\(\int \frac {1}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\) [984]

Optimal result

Integrand size = 21, antiderivative size = 122 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {d x}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac {d (5 b c-2 a d) x}{3 c^2 (b c-a d)^2 \sqrt {c+d x^2}}+\frac {b^2 \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} (b c-a d)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {d x \left (a d \left (3 c+2 d x^2\right )-b c \left (6 c+5 d x^2\right )\right )}{3 c^2 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {b^2 \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{\sqrt {a} (b c-a d)^{5/2}} \] Input:

Integrate[1/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

(d*x*(a*d*(3*c + 2*d*x^2) - b*c*(6*c + 5*d*x^2)))/(3*c^2*(b*c - a*d)^2*(c 
+ d*x^2)^(3/2)) - (b^2*ArcTan[(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c + d*x^2 
]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(Sqrt[a]*(b*c - a*d)^(5/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {316, 402, 27, 291, 218}

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.

\(\displaystyle \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\int \frac {3 b^2 c^2}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{c (b c-a d)}-\frac {d x (5 b c-2 a d)}{c \sqrt {c+d x^2} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 b^2 c \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b c-a d}-\frac {d x (5 b c-2 a d)}{c \sqrt {c+d x^2} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {3 b^2 c \int \frac {1}{a-\frac {(a d-b c) x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{b c-a d}-\frac {d x (5 b c-2 a d)}{c \sqrt {c+d x^2} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {3 b^2 c \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} (b c-a d)^{3/2}}-\frac {d x (5 b c-2 a d)}{c \sqrt {c+d x^2} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}\)

Input:

Int[1/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

-1/3*(d*x)/(c*(b*c - a*d)*(c + d*x^2)^(3/2)) + (-((d*(5*b*c - 2*a*d)*x)/(c 
*(b*c - a*d)*Sqrt[c + d*x^2])) + (3*b^2*c*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt 
[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*(b*c - a*d)^(3/2)))/(3*c*(b*c - a*d))
 

Defintions of rubi rules used

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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 316

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

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 = 0.76 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {b^{2} c^{2} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right ) \left (x^{2} d +c \right )^{\frac {3}{2}}+\left (-2 b \,c^{2}+d \left (-\frac {5 b \,x^{2}}{3}+a \right ) c +\frac {2 a \,d^{2} x^{2}}{3}\right ) d \sqrt {\left (a d -b c \right ) a}\, x}{\sqrt {\left (a d -b c \right ) a}\, \left (x^{2} d +c \right )^{\frac {3}{2}} \left (a d -b c \right )^{2} c^{2}}\)	\(122\)
default	\(\text {Expression too large to display}\)	\(1396\)

Input:

int(1/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (104) = 208\).

Time = 0.30 (sec) , antiderivative size = 764, normalized size of antiderivative = 6.26 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\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 ) + 4 \, {\left ({\left (5 \, a b^{2} c^{2} d^{2} - 7 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{3} + 3 \, {\left (2 \, a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{12 \, {\left (a b^{3} c^{7} - 3 \, a^{2} b^{2} c^{6} d + 3 \, a^{3} b c^{5} d^{2} - a^{4} c^{4} d^{3} + {\left (a b^{3} c^{5} d^{2} - 3 \, a^{2} b^{2} c^{4} d^{3} + 3 \, a^{3} b c^{3} d^{4} - a^{4} c^{2} d^{5}\right )} x^{4} + 2 \, {\left (a b^{3} c^{6} d - 3 \, a^{2} b^{2} c^{5} d^{2} + 3 \, a^{3} b c^{4} d^{3} - a^{4} c^{3} d^{4}\right )} x^{2}\right )}}, \frac {3 \, {\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\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 ) - 2 \, {\left ({\left (5 \, a b^{2} c^{2} d^{2} - 7 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{3} + 3 \, {\left (2 \, a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{6 \, {\left (a b^{3} c^{7} - 3 \, a^{2} b^{2} c^{6} d + 3 \, a^{3} b c^{5} d^{2} - a^{4} c^{4} d^{3} + {\left (a b^{3} c^{5} d^{2} - 3 \, a^{2} b^{2} c^{4} d^{3} + 3 \, a^{3} b c^{3} d^{4} - a^{4} c^{2} d^{5}\right )} x^{4} + 2 \, {\left (a b^{3} c^{6} d - 3 \, a^{2} b^{2} c^{5} d^{2} + 3 \, a^{3} b c^{4} d^{3} - a^{4} c^{3} d^{4}\right )} x^{2}\right )}}\right ] \] Input:

integrate(1/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*(b^2*c^2*d^2*x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*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)*sqrt(d* 
x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*((5*a*b^2*c^2*d^2 - 7*a^2*b*c*d 
^3 + 2*a^3*d^4)*x^3 + 3*(2*a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + a^3*c*d^3)*x)*s 
qrt(d*x^2 + c))/(a*b^3*c^7 - 3*a^2*b^2*c^6*d + 3*a^3*b*c^5*d^2 - a^4*c^4*d 
^3 + (a*b^3*c^5*d^2 - 3*a^2*b^2*c^4*d^3 + 3*a^3*b*c^3*d^4 - a^4*c^2*d^5)*x 
^4 + 2*(a*b^3*c^6*d - 3*a^2*b^2*c^5*d^2 + 3*a^3*b*c^4*d^3 - a^4*c^3*d^4)*x 
^2), 1/6*(3*(b^2*c^2*d^2*x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*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)) - 2*((5*a*b^2*c^2*d^ 
2 - 7*a^2*b*c*d^3 + 2*a^3*d^4)*x^3 + 3*(2*a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + 
a^3*c*d^3)*x)*sqrt(d*x^2 + c))/(a*b^3*c^7 - 3*a^2*b^2*c^6*d + 3*a^3*b*c^5* 
d^2 - a^4*c^4*d^3 + (a*b^3*c^5*d^2 - 3*a^2*b^2*c^4*d^3 + 3*a^3*b*c^3*d^4 - 
 a^4*c^2*d^5)*x^4 + 2*(a*b^3*c^6*d - 3*a^2*b^2*c^5*d^2 + 3*a^3*b*c^4*d^3 - 
 a^4*c^3*d^4)*x^2)]
 

Sympy [F]

\[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(b*x**2+a)/(d*x**2+c)**(5/2),x)
 

Output:

Integral(1/((a + b*x**2)*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (104) = 208\).

Time = 0.13 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.63 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {b^{2} \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\frac {{\left (5 \, b^{3} c^{3} d^{3} - 12 \, a b^{2} c^{2} d^{4} + 9 \, a^{2} b c d^{5} - 2 \, a^{3} d^{6}\right )} x^{2}}{b^{4} c^{6} d - 4 \, a b^{3} c^{5} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{3} - 4 \, a^{3} b c^{3} d^{4} + a^{4} c^{2} d^{5}} + \frac {3 \, {\left (2 \, b^{3} c^{4} d^{2} - 5 \, a b^{2} c^{3} d^{3} + 4 \, a^{2} b c^{2} d^{4} - a^{3} c d^{5}\right )}}{b^{4} c^{6} d - 4 \, a b^{3} c^{5} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{3} - 4 \, a^{3} b c^{3} d^{4} + a^{4} c^{2} d^{5}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \] Input:

integrate(1/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

-b^2*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))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d - a 
^2*d^2)) - 1/3*((5*b^3*c^3*d^3 - 12*a*b^2*c^2*d^4 + 9*a^2*b*c*d^5 - 2*a^3* 
d^6)*x^2/(b^4*c^6*d - 4*a*b^3*c^5*d^2 + 6*a^2*b^2*c^4*d^3 - 4*a^3*b*c^3*d^ 
4 + a^4*c^2*d^5) + 3*(2*b^3*c^4*d^2 - 5*a*b^2*c^3*d^3 + 4*a^2*b*c^2*d^4 - 
a^3*c*d^5)/(b^4*c^6*d - 4*a*b^3*c^5*d^2 + 6*a^2*b^2*c^4*d^3 - 4*a^3*b*c^3* 
d^4 + a^4*c^2*d^5))*x/(d*x^2 + c)^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(1/((a + b*x^2)*(c + d*x^2)^(5/2)),x)
 

Output:

int(1/((a + b*x^2)*(c + d*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 997, normalized size of antiderivative = 8.17 \[ \int \frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(1/(b*x^2+a)/(d*x^2+c)^(5/2),x)
 

Output:

(3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 
 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b**2*c**4 + 
6*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 
2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b**2*c**3*d*x 
**2 + 3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b 
*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b**2*c* 
*2*d**2*x**4 + 3*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a 
*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b 
**2*c**4 + 6*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - 
 b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b**2* 
c**3*d*x**2 + 3*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a* 
d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b* 
*2*c**2*d**2*x**4 - 3*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a 
*d - b*c) + 2*sqrt(d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*b**2*c**4 
 - 6*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqr 
t(d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*b**2*c**3*d*x**2 - 3*sqrt( 
a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqrt(d)*sqrt( 
c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*b**2*c**2*d**2*x**4 + 6*sqrt(c + d*x 
**2)*a**3*c*d**3*x + 4*sqrt(c + d*x**2)*a**3*d**4*x**3 - 18*sqrt(c + d*x** 
2)*a**2*b*c**2*d**2*x - 14*sqrt(c + d*x**2)*a**2*b*c*d**3*x**3 + 12*sqr...