\(\int \frac {x^4 (a+b x^2)^2}{\sqrt {c+d x^2}} \, dx\) [636]

Optimal result

Integrand size = 24, antiderivative size = 194 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=-\frac {c \left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) x \sqrt {c+d x^2}}{128 d^4}+\frac {\left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) x^3 \sqrt {c+d x^2}}{192 d^3}-\frac {b (7 b c-16 a d) x^5 \sqrt {c+d x^2}}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}+\frac {c^2 \left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{9/2}} \]

[Out]

1/128*c^2*(48*a^2*d^2+5*b*c*(-16*a*d+7*b*c))*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(9/2)-1/128*c*(48*a^2*d^2+5*
b*c*(-16*a*d+7*b*c))*x*(d*x^2+c)^(1/2)/d^4+1/192*(48*a^2*d^2+5*b*c*(-16*a*d+7*b*c))*x^3*(d*x^2+c)^(1/2)/d^3-1/
48*b*(-16*a*d+7*b*c)*x^5*(d*x^2+c)^(1/2)/d^2+1/8*b^2*x^7*(d*x^2+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {475, 470, 327, 223, 212} \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {c^2 \left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{9/2}}+\frac {x^3 \sqrt {c+d x^2} \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right )}{192 d}-\frac {c x \sqrt {c+d x^2} \left (48 a^2 d^2+5 b c (7 b c-16 a d)\right )}{128 d^4}-\frac {b x^5 \sqrt {c+d x^2} (7 b c-16 a d)}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d} \]

[In]

Int[(x^4*(a + b*x^2)^2)/Sqrt[c + d*x^2],x]

[Out]

-1/128*(c*(48*a^2*d^2 + 5*b*c*(7*b*c - 16*a*d))*x*Sqrt[c + d*x^2])/d^4 + ((48*a^2 + (5*b*c*(7*b*c - 16*a*d))/d
^2)*x^3*Sqrt[c + d*x^2])/(192*d) - (b*(7*b*c - 16*a*d)*x^5*Sqrt[c + d*x^2])/(48*d^2) + (b^2*x^7*Sqrt[c + d*x^2
])/(8*d) + (c^2*(48*a^2*d^2 + 5*b*c*(7*b*c - 16*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(128*d^(9/2))

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^7 \sqrt {c+d x^2}}{8 d}+\frac {\int \frac {x^4 \left (8 a^2 d-b (7 b c-16 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{8 d} \\ & = -\frac {b (7 b c-16 a d) x^5 \sqrt {c+d x^2}}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}-\frac {1}{48} \left (-48 a^2-\frac {5 b c (7 b c-16 a d)}{d^2}\right ) \int \frac {x^4}{\sqrt {c+d x^2}} \, dx \\ & = \frac {\left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}}{192 d}-\frac {b (7 b c-16 a d) x^5 \sqrt {c+d x^2}}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}-\frac {\left (c \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right )\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx}{64 d} \\ & = -\frac {c \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d^2}+\frac {\left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}}{192 d}-\frac {b (7 b c-16 a d) x^5 \sqrt {c+d x^2}}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}+\frac {\left (c^2 \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{128 d^2} \\ & = -\frac {c \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d^2}+\frac {\left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}}{192 d}-\frac {b (7 b c-16 a d) x^5 \sqrt {c+d x^2}}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}+\frac {\left (c^2 \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right )\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{128 d^2} \\ & = -\frac {c \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d^2}+\frac {\left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}}{192 d}-\frac {b (7 b c-16 a d) x^5 \sqrt {c+d x^2}}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}+\frac {c^2 \left (48 a^2+\frac {5 b c (7 b c-16 a d)}{d^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (48 a^2 d^2 \left (-3 c+2 d x^2\right )+16 a b d \left (15 c^2-10 c d x^2+8 d^2 x^4\right )+b^2 \left (-105 c^3+70 c^2 d x^2-56 c d^2 x^4+48 d^3 x^6\right )\right )+6 c^2 \left (35 b^2 c^2-80 a b c d+48 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{384 d^{9/2}} \]

[In]

Integrate[(x^4*(a + b*x^2)^2)/Sqrt[c + d*x^2],x]

[Out]

(Sqrt[d]*x*Sqrt[c + d*x^2]*(48*a^2*d^2*(-3*c + 2*d*x^2) + 16*a*b*d*(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4) + b^2*(-1
05*c^3 + 70*c^2*d*x^2 - 56*c*d^2*x^4 + 48*d^3*x^6)) + 6*c^2*(35*b^2*c^2 - 80*a*b*c*d + 48*a^2*d^2)*ArcTanh[(Sq
rt[d]*x)/(-Sqrt[c] + Sqrt[c + d*x^2])])/(384*d^(9/2))

Maple [A] (verified)

Time = 2.99 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(-\frac {3 \left (\left (-a^{2} c^{2} d^{2}+\frac {5}{3} a b \,c^{3} d -\frac {35}{48} b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+\left (c \left (\frac {7}{18} b^{2} x^{4}+\frac {10}{9} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {\left (-b^{2} x^{6}-\frac {8}{3} a b \,x^{4}-2 a^{2} x^{2}\right ) d^{\frac {7}{2}}}{3}-\frac {5 b \,c^{2} \left (\left (\frac {7 b \,x^{2}}{24}+a \right ) d^{\frac {3}{2}}-\frac {7 b \sqrt {d}\, c}{16}\right )}{3}\right ) x \sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {9}{2}}}\)	\(144\)
risch	\(-\frac {x \left (-48 b^{2} d^{3} x^{6}-128 a b \,d^{3} x^{4}+56 b^{2} c \,d^{2} x^{4}-96 a^{2} d^{3} x^{2}+160 a b c \,d^{2} x^{2}-70 b^{2} c^{2} d \,x^{2}+144 c \,a^{2} d^{2}-240 a b \,c^{2} d +105 b^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{384 d^{4}}+\frac {c^{2} \left (48 a^{2} d^{2}-80 a b c d +35 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {9}{2}}}\)	\(157\)
default	\(b^{2} \left (\frac {x^{7} \sqrt {d \,x^{2}+c}}{8 d}-\frac {7 c \left (\frac {x^{5} \sqrt {d \,x^{2}+c}}{6 d}-\frac {5 c \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )}{6 d}\right )}{8 d}\right )+a^{2} \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )+2 a b \left (\frac {x^{5} \sqrt {d \,x^{2}+c}}{6 d}-\frac {5 c \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )}{6 d}\right )\)	\(272\)

[In]

int(x^4*(b*x^2+a)^2/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-3/8/d^(9/2)*((-a^2*c^2*d^2+5/3*a*b*c^3*d-35/48*b^2*c^4)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))+(c*(7/18*b^2*x^4+1
0/9*a*b*x^2+a^2)*d^(5/2)+1/3*(-b^2*x^6-8/3*a*b*x^4-2*a^2*x^2)*d^(7/2)-5/3*b*c^2*((7/24*b*x^2+a)*d^(3/2)-7/16*b
*d^(1/2)*c))*x*(d*x^2+c)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.77 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 80 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (48 \, b^{2} d^{4} x^{7} - 8 \, {\left (7 \, b^{2} c d^{3} - 16 \, a b d^{4}\right )} x^{5} + 2 \, {\left (35 \, b^{2} c^{2} d^{2} - 80 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (35 \, b^{2} c^{3} d - 80 \, a b c^{2} d^{2} + 48 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{768 \, d^{5}}, -\frac {3 \, {\left (35 \, b^{2} c^{4} - 80 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (48 \, b^{2} d^{4} x^{7} - 8 \, {\left (7 \, b^{2} c d^{3} - 16 \, a b d^{4}\right )} x^{5} + 2 \, {\left (35 \, b^{2} c^{2} d^{2} - 80 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (35 \, b^{2} c^{3} d - 80 \, a b c^{2} d^{2} + 48 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{384 \, d^{5}}\right ] \]

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(35*b^2*c^4 - 80*a*b*c^3*d + 48*a^2*c^2*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c)
+ 2*(48*b^2*d^4*x^7 - 8*(7*b^2*c*d^3 - 16*a*b*d^4)*x^5 + 2*(35*b^2*c^2*d^2 - 80*a*b*c*d^3 + 48*a^2*d^4)*x^3 -
3*(35*b^2*c^3*d - 80*a*b*c^2*d^2 + 48*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^5, -1/384*(3*(35*b^2*c^4 - 80*a*b*c^3*
d + 48*a^2*c^2*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (48*b^2*d^4*x^7 - 8*(7*b^2*c*d^3 - 16*a*b*d^
4)*x^5 + 2*(35*b^2*c^2*d^2 - 80*a*b*c*d^3 + 48*a^2*d^4)*x^3 - 3*(35*b^2*c^3*d - 80*a*b*c^2*d^2 + 48*a^2*c*d^3)
*x)*sqrt(d*x^2 + c))/d^5]

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.13 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\begin {cases} \frac {3 c^{2} \left (a^{2} - \frac {5 c \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d^{2}} + \sqrt {c + d x^{2}} \left (\frac {b^{2} x^{7}}{8 d} - \frac {3 c x \left (a^{2} - \frac {5 c \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d}\right )}{8 d^{2}} + \frac {x^{5} \cdot \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d} + \frac {x^{3} \left (a^{2} - \frac {5 c \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d}\right )}{4 d}\right ) & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} x^{5}}{5} + \frac {2 a b x^{7}}{7} + \frac {b^{2} x^{9}}{9}}{\sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**4*(b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

Piecewise((3*c**2*(a**2 - 5*c*(2*a*b - 7*b**2*c/(8*d))/(6*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))/(8*d**2) + sqrt(c + d*x**2)*(b**2*x**7/(8*d) - 3*c*x*(a*
*2 - 5*c*(2*a*b - 7*b**2*c/(8*d))/(6*d))/(8*d**2) + x**5*(2*a*b - 7*b**2*c/(8*d))/(6*d) + x**3*(a**2 - 5*c*(2*
a*b - 7*b**2*c/(8*d))/(6*d))/(4*d)), Ne(d, 0)), ((a**2*x**5/5 + 2*a*b*x**7/7 + b**2*x**9/9)/sqrt(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.25 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x^{7}}{8 \, d} - \frac {7 \, \sqrt {d x^{2} + c} b^{2} c x^{5}}{48 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b x^{5}}{3 \, d} + \frac {35 \, \sqrt {d x^{2} + c} b^{2} c^{2} x^{3}}{192 \, d^{3}} - \frac {5 \, \sqrt {d x^{2} + c} a b c x^{3}}{12 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} x^{3}}{4 \, d} - \frac {35 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d^{4}} + \frac {5 \, \sqrt {d x^{2} + c} a b c^{2} x}{8 \, d^{3}} - \frac {3 \, \sqrt {d x^{2} + c} a^{2} c x}{8 \, d^{2}} + \frac {35 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {9}{2}}} - \frac {5 \, a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {7}{2}}} + \frac {3 \, a^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {5}{2}}} \]

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

1/8*sqrt(d*x^2 + c)*b^2*x^7/d - 7/48*sqrt(d*x^2 + c)*b^2*c*x^5/d^2 + 1/3*sqrt(d*x^2 + c)*a*b*x^5/d + 35/192*sq
rt(d*x^2 + c)*b^2*c^2*x^3/d^3 - 5/12*sqrt(d*x^2 + c)*a*b*c*x^3/d^2 + 1/4*sqrt(d*x^2 + c)*a^2*x^3/d - 35/128*sq
rt(d*x^2 + c)*b^2*c^3*x/d^4 + 5/8*sqrt(d*x^2 + c)*a*b*c^2*x/d^3 - 3/8*sqrt(d*x^2 + c)*a^2*c*x/d^2 + 35/128*b^2
*c^4*arcsinh(d*x/sqrt(c*d))/d^(9/2) - 5/8*a*b*c^3*arcsinh(d*x/sqrt(c*d))/d^(7/2) + 3/8*a^2*c^2*arcsinh(d*x/sqr
t(c*d))/d^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.92 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (\frac {6 \, b^{2} x^{2}}{d} - \frac {7 \, b^{2} c d^{5} - 16 \, a b d^{6}}{d^{7}}\right )} x^{2} + \frac {35 \, b^{2} c^{2} d^{4} - 80 \, a b c d^{5} + 48 \, a^{2} d^{6}}{d^{7}}\right )} x^{2} - \frac {3 \, {\left (35 \, b^{2} c^{3} d^{3} - 80 \, a b c^{2} d^{4} + 48 \, a^{2} c d^{5}\right )}}{d^{7}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (35 \, b^{2} c^{4} - 80 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{128 \, d^{\frac {9}{2}}} \]

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/384*(2*(4*(6*b^2*x^2/d - (7*b^2*c*d^5 - 16*a*b*d^6)/d^7)*x^2 + (35*b^2*c^2*d^4 - 80*a*b*c*d^5 + 48*a^2*d^6)/
d^7)*x^2 - 3*(35*b^2*c^3*d^3 - 80*a*b*c^2*d^4 + 48*a^2*c*d^5)/d^7)*sqrt(d*x^2 + c)*x - 1/128*(35*b^2*c^4 - 80*
a*b*c^3*d + 48*a^2*c^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(9/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\int \frac {x^4\,{\left (b\,x^2+a\right )}^2}{\sqrt {d\,x^2+c}} \,d x \]

[In]

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(1/2),x)

[Out]

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(1/2), x)