3.57 \(\int \frac{(a+b x^2) \sqrt{c+d x^2}}{e+f x^2} \, dx\)
Optimal. Leaf size=128 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right ) (-2 a d f-b c f+2 b d e)}{2 \sqrt{d} f^2}+\frac{(b e-a f) \sqrt{d e-c f} \tanh ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{e} \sqrt{c+d x^2}}\right )}{\sqrt{e} f^2}+\frac{b x \sqrt{c+d x^2}}{2 f} \]
[Out]
(b*x*Sqrt[c + d*x^2])/(2*f) - ((2*b*d*e - b*c*f - 2*a*d*f)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*Sqrt[d]*f^
2) + ((b*e - a*f)*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d*e - c*f]*x)/(Sqrt[e]*Sqrt[c + d*x^2])])/(Sqrt[e]*f^2)
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Rubi [A] time = 0.142179, antiderivative size = 128, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used =
{528, 523, 217, 206, 377, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right ) (-2 a d f-b c f+2 b d e)}{2 \sqrt{d} f^2}+\frac{(b e-a f) \sqrt{d e-c f} \tanh ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{e} \sqrt{c+d x^2}}\right )}{\sqrt{e} f^2}+\frac{b x \sqrt{c+d x^2}}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*x^2)*Sqrt[c + d*x^2])/(e + f*x^2),x]
[Out]
(b*x*Sqrt[c + d*x^2])/(2*f) - ((2*b*d*e - b*c*f - 2*a*d*f)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*Sqrt[d]*f^
2) + ((b*e - a*f)*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d*e - c*f]*x)/(Sqrt[e]*Sqrt[c + d*x^2])])/(Sqrt[e]*f^2)
Rule 528
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
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 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]
Rubi steps
\begin{align*} \int \frac{\left (a+b x^2\right ) \sqrt{c+d x^2}}{e+f x^2} \, dx &=\frac{b x \sqrt{c+d x^2}}{2 f}+\frac{\int \frac{-c (b e-2 a f)+(-2 b d e+b c f+2 a d f) x^2}{\sqrt{c+d x^2} \left (e+f x^2\right )} \, dx}{2 f}\\ &=\frac{b x \sqrt{c+d x^2}}{2 f}+\frac{((b e-a f) (d e-c f)) \int \frac{1}{\sqrt{c+d x^2} \left (e+f x^2\right )} \, dx}{f^2}-\frac{(2 b d e-b c f-2 a d f) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{2 f^2}\\ &=\frac{b x \sqrt{c+d x^2}}{2 f}+\frac{((b e-a f) (d e-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 )}{f^2}-\frac{(2 b d e-b c f-2 a d f) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{2 f^2}\\ &=\frac{b x \sqrt{c+d x^2}}{2 f}-\frac{(2 b d e-b c f-2 a d f) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 \sqrt{d} f^2}+\frac{(b e-a f) \sqrt{d e-c f} \tanh ^{-1}\left (\frac{\sqrt{d e-c f} x}{\sqrt{e} \sqrt{c+d x^2}}\right )}{\sqrt{e} f^2}\\ \end{align*}
Mathematica [A] time = 0.245454, size = 124, normalized size = 0.97 \[ \frac{\frac{\log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right ) (2 a d f+b c f-2 b d e)}{\sqrt{d}}-\frac{2 (b e-a f) \sqrt{c f-d e} \tan ^{-1}\left (\frac{x \sqrt{c f-d e}}{\sqrt{e} \sqrt{c+d x^2}}\right )}{\sqrt{e}}+b f x \sqrt{c+d x^2}}{2 f^2} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*x^2)*Sqrt[c + d*x^2])/(e + f*x^2),x]
[Out]
(b*f*x*Sqrt[c + d*x^2] - (2*(b*e - a*f)*Sqrt[-(d*e) + c*f]*ArcTan[(Sqrt[-(d*e) + c*f]*x)/(Sqrt[e]*Sqrt[c + d*x
^2])])/Sqrt[e] + ((-2*b*d*e + b*c*f + 2*a*d*f)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/(2*f^2)
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Maple [B] time = 0.042, size = 1942, normalized size = 15.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x^2+a)*(d*x^2+c)^(1/2)/(f*x^2+e),x)
[Out]
1/2*b*x*(d*x^2+c)^(1/2)/f+1/2*b/f*c/d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/2/(-e*f)^(1/2)*((x-(-e*f)^(1/2)/f)
^2*d+2*d*(-e*f)^(1/2)/f*(x-(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2)*a-1/2/(-e*f)^(1/2)/f*((x-(-e*f)^(1/2)/f)^2*d+2*d
*(-e*f)^(1/2)/f*(x-(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2)*b*e+1/2/f*d^(1/2)*ln((d*(-e*f)^(1/2)/f+(x-(-e*f)^(1/2)/f
)*d)/d^(1/2)+((x-(-e*f)^(1/2)/f)^2*d+2*d*(-e*f)^(1/2)/f*(x-(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2))*a-1/2/f^2*d^(1/
2)*ln((d*(-e*f)^(1/2)/f+(x-(-e*f)^(1/2)/f)*d)/d^(1/2)+((x-(-e*f)^(1/2)/f)^2*d+2*d*(-e*f)^(1/2)/f*(x-(-e*f)^(1/
2)/f)+(c*f-d*e)/f)^(1/2))*b*e-1/2/(-e*f)^(1/2)/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f+2*d*(-e*f)^(1/2)/f*(x-(-e
*f)^(1/2)/f)+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*d*(-e*f)^(1/2)/f*(x-(-e*f)^(1/2)/f)+(c*f-d*e)/f)^
(1/2))/(x-(-e*f)^(1/2)/f))*c*a+1/2/(-e*f)^(1/2)/f/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f+2*d*(-e*f)^(1/2)/f*(x-
(-e*f)^(1/2)/f)+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*d*(-e*f)^(1/2)/f*(x-(-e*f)^(1/2)/f)+(c*f-d*e)/
f)^(1/2))/(x-(-e*f)^(1/2)/f))*c*b*e+1/2/(-e*f)^(1/2)/f/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f+2*d*(-e*f)^(1/2)/
f*(x-(-e*f)^(1/2)/f)+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*d*(-e*f)^(1/2)/f*(x-(-e*f)^(1/2)/f)+(c*f-
d*e)/f)^(1/2))/(x-(-e*f)^(1/2)/f))*d*e*a-1/2/(-e*f)^(1/2)/f^2/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f+2*d*(-e*f)
^(1/2)/f*(x-(-e*f)^(1/2)/f)+2*((c*f-d*e)/f)^(1/2)*((x-(-e*f)^(1/2)/f)^2*d+2*d*(-e*f)^(1/2)/f*(x-(-e*f)^(1/2)/f
)+(c*f-d*e)/f)^(1/2))/(x-(-e*f)^(1/2)/f))*d*e^2*b-1/2/(-e*f)^(1/2)*((x+(-e*f)^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*
(x+(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2)*a+1/2/(-e*f)^(1/2)/f*((x+(-e*f)^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*(x+(-e*f
)^(1/2)/f)+(c*f-d*e)/f)^(1/2)*b*e+1/2/f*d^(1/2)*ln((-d*(-e*f)^(1/2)/f+(x+(-e*f)^(1/2)/f)*d)/d^(1/2)+((x+(-e*f)
^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2))*a-1/2/f^2*d^(1/2)*ln((-d*(-e*f)^(1/2)/
f+(x+(-e*f)^(1/2)/f)*d)/d^(1/2)+((x+(-e*f)^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/
2))*b*e+1/2/(-e*f)^(1/2)/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+2*((c*f-d
*e)/f)^(1/2)*((x+(-e*f)^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2))/(x+(-e*f)^(1/2)
/f))*c*a-1/2/(-e*f)^(1/2)/f/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+2*((c*
f-d*e)/f)^(1/2)*((x+(-e*f)^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2))/(x+(-e*f)^(1
/2)/f))*c*b*e-1/2/(-e*f)^(1/2)/f/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+2
*((c*f-d*e)/f)^(1/2)*((x+(-e*f)^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2))/(x+(-e*
f)^(1/2)/f))*d*e*a+1/2/(-e*f)^(1/2)/f^2/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/
2)/f)+2*((c*f-d*e)/f)^(1/2)*((x+(-e*f)^(1/2)/f)^2*d-2*d*(-e*f)^(1/2)/f*(x+(-e*f)^(1/2)/f)+(c*f-d*e)/f)^(1/2))/
(x+(-e*f)^(1/2)/f))*d*e^2*b
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x^2+a)*(d*x^2+c)^(1/2)/(f*x^2+e),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 7.51925, size = 1701, normalized size = 13.29 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x^2+a)*(d*x^2+c)^(1/2)/(f*x^2+e),x, algorithm="fricas")
[Out]
[1/4*(2*sqrt(d*x^2 + c)*b*d*f*x - (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)
*x - c) - (b*d*e - a*d*f)*sqrt((d*e - c*f)/e)*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*(c*e^2*x + (2*d*e^2 - c*e*f)*x^3)*sqrt(d*x^2 + c)*sqrt((d*e - c*f)/e))/(f^2*x^4 + 2*e
*f*x^2 + e^2)))/(d*f^2), 1/4*(2*sqrt(d*x^2 + c)*b*d*f*x + 2*(2*b*d*e - (b*c + 2*a*d)*f)*sqrt(-d)*arctan(sqrt(-
d)*x/sqrt(d*x^2 + c)) - (b*d*e - a*d*f)*sqrt((d*e - c*f)/e)*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*(c*e^2*x + (2*d*e^2 - c*e*f)*x^3)*sqrt(d*x^2 + c)*sqrt((d*e - c*f)/e))/
(f^2*x^4 + 2*e*f*x^2 + e^2)))/(d*f^2), 1/4*(2*sqrt(d*x^2 + c)*b*d*f*x - 2*(b*d*e - a*d*f)*sqrt(-(d*e - c*f)/e)
*arctan(1/2*((2*d*e - c*f)*x^2 + c*e)*sqrt(d*x^2 + c)*sqrt(-(d*e - c*f)/e)/((d^2*e - c*d*f)*x^3 + (c*d*e - c^2
*f)*x)) - (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c))/(d*f^2), 1/2*(s
qrt(d*x^2 + c)*b*d*f*x + (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (b*d*e - a*
d*f)*sqrt(-(d*e - c*f)/e)*arctan(1/2*((2*d*e - c*f)*x^2 + c*e)*sqrt(d*x^2 + c)*sqrt(-(d*e - c*f)/e)/((d^2*e -
c*d*f)*x^3 + (c*d*e - c^2*f)*x)))/(d*f^2)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right ) \sqrt{c + d x^{2}}}{e + f x^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x**2+a)*(d*x**2+c)**(1/2)/(f*x**2+e),x)
[Out]
Integral((a + b*x**2)*sqrt(c + d*x**2)/(e + f*x**2), x)
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Giac [A] time = 1.47179, size = 227, normalized size = 1.77 \begin{align*} \frac{\sqrt{d x^{2} + c} b x}{2 \, f} - \frac{{\left (a c \sqrt{d} f^{2} - b c \sqrt{d} f e - a d^{\frac{3}{2}} f e + b d^{\frac{3}{2}} e^{2}\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 )}{\sqrt{c d f e - d^{2} e^{2}} f^{2}} - \frac{{\left (b c f + 2 \, a d f - 2 \, b d e\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{4 \, \sqrt{d} f^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x^2+a)*(d*x^2+c)^(1/2)/(f*x^2+e),x, algorithm="giac")
[Out]
1/2*sqrt(d*x^2 + c)*b*x/f - (a*c*sqrt(d)*f^2 - b*c*sqrt(d)*f*e - a*d^(3/2)*f*e + b*d^(3/2)*e^2)*arctan(1/2*((s
qrt(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)*f^2) - 1/4*(b
*c*f + 2*a*d*f - 2*b*d*e)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/(sqrt(d)*f^2)