\(\int \frac {(a+b x^2)^2}{\sqrt {c+d x^2}} \, dx\) [640]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=-\frac {3 b (b c-2 a d) x \sqrt {c+d x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}+\frac {\left (3 b^2 c^2-8 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{5/2}} \]

[Out]

1/8*(8*a^2*d^2-8*a*b*c*d+3*b^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(5/2)-3/8*b*(-2*a*d+b*c)*x*(d*x^2+c)^
(1/2)/d^2+1/4*b*x*(b*x^2+a)*(d*x^2+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {427, 396, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\left (8 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{5/2}}-\frac {3 b x \sqrt {c+d x^2} (b c-2 a d)}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d} \]

[In]

Int[(a + b*x^2)^2/Sqrt[c + d*x^2],x]

[Out]

(-3*b*(b*c - 2*a*d)*x*Sqrt[c + d*x^2])/(8*d^2) + (b*x*(a + b*x^2)*Sqrt[c + d*x^2])/(4*d) + ((3*b^2*c^2 - 8*a*b
*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*d^(5/2))

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 396

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

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps \begin{align*} \text {integral}& = \frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}+\frac {\int \frac {-a (b c-4 a d)-3 b (b c-2 a d) x^2}{\sqrt {c+d x^2}} \, dx}{4 d} \\ & = -\frac {3 b (b c-2 a d) x \sqrt {c+d x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}-\frac {(2 a d (b c-4 a d)-3 b c (b c-2 a d)) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 d^2} \\ & = -\frac {3 b (b c-2 a d) x \sqrt {c+d x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}-\frac {(2 a d (b c-4 a d)-3 b c (b c-2 a d)) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 d^2} \\ & = -\frac {3 b (b c-2 a d) x \sqrt {c+d x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}+\frac {\left (3 b^2 c^2-8 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {b x \sqrt {c+d x^2} \left (-3 b c+8 a d+2 b d x^2\right )}{8 d^2}+\frac {\left (-3 b^2 c^2+8 a b c d-8 a^2 d^2\right ) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{8 d^{5/2}} \]

[In]

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

[Out]

(b*x*Sqrt[c + d*x^2]*(-3*b*c + 8*a*d + 2*b*d*x^2))/(8*d^2) + ((-3*b^2*c^2 + 8*a*b*c*d - 8*a^2*d^2)*Log[-(Sqrt[
d]*x) + Sqrt[c + d*x^2]])/(8*d^(5/2))

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {b x \left (2 b d \,x^{2}+8 a d -3 b c \right ) \sqrt {d \,x^{2}+c}}{8 d^{2}}+\frac {\left (8 a^{2} d^{2}-8 a b c d +3 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {5}{2}}}\)	\(78\)
pseudoelliptic	\(\frac {\left (a^{2} d^{2}-a b c d +\frac {3}{8} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x b \sqrt {d \,x^{2}+c}\, \left (\left (\frac {b \,x^{2}}{4}+a \right ) d^{\frac {3}{2}}-\frac {3 b \sqrt {d}\, c}{8}\right )}{d^{\frac {5}{2}}}\)	\(78\)
default	\(\frac {a^{2} \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{\sqrt {d}}+b^{2} \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )+2 a b \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )\)	\(133\)

[In]

int((b*x^2+a)^2/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8*b*x*(2*b*d*x^2+8*a*d-3*b*c)*(d*x^2+c)^(1/2)/d^2+1/8*(8*a^2*d^2-8*a*b*c*d+3*b^2*c^2)/d^(5/2)*ln(x*d^(1/2)+(
d*x^2+c)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.81 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\left [\frac {{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (2 \, b^{2} d^{2} x^{3} - {\left (3 \, b^{2} c d - 8 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, d^{3}}, -\frac {{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} - {\left (3 \, b^{2} c d - 8 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, d^{3}}\right ] \]

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*((3*b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*b^2*
d^2*x^3 - (3*b^2*c*d - 8*a*b*d^2)*x)*sqrt(d*x^2 + c))/d^3, -1/8*((3*b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*sqrt(-d)*
arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (2*b^2*d^2*x^3 - (3*b^2*c*d - 8*a*b*d^2)*x)*sqrt(d*x^2 + c))/d^3]

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\begin {cases} \left (a^{2} - \frac {c \left (2 a b - \frac {3 b^{2} c}{4 d}\right )}{2 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {c + d x^{2}} \left (\frac {b^{2} x^{3}}{4 d} + \frac {x \left (2 a b - \frac {3 b^{2} c}{4 d}\right )}{2 d}\right ) & \text {for}\: d \neq 0 \\\frac {a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5}}{\sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

integrate((b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x^{3}}{4 \, d} - \frac {3 \, \sqrt {d x^{2} + c} b^{2} c x}{8 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b x}{d} + \frac {3 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {5}{2}}} - \frac {a b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} + \frac {a^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} \]

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(d*x^2 + c)*b^2*x^3/d - 3/8*sqrt(d*x^2 + c)*b^2*c*x/d^2 + sqrt(d*x^2 + c)*a*b*x/d + 3/8*b^2*c^2*arcsin
h(d*x/sqrt(c*d))/d^(5/2) - a*b*c*arcsinh(d*x/sqrt(c*d))/d^(3/2) + a^2*arcsinh(d*x/sqrt(c*d))/sqrt(d)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {1}{8} \, {\left (\frac {2 \, b^{2} x^{2}}{d} - \frac {3 \, b^{2} c d - 8 \, a b d^{2}}{d^{3}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{8 \, d^{\frac {5}{2}}} \]

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/8*(2*b^2*x^2/d - (3*b^2*c*d - 8*a*b*d^2)/d^3)*sqrt(d*x^2 + c)*x - 1/8*(3*b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*lo
g(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)

Mupad [F(-1)]

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

[In]

int((a + b*x^2)^2/(c + d*x^2)^(1/2),x)

[Out]

int((a + b*x^2)^2/(c + d*x^2)^(1/2), x)