\(\int \frac {\sqrt {c+d x^2}}{(a+b x^2)^2} \, dx\) [735]

Optimal result

Integrand size = 21, antiderivative size = 82 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=\frac {x \sqrt {c+d x^2}}{2 a \left (a+b x^2\right )}+\frac {c \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} \sqrt {b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {386, 385, 211} \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=\frac {c \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} \sqrt {b c-a d}}+\frac {x \sqrt {c+d x^2}}{2 a \left (a+b x^2\right )} \]

[In]

Int[Sqrt[c + d*x^2]/(a + b*x^2)^2,x]

[Out]

(x*Sqrt[c + d*x^2])/(2*a*(a + b*x^2)) + (c*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(3/2)*S
qrt[b*c - a*d])

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]

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=\frac {x \sqrt {c+d x^2}}{2 a^2+2 a b x^2}-\frac {c \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 a^{3/2} \sqrt {b c-a d}} \]

[In]

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

[Out]

(x*Sqrt[c + d*x^2])/(2*a^2 + 2*a*b*x^2) - (c*ArcTan[(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c + d*x^2]))/(Sqrt[a]*S
qrt[b*c - a*d])])/(2*a^(3/2)*Sqrt[b*c - a*d])

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(-\frac {c \left (-\frac {\sqrt {d \,x^{2}+c}\, x}{c \left (b \,x^{2}+a \right )}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{2 a}\)	\(73\)
default	\(\text {Expression too large to display}\)	\(1965\)

[In]

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

[Out]

-1/2*c/a*(-(d*x^2+c)^(1/2)*x/c/(b*x^2+a)-1/((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. 164 vs. \(2 (66) = 132\).

Time = 0.31 (sec) , antiderivative size = 369, normalized size of antiderivative = 4.50 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=\left [\frac {4 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c} x - {\left (b c x^{2} + a c\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{8 \, {\left (a^{3} b c - a^{4} d + {\left (a^{2} b^{2} c - a^{3} b d\right )} x^{2}\right )}}, \frac {2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x^{2} + c} x + {\left (b c x^{2} + a c\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right )}{4 \, {\left (a^{3} b c - a^{4} d + {\left (a^{2} b^{2} c - a^{3} b d\right )} x^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/8*(4*(a*b*c - a^2*d)*sqrt(d*x^2 + c)*x - (b*c*x^2 + a*c)*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)*s
qrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a^3*b*c - a^4*d + (a^2*b^2*c - a^3*b*d)*x^2), 1/4*(2*(a*b*c - a
^2*d)*sqrt(d*x^2 + c)*x + (b*c*x^2 + a*c)*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)))/(a^3*b*c - a^4*d + (a^2*b^2*c - a
^3*b*d)*x^2)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (66) = 132\).

Time = 0.82 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.66 \[ \int \frac {\sqrt {c+d x^2}}{\left (a+b x^2\right )^2} \, dx=-\frac {c \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} a} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d^{\frac {3}{2}} - b c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} a b} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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