\(\int \frac {x^2 (a+b x^2)^2}{(c+d x^2)^{3/2}} \, dx\) [650]

Optimal result

Integrand size = 24, antiderivative size = 152 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^3}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}+\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{7/2}} \]

[Out]

1/8*(8*a^2*d^2-24*a*b*c*d+15*b^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(7/2)+(-a*d+b*c)^2*x^3/c/d^2/(d*x^2
+c)^(1/2)-1/8*(8*a^2*d^2-24*a*b*c*d+15*b^2*c^2)*x*(d*x^2+c)^(1/2)/c/d^3+1/4*b^2*x^3*(d*x^2+c)^(1/2)/d^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 152, 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 = {474, 470, 327, 223, 212} \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {\left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{7/2}}-\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right )}{8 c d^3}+\frac {x^3 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2} \]

[In]

Int[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

((b*c - a*d)^2*x^3)/(c*d^2*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(8*c*d
^3) + (b^2*x^3*Sqrt[c + d*x^2])/(4*d^2) + ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(8*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 474

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

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\int \frac {x^2 \left (-a^2 d^2+3 (b c-a d)^2-b^2 c d x^2\right )}{\sqrt {c+d x^2}} \, dx}{c d^2} \\ & = \frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx}{4 c d^2} \\ & = \frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^3}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}+\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 d^3} \\ & = \frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^3}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}+\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 d^3} \\ & = \frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^3}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}+\frac {\left (15 b^2 c^2-24 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^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.85 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {x \left (-8 a^2 d^2+8 a b d \left (3 c+d x^2\right )+b^2 \left (-15 c^2-5 c d x^2+2 d^2 x^4\right )\right )}{8 d^3 \sqrt {c+d x^2}}+\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{4 d^{7/2}} \]

[In]

Integrate[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

(x*(-8*a^2*d^2 + 8*a*b*d*(3*c + d*x^2) + b^2*(-15*c^2 - 5*c*d*x^2 + 2*d^2*x^4)))/(8*d^3*Sqrt[c + d*x^2]) + ((1
5*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/(-Sqrt[c] + Sqrt[c + d*x^2])])/(4*d^(7/2))

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(\frac {3 \left (-\frac {5 b \,x^{2}}{24}+a \right ) x b c \,d^{\frac {3}{2}}+\left (\frac {1}{4} b^{2} x^{5}+a b \,x^{3}-a^{2} x \right ) d^{\frac {5}{2}}-\frac {15 \sqrt {d}\, b^{2} c^{2} x}{8}+\left (a^{2} d^{2}-3 a b c d +\frac {15}{8} b^{2} c^{2}\right ) \sqrt {d \,x^{2}+c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{\sqrt {d \,x^{2}+c}\, d^{\frac {7}{2}}}\)	\(117\)
risch	\(\frac {b x \left (2 b d \,x^{2}+8 a d -7 b c \right ) \sqrt {d \,x^{2}+c}}{8 d^{3}}+\frac {\frac {7 b^{2} c^{2} x}{\sqrt {d \,x^{2}+c}}-\frac {8 a b c d x}{\sqrt {d \,x^{2}+c}}+\left (8 a^{2} d^{3}-24 a b c \,d^{2}+15 b^{2} c^{2} d \right ) \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{8 d^{3}}\)	\(137\)
default	\(b^{2} \left (\frac {x^{5}}{4 d \sqrt {d \,x^{2}+c}}-\frac {5 c \left (\frac {x^{3}}{2 d \sqrt {d \,x^{2}+c}}-\frac {3 c \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )}{4 d}\right )+a^{2} \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )+2 a b \left (\frac {x^{3}}{2 d \sqrt {d \,x^{2}+c}}-\frac {3 c \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )\)	\(194\)

[In]

int(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/(d*x^2+c)^(1/2)/d^(7/2)*(3*(-5/24*b*x^2+a)*x*b*c*d^(3/2)+(1/4*b^2*x^5+a*b*x^3-a^2*x)*d^(5/2)-15/8*d^(1/2)*b^
2*c^2*x+(a^2*d^2-3*a*b*c*d+15/8*b^2*c^2)*(d*x^2+c)^(1/2)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.30 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {{\left (15 \, b^{2} c^{3} - 24 \, a b c^{2} d + 8 \, a^{2} c d^{2} + {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (2 \, b^{2} d^{3} x^{5} - {\left (5 \, b^{2} c d^{2} - 8 \, a b d^{3}\right )} x^{3} - {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, {\left (d^{5} x^{2} + c d^{4}\right )}}, -\frac {{\left (15 \, b^{2} c^{3} - 24 \, a b c^{2} d + 8 \, a^{2} c d^{2} + {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b^{2} d^{3} x^{5} - {\left (5 \, b^{2} c d^{2} - 8 \, a b d^{3}\right )} x^{3} - {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, {\left (d^{5} x^{2} + c d^{4}\right )}}\right ] \]

[In]

integrate(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*((15*b^2*c^3 - 24*a*b*c^2*d + 8*a^2*c*d^2 + (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x^2)*sqrt(d)*log(-
2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*b^2*d^3*x^5 - (5*b^2*c*d^2 - 8*a*b*d^3)*x^3 - (15*b^2*c^2*d
- 24*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/(d^5*x^2 + c*d^4), -1/8*((15*b^2*c^3 - 24*a*b*c^2*d + 8*a^2*c*
d^2 + (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (2*b^2*d^3*
x^5 - (5*b^2*c*d^2 - 8*a*b*d^3)*x^3 - (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/(d^5*x^2 +
c*d^4)]

Sympy [F]

\[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**2*(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**2*(a + b*x**2)**2/(c + d*x**2)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {b^{2} x^{5}}{4 \, \sqrt {d x^{2} + c} d} - \frac {5 \, b^{2} c x^{3}}{8 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a b x^{3}}{\sqrt {d x^{2} + c} d} - \frac {15 \, b^{2} c^{2} x}{8 \, \sqrt {d x^{2} + c} d^{3}} + \frac {3 \, a b c x}{\sqrt {d x^{2} + c} d^{2}} - \frac {a^{2} x}{\sqrt {d x^{2} + c} d} + \frac {15 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {7}{2}}} - \frac {3 \, a b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {5}{2}}} + \frac {a^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} \]

[In]

integrate(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

1/4*b^2*x^5/(sqrt(d*x^2 + c)*d) - 5/8*b^2*c*x^3/(sqrt(d*x^2 + c)*d^2) + a*b*x^3/(sqrt(d*x^2 + c)*d) - 15/8*b^2
*c^2*x/(sqrt(d*x^2 + c)*d^3) + 3*a*b*c*x/(sqrt(d*x^2 + c)*d^2) - a^2*x/(sqrt(d*x^2 + c)*d) + 15/8*b^2*c^2*arcs
inh(d*x/sqrt(c*d))/d^(7/2) - 3*a*b*c*arcsinh(d*x/sqrt(c*d))/d^(5/2) + a^2*arcsinh(d*x/sqrt(c*d))/d^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.86 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {2 \, b^{2} x^{2}}{d} - \frac {5 \, b^{2} c d^{3} - 8 \, a b d^{4}}{d^{5}}\right )} x^{2} - \frac {15 \, b^{2} c^{2} d^{2} - 24 \, a b c d^{3} + 8 \, a^{2} d^{4}}{d^{5}}\right )} x}{8 \, \sqrt {d x^{2} + c}} - \frac {{\left (15 \, b^{2} c^{2} - 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{8 \, d^{\frac {7}{2}}} \]

[In]

integrate(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

1/8*((2*b^2*x^2/d - (5*b^2*c*d^3 - 8*a*b*d^4)/d^5)*x^2 - (15*b^2*c^2*d^2 - 24*a*b*c*d^3 + 8*a^2*d^4)/d^5)*x/sq
rt(d*x^2 + c) - 1/8*(15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\int \frac {x^2\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

[In]

int((x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x)

[Out]

int((x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2), x)