\(\int \frac {x (c+d x^2)^{5/2}}{a+b x^2} \, dx\) [697]

Optimal result

Integrand size = 22, antiderivative size = 119 \[ \int \frac {x \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {(b c-a d)^2 \sqrt {c+d x^2}}{b^3}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b}-\frac {(b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {455, 52, 65, 214} \[ \int \frac {x \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=-\frac {(b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {\sqrt {c+d x^2} (b c-a d)^2}{b^3}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b} \]

[In]

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

[Out]

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

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 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 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97 \[ \int \frac {x \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2} \left (15 a^2 d^2-5 a b d \left (7 c+d x^2\right )+b^2 \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{15 b^3}-\frac {(-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{7/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(-\frac {\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\sqrt {d \,x^{2}+c}\, \left (\frac {\left (d^{2} x^{4}+\frac {11}{3} c d \,x^{2}+\frac {23}{3} c^{2}\right ) b^{2}}{5}-\frac {7 \left (\frac {d \,x^{2}}{7}+c \right ) d a b}{3}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, b^{3}}\)	\(123\)
risch	\(\frac {\left (3 b^{2} d^{2} x^{4}-5 x^{2} a b \,d^{2}+11 x^{2} b^{2} c d +15 a^{2} d^{2}-35 a b c d +23 b^{2} c^{2}\right ) \sqrt {d \,x^{2}+c}}{15 b^{3}}-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}\right )}{b^{3}}\)	\(410\)
default	\(\text {Expression too large to display}\)	\(2068\)

[In]

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

[Out]

-((a*d-b*c)^3*arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2))-(d*x^2+c)^(1/2)*(1/5*(d^2*x^4+11/3*c*d*x^2+23/3*c^
2)*b^2-7/3*(1/7*d*x^2+c)*d*a*b+a^2*d^2)*((a*d-b*c)*b)^(1/2))/((a*d-b*c)*b)^(1/2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 405, normalized size of antiderivative = 3.40 \[ \int \frac {x \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (3 \, b^{2} d^{2} x^{4} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, b^{3}}, -\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (3 \, b^{2} d^{2} x^{4} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, b^{3}}\right ] \]

[In]

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

[Out]

[1/60*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*(3*b^2*d^2*x^4 + 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d - 5*a*b*d^2)*
x^2)*sqrt(d*x^2 + c))/b^3, -1/30*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*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*(3*b^2*d^2*x^4
+ 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d - 5*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/b^3]

Sympy [A] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.31 \[ \int \frac {x \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\begin {cases} \frac {2 \left (\frac {d \left (c + d x^{2}\right )^{\frac {5}{2}}}{10 b} + \frac {\left (c + d x^{2}\right )^{\frac {3}{2}} \left (- a d^{2} + b c d\right )}{6 b^{2}} + \frac {\sqrt {c + d x^{2}} \left (a^{2} d^{3} - 2 a b c d^{2} + b^{2} c^{2} d\right )}{2 b^{3}} - \frac {d \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 b^{4} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (\begin {cases} \frac {x^{2}}{2 a} & \text {for}\: b = 0 \\\frac {\log {\left (2 a + 2 b x^{2} \right )}}{2 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(x*(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 = 184, normalized size of antiderivative = 1.55 \[ \int \frac {x \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{4} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{4} c + 15 \, \sqrt {d x^{2} + c} b^{4} c^{2} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b^{3} d - 30 \, \sqrt {d x^{2} + c} a b^{3} c d + 15 \, \sqrt {d x^{2} + c} a^{2} b^{2} d^{2}}{15 \, b^{5}} \]

[In]

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

[Out]

(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*
c + a*b*d)*b^3) + 1/15*(3*(d*x^2 + c)^(5/2)*b^4 + 5*(d*x^2 + c)^(3/2)*b^4*c + 15*sqrt(d*x^2 + c)*b^4*c^2 - 5*(
d*x^2 + c)^(3/2)*a*b^3*d - 30*sqrt(d*x^2 + c)*a*b^3*c*d + 15*sqrt(d*x^2 + c)*a^2*b^2*d^2)/b^5

Mupad [B] (verification not implemented)

Time = 5.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15 \[ \int \frac {x \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx=\frac {{\left (d\,x^2+c\right )}^{5/2}}{5\,b}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{5/2}}{a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{5/2}}{b^{7/2}}-\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (a\,d-b\,c\right )}{3\,b^2}+\frac {\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^2}{b^3} \]

[In]

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

[Out]

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