\(\int \frac {x^2 (a+b x^2)^2}{\sqrt {c+d x^2}} \, dx\) [638]

Optimal result

Integrand size = 24, antiderivative size = 146 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\left (8 a^2 d^2+b c (5 b c-12 a d)\right ) x \sqrt {c+d x^2}}{16 d^3}-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}-\frac {c \left (8 a^2 d^2+b c (5 b c-12 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{7/2}} \]

[Out]

-1/16*c*(8*a^2*d^2+b*c*(-12*a*d+5*b*c))*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(7/2)+1/16*(8*a^2*d^2+b*c*(-12*a*
d+5*b*c))*x*(d*x^2+c)^(1/2)/d^3-1/24*b*(-12*a*d+5*b*c)*x^3*(d*x^2+c)^(1/2)/d^2+1/6*b^2*x^5*(d*x^2+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {475, 470, 327, 223, 212} \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=-\frac {c \left (8 a^2 d^2+b c (5 b c-12 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{7/2}}+\frac {x \sqrt {c+d x^2} \left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right )}{16 d}-\frac {b x^3 \sqrt {c+d x^2} (5 b c-12 a d)}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d} \]

[In]

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

[Out]

((8*a^2 + (b*c*(5*b*c - 12*a*d))/d^2)*x*Sqrt[c + d*x^2])/(16*d) - (b*(5*b*c - 12*a*d)*x^3*Sqrt[c + d*x^2])/(24
*d^2) + (b^2*x^5*Sqrt[c + d*x^2])/(6*d) - (c*(8*a^2*d^2 + b*c*(5*b*c - 12*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d
*x^2]])/(16*d^(7/2))

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 327

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

Rule 470

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

Rule 475

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[d^2*(e*x)^(
m + n + 1)*((a + b*x^n)^(p + 1)/(b*e^(n + 1)*(m + n*(p + 2) + 1))), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}+\frac {\int \frac {x^2 \left (6 a^2 d-b (5 b c-12 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{6 d} \\ & = -\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}+\frac {1}{8} \left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx \\ & = \frac {\left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{16 d}-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}-\frac {\left (c \left (5 b^2 c^2-12 a b c d+8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 d^3} \\ & = \frac {\left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{16 d}-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}-\frac {\left (c \left (5 b^2 c^2-12 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^3} \\ & = \frac {\left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{16 d}-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}-\frac {c \left (5 b^2 c^2-12 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^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {x \sqrt {c+d x^2} \left (24 a^2 d^2+12 a b d \left (-3 c+2 d x^2\right )+b^2 \left (15 c^2-10 c d x^2+8 d^2 x^4\right )\right )}{48 d^3}+\frac {c \left (5 b^2 c^2-12 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+d x^2}}\right )}{8 d^{7/2}} \]

[In]

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

[Out]

(x*Sqrt[c + d*x^2]*(24*a^2*d^2 + 12*a*b*d*(-3*c + 2*d*x^2) + b^2*(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4)))/(48*d^3)
+ (c*(5*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/(Sqrt[c] - Sqrt[c + d*x^2])])/(8*d^(7/2))

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\left (-c \,a^{2} d^{2}+\frac {3}{2} a b \,c^{2} d -\frac {5}{8} b^{2} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x \sqrt {d \,x^{2}+c}\, \left (\left (\frac {1}{3} b^{2} x^{4}+a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}-\frac {3 b c \left (\left (\frac {5 b \,x^{2}}{18}+a \right ) d^{\frac {3}{2}}-\frac {5 b \sqrt {d}\, c}{12}\right )}{2}\right )}{2 d^{\frac {7}{2}}}\)	\(109\)
risch	\(\frac {x \left (8 b^{2} d^{2} x^{4}+24 x^{2} a b \,d^{2}-10 x^{2} b^{2} c d +24 a^{2} d^{2}-36 a b c d +15 b^{2} c^{2}\right ) \sqrt {d \,x^{2}+c}}{48 d^{3}}-\frac {c \left (8 a^{2} d^{2}-12 a b c d +5 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{16 d^{\frac {7}{2}}}\)	\(116\)
default	\(b^{2} \left (\frac {x^{5} \sqrt {d \,x^{2}+c}}{6 d}-\frac {5 c \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )}{6 d}\right )+a^{2} \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )+2 a b \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )\)	\(200\)

[In]

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

[Out]

1/2/d^(7/2)*((-c*a^2*d^2+3/2*a*b*c^2*d-5/8*b^2*c^3)*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)-3/2*b*c*((5/18*b*x^2+a)*d^(3/2)-5/12*b*d^(1/2)*c)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.83 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{3} - 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{3} x^{5} - 2 \, {\left (5 \, b^{2} c d^{2} - 12 \, a b d^{3}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{96 \, d^{4}}, \frac {3 \, {\left (5 \, b^{2} c^{3} - 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} d^{3} x^{5} - 2 \, {\left (5 \, b^{2} c d^{2} - 12 \, a b d^{3}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, d^{4}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\begin {cases} - \frac {c \left (a^{2} - \frac {3 c \left (2 a b - \frac {5 b^{2} c}{6 d}\right )}{4 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2 d} + \sqrt {c + d x^{2}} \left (\frac {b^{2} x^{5}}{6 d} + \frac {x^{3} \cdot \left (2 a b - \frac {5 b^{2} c}{6 d}\right )}{4 d} + \frac {x \left (a^{2} - \frac {3 c \left (2 a b - \frac {5 b^{2} c}{6 d}\right )}{4 d}\right )}{2 d}\right ) & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} x^{3}}{3} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{7}}{7}}{\sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-c*(a**2 - 3*c*(2*a*b - 5*b**2*c/(6*d))/(4*d))*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/s
qrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/(2*d) + sqrt(c + d*x**2)*(b**2*x**5/(6*d) + x**3*(2*a*b - 5*
b**2*c/(6*d))/(4*d) + x*(a**2 - 3*c*(2*a*b - 5*b**2*c/(6*d))/(4*d))/(2*d)), Ne(d, 0)), ((a**2*x**3/3 + 2*a*b*x
**5/5 + b**2*x**7/7)/sqrt(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.20 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x^{5}}{6 \, d} - \frac {5 \, \sqrt {d x^{2} + c} b^{2} c x^{3}}{24 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b x^{3}}{2 \, d} + \frac {5 \, \sqrt {d x^{2} + c} b^{2} c^{2} x}{16 \, d^{3}} - \frac {3 \, \sqrt {d x^{2} + c} a b c x}{4 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} x}{2 \, d} - \frac {5 \, b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {7}{2}}} + \frac {3 \, a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, d^{\frac {5}{2}}} - \frac {a^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {3}{2}}} \]

[In]

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

[Out]

1/6*sqrt(d*x^2 + c)*b^2*x^5/d - 5/24*sqrt(d*x^2 + c)*b^2*c*x^3/d^2 + 1/2*sqrt(d*x^2 + c)*a*b*x^3/d + 5/16*sqrt
(d*x^2 + c)*b^2*c^2*x/d^3 - 3/4*sqrt(d*x^2 + c)*a*b*c*x/d^2 + 1/2*sqrt(d*x^2 + c)*a^2*x/d - 5/16*b^2*c^3*arcsi
nh(d*x/sqrt(c*d))/d^(7/2) + 3/4*a*b*c^2*arcsinh(d*x/sqrt(c*d))/d^(5/2) - 1/2*a^2*c*arcsinh(d*x/sqrt(c*d))/d^(3
/2)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, b^{2} x^{2}}{d} - \frac {5 \, b^{2} c d^{3} - 12 \, a b d^{4}}{d^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{2} d^{2} - 12 \, a b c d^{3} + 8 \, a^{2} d^{4}\right )}}{d^{5}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (5 \, b^{2} c^{3} - 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{16 \, d^{\frac {7}{2}}} \]

[In]

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

[Out]

1/48*(2*(4*b^2*x^2/d - (5*b^2*c*d^3 - 12*a*b*d^4)/d^5)*x^2 + 3*(5*b^2*c^2*d^2 - 12*a*b*c*d^3 + 8*a^2*d^4)/d^5)
*sqrt(d*x^2 + c)*x + 1/16*(5*b^2*c^3 - 12*a*b*c^2*d + 8*a^2*c*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7
/2)

Mupad [F(-1)]

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

[In]

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

[Out]

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