\(\int \frac {x^2}{(a+b x^2)^2 (c+d x^2)^{5/2}} \, dx\) [778]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {5 d x}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {d (13 b c+2 a d) x}{6 c (b c-a d)^3 \sqrt {c+d x^2}}+\frac {b (b c+4 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} (b c-a d)^{7/2}} \]

[Out]

-5/6*d*x/(-a*d+b*c)^2/(d*x^2+c)^(3/2)-1/2*x/(-a*d+b*c)/(b*x^2+a)/(d*x^2+c)^(3/2)+1/2*b*(4*a*d+b*c)*arctan(x*(-
a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))/(-a*d+b*c)^(7/2)/a^(1/2)-1/6*d*(2*a*d+13*b*c)*x/c/(-a*d+b*c)^3/(d*x^2+
c)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 163, 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 = {482, 541, 12, 385, 211} \[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {b (4 a d+b c) \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} (b c-a d)^{7/2}}-\frac {d x (2 a d+13 b c)}{6 c \sqrt {c+d x^2} (b c-a d)^3}-\frac {x}{2 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {5 d x}{6 \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

[In]

Int[x^2/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x]

[Out]

(-5*d*x)/(6*(b*c - a*d)^2*(c + d*x^2)^(3/2)) - x/(2*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) - (d*(13*b*c +
2*a*d)*x)/(6*c*(b*c - a*d)^3*Sqrt[c + d*x^2]) + (b*(b*c + 4*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c +
d*x^2])])/(2*Sqrt[a]*(b*c - a*d)^(7/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 385

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

Rule 482

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

Rule 541

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

Rubi steps \begin{align*} \text {integral}& = -\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {c-4 d x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx}{2 (b c-a d)} \\ & = -\frac {5 d x}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {c (3 b c+2 a d)-10 b c d x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{6 c (b c-a d)^2} \\ & = -\frac {5 d x}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {d (13 b c+2 a d) x}{6 c (b c-a d)^3 \sqrt {c+d x^2}}+\frac {\int \frac {3 b c^2 (b c+4 a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 c^2 (b c-a d)^3} \\ & = -\frac {5 d x}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {d (13 b c+2 a d) x}{6 c (b c-a d)^3 \sqrt {c+d x^2}}+\frac {(b (b c+4 a d)) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 (b c-a d)^3} \\ & = -\frac {5 d x}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {d (13 b c+2 a d) x}{6 c (b c-a d)^3 \sqrt {c+d x^2}}+\frac {(b (b c+4 a d)) \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 (b c-a d)^3} \\ & = -\frac {5 d x}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {x}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {d (13 b c+2 a d) x}{6 c (b c-a d)^3 \sqrt {c+d x^2}}+\frac {b (b c+4 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 \sqrt {a} (b c-a d)^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {x \left (2 a^2 d^3 x^2+2 a b d \left (6 c^2+5 c d x^2+d^2 x^4\right )+b^2 c \left (3 c^2+18 c d x^2+13 d^2 x^4\right )\right )}{6 c (b c-a d)^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {b (b c+4 a d) \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 \sqrt {a} (b c-a d)^{7/2}} \]

[In]

Integrate[x^2/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x]

[Out]

-1/6*(x*(2*a^2*d^3*x^2 + 2*a*b*d*(6*c^2 + 5*c*d*x^2 + d^2*x^4) + b^2*c*(3*c^2 + 18*c*d*x^2 + 13*d^2*x^4)))/(c*
(b*c - a*d)^3*(a + b*x^2)*(c + d*x^2)^(3/2)) - (b*(b*c + 4*a*d)*ArcTan[(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c +
d*x^2]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(2*Sqrt[a]*(b*c - a*d)^(7/2))

Maple [A] (verified)

Time = 3.21 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {\frac {b c \left (\frac {b \sqrt {d \,x^{2}+c}\, x}{b \,x^{2}+a}-\frac {\left (4 a d +b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{2 \left (a d -b c \right )^{3}}+\frac {d^{2} x^{3}}{3 \left (a d -b c \right )^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 b c d x}{\left (a d -b c \right )^{3} \sqrt {d \,x^{2}+c}}}{c}\)	\(143\)
default	\(\text {Expression too large to display}\)	\(3483\)

[In]

int(x^2/(b*x^2+a)^2/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 626 vs. \(2 (139) = 278\).

Time = 1.13 (sec) , antiderivative size = 1292, normalized size of antiderivative = 7.93 \[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(x^2/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(3*(a*b^2*c^4 + 4*a^2*b*c^3*d + (b^3*c^2*d^2 + 4*a*b^2*c*d^3)*x^6 + (2*b^3*c^3*d + 9*a*b^2*c^2*d^2 + 4*a
^2*b*c*d^3)*x^4 + (b^3*c^4 + 6*a*b^2*c^3*d + 8*a^2*b*c^2*d^2)*x^2)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*
c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 + 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a
^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*((13*a*b^3*c^2*d^2 - 11*a^2*b^2*c*d^3 - 2*a^3*b*d^4)*x
^5 + 2*(9*a*b^3*c^3*d - 4*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4)*x^3 + 3*(a*b^3*c^4 + 3*a^2*b^2*c^3*d - 4*
a^3*b*c^2*d^2)*x)*sqrt(d*x^2 + c))/(a^2*b^4*c^7 - 4*a^3*b^3*c^6*d + 6*a^4*b^2*c^5*d^2 - 4*a^5*b*c^4*d^3 + a^6*
c^3*d^4 + (a*b^5*c^5*d^2 - 4*a^2*b^4*c^4*d^3 + 6*a^3*b^3*c^3*d^4 - 4*a^4*b^2*c^2*d^5 + a^5*b*c*d^6)*x^6 + (2*a
*b^5*c^6*d - 7*a^2*b^4*c^5*d^2 + 8*a^3*b^3*c^4*d^3 - 2*a^4*b^2*c^3*d^4 - 2*a^5*b*c^2*d^5 + a^6*c*d^6)*x^4 + (a
*b^5*c^7 - 2*a^2*b^4*c^6*d - 2*a^3*b^3*c^5*d^2 + 8*a^4*b^2*c^4*d^3 - 7*a^5*b*c^3*d^4 + 2*a^6*c^2*d^5)*x^2), 1/
12*(3*(a*b^2*c^4 + 4*a^2*b*c^3*d + (b^3*c^2*d^2 + 4*a*b^2*c*d^3)*x^6 + (2*b^3*c^3*d + 9*a*b^2*c^2*d^2 + 4*a^2*
b*c*d^3)*x^4 + (b^3*c^4 + 6*a*b^2*c^3*d + 8*a^2*b*c^2*d^2)*x^2)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^
2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - 2*((13*a*b
^3*c^2*d^2 - 11*a^2*b^2*c*d^3 - 2*a^3*b*d^4)*x^5 + 2*(9*a*b^3*c^3*d - 4*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*
d^4)*x^3 + 3*(a*b^3*c^4 + 3*a^2*b^2*c^3*d - 4*a^3*b*c^2*d^2)*x)*sqrt(d*x^2 + c))/(a^2*b^4*c^7 - 4*a^3*b^3*c^6*
d + 6*a^4*b^2*c^5*d^2 - 4*a^5*b*c^4*d^3 + a^6*c^3*d^4 + (a*b^5*c^5*d^2 - 4*a^2*b^4*c^4*d^3 + 6*a^3*b^3*c^3*d^4
- 4*a^4*b^2*c^2*d^5 + a^5*b*c*d^6)*x^6 + (2*a*b^5*c^6*d - 7*a^2*b^4*c^5*d^2 + 8*a^3*b^3*c^4*d^3 - 2*a^4*b^2*c
^3*d^4 - 2*a^5*b*c^2*d^5 + a^6*c*d^6)*x^4 + (a*b^5*c^7 - 2*a^2*b^4*c^6*d - 2*a^3*b^3*c^5*d^2 + 8*a^4*b^2*c^4*d
^3 - 7*a^5*b*c^3*d^4 + 2*a^6*c^2*d^5)*x^2)]

Sympy [F]

\[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^{2}}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**2/(b*x**2+a)**2/(d*x**2+c)**(5/2),x)

[Out]

Integral(x**2/((a + b*x**2)**2*(c + d*x**2)**(5/2)), x)

Maxima [F]

\[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^2/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^2/((b*x^2 + a)^2*(d*x^2 + c)^(5/2)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 595 vs. \(2 (139) = 278\).

Time = 0.94 (sec) , antiderivative size = 595, normalized size of antiderivative = 3.65 \[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {{\left (\frac {{\left (5 \, b^{4} c^{4} d^{3} - 14 \, a b^{3} c^{3} d^{4} + 12 \, a^{2} b^{2} c^{2} d^{5} - 2 \, a^{3} b c d^{6} - a^{4} d^{7}\right )} x^{2}}{b^{6} c^{7} d - 6 \, a b^{5} c^{6} d^{2} + 15 \, a^{2} b^{4} c^{5} d^{3} - 20 \, a^{3} b^{3} c^{4} d^{4} + 15 \, a^{4} b^{2} c^{3} d^{5} - 6 \, a^{5} b c^{2} d^{6} + a^{6} c d^{7}} + \frac {6 \, {\left (b^{4} c^{5} d^{2} - 3 \, a b^{3} c^{4} d^{3} + 3 \, a^{2} b^{2} c^{3} d^{4} - a^{3} b c^{2} d^{5}\right )}}{b^{6} c^{7} d - 6 \, a b^{5} c^{6} d^{2} + 15 \, a^{2} b^{4} c^{5} d^{3} - 20 \, a^{3} b^{3} c^{4} d^{4} + 15 \, a^{4} b^{2} c^{3} d^{5} - 6 \, a^{5} b c^{2} d^{6} + a^{6} c d^{7}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} - \frac {{\left (b^{2} c \sqrt {d} + 4 \, a b d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b d^{\frac {3}{2}} - b^{2} c^{2} \sqrt {d}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} \]

[In]

integrate(x^2/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

-1/3*((5*b^4*c^4*d^3 - 14*a*b^3*c^3*d^4 + 12*a^2*b^2*c^2*d^5 - 2*a^3*b*c*d^6 - a^4*d^7)*x^2/(b^6*c^7*d - 6*a*b
^5*c^6*d^2 + 15*a^2*b^4*c^5*d^3 - 20*a^3*b^3*c^4*d^4 + 15*a^4*b^2*c^3*d^5 - 6*a^5*b*c^2*d^6 + a^6*c*d^7) + 6*(
b^4*c^5*d^2 - 3*a*b^3*c^4*d^3 + 3*a^2*b^2*c^3*d^4 - a^3*b*c^2*d^5)/(b^6*c^7*d - 6*a*b^5*c^6*d^2 + 15*a^2*b^4*c
^5*d^3 - 20*a^3*b^3*c^4*d^4 + 15*a^4*b^2*c^3*d^5 - 6*a^5*b*c^2*d^6 + a^6*c*d^7))*x/(d*x^2 + c)^(3/2) - 1/2*(b^
2*c*sqrt(d) + 4*a*b*d^(3/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d
^2))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a*b*c*d - a^2*d^2)) + ((sqrt(d)*x - sqrt(d*x^2
+ c))^2*b^2*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*d^(3/2) - b^2*c^2*sqrt(d))/((b^3*c^3 - 3*a*b^2*c
^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4
*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^2}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \]

[In]

int(x^2/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x)

[Out]

int(x^2/((a + b*x^2)^2*(c + d*x^2)^(5/2)), x)