\(\int \frac {x (c+d x^2)^{3/2}}{a+b x^2} \, dx\) [688]

Optimal result

Integrand size = 22, antiderivative size = 91 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {(b c-a d) \sqrt {c+d x^2}}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b}-\frac {(b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {455, 52, 65, 214} \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=-\frac {(b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}}+\frac {\sqrt {c+d x^2} (b c-a d)}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2} \left (4 b c-3 a d+b d x^2\right )}{3 b^2}+\frac {(-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{5/2}} \]

[In]

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

[Out]

(Sqrt[c + d*x^2]*(4*b*c - 3*a*d + b*d*x^2))/(3*b^2) + ((-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/S
qrt[-(b*c) + a*d]])/b^(5/2)

Maple [A] (verified)

Time = 2.96 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(-\frac {-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {d \,x^{2}+c}\, \left (\frac {\left (-d \,x^{2}-4 c \right ) b}{3}+a d \right ) \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, b^{2}}\)	\(94\)
risch	\(-\frac {\left (-b d \,x^{2}+3 a d -4 b c \right ) \sqrt {d \,x^{2}+c}}{3 b^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}\right )}{b^{2}}\)	\(356\)
default	\(\text {Expression too large to display}\)	\(1237\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 303, normalized size of antiderivative = 3.33 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (b d x^{2} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{12 \, b^{2}}, -\frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (b d x^{2} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{6 \, b^{2}}\right ] \]

[In]

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

[Out]

[-1/12*(3*(b*c - a*d)*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*(b*d*x^2 + 4*b*c - 3*a*d)*sqrt(d*x^2 + c))/b^2, -1/6*(3*(b*c - a*d)*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*(b*d*x
^2 + 4*b*c - 3*a*d)*sqrt(d*x^2 + c))/b^2]

Sympy [A] (verification not implemented)

Time = 5.16 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\begin {cases} \frac {2 \left (\frac {d \left (c + d x^{2}\right )^{\frac {3}{2}}}{6 b} + \frac {\sqrt {c + d x^{2}} \left (- a d^{2} + b c d\right )}{2 b^{2}} + \frac {d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 b^{3} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\begin {cases} \frac {x^{2}}{2 a} & \text {for}\: b = 0 \\\frac {\log {\left (2 a + 2 b x^{2} \right )}}{2 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(d*(c + d*x**2)**(3/2)/(6*b) + sqrt(c + d*x**2)*(-a*d**2 + b*c*d)/(2*b**2) + d*(a*d - b*c)**2*ata
n(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*b**3*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), (c**(3/2)*Piecewise((x**2/
(2*a), Eq(b, 0)), (log(2*a + 2*b*x**2)/(2*b), True)), True))

Maxima [F(-2)]

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

[In]

integrate(x*(d*x^2+c)^(3/2)/(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 = 112, normalized size of antiderivative = 1.23 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {d x^{2} + c} b^{2} c - 3 \, \sqrt {d x^{2} + c} a b d}{3 \, b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.08 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx=\frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b}-\frac {\sqrt {d\,x^2+c}\,\left (a\,d-b\,c\right )}{b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{3/2}}{a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{b^{5/2}} \]

[In]

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

[Out]

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