\(\int (a+b x^2)^2 \sqrt {c+d x^2} \, dx\) [606]

Optimal result

Integrand size = 21, antiderivative size = 149 \[ \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^2}-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{5/2}} \]

[Out]

-1/24*b*(-8*a*d+3*b*c)*x*(d*x^2+c)^(3/2)/d^2+1/6*b*x*(b*x^2+a)*(d*x^2+c)^(3/2)/d+1/16*c*(8*a^2*d^2-4*a*b*c*d+b
^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(5/2)+1/16*(8*a^2*d^2-4*a*b*c*d+b^2*c^2)*x*(d*x^2+c)^(1/2)/d^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {427, 396, 201, 223, 212} \[ \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {c \left (8 a^2 d^2-4 a b c d+b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{5/2}}+\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-4 a b c d+b^2 c^2\right )}{16 d^2}-\frac {b x \left (c+d x^2\right )^{3/2} (3 b c-8 a d)}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d} \]

[In]

Int[(a + b*x^2)^2*Sqrt[c + d*x^2],x]

[Out]

((b^2*c^2 - 4*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*d^2) - (b*(3*b*c - 8*a*d)*x*(c + d*x^2)^(3/2))/(24*d
^2) + (b*x*(a + b*x^2)*(c + d*x^2)^(3/2))/(6*d) + (c*(b^2*c^2 - 4*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqr
t[c + d*x^2]])/(16*d^(5/2))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\int \sqrt {c+d x^2} \left (-a (b c-6 a d)-b (3 b c-8 a d) x^2\right ) \, dx}{6 d} \\ & = -\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) \int \sqrt {c+d x^2} \, dx}{8 d^2} \\ & = \frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^2}-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\left (c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 d^2} \\ & = \frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^2}-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\left (c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 d^2} \\ & = \frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^2}-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.81 \[ \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (24 a^2 d^2+12 a b d \left (c+2 d x^2\right )+b^2 \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )\right )-3 c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{48 d^{5/2}} \]

[In]

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

[Out]

(Sqrt[d]*x*Sqrt[c + d*x^2]*(24*a^2*d^2 + 12*a*b*d*(c + 2*d*x^2) + b^2*(-3*c^2 + 2*c*d*x^2 + 8*d^2*x^4)) - 3*c*
(b^2*c^2 - 4*a*b*c*d + 8*a^2*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(48*d^(5/2))

Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(\frac {c \left (a^{2} d^{2}-\frac {1}{2} a b c d +\frac {1}{8} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x \sqrt {d \,x^{2}+c}\, \left (\left (\frac {1}{3} b^{2} x^{4}+a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {b c \left (\left (\frac {b \,x^{2}}{6}+a \right ) d^{\frac {3}{2}}-\frac {b \sqrt {d}\, c}{4}\right )}{2}\right )}{2 d^{\frac {5}{2}}}\)	\(106\)
risch	\(\frac {x \left (8 b^{2} d^{2} x^{4}+24 x^{2} a b \,d^{2}+2 x^{2} b^{2} c d +24 a^{2} d^{2}+12 a b c d -3 b^{2} c^{2}\right ) \sqrt {d \,x^{2}+c}}{48 d^{2}}+\frac {c \left (8 a^{2} d^{2}-4 a b c d +b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{16 d^{\frac {5}{2}}}\)	\(115\)
default	\(a^{2} \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )+b^{2} \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )}{2 d}\right )+2 a b \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )\)	\(187\)

[In]

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

[Out]

1/2/d^(5/2)*(c*(a^2*d^2-1/2*a*b*c*d+1/8*b^2*c^2)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))+x*(d*x^2+c)^(1/2)*((1/3*b^
2*x^4+a*b*x^2+a^2)*d^(5/2)+1/2*b*c*((1/6*b*x^2+a)*d^(3/2)-1/4*b*d^(1/2)*c)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.76 \[ \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\left [\frac {3 \, {\left (b^{2} c^{3} - 4 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{3} x^{5} + 2 \, {\left (b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{3} - 3 \, {\left (b^{2} c^{2} d - 4 \, a b c d^{2} - 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{96 \, d^{3}}, -\frac {3 \, {\left (b^{2} c^{3} - 4 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (8 \, b^{2} d^{3} x^{5} + 2 \, {\left (b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{3} - 3 \, {\left (b^{2} c^{2} d - 4 \, a b c d^{2} - 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, d^{3}}\right ] \]

[In]

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

[Out]

[1/96*(3*(b^2*c^3 - 4*a*b*c^2*d + 8*a^2*c*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(8*
b^2*d^3*x^5 + 2*(b^2*c*d^2 + 12*a*b*d^3)*x^3 - 3*(b^2*c^2*d - 4*a*b*c*d^2 - 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/d^3
, -1/48*(3*(b^2*c^3 - 4*a*b*c^2*d + 8*a^2*c*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b^2*d^3*x^5
+ 2*(b^2*c*d^2 + 12*a*b*d^3)*x^3 - 3*(b^2*c^2*d - 4*a*b*c*d^2 - 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/d^3]

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.27 \[ \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\begin {cases} \sqrt {c + d x^{2}} \left (\frac {b^{2} x^{5}}{6} + \frac {x^{3} \cdot \left (2 a b d + \frac {b^{2} c}{6}\right )}{4 d} + \frac {x \left (a^{2} d + 2 a b c - \frac {3 c \left (2 a b d + \frac {b^{2} c}{6}\right )}{4 d}\right )}{2 d}\right ) + \left (a^{2} c - \frac {c \left (a^{2} d + 2 a b c - \frac {3 c \left (2 a b d + \frac {b^{2} c}{6}\right )}{4 d}\right )}{2 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d \neq 0 \\\sqrt {c} \left (a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.13 \[ \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{3}}{6 \, d} + \frac {1}{2} \, \sqrt {d x^{2} + c} a^{2} x - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x}{8 \, d^{2}} + \frac {\sqrt {d x^{2} + c} b^{2} c^{2} x}{16 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x}{2 \, d} - \frac {\sqrt {d x^{2} + c} a b c x}{4 \, d} + \frac {b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {5}{2}}} - \frac {a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, d^{\frac {3}{2}}} + \frac {a^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} \]

[In]

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

[Out]

1/6*(d*x^2 + c)^(3/2)*b^2*x^3/d + 1/2*sqrt(d*x^2 + c)*a^2*x - 1/8*(d*x^2 + c)^(3/2)*b^2*c*x/d^2 + 1/16*sqrt(d*
x^2 + c)*b^2*c^2*x/d^2 + 1/2*(d*x^2 + c)^(3/2)*a*b*x/d - 1/4*sqrt(d*x^2 + c)*a*b*c*x/d + 1/16*b^2*c^3*arcsinh(
d*x/sqrt(c*d))/d^(5/2) - 1/4*a*b*c^2*arcsinh(d*x/sqrt(c*d))/d^(3/2) + 1/2*a^2*c*arcsinh(d*x/sqrt(c*d))/sqrt(d)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86 \[ \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, b^{2} x^{2} + \frac {b^{2} c d^{3} + 12 \, a b d^{4}}{d^{4}}\right )} x^{2} - \frac {3 \, {\left (b^{2} c^{2} d^{2} - 4 \, a b c d^{3} - 8 \, a^{2} d^{4}\right )}}{d^{4}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (b^{2} c^{3} - 4 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{16 \, d^{\frac {5}{2}}} \]

[In]

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

[Out]

1/48*(2*(4*b^2*x^2 + (b^2*c*d^3 + 12*a*b*d^4)/d^4)*x^2 - 3*(b^2*c^2*d^2 - 4*a*b*c*d^3 - 8*a^2*d^4)/d^4)*sqrt(d
*x^2 + c)*x - 1/16*(b^2*c^3 - 4*a*b*c^2*d + 8*a^2*c*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)

Mupad [F(-1)]

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

[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)