time = 12.50 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.30 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\begin {cases} \frac {2 \left (- \frac {a d \left (c + d x^{2}\right )^{\frac {5}{2}}}{10 b^{2}} + \frac {a d \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 b^{5} \sqrt {\frac {a d - b c}{b}}} + \frac {\left (c + d x^{2}\right )^{\frac {7}{2}}}{14 b} + \frac {\left (c + d x^{2}\right )^{\frac {3}{2}} \left (a^{2} d^{2} - a b c d\right )}{6 b^{3}} + \frac {\sqrt {c + d x^{2}} \left (- a^{3} d^{3} + 2 a^{2} b c d^{2} - a b^{2} c^{2} d\right )}{2 b^{4}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (- \frac {a \left (\begin {cases} \frac {x^{2}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{2} \right )}}{b} & \text {otherwise} \end {cases}\right )}{2 b} + \frac {x^{2}}{2 b}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(d*x**2+c)**(5/2)/(b*x**2+a),x)

[Out]

Piecewise((2*(-a*d*(c + d*x**2)**(5/2)/(10*b**2) + a*d*(a*d - b*c)**3*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b
))/(2*b**5*sqrt((a*d - b*c)/b)) + (c + d*x**2)**(7/2)/(14*b) + (c + d*x**2)**(3/2)*(a**2*d**2 - a*b*c*d)/(6*b*
*3) + sqrt(c + d*x**2)*(-a**3*d**3 + 2*a**2*b*c*d**2 - a*b**2*c**2*d)/(2*b**4))/d, Ne(d, 0)), (c**(5/2)*(-a*Pi
ecewise((x**2/a, Eq(b, 0)), (log(a + b*x**2)/b, True))/(2*b) + x**2/(2*b)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^3*(d*x^2+c)^(5/2)/(b*x^2+a),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 detail

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.58 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=-\frac {{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{4}} + \frac {15 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{6} d^{6} - 21 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b^{5} d^{7} - 35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b^{5} c d^{7} - 105 \, \sqrt {d x^{2} + c} a b^{5} c^{2} d^{7} + 35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} b^{4} d^{8} + 210 \, \sqrt {d x^{2} + c} a^{2} b^{4} c d^{8} - 105 \, \sqrt {d x^{2} + c} a^{3} b^{3} d^{9}}{105 \, b^{7} d^{7}} \]

[In]

integrate(x^3*(d*x^2+c)^(5/2)/(b*x^2+a),x, algorithm="giac")

[Out]

-(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(
-b^2*c + a*b*d)*b^4) + 1/105*(15*(d*x^2 + c)^(7/2)*b^6*d^6 - 21*(d*x^2 + c)^(5/2)*a*b^5*d^7 - 35*(d*x^2 + c)^(
3/2)*a*b^5*c*d^7 - 105*sqrt(d*x^2 + c)*a*b^5*c^2*d^7 + 35*(d*x^2 + c)^(3/2)*a^2*b^4*d^8 + 210*sqrt(d*x^2 + c)*
a^2*b^4*c*d^8 - 105*sqrt(d*x^2 + c)*a^3*b^3*d^9)/(b^7*d^7)

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.74 \[ \int \frac {x^3 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {{\left (d\,x^2+c\right )}^{7/2}}{7\,b\,d}-{\left (d\,x^2+c\right )}^{5/2}\,\left (\frac {c}{5\,b\,d}+\frac {a\,d^2-b\,c\,d}{5\,b^2\,d^2}\right )+\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{5/2}}{a^4\,d^3-3\,a^3\,b\,c\,d^2+3\,a^2\,b^2\,c^2\,d-a\,b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{5/2}}{b^{9/2}}+\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (a\,d^2-b\,c\,d\right )\,\left (\frac {c}{b\,d}+\frac {a\,d^2-b\,c\,d}{b^2\,d^2}\right )}{3\,b\,d}-\frac {\sqrt {d\,x^2+c}\,{\left (a\,d^2-b\,c\,d\right )}^2\,\left (\frac {c}{b\,d}+\frac {a\,d^2-b\,c\,d}{b^2\,d^2}\right )}{b^2\,d^2} \]

[In]

int((x^3*(c + d*x^2)^(5/2))/(a + b*x^2),x)

[Out]

(c + d*x^2)^(7/2)/(7*b*d) - (c + d*x^2)^(5/2)*(c/(5*b*d) + (a*d^2 - b*c*d)/(5*b^2*d^2)) + (a*atan((a*b^(1/2)*(
c + d*x^2)^(1/2)*(a*d - b*c)^(5/2))/(a^4*d^3 - a*b^3*c^3 + 3*a^2*b^2*c^2*d - 3*a^3*b*c*d^2))*(a*d - b*c)^(5/2)
)/b^(9/2) + ((c + d*x^2)^(3/2)*(a*d^2 - b*c*d)*(c/(b*d) + (a*d^2 - b*c*d)/(b^2*d^2)))/(3*b*d) - ((c + d*x^2)^(
1/2)*(a*d^2 - b*c*d)^2*(c/(b*d) + (a*d^2 - b*c*d)/(b^2*d^2)))/(b^2*d^2)