3.1.30 \(\int \frac {a+b x^2}{\sqrt {c+d x^2} (e+f x^2)^2} \, dx\)

Optimal. Leaf size=113 \[ \frac {x \sqrt {c+d x^2} (b e-a f)}{2 e \left (e+f x^2\right ) (d e-c f)}-\frac {(a c f-2 a d e+b c e) \tanh ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {e} \sqrt {c+d x^2}}\right )}{2 e^{3/2} (d e-c f)^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {527, 12, 377, 208} \begin {gather*} \frac {x \sqrt {c+d x^2} (b e-a f)}{2 e \left (e+f x^2\right ) (d e-c f)}-\frac {(a c f-2 a d e+b c e) \tanh ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {e} \sqrt {c+d x^2}}\right )}{2 e^{3/2} (d e-c f)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*e - a*f)*x*Sqrt[c + d*x^2])/(2*e*(d*e - c*f)*(e + f*x^2)) - ((b*c*e - 2*a*d*e + a*c*f)*ArcTanh[(Sqrt[d*e -
c*f]*x)/(Sqrt[e]*Sqrt[c + d*x^2])])/(2*e^(3/2)*(d*e - c*f)^(3/2))

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 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 {a+b x^2}{\sqrt {c+d x^2} \left (e+f x^2\right )^2} \, dx &=\frac {(b e-a f) x \sqrt {c+d x^2}}{2 e (d e-c f) \left (e+f x^2\right )}+\frac {\int \frac {-b c e+2 a d e-a c f}{\sqrt {c+d x^2} \left (e+f x^2\right )} \, dx}{2 e (d e-c f)}\\ &=\frac {(b e-a f) x \sqrt {c+d x^2}}{2 e (d e-c f) \left (e+f x^2\right )}-\frac {(b c e-2 a d e+a c f) \int \frac {1}{\sqrt {c+d x^2} \left (e+f x^2\right )} \, dx}{2 e (d e-c f)}\\ &=\frac {(b e-a f) x \sqrt {c+d x^2}}{2 e (d e-c f) \left (e+f x^2\right )}-\frac {(b c e-2 a d e+a c f) \operatorname {Subst}\left (\int \frac {1}{e-(d e-c f) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 e (d e-c f)}\\ &=\frac {(b e-a f) x \sqrt {c+d x^2}}{2 e (d e-c f) \left (e+f x^2\right )}-\frac {(b c e-2 a d e+a c f) \tanh ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right )}{2 e^{3/2} (d e-c f)^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.81, size = 191, normalized size = 1.69 \begin {gather*} \frac {x (b e-a f) \left (f \left (c+d x^2\right )-\frac {\left (e+f x^2\right ) (2 d e-c f) \tanh ^{-1}\left (\sqrt {\frac {x^2 (d e-c f)}{e \left (c+d x^2\right )}}\right )}{e \sqrt {\frac {x^2 (d e-c f)}{e \left (c+d x^2\right )}}}\right )}{2 e f \sqrt {c+d x^2} \left (e+f x^2\right ) (d e-c f)}+\frac {b \tanh ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {e} \sqrt {c+d x^2}}\right )}{\sqrt {e} f \sqrt {d e-c f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*e - a*f)*x*(f*(c + d*x^2) - ((2*d*e - c*f)*(e + f*x^2)*ArcTanh[Sqrt[((d*e - c*f)*x^2)/(e*(c + d*x^2))]])/(
e*Sqrt[((d*e - c*f)*x^2)/(e*(c + d*x^2))])))/(2*e*f*(d*e - c*f)*Sqrt[c + d*x^2]*(e + f*x^2)) + (b*ArcTanh[(Sqr
t[d*e - c*f]*x)/(Sqrt[e]*Sqrt[c + d*x^2])])/(Sqrt[e]*f*Sqrt[d*e - c*f])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.62, size = 146, normalized size = 1.29 \begin {gather*} \frac {\sqrt {c f-d e} (-a c f+2 a d e-b c e) \tan ^{-1}\left (\frac {-f x \sqrt {c+d x^2}+\sqrt {d} e+\sqrt {d} f x^2}{\sqrt {e} \sqrt {c f-d e}}\right )}{2 e^{3/2} (d e-c f)^2}+\frac {x \sqrt {c+d x^2} (b e-a f)}{2 e \left (e+f x^2\right ) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 5.77, size = 513, normalized size = 4.54 \begin {gather*} \left [\frac {4 \, {\left (b d e^{3} + a c e f^{2} - {\left (b c + a d\right )} e^{2} f\right )} \sqrt {d x^{2} + c} x - {\left (a c e f + {\left (b c - 2 \, a d\right )} e^{2} + {\left (a c f^{2} + {\left (b c - 2 \, a d\right )} e f\right )} x^{2}\right )} \sqrt {d e^{2} - c e f} \log \left (\frac {{\left (8 \, d^{2} e^{2} - 8 \, c d e f + c^{2} f^{2}\right )} x^{4} + c^{2} e^{2} + 2 \, {\left (4 \, c d e^{2} - 3 \, c^{2} e f\right )} x^{2} + 4 \, {\left ({\left (2 \, d e - c f\right )} x^{3} + c e x\right )} \sqrt {d e^{2} - c e f} \sqrt {d x^{2} + c}}{f^{2} x^{4} + 2 \, e f x^{2} + e^{2}}\right )}{8 \, {\left (d^{2} e^{5} - 2 \, c d e^{4} f + c^{2} e^{3} f^{2} + {\left (d^{2} e^{4} f - 2 \, c d e^{3} f^{2} + c^{2} e^{2} f^{3}\right )} x^{2}\right )}}, \frac {2 \, {\left (b d e^{3} + a c e f^{2} - {\left (b c + a d\right )} e^{2} f\right )} \sqrt {d x^{2} + c} x + {\left (a c e f + {\left (b c - 2 \, a d\right )} e^{2} + {\left (a c f^{2} + {\left (b c - 2 \, a d\right )} e f\right )} x^{2}\right )} \sqrt {-d e^{2} + c e f} \arctan \left (\frac {\sqrt {-d e^{2} + c e f} {\left ({\left (2 \, d e - c f\right )} x^{2} + c e\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (d^{2} e^{2} - c d e f\right )} x^{3} + {\left (c d e^{2} - c^{2} e f\right )} x\right )}}\right )}{4 \, {\left (d^{2} e^{5} - 2 \, c d e^{4} f + c^{2} e^{3} f^{2} + {\left (d^{2} e^{4} f - 2 \, c d e^{3} f^{2} + c^{2} e^{2} f^{3}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 3.40, size = 336, normalized size = 2.97 \begin {gather*} -\frac {{\left (a c \sqrt {d} f + b c \sqrt {d} e - 2 \, a d^{\frac {3}{2}} e\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} f - c f + 2 \, d e}{2 \, \sqrt {c d f e - d^{2} e^{2}}}\right )}{2 \, \sqrt {c d f e - d^{2} e^{2}} {\left (c f e - d e^{2}\right )}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c \sqrt {d} f^{2} - {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c \sqrt {d} f e - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d^{\frac {3}{2}} f e - a c^{2} \sqrt {d} f^{2} + 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b d^{\frac {3}{2}} e^{2} + b c^{2} \sqrt {d} f e}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} f - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} c f + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} d e + c^{2} f\right )} {\left (c f^{2} e - d f e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*c*sqrt(d)*f + b*c*sqrt(d)*e - 2*a*d^(3/2)*e)*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*f - c*f + 2*d
*e)/sqrt(c*d*f*e - d^2*e^2))/(sqrt(c*d*f*e - d^2*e^2)*(c*f*e - d*e^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*
sqrt(d)*f^2 - (sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c*sqrt(d)*f*e - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d^(3/2)*f*
e - a*c^2*sqrt(d)*f^2 + 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*d^(3/2)*e^2 + b*c^2*sqrt(d)*f*e)/(((sqrt(d)*x - sq
rt(d*x^2 + c))^4*f - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*c*f + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*d*e + c^2*f)*(c
*f^2*e - d*f*e^2))

________________________________________________________________________________________

maple [B]  time = 0.03, size = 1622, normalized size = 14.35

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/e/(c*f-d*e)/(x-(-e*f)^(1/2)/f)*((x-(-e*f)^(1/2)/f)^2*d+2*(-e*f)^(1/2)*(x-(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f)^
(1/2)*a-1/4/f/(c*f-d*e)/(x-(-e*f)^(1/2)/f)*((x-(-e*f)^(1/2)/f)^2*d+2*(-e*f)^(1/2)*(x-(-e*f)^(1/2)/f)*d/f+(c*f-
d*e)/f)^(1/2)*b-1/4/e/f*(-e*f)^(1/2)*d/(c*f-d*e)/((c*f-d*e)/f)^(1/2)*ln((2*(-e*f)^(1/2)*(x-(-e*f)^(1/2)/f)*d/f
+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*(-e*f)^(1/2)*(x-(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f
)^(1/2))/(x-(-e*f)^(1/2)/f))*a+1/4/f^2*(-e*f)^(1/2)*d/(c*f-d*e)/((c*f-d*e)/f)^(1/2)*ln((2*(-e*f)^(1/2)*(x-(-e*
f)^(1/2)/f)*d/f+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*(-e*f)^(1/2)*(x-(-e*f)^(1/2)/f)*
d/f+(c*f-d*e)/f)^(1/2))/(x-(-e*f)^(1/2)/f))*b-1/4/e/(-e*f)^(1/2)/((c*f-d*e)/f)^(1/2)*ln((2*(-e*f)^(1/2)*(x-(-e
*f)^(1/2)/f)*d/f+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*(-e*f)^(1/2)*(x-(-e*f)^(1/2)/f)
*d/f+(c*f-d*e)/f)^(1/2))/(x-(-e*f)^(1/2)/f))*a-1/4/(-e*f)^(1/2)/f/((c*f-d*e)/f)^(1/2)*ln((2*(-e*f)^(1/2)*(x-(-
e*f)^(1/2)/f)*d/f+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*(-e*f)^(1/2)*(x-(-e*f)^(1/2)/f
)*d/f+(c*f-d*e)/f)^(1/2))/(x-(-e*f)^(1/2)/f))*b+1/4/e/(c*f-d*e)/(x+(-e*f)^(1/2)/f)*((x+(-e*f)^(1/2)/f)^2*d-2*(
-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f)^(1/2)*a-1/4/f/(c*f-d*e)/(x+(-e*f)^(1/2)/f)*((x+(-e*f)^(1/2)/f)
^2*d-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f)^(1/2)*b+1/4/e/f*(-e*f)^(1/2)*d/(c*f-d*e)/((c*f-d*e)/f)
^(1/2)*ln((-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x+(-e*f)^(1/2)/f)^2*d-
2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f)^(1/2))/(x+(-e*f)^(1/2)/f))*a-1/4/f^2*(-e*f)^(1/2)*d/(c*f-d*
e)/((c*f-d*e)/f)^(1/2)*ln((-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x+(-e*
f)^(1/2)/f)^2*d-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f)^(1/2))/(x+(-e*f)^(1/2)/f))*b+1/4/e/(-e*f)^(
1/2)/((c*f-d*e)/f)^(1/2)*ln((-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x+(-
e*f)^(1/2)/f)^2*d-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f)^(1/2))/(x+(-e*f)^(1/2)/f))*a+1/4/(-e*f)^(
1/2)/f/((c*f-d*e)/f)^(1/2)*ln((-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+2*(c*f-d*e)/f+2*((c*f-d*e)/f)^(1/2)*((x+
(-e*f)^(1/2)/f)^2*d-2*(-e*f)^(1/2)*(x+(-e*f)^(1/2)/f)*d/f+(c*f-d*e)/f)^(1/2))/(x+(-e*f)^(1/2)/f))*b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {b\,x^2+a}{\sqrt {d\,x^2+c}\,{\left (f\,x^2+e\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b x^{2}}{\sqrt {c + d x^{2}} \left (e + f x^{2}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________