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

Optimal. Leaf size=208 \[ \frac {a \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{16 c^{7/2} (b c-a d)^{5/2}}+\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{48 c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac {x \sqrt {a+b x^2} (4 b c-5 a d)}{24 c^2 \left (c+d x^2\right )^2 (b c-a d)}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3} \]

[Out]

1/16*a*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*arctanh(x*(-a*d+b*c)^(1/2)/c^(1/2)/(b*x^2+a)^(1/2))/c^(7/2)/(-a*d+b*c)
^(5/2)+1/6*x*(b*x^2+a)^(1/2)/c/(d*x^2+c)^3+1/24*(-5*a*d+4*b*c)*x*(b*x^2+a)^(1/2)/c^2/(-a*d+b*c)/(d*x^2+c)^2+1/
48*(-5*a*d+2*b*c)*(-3*a*d+4*b*c)*x*(b*x^2+a)^(1/2)/c^3/(-a*d+b*c)^2/(d*x^2+c)

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {412, 527, 12, 377, 208} \[ \frac {a \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{16 c^{7/2} (b c-a d)^{5/2}}+\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{48 c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac {x \sqrt {a+b x^2} (4 b c-5 a d)}{24 c^2 \left (c+d x^2\right )^2 (b c-a d)}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/(c + d*x^2)^4,x]

[Out]

(x*Sqrt[a + b*x^2])/(6*c*(c + d*x^2)^3) + ((4*b*c - 5*a*d)*x*Sqrt[a + b*x^2])/(24*c^2*(b*c - a*d)*(c + d*x^2)^
2) + ((2*b*c - 5*a*d)*(4*b*c - 3*a*d)*x*Sqrt[a + b*x^2])/(48*c^3*(b*c - a*d)^2*(c + d*x^2)) + (a*(8*b^2*c^2 -
12*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(16*c^(7/2)*(b*c - a*d)^(5/2))

Rule 12

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

Rule 208

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

Rule 377

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 412

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

Rule 527

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*} \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx &=\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}-\frac {\int \frac {-5 a-4 b x^2}{\sqrt {a+b x^2} \left (c+d x^2\right )^3} \, dx}{6 c}\\ &=\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}+\frac {(4 b c-5 a d) x \sqrt {a+b x^2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}-\frac {\int \frac {-a (16 b c-15 a d)-2 b (4 b c-5 a d) x^2}{\sqrt {a+b x^2} \left (c+d x^2\right )^2} \, dx}{24 c^2 (b c-a d)}\\ &=\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}+\frac {(4 b c-5 a d) x \sqrt {a+b x^2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac {(2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{48 c^3 (b c-a d)^2 \left (c+d x^2\right )}-\frac {\int -\frac {3 a \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right )}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx}{48 c^3 (b c-a d)^2}\\ &=\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}+\frac {(4 b c-5 a d) x \sqrt {a+b x^2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac {(2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{48 c^3 (b c-a d)^2 \left (c+d x^2\right )}+\frac {\left (a \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx}{16 c^3 (b c-a d)^2}\\ &=\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}+\frac {(4 b c-5 a d) x \sqrt {a+b x^2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac {(2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{48 c^3 (b c-a d)^2 \left (c+d x^2\right )}+\frac {\left (a \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{c-(b c-a d) x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{16 c^3 (b c-a d)^2}\\ &=\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}+\frac {(4 b c-5 a d) x \sqrt {a+b x^2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac {(2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{48 c^3 (b c-a d)^2 \left (c+d x^2\right )}+\frac {a \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{16 c^{7/2} (b c-a d)^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.99, size = 227, normalized size = 1.09 \[ \frac {x \sqrt {a+b x^2} \left (\frac {3 a \left (c+d x^2\right )^3 \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \sqrt {\frac {x^2 (b c-a d)}{c \left (a+b x^2\right )}} \tanh ^{-1}\left (\sqrt {\frac {x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{x^2}+(b c-a d) \left (a^2 d^2 \left (33 c^2+40 c d x^2+15 d^2 x^4\right )-2 a b c d \left (30 c^2+35 c d x^2+13 d^2 x^4\right )+8 b^2 c^2 \left (3 c^2+3 c d x^2+d^2 x^4\right )\right )\right )}{48 c^3 \left (c+d x^2\right )^3 (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*((b*c - a*d)*(8*b^2*c^2*(3*c^2 + 3*c*d*x^2 + d^2*x^4) - 2*a*b*c*d*(30*c^2 + 35*c*d*x^2 + 13
*d^2*x^4) + a^2*d^2*(33*c^2 + 40*c*d*x^2 + 15*d^2*x^4)) + (3*a*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Sqrt[((b*c
- a*d)*x^2)/(c*(a + b*x^2))]*(c + d*x^2)^3*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/x^2))/(48*c^3*(b
*c - a*d)^3*(c + d*x^2)^3)

________________________________________________________________________________________

fricas [B]  time = 1.64, size = 1220, normalized size = 5.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(3*(8*a*b^2*c^5 - 12*a^2*b*c^4*d + 5*a^3*c^3*d^2 + (8*a*b^2*c^2*d^3 - 12*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 +
3*(8*a*b^2*c^3*d^2 - 12*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^4 + 3*(8*a*b^2*c^4*d - 12*a^2*b*c^3*d^2 + 5*a^3*c^2*d^
3)*x^2)*sqrt(b*c^2 - a*c*d)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x
^2 + 4*((2*b*c - a*d)*x^3 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*((8*b
^3*c^4*d^2 - 34*a*b^2*c^3*d^3 + 41*a^2*b*c^2*d^4 - 15*a^3*c*d^5)*x^5 + 2*(12*b^3*c^5*d - 47*a*b^2*c^4*d^2 + 55
*a^2*b*c^3*d^3 - 20*a^3*c^2*d^4)*x^3 + 3*(8*b^3*c^6 - 28*a*b^2*c^5*d + 31*a^2*b*c^4*d^2 - 11*a^3*c^3*d^3)*x)*s
qrt(b*x^2 + a))/(b^3*c^10 - 3*a*b^2*c^9*d + 3*a^2*b*c^8*d^2 - a^3*c^7*d^3 + (b^3*c^7*d^3 - 3*a*b^2*c^6*d^4 + 3
*a^2*b*c^5*d^5 - a^3*c^4*d^6)*x^6 + 3*(b^3*c^8*d^2 - 3*a*b^2*c^7*d^3 + 3*a^2*b*c^6*d^4 - a^3*c^5*d^5)*x^4 + 3*
(b^3*c^9*d - 3*a*b^2*c^8*d^2 + 3*a^2*b*c^7*d^3 - a^3*c^6*d^4)*x^2), -1/96*(3*(8*a*b^2*c^5 - 12*a^2*b*c^4*d + 5
*a^3*c^3*d^2 + (8*a*b^2*c^2*d^3 - 12*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 3*(8*a*b^2*c^3*d^2 - 12*a^2*b*c^2*d^3 + 5*
a^3*c*d^4)*x^4 + 3*(8*a*b^2*c^4*d - 12*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x^2)*sqrt(-b*c^2 + a*c*d)*arctan(1/2*sqr
t(-b*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c*d)*x^3 + (a*b*c^2 - a^2*c*d)*x))
- 2*((8*b^3*c^4*d^2 - 34*a*b^2*c^3*d^3 + 41*a^2*b*c^2*d^4 - 15*a^3*c*d^5)*x^5 + 2*(12*b^3*c^5*d - 47*a*b^2*c^
4*d^2 + 55*a^2*b*c^3*d^3 - 20*a^3*c^2*d^4)*x^3 + 3*(8*b^3*c^6 - 28*a*b^2*c^5*d + 31*a^2*b*c^4*d^2 - 11*a^3*c^3
*d^3)*x)*sqrt(b*x^2 + a))/(b^3*c^10 - 3*a*b^2*c^9*d + 3*a^2*b*c^8*d^2 - a^3*c^7*d^3 + (b^3*c^7*d^3 - 3*a*b^2*c
^6*d^4 + 3*a^2*b*c^5*d^5 - a^3*c^4*d^6)*x^6 + 3*(b^3*c^8*d^2 - 3*a*b^2*c^7*d^3 + 3*a^2*b*c^6*d^4 - a^3*c^5*d^5
)*x^4 + 3*(b^3*c^9*d - 3*a*b^2*c^8*d^2 + 3*a^2*b*c^7*d^3 - a^3*c^6*d^4)*x^2)]

________________________________________________________________________________________

giac [B]  time = 2.87, size = 958, normalized size = 4.61 \[ -\frac {{\left (8 \, a b^{\frac {5}{2}} c^{2} - 12 \, a^{2} b^{\frac {3}{2}} c d + 5 \, a^{3} \sqrt {b} d^{2}\right )} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{16 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt {-b^{2} c^{2} + a b c d}} - \frac {24 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} a b^{\frac {5}{2}} c^{2} d^{3} - 36 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} a^{2} b^{\frac {3}{2}} c d^{4} + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} a^{3} \sqrt {b} d^{5} + 240 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a b^{\frac {7}{2}} c^{3} d^{2} - 480 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a^{2} b^{\frac {5}{2}} c^{2} d^{3} + 330 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a^{3} b^{\frac {3}{2}} c d^{4} - 75 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a^{4} \sqrt {b} d^{5} - 256 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} b^{\frac {11}{2}} c^{5} + 1216 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a b^{\frac {9}{2}} c^{4} d - 2016 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{2} b^{\frac {7}{2}} c^{3} d^{2} + 1736 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{3} b^{\frac {5}{2}} c^{2} d^{3} - 800 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{4} b^{\frac {3}{2}} c d^{4} + 150 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{5} \sqrt {b} d^{5} - 384 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} b^{\frac {9}{2}} c^{4} d + 1392 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{3} b^{\frac {7}{2}} c^{3} d^{2} - 1608 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{4} b^{\frac {5}{2}} c^{2} d^{3} + 780 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{5} b^{\frac {3}{2}} c d^{4} - 150 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{6} \sqrt {b} d^{5} - 96 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{4} b^{\frac {7}{2}} c^{3} d^{2} + 336 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{5} b^{\frac {5}{2}} c^{2} d^{3} - 300 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{6} b^{\frac {3}{2}} c d^{4} + 75 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{7} \sqrt {b} d^{5} - 8 \, a^{6} b^{\frac {5}{2}} c^{2} d^{3} + 26 \, a^{7} b^{\frac {3}{2}} c d^{4} - 15 \, a^{8} \sqrt {b} d^{5}}{24 \, {\left (b^{2} c^{5} d - 2 \, a b c^{4} d^{2} + a^{2} c^{3} d^{3}\right )} {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} d + 4 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b c - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a d + a^{2} d\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(8*a*b^(5/2)*c^2 - 12*a^2*b^(3/2)*c*d + 5*a^3*sqrt(b)*d^2)*arctan(1/2*((sqrt(b)*x - sqrt(b*x^2 + a))^2*d
+ 2*b*c - a*d)/sqrt(-b^2*c^2 + a*b*c*d))/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2)*sqrt(-b^2*c^2 + a*b*c*d)) - 1
/24*(24*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a*b^(5/2)*c^2*d^3 - 36*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^2*b^(3/2)*c
*d^4 + 15*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^3*sqrt(b)*d^5 + 240*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a*b^(7/2)*c^3
*d^2 - 480*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^2*b^(5/2)*c^2*d^3 + 330*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^3*b^(3/
2)*c*d^4 - 75*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^4*sqrt(b)*d^5 - 256*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(11/2)*c
^5 + 1216*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(9/2)*c^4*d - 2016*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^2*b^(7/2)*c
^3*d^2 + 1736*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^3*b^(5/2)*c^2*d^3 - 800*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^4*b^
(3/2)*c*d^4 + 150*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^5*sqrt(b)*d^5 - 384*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^
(9/2)*c^4*d + 1392*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*b^(7/2)*c^3*d^2 - 1608*(sqrt(b)*x - sqrt(b*x^2 + a))^4*
a^4*b^(5/2)*c^2*d^3 + 780*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^5*b^(3/2)*c*d^4 - 150*(sqrt(b)*x - sqrt(b*x^2 + a)
)^4*a^6*sqrt(b)*d^5 - 96*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^4*b^(7/2)*c^3*d^2 + 336*(sqrt(b)*x - sqrt(b*x^2 + a
))^2*a^5*b^(5/2)*c^2*d^3 - 300*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^6*b^(3/2)*c*d^4 + 75*(sqrt(b)*x - sqrt(b*x^2
+ a))^2*a^7*sqrt(b)*d^5 - 8*a^6*b^(5/2)*c^2*d^3 + 26*a^7*b^(3/2)*c*d^4 - 15*a^8*sqrt(b)*d^5)/((b^2*c^5*d - 2*a
*b*c^4*d^2 + a^2*c^3*d^3)*((sqrt(b)*x - sqrt(b*x^2 + a))^4*d + 4*(sqrt(b)*x - sqrt(b*x^2 + a))^2*b*c - 2*(sqrt
(b)*x - sqrt(b*x^2 + a))^2*a*d + a^2*d)^3)

________________________________________________________________________________________

maple [B]  time = 0.03, size = 7922, normalized size = 38.09 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x^{2} + a}}{{\left (d x^{2} + c\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {b\,x^2+a}}{{\left (d\,x^2+c\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________