\(\int \frac {(c+d x^2)^2}{\sqrt {a+b x^2}} \, dx\) [75]

Optimal result

Integrand size = 21, antiderivative size = 108 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {3 d (2 b c-a d) x \sqrt {a+b x^2}}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b}+\frac {\left (8 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 108, 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} \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a^2 d^2-8 a b c d+8 b^2 c^2\right )}{8 b^{5/2}}+\frac {3 d x \sqrt {a+b x^2} (2 b c-a d)}{8 b^2}+\frac {d x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 b} \]

[In]

Int[(c + d*x^2)^2/Sqrt[a + b*x^2],x]

[Out]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {d x \sqrt {a+b x^2} \left (8 b c-3 a d+2 b d x^2\right )}{8 b^2}+\frac {\left (-8 b^2 c^2+8 a b c d-3 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.35 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {d x \left (-2 b d \,x^{2}+3 a d -8 b c \right ) \sqrt {b \,x^{2}+a}}{8 b^{2}}+\frac {\left (3 a^{2} d^{2}-8 a b c d +8 b^{2} c^{2}\right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\)	\(78\)
pseudoelliptic	\(\frac {\frac {3 \left (a^{2} d^{2}-\frac {8}{3} a b c d +\frac {8}{3} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{8}-\frac {3 x \sqrt {b \,x^{2}+a}\, \left (\frac {2 \left (-d \,x^{2}-4 c \right ) b^{\frac {3}{2}}}{3}+a d \sqrt {b}\right ) d}{8}}{b^{\frac {5}{2}}}\)	\(82\)
default	\(\frac {c^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+d^{2} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+2 c d \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\)	\(133\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.78 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\left [\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, b^{2} d^{2} x^{3} + {\left (8 \, b^{2} c d - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, -\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} + {\left (8 \, b^{2} c d - 3 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.24 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {d^{2} x^{3}}{4 b} + \frac {x \left (- \frac {3 a d^{2}}{4 b} + 2 c d\right )}{2 b}\right ) + \left (- \frac {a \left (- \frac {3 a d^{2}}{4 b} + 2 c d\right )}{2 b} + c^{2}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {c^{2} x + \frac {2 c d x^{3}}{3} + \frac {d^{2} x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} d^{2} x^{3}}{4 \, b} + \frac {\sqrt {b x^{2} + a} c d x}{b} - \frac {3 \, \sqrt {b x^{2} + a} a d^{2} x}{8 \, b^{2}} + \frac {c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {a c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {1}{8} \, \sqrt {b x^{2} + a} {\left (\frac {2 \, d^{2} x^{2}}{b} + \frac {8 \, b^{2} c d - 3 \, a b d^{2}}{b^{3}}\right )} x - \frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + 3 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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