\(\int \frac {\sqrt {d+e x^2} (A+B x^2+C x^4)}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx\) [148]

Optimal result

Integrand size = 49, antiderivative size = 173 \[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {(4 b B-3 a C) x \sqrt {a d+(b d+a e) x^2+b e x^4}}{8 b^2 \sqrt {d+e x^2}}+\frac {C x^3 \sqrt {a d+(b d+a e) x^2+b e x^4}}{4 b \sqrt {d+e x^2}}+\frac {\left (8 A b^2-a (4 b B-3 a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} x \sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}}\right )}{8 b^{5/2}} \] Output:

1/8*(4*B*b-3*C*a)*x*(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2)/b^2/(e*x^2+d)^(1/2)+ 
1/4*C*x^3*(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2)/b/(e*x^2+d)^(1/2)+1/8*(8*A*b^2 
-a*(4*B*b-3*C*a))*arctanh(b^(1/2)*x*(e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e 
*x^4)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {\sqrt {d+e x^2} \left (\sqrt {b} x \left (a+b x^2\right ) \left (4 b B-3 a C+2 b C x^2\right )+2 \left (8 A b^2+a (-4 b B+3 a C)\right ) \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )\right )}{8 b^{5/2} \sqrt {\left (a+b x^2\right ) \left (d+e x^2\right )}} \] Input:

Integrate[(Sqrt[d + e*x^2]*(A + B*x^2 + C*x^4))/Sqrt[a*d + (b*d + a*e)*x^2 
 + b*e*x^4],x]
 

Output:

(Sqrt[d + e*x^2]*(Sqrt[b]*x*(a + b*x^2)*(4*b*B - 3*a*C + 2*b*C*x^2) + 2*(8 
*A*b^2 + a*(-4*b*B + 3*a*C))*Sqrt[a + b*x^2]*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] 
 + Sqrt[a + b*x^2])]))/(8*b^(5/2)*Sqrt[(a + b*x^2)*(d + e*x^2)])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {1395, 1473, 299, 224, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}} \, dx\)

\(\Big \downarrow \) 1395

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \int \frac {C x^4+B x^2+A}{\sqrt {b x^2+a}}dx}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\int \frac {(4 b B-3 a C) x^2+4 A b}{\sqrt {b x^2+a}}dx}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {\left (8 A b^2-a (4 b B-3 a C)\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {x \sqrt {a+b x^2} (4 b B-3 a C)}{2 b}}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {\left (8 A b^2-a (4 b B-3 a C)\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {x \sqrt {a+b x^2} (4 b B-3 a C)}{2 b}}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (8 A b^2-a (4 b B-3 a C)\right )}{2 b^{3/2}}+\frac {x \sqrt {a+b x^2} (4 b B-3 a C)}{2 b}}{4 b}+\frac {C x^3 \sqrt {a+b x^2}}{4 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

Input:

Int[(Sqrt[d + e*x^2]*(A + B*x^2 + C*x^4))/Sqrt[a*d + (b*d + a*e)*x^2 + b*e 
*x^4],x]
 

Output:

(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*((C*x^3*Sqrt[a + b*x^2])/(4*b) + (((4*b*B 
 - 3*a*C)*x*Sqrt[a + b*x^2])/(2*b) + ((8*A*b^2 - a*(4*b*B - 3*a*C))*ArcTan 
h[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/(4*b)))/Sqrt[a*d + (b*d + a*e 
)*x^2 + b*e*x^4]
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 1395

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( 
x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d 
+ e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p 
+ q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E 
qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 
1] && EqQ[n, 2])
 

rule 1473

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) 
, x] + Simp[1/(e*(4*p + 2*q + 1))   Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 
2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 
 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[q, -1]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {x \left (2 C \,x^{2} b +4 B b -3 C a \right ) \left (b \,x^{2}+a \right ) \sqrt {e \,x^{2}+d}}{8 b^{2} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}+\frac {\left (8 A \,b^{2}-4 a b B +3 C \,a^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {e \,x^{2}+d}}{8 b^{\frac {5}{2}} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}\)	\(131\)
default	\(\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, \left (2 C \,b^{\frac {3}{2}} x^{3} \sqrt {b \,x^{2}+a}+4 B \,b^{\frac {3}{2}} \sqrt {b \,x^{2}+a}\, x -3 C a x \sqrt {b}\, \sqrt {b \,x^{2}+a}+8 A \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) b^{2}-4 B \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a b +3 C \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{2}\right )}{8 b^{\frac {5}{2}} \sqrt {e \,x^{2}+d}\, \sqrt {b \,x^{2}+a}}\)	\(158\)

Input:

int((e*x^2+d)^(1/2)*(C*x^4+B*x^2+A)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

1/8*x*(2*C*b*x^2+4*B*b-3*C*a)*(b*x^2+a)/b^2/((e*x^2+d)*(b*x^2+a))^(1/2)*(e 
*x^2+d)^(1/2)+1/8*(8*A*b^2-4*B*a*b+3*C*a^2)/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a) 
^(1/2))*(b*x^2+a)^(1/2)/((e*x^2+d)*(b*x^2+a))^(1/2)*(e*x^2+d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.12 \[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\left [\frac {{\left ({\left (3 \, C a^{2} - 4 \, B a b + 8 \, A b^{2}\right )} e x^{2} + {\left (3 \, C a^{2} - 4 \, B a b + 8 \, A b^{2}\right )} d\right )} \sqrt {b} \log \left (\frac {2 \, b e x^{4} + {\left (2 \, b d + a e\right )} x^{2} + 2 \, \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d} \sqrt {b} x + a d}{e x^{2} + d}\right ) + 2 \, {\left (2 \, C b^{2} x^{3} - {\left (3 \, C a b - 4 \, B b^{2}\right )} x\right )} \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d}}{16 \, {\left (b^{3} e x^{2} + b^{3} d\right )}}, -\frac {{\left ({\left (3 \, C a^{2} - 4 \, B a b + 8 \, A b^{2}\right )} e x^{2} + {\left (3 \, C a^{2} - 4 \, B a b + 8 \, A b^{2}\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {e x^{2} + d} \sqrt {-b} x}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}}\right ) - {\left (2 \, C b^{2} x^{3} - {\left (3 \, C a b - 4 \, B b^{2}\right )} x\right )} \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d}}{8 \, {\left (b^{3} e x^{2} + b^{3} d\right )}}\right ] \] Input:

integrate((e*x^2+d)^(1/2)*(C*x^4+B*x^2+A)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2 
),x, algorithm="fricas")
 

Output:

[1/16*(((3*C*a^2 - 4*B*a*b + 8*A*b^2)*e*x^2 + (3*C*a^2 - 4*B*a*b + 8*A*b^2 
)*d)*sqrt(b)*log((2*b*e*x^4 + (2*b*d + a*e)*x^2 + 2*sqrt(b*e*x^4 + (b*d + 
a*e)*x^2 + a*d)*sqrt(e*x^2 + d)*sqrt(b)*x + a*d)/(e*x^2 + d)) + 2*(2*C*b^2 
*x^3 - (3*C*a*b - 4*B*b^2)*x)*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(e 
*x^2 + d))/(b^3*e*x^2 + b^3*d), -1/8*(((3*C*a^2 - 4*B*a*b + 8*A*b^2)*e*x^2 
 + (3*C*a^2 - 4*B*a*b + 8*A*b^2)*d)*sqrt(-b)*arctan(sqrt(e*x^2 + d)*sqrt(- 
b)*x/sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)) - (2*C*b^2*x^3 - (3*C*a*b - 4* 
B*b^2)*x)*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(e*x^2 + d))/(b^3*e*x^ 
2 + b^3*d)]
 

Sympy [F]

\[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int \frac {\sqrt {d + e x^{2}} \left (A + B x^{2} + C x^{4}\right )}{\sqrt {\left (a + b x^{2}\right ) \left (d + e x^{2}\right )}}\, dx \] Input:

integrate((e*x**2+d)**(1/2)*(C*x**4+B*x**2+A)/(a*d+(a*e+b*d)*x**2+b*e*x**4 
)**(1/2),x)
 

Output:

Integral(sqrt(d + e*x**2)*(A + B*x**2 + C*x**4)/sqrt((a + b*x**2)*(d + e*x 
**2)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} \sqrt {e x^{2} + d}}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}} \,d x } \] Input:

integrate((e*x^2+d)^(1/2)*(C*x^4+B*x^2+A)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2 
),x, algorithm="maxima")
 

Output:

integrate((C*x^4 + B*x^2 + A)*sqrt(e*x^2 + d)/sqrt(b*e*x^4 + (b*d + a*e)*x 
^2 + a*d), x)
 

Giac [F]

\[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} \sqrt {e x^{2} + d}}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}} \,d x } \] Input:

integrate((e*x^2+d)^(1/2)*(C*x^4+B*x^2+A)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2 
),x, algorithm="giac")
 

Output:

integrate((C*x^4 + B*x^2 + A)*sqrt(e*x^2 + d)/sqrt(b*e*x^4 + (b*d + a*e)*x 
^2 + a*d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int \frac {\sqrt {e\,x^2+d}\,\left (C\,x^4+B\,x^2+A\right )}{\sqrt {b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d}} \,d x \] Input:

int(((d + e*x^2)^(1/2)*(A + B*x^2 + C*x^4))/(a*d + x^2*(a*e + b*d) + b*e*x 
^4)^(1/2),x)
 

Output:

int(((d + e*x^2)^(1/2)*(A + B*x^2 + C*x^4))/(a*d + x^2*(a*e + b*d) + b*e*x 
^4)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {d+e x^2} \left (A+B x^2+C x^4\right )}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a b c x +4 \sqrt {b \,x^{2}+a}\, b^{3} x +2 \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c +4 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2}}{8 b^{3}} \] Input:

int((e*x^2+d)^(1/2)*(C*x^4+B*x^2+A)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x)
 

Output:

( - 3*sqrt(a + b*x**2)*a*b*c*x + 4*sqrt(a + b*x**2)*b**3*x + 2*sqrt(a + b* 
x**2)*b**2*c*x**3 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))* 
a**2*c + 4*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2)/(8* 
b**3)