\(\int \frac {c+d x^2}{(a+b x^2)^{3/2} (e+f x^2)^2} \, dx\) [341]

Optimal result

Integrand size = 28, antiderivative size = 170 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx=\frac {b (2 b c e-3 a d e+a c f) x}{2 a e (b e-a f)^2 \sqrt {a+b x^2}}+\frac {(d e-c f) x}{2 e (b e-a f) \sqrt {a+b x^2} \left (e+f x^2\right )}+\frac {(2 b e (d e-2 c f)+a f (d e+c f)) \text {arctanh}\left (\frac {\sqrt {b e-a f} x}{\sqrt {e} \sqrt {a+b x^2}}\right )}{2 e^{3/2} (b e-a f)^{5/2}} \] Output:

1/2*b*(a*c*f-3*a*d*e+2*b*c*e)*x/a/e/(-a*f+b*e)^2/(b*x^2+a)^(1/2)+1/2*(-c*f 
+d*e)*x/e/(-a*f+b*e)/(b*x^2+a)^(1/2)/(f*x^2+e)+1/2*(2*b*e*(-2*c*f+d*e)+a*f 
*(c*f+d*e))*arctanh((-a*f+b*e)^(1/2)*x/e^(1/2)/(b*x^2+a)^(1/2))/e^(3/2)/(- 
a*f+b*e)^(5/2)
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.11 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx=\frac {x \left (a^2 f (-d e+c f)+2 b^2 c e \left (e+f x^2\right )+a b \left (c f^2 x^2-d e \left (2 e+3 f x^2\right )\right )\right )}{2 a e (b e-a f)^2 \sqrt {a+b x^2} \left (e+f x^2\right )}-\frac {(2 b e (d e-2 c f)+a f (d e+c f)) \arctan \left (\frac {-f x \sqrt {a+b x^2}+\sqrt {b} \left (e+f x^2\right )}{\sqrt {e} \sqrt {-b e+a f}}\right )}{2 e^{3/2} (-b e+a f)^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {402, 25, 402, 27, 291, 221}

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.

\(\displaystyle \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {x (b c-a d)}{a \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f)}-\frac {\int -\frac {2 (b c-a d) f x^2+a (d e-c f)}{\sqrt {b x^2+a} \left (f x^2+e\right )^2}dx}{a (b e-a f)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {2 (b c-a d) f x^2+a (d e-c f)}{\sqrt {b x^2+a} \left (f x^2+e\right )^2}dx}{a (b e-a f)}+\frac {x (b c-a d)}{a \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\int \frac {a (2 b e (d e-2 c f)+a f (d e+c f))}{\sqrt {b x^2+a} \left (f x^2+e\right )}dx}{2 e (b e-a f)}+\frac {f x \sqrt {a+b x^2} (a c f-3 a d e+2 b c e)}{2 e \left (e+f x^2\right ) (b e-a f)}}{a (b e-a f)}+\frac {x (b c-a d)}{a \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a (a f (c f+d e)+2 b e (d e-2 c f)) \int \frac {1}{\sqrt {b x^2+a} \left (f x^2+e\right )}dx}{2 e (b e-a f)}+\frac {f x \sqrt {a+b x^2} (a c f-3 a d e+2 b c e)}{2 e \left (e+f x^2\right ) (b e-a f)}}{a (b e-a f)}+\frac {x (b c-a d)}{a \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {a (a f (c f+d e)+2 b e (d e-2 c f)) \int \frac {1}{e-\frac {(b e-a f) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 e (b e-a f)}+\frac {f x \sqrt {a+b x^2} (a c f-3 a d e+2 b c e)}{2 e \left (e+f x^2\right ) (b e-a f)}}{a (b e-a f)}+\frac {x (b c-a d)}{a \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {a \text {arctanh}\left (\frac {x \sqrt {b e-a f}}{\sqrt {e} \sqrt {a+b x^2}}\right ) (a f (c f+d e)+2 b e (d e-2 c f))}{2 e^{3/2} (b e-a f)^{3/2}}+\frac {f x \sqrt {a+b x^2} (a c f-3 a d e+2 b c e)}{2 e \left (e+f x^2\right ) (b e-a f)}}{a (b e-a f)}+\frac {x (b c-a d)}{a \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f)}\)

Input:

Int[(c + d*x^2)/((a + b*x^2)^(3/2)*(e + f*x^2)^2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

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

rule 291

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

rule 402

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.13

method	result	size
pseudoelliptic	\(\frac {-a \sqrt {b \,x^{2}+a}\, \left (2 b d \,e^{2}+f \left (a d -4 b c \right ) e +a c \,f^{2}\right ) \left (f \,x^{2}+e \right ) \arctan \left (\frac {e \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a f -b e \right ) e}}\right )+\left (\left (-2 a b d +2 b^{2} c \right ) e^{2}-f \left (3 a b d \,x^{2}-2 b^{2} c \,x^{2}+d \,a^{2}\right ) e +a c \,f^{2} \left (b \,x^{2}+a \right )\right ) \sqrt {\left (a f -b e \right ) e}\, x}{2 \sqrt {\left (a f -b e \right ) e}\, \sqrt {b \,x^{2}+a}\, e \left (f \,x^{2}+e \right ) \left (a f -b e \right )^{2} a}\)	\(192\)
default	\(\text {Expression too large to display}\)	\(1942\)

Input:

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

Output:

1/2*(-a*(b*x^2+a)^(1/2)*(2*b*d*e^2+f*(a*d-4*b*c)*e+a*c*f^2)*(f*x^2+e)*arct 
an(e*(b*x^2+a)^(1/2)/x/((a*f-b*e)*e)^(1/2))+((-2*a*b*d+2*b^2*c)*e^2-f*(3*a 
*b*d*x^2-2*b^2*c*x^2+a^2*d)*e+a*c*f^2*(b*x^2+a))*((a*f-b*e)*e)^(1/2)*x)/(( 
a*f-b*e)*e)^(1/2)/(b*x^2+a)^(1/2)/e/(f*x^2+e)/(a*f-b*e)^2/a
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (150) = 300\).

Time = 3.27 (sec) , antiderivative size = 1108, normalized size of antiderivative = 6.52 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/8*((2*a^2*b*d*e^3 + a^3*c*e*f^2 + (2*a*b^2*d*e^2*f + a^2*b*c*f^3 - (4*a 
*b^2*c - a^2*b*d)*e*f^2)*x^4 - (4*a^2*b*c - a^3*d)*e^2*f + (2*a*b^2*d*e^3 
+ a^3*c*f^3 - (4*a*b^2*c - 3*a^2*b*d)*e^2*f - (3*a^2*b*c - a^3*d)*e*f^2)*x 
^2)*sqrt(b*e^2 - a*e*f)*log(((8*b^2*e^2 - 8*a*b*e*f + a^2*f^2)*x^4 + a^2*e 
^2 + 2*(4*a*b*e^2 - 3*a^2*e*f)*x^2 + 4*((2*b*e - a*f)*x^3 + a*e*x)*sqrt(b* 
e^2 - a*e*f)*sqrt(b*x^2 + a))/(f^2*x^4 + 2*e*f*x^2 + e^2)) - 4*((a^2*b*c*e 
*f^3 - (2*b^3*c - 3*a*b^2*d)*e^3*f + (a*b^2*c - 3*a^2*b*d)*e^2*f^2)*x^3 + 
(a^3*c*e*f^3 - 2*(b^3*c - a*b^2*d)*e^4 + (2*a*b^2*c - a^2*b*d)*e^3*f - (a^ 
2*b*c + a^3*d)*e^2*f^2)*x)*sqrt(b*x^2 + a))/(a^2*b^3*e^6 - 3*a^3*b^2*e^5*f 
 + 3*a^4*b*e^4*f^2 - a^5*e^3*f^3 + (a*b^4*e^5*f - 3*a^2*b^3*e^4*f^2 + 3*a^ 
3*b^2*e^3*f^3 - a^4*b*e^2*f^4)*x^4 + (a*b^4*e^6 - 2*a^2*b^3*e^5*f + 2*a^4* 
b*e^3*f^3 - a^5*e^2*f^4)*x^2), -1/4*((2*a^2*b*d*e^3 + a^3*c*e*f^2 + (2*a*b 
^2*d*e^2*f + a^2*b*c*f^3 - (4*a*b^2*c - a^2*b*d)*e*f^2)*x^4 - (4*a^2*b*c - 
 a^3*d)*e^2*f + (2*a*b^2*d*e^3 + a^3*c*f^3 - (4*a*b^2*c - 3*a^2*b*d)*e^2*f 
 - (3*a^2*b*c - a^3*d)*e*f^2)*x^2)*sqrt(-b*e^2 + a*e*f)*arctan(1/2*sqrt(-b 
*e^2 + a*e*f)*((2*b*e - a*f)*x^2 + a*e)*sqrt(b*x^2 + a)/((b^2*e^2 - a*b*e* 
f)*x^3 + (a*b*e^2 - a^2*e*f)*x)) + 2*((a^2*b*c*e*f^3 - (2*b^3*c - 3*a*b^2* 
d)*e^3*f + (a*b^2*c - 3*a^2*b*d)*e^2*f^2)*x^3 + (a^3*c*e*f^3 - 2*(b^3*c - 
a*b^2*d)*e^4 + (2*a*b^2*c - a^2*b*d)*e^3*f - (a^2*b*c + a^3*d)*e^2*f^2)*x) 
*sqrt(b*x^2 + a))/(a^2*b^3*e^6 - 3*a^3*b^2*e^5*f + 3*a^4*b*e^4*f^2 - a^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (f x^{2} + e\right )}^{2}} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (150) = 300\).

Time = 0.44 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.44 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx=\frac {{\left (b^{2} c - a b d\right )} x}{{\left (a b^{2} e^{2} - 2 \, a^{2} b e f + a^{3} f^{2}\right )} \sqrt {b x^{2} + a}} - \frac {{\left (2 \, b^{\frac {3}{2}} d e^{2} - 4 \, b^{\frac {3}{2}} c e f + a \sqrt {b} d e f + a \sqrt {b} c f^{2}\right )} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} f + 2 \, b e - a f}{2 \, \sqrt {-b^{2} e^{2} + a b e f}}\right )}{2 \, {\left (b^{2} e^{3} - 2 \, a b e^{2} f + a^{2} e f^{2}\right )} \sqrt {-b^{2} e^{2} + a b e f}} - \frac {2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b^{\frac {3}{2}} d e^{2} - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b^{\frac {3}{2}} c e f - {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a \sqrt {b} d e f + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a \sqrt {b} c f^{2} + a^{2} \sqrt {b} d e f - a^{2} \sqrt {b} c f^{2}}{{\left (b^{2} e^{3} - 2 \, a b e^{2} f + a^{2} e f^{2}\right )} {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} f + 4 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b e - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a f + a^{2} f\right )}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 3450, normalized size of antiderivative = 20.29 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{3/2} \left (e+f x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x** 
2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**4*c*e*f**3 - sqrt(e)*sqrt(a* 
f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b 
)*x)/(sqrt(e)*sqrt(b)))*a**4*c*f**4*x**2 - sqrt(e)*sqrt(a*f - b*e)*atan((s 
qrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sq 
rt(b)))*a**4*d*e**2*f**2 - sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - 
 sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**4*d*e 
*f**3*x**2 + 8*sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqr 
t(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*b*c*e**2*f**2 + 
 7*sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2 
) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*b*c*e*f**3*x**2 - sqrt(e)*s 
qrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)* 
sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*b*c*f**4*x**4 + 2*sqrt(e)*sqrt(a*f - b* 
e)*atan((sqrt(a*f - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/( 
sqrt(e)*sqrt(b)))*a**3*b*d*e**3*f + sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f 
 - b*e) - sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b))) 
*a**3*b*d*e**2*f**2*x**2 - sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - 
 sqrt(f)*sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**3*b*d 
*e*f**3*x**4 - 16*sqrt(e)*sqrt(a*f - b*e)*atan((sqrt(a*f - b*e) - sqrt(f)* 
sqrt(a + b*x**2) - sqrt(f)*sqrt(b)*x)/(sqrt(e)*sqrt(b)))*a**2*b**2*c*e*...