\(\int \frac {\sqrt {c+d x^2}}{(a+b x^2)^3} \, dx\) [103]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 21, antiderivative size = 149 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^3} \, dx=\frac {(3 b c-4 a d) x \sqrt {c+d x^2}}{8 a^2 (b c-a d) \left (a+b x^2\right )}+\frac {b x \left (c+d x^2\right )^{3/2}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {c (3 b c-4 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{5/2} (b c-a d)^{3/2}} \]

[Out]

1/4*b*x*(d*x^2+c)^(3/2)/a/(-a*d+b*c)/(b*x^2+a)^2+1/8*c*(-4*a*d+3*b*c)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2
+c)^(1/2))/a^(5/2)/(-a*d+b*c)^(3/2)+1/8*(-4*a*d+3*b*c)*x*(d*x^2+c)^(1/2)/a^2/(-a*d+b*c)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {390, 386, 385, 211} \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^3} \, dx=\frac {c (3 b c-4 a d) \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{5/2} (b c-a d)^{3/2}}+\frac {x \sqrt {c+d x^2} (3 b c-4 a d)}{8 a^2 \left (a+b x^2\right ) (b c-a d)}+\frac {b x \left (c+d x^2\right )^{3/2}}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]

[In]

Int[Sqrt[c + d*x^2]/(a + b*x^2)^3,x]

[Out]

((3*b*c - 4*a*d)*x*Sqrt[c + d*x^2])/(8*a^2*(b*c - a*d)*(a + b*x^2)) + (b*x*(c + d*x^2)^(3/2))/(4*a*(b*c - a*d)
*(a + b*x^2)^2) + (c*(3*b*c - 4*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*a^(5/2)*(b*c -
a*d)^(3/2))

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 386

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[c*(q/(a*(p + 1))), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x]
 /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {b x \left (c+d x^2\right )^{3/2}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {(3 b c-4 a d) \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx}{4 a (b c-a d)} \\ & = \frac {(3 b c-4 a d) x \sqrt {c+d x^2}}{8 a^2 (b c-a d) \left (a+b x^2\right )}+\frac {b x \left (c+d x^2\right )^{3/2}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {(c (3 b c-4 a d)) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{8 a^2 (b c-a d)} \\ & = \frac {(3 b c-4 a d) x \sqrt {c+d x^2}}{8 a^2 (b c-a d) \left (a+b x^2\right )}+\frac {b x \left (c+d x^2\right )^{3/2}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {(c (3 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 )}{8 a^2 (b c-a d)} \\ & = \frac {(3 b c-4 a d) x \sqrt {c+d x^2}}{8 a^2 (b c-a d) \left (a+b x^2\right )}+\frac {b x \left (c+d x^2\right )^{3/2}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {c (3 b c-4 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{5/2} (b c-a d)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^3} \, dx=\frac {x \sqrt {c+d x^2} \left (-5 a b c+4 a^2 d-3 b^2 c x^2+2 a b d x^2\right )}{8 a^2 (-b c+a d) \left (a+b x^2\right )^2}-\frac {c (3 b c-4 a d) \arctan \left (\frac {a \sqrt {d}+b \sqrt {d} x^2-b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}\right )}{8 a^{5/2} (b c-a d)^{3/2}} \]

[In]

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

[Out]

(x*Sqrt[c + d*x^2]*(-5*a*b*c + 4*a^2*d - 3*b^2*c*x^2 + 2*a*b*d*x^2))/(8*a^2*(-(b*c) + a*d)*(a + b*x^2)^2) - (c
*(3*b*c - 4*a*d)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^2 - b*x*Sqrt[c + d*x^2])/(Sqrt[a]*Sqrt[b*c - a*d])])/(8*a^(5/
2)*(b*c - a*d)^(3/2))

Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.81

method result size
pseudoelliptic \(-\frac {c \left (-\frac {\sqrt {d \,x^{2}+c}\, \left (2 x^{2} a b d -3 b^{2} c \,x^{2}+4 a^{2} d -5 a b c \right ) x}{c \left (b \,x^{2}+a \right )^{2}}-\frac {\left (4 a d -3 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{8 \left (a d -b c \right ) a^{2}}\) \(121\)
default \(\text {Expression too large to display}\) \(4155\)

[In]

int((d*x^2+c)^(1/2)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (129) = 258\).

Time = 0.38 (sec) , antiderivative size = 698, normalized size of antiderivative = 4.68 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^3} \, dx=\left [-\frac {{\left (3 \, a^{2} b c^{2} - 4 \, a^{3} c d + {\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{4} + 2 \, {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{2}\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{3} + {\left (5 \, a^{2} b^{2} c^{2} - 9 \, a^{3} b c d + 4 \, a^{4} d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{32 \, {\left (a^{5} b^{2} c^{2} - 2 \, a^{6} b c d + a^{7} d^{2} + {\left (a^{3} b^{4} c^{2} - 2 \, a^{4} b^{3} c d + a^{5} b^{2} d^{2}\right )} x^{4} + 2 \, {\left (a^{4} b^{3} c^{2} - 2 \, a^{5} b^{2} c d + a^{6} b d^{2}\right )} x^{2}\right )}}, \frac {{\left (3 \, a^{2} b c^{2} - 4 \, a^{3} c d + {\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{4} + 2 \, {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{2}\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{3} + {\left (5 \, a^{2} b^{2} c^{2} - 9 \, a^{3} b c d + 4 \, a^{4} d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, {\left (a^{5} b^{2} c^{2} - 2 \, a^{6} b c d + a^{7} d^{2} + {\left (a^{3} b^{4} c^{2} - 2 \, a^{4} b^{3} c d + a^{5} b^{2} d^{2}\right )} x^{4} + 2 \, {\left (a^{4} b^{3} c^{2} - 2 \, a^{5} b^{2} c d + a^{6} b d^{2}\right )} x^{2}\right )}}\right ] \]

[In]

integrate((d*x^2+c)^(1/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[-1/32*((3*a^2*b*c^2 - 4*a^3*c*d + (3*b^3*c^2 - 4*a*b^2*c*d)*x^4 + 2*(3*a*b^2*c^2 - 4*a^2*b*c*d)*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*((3*a*b^3*c^2 - 5*a
^2*b^2*c*d + 2*a^3*b*d^2)*x^3 + (5*a^2*b^2*c^2 - 9*a^3*b*c*d + 4*a^4*d^2)*x)*sqrt(d*x^2 + c))/(a^5*b^2*c^2 - 2
*a^6*b*c*d + a^7*d^2 + (a^3*b^4*c^2 - 2*a^4*b^3*c*d + a^5*b^2*d^2)*x^4 + 2*(a^4*b^3*c^2 - 2*a^5*b^2*c*d + a^6*
b*d^2)*x^2), 1/16*((3*a^2*b*c^2 - 4*a^3*c*d + (3*b^3*c^2 - 4*a*b^2*c*d)*x^4 + 2*(3*a*b^2*c^2 - 4*a^2*b*c*d)*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*((3*a*b^3*c^2 - 5*a^2*b^2*c*d + 2*a^3*b*d^2)*x^3 + (5*a^2*b^2*c^2 - 9
*a^3*b*c*d + 4*a^4*d^2)*x)*sqrt(d*x^2 + c))/(a^5*b^2*c^2 - 2*a^6*b*c*d + a^7*d^2 + (a^3*b^4*c^2 - 2*a^4*b^3*c*
d + a^5*b^2*d^2)*x^4 + 2*(a^4*b^3*c^2 - 2*a^5*b^2*c*d + a^6*b*d^2)*x^2)]

Sympy [F]

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

[In]

integrate((d*x**2+c)**(1/2)/(b*x**2+a)**3,x)

[Out]

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

Maxima [F]

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

[In]

integrate((d*x^2+c)^(1/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 487 vs. \(2 (129) = 258\).

Time = 1.66 (sec) , antiderivative size = 487, normalized size of antiderivative = 3.27 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^3} \, dx=-\frac {{\left (3 \, b c^{2} \sqrt {d} - 4 \, a c d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{8 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{3} c^{2} \sqrt {d} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b^{2} c d^{\frac {3}{2}} - 9 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{3} c^{3} \sqrt {d} + 30 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b^{2} c^{2} d^{\frac {3}{2}} - 40 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} b c d^{\frac {5}{2}} + 16 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{3} d^{\frac {7}{2}} + 9 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c^{4} \sqrt {d} - 28 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} c^{3} d^{\frac {3}{2}} + 16 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b c^{2} d^{\frac {5}{2}} - 3 \, b^{3} c^{5} \sqrt {d} + 2 \, a b^{2} c^{4} d^{\frac {3}{2}}}{4 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}^{2} {\left (a^{2} b^{2} c - a^{3} b d\right )}} \]

[In]

integrate((d*x^2+c)^(1/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-1/8*(3*b*c^2*sqrt(d) - 4*a*c*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))/((a^2*b*c - a^3*d)*sqrt(a*b*c*d - a^2*d^2)) - 1/4*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^3*c^2*sq
rt(d) - 4*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b^2*c*d^(3/2) - 9*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^3*c^3*sqrt(d)
+ 30*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b^2*c^2*d^(3/2) - 40*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*b*c*d^(5/2) +
16*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^3*d^(7/2) + 9*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c^4*sqrt(d) - 28*(sqrt(
d)*x - sqrt(d*x^2 + c))^2*a*b^2*c^3*d^(3/2) + 16*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*b*c^2*d^(5/2) - 3*b^3*c^5
*sqrt(d) + 2*a*b^2*c^4*d^(3/2))/(((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)^2*(a^2*b^2*c - a^3*b*d))

Mupad [F(-1)]

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

[In]

int((c + d*x^2)^(1/2)/(a + b*x^2)^3,x)

[Out]

int((c + d*x^2)^(1/2)/(a + b*x^2)^3, x)