\(\int \frac {1}{(a+b x^2)^{5/2} (c+d x^2)^3} \, dx\) [110]

Optimal result

Integrand size = 21, antiderivative size = 292 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx=\frac {b \left (8 b^2 c^2+36 a b c d-9 a^2 d^2\right ) x}{24 a c^2 (b c-a d)^3 \left (a+b x^2\right )^{3/2}}+\frac {b \left (16 b^3 c^3-88 a b^2 c^2 d-42 a^2 b c d^2+9 a^3 d^3\right ) x}{24 a^2 c^2 (b c-a d)^4 \sqrt {a+b x^2}}-\frac {d x}{4 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}-\frac {d (10 b c-3 a d) x}{8 c^2 (b c-a d)^2 \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac {d^2 \left (48 b^2 c^2-16 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{8 c^{5/2} (b c-a d)^{9/2}} \] Output:

1/24*b*(-9*a^2*d^2+36*a*b*c*d+8*b^2*c^2)*x/a/c^2/(-a*d+b*c)^3/(b*x^2+a)^(3 
/2)+1/24*b*(9*a^3*d^3-42*a^2*b*c*d^2-88*a*b^2*c^2*d+16*b^3*c^3)*x/a^2/c^2/ 
(-a*d+b*c)^4/(b*x^2+a)^(1/2)-1/4*d*x/c/(-a*d+b*c)/(b*x^2+a)^(3/2)/(d*x^2+c 
)^2-1/8*d*(-3*a*d+10*b*c)*x/c^2/(-a*d+b*c)^2/(b*x^2+a)^(3/2)/(d*x^2+c)+1/8 
*d^2*(3*a^2*d^2-16*a*b*c*d+48*b^2*c^2)*arctanh((-a*d+b*c)^(1/2)*x/c^(1/2)/ 
(b*x^2+a)^(1/2))/c^(5/2)/(-a*d+b*c)^(9/2)
 

Mathematica [A] (verified)

Time = 2.89 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx=\frac {\frac {\sqrt {c} x \left (16 b^5 c^3 x^2 \left (c+d x^2\right )^2+8 a b^4 c^2 \left (3 c-11 d x^2\right ) \left (c+d x^2\right )^2+3 a^5 d^4 \left (5 c+3 d x^2\right )+3 a^3 b^2 d^3 x^2 \left (-32 c^2-23 c d x^2+3 d^2 x^4\right )+6 a^4 b d^3 \left (-8 c^2-2 c d x^2+3 d^2 x^4\right )-6 a^2 b^3 c d \left (16 c^3+32 c^2 d x^2+24 c d^2 x^4+7 d^3 x^6\right )\right )}{a^2 (b c-a d)^4 \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2}-\frac {9 d^2 (-4 b c+a d)^2 \arctan \left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{(-b c+a d)^{9/2}}+\frac {24 a b c d^3 \text {arctanh}\left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {b c-a d}}\right )}{(b c-a d)^{9/2}}}{24 c^{5/2}} \] Input:

Integrate[1/((a + b*x^2)^(5/2)*(c + d*x^2)^3),x]
 

Output:

((Sqrt[c]*x*(16*b^5*c^3*x^2*(c + d*x^2)^2 + 8*a*b^4*c^2*(3*c - 11*d*x^2)*( 
c + d*x^2)^2 + 3*a^5*d^4*(5*c + 3*d*x^2) + 3*a^3*b^2*d^3*x^2*(-32*c^2 - 23 
*c*d*x^2 + 3*d^2*x^4) + 6*a^4*b*d^3*(-8*c^2 - 2*c*d*x^2 + 3*d^2*x^4) - 6*a 
^2*b^3*c*d*(16*c^3 + 32*c^2*d*x^2 + 24*c*d^2*x^4 + 7*d^3*x^6)))/(a^2*(b*c 
- a*d)^4*(a + b*x^2)^(3/2)*(c + d*x^2)^2) - (9*d^2*(-4*b*c + a*d)^2*ArcTan 
[(-(d*x*Sqrt[a + b*x^2]) + Sqrt[b]*(c + d*x^2))/(Sqrt[c]*Sqrt[-(b*c) + a*d 
])])/(-(b*c) + a*d)^(9/2) + (24*a*b*c*d^3*ArcTanh[(-(d*x*Sqrt[a + b*x^2]) 
+ Sqrt[b]*(c + d*x^2))/(Sqrt[c]*Sqrt[b*c - a*d])])/(b*c - a*d)^(9/2))/(24* 
c^(5/2))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {316, 402, 25, 402, 25, 27, 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.

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

\(\Big \downarrow \) 316

\(\displaystyle \frac {\int \frac {-6 b d x^2+4 b c-3 a d}{\left (b x^2+a\right )^{5/2} \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}-\frac {\int -\frac {8 b^2 c^2-24 a b d c+9 a^2 d^2+4 b d (4 b c+3 a d) x^2}{\left (b x^2+a\right )^{3/2} \left (d x^2+c\right )^2}dx}{3 a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {8 b^2 c^2-24 a b d c+9 a^2 d^2+4 b d (4 b c+3 a d) x^2}{\left (b x^2+a\right )^{3/2} \left (d x^2+c\right )^2}dx}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {b x \left (-3 a^2 d^2-40 a b c d+8 b^2 c^2\right )}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}-\frac {\int -\frac {d \left (2 b \left (8 b^2 c^2-40 a b d c-3 a^2 d^2\right ) x^2+a \left (8 b^2 c^2+36 a b d c-9 a^2 d^2\right )\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )^2}dx}{a (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {d \left (2 b \left (8 b^2 c^2-40 a b d c-3 a^2 d^2\right ) x^2+a \left (8 b^2 c^2+36 a b d c-9 a^2 d^2\right )\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )^2}dx}{a (b c-a d)}+\frac {b x \left (-3 a^2 d^2-40 a b c d+8 b^2 c^2\right )}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {d \int \frac {2 b \left (8 b^2 c^2-40 a b d c-3 a^2 d^2\right ) x^2+a \left (8 b^2 c^2+36 a b d c-9 a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )^2}dx}{a (b c-a d)}+\frac {b x \left (-3 a^2 d^2-40 a b c d+8 b^2 c^2\right )}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {d \left (\frac {\int \frac {3 a^2 d \left (48 b^2 c^2-16 a b d c+3 a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (9 a^3 d^3-42 a^2 b c d^2-88 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}+\frac {b x \left (-3 a^2 d^2-40 a b c d+8 b^2 c^2\right )}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {d \left (\frac {3 a^2 d \left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (9 a^3 d^3-42 a^2 b c d^2-88 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}+\frac {b x \left (-3 a^2 d^2-40 a b c d+8 b^2 c^2\right )}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {d \left (\frac {3 a^2 d \left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \int \frac {1}{c-\frac {(b c-a d) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (9 a^3 d^3-42 a^2 b c d^2-88 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}+\frac {b x \left (-3 a^2 d^2-40 a b c d+8 b^2 c^2\right )}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {b x \left (-3 a^2 d^2-40 a b c d+8 b^2 c^2\right )}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}+\frac {d \left (\frac {3 a^2 d \left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{3/2}}+\frac {x \sqrt {a+b x^2} \left (9 a^3 d^3-42 a^2 b c d^2-88 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+4 b c)}{3 a \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 (b c-a d)}\)

Input:

Int[1/((a + b*x^2)^(5/2)*(c + d*x^2)^3),x]
 

Output:

-1/4*(d*x)/(c*(b*c - a*d)*(a + b*x^2)^(3/2)*(c + d*x^2)^2) + ((b*(4*b*c + 
3*a*d)*x)/(3*a*(b*c - a*d)*(a + b*x^2)^(3/2)*(c + d*x^2)) + ((b*(8*b^2*c^2 
 - 40*a*b*c*d - 3*a^2*d^2)*x)/(a*(b*c - a*d)*Sqrt[a + b*x^2]*(c + d*x^2)) 
+ (d*(((16*b^3*c^3 - 88*a*b^2*c^2*d - 42*a^2*b*c*d^2 + 9*a^3*d^3)*x*Sqrt[a 
 + b*x^2])/(2*c*(b*c - a*d)*(c + d*x^2)) + (3*a^2*d*(48*b^2*c^2 - 16*a*b*c 
*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(2 
*c^(3/2)*(b*c - a*d)^(3/2))))/(a*(b*c - a*d)))/(3*a*(b*c - a*d)))/(4*c*(b* 
c - a*d))
 

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 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 = 0.84 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(-\frac {3 \left (a^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}} d^{2} \left (x^{2} d +c \right )^{2} \left (a^{2} d^{2}-\frac {16}{3} a b c d +16 b^{2} c^{2}\right ) \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )-\frac {5 \sqrt {\left (a d -b c \right ) c}\, \left (\frac {3 a^{3} x^{2} \left (b \,x^{2}+a \right )^{2} d^{5}}{5}+a^{2} \left (-\frac {14 b \,x^{2}}{5}+a \right ) \left (b \,x^{2}+a \right )^{2} c \,d^{4}-\frac {16 a \,c^{2} \left (\frac {11}{6} b^{3} x^{6}+3 a \,b^{2} x^{4}+2 a^{2} b \,x^{2}+a^{3}\right ) b \,d^{3}}{5}-\frac {64 \left (-\frac {1}{12} b^{2} x^{4}+\frac {19}{24} a b \,x^{2}+a^{2}\right ) c^{3} b^{3} x^{2} d^{2}}{5}-\frac {32 \left (-\frac {1}{3} b^{2} x^{4}+\frac {5}{12} a b \,x^{2}+a^{2}\right ) c^{4} b^{3} d}{5}+\frac {8 c^{5} \left (\frac {2 b \,x^{2}}{3}+a \right ) b^{4}}{5}\right ) x}{3}\right )}{8 \sqrt {\left (a d -b c \right ) c}\, \left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (x^{2} d +c \right )^{2} c^{2} \left (a d -b c \right )^{4} a^{2}}\)	\(302\)
default	\(\text {Expression too large to display}\)	\(7121\)

Input:

int(1/(b*x^2+a)^(5/2)/(d*x^2+c)^3,x,method=_RETURNVERBOSE)
 

Output:

-3/8*(a^2*(b*x^2+a)^(3/2)*d^2*(d*x^2+c)^2*(a^2*d^2-16/3*a*b*c*d+16*b^2*c^2 
)*arctan(c*(b*x^2+a)^(1/2)/x/((a*d-b*c)*c)^(1/2))-5/3*((a*d-b*c)*c)^(1/2)* 
(3/5*a^3*x^2*(b*x^2+a)^2*d^5+a^2*(-14/5*b*x^2+a)*(b*x^2+a)^2*c*d^4-16/5*a* 
c^2*(11/6*b^3*x^6+3*a*b^2*x^4+2*a^2*b*x^2+a^3)*b*d^3-64/5*(-1/12*b^2*x^4+1 
9/24*a*b*x^2+a^2)*c^3*b^3*x^2*d^2-32/5*(-1/3*b^2*x^4+5/12*a*b*x^2+a^2)*c^4 
*b^3*d+8/5*c^5*(2/3*b*x^2+a)*b^4)*x)/((a*d-b*c)*c)^(1/2)/(b*x^2+a)^(3/2)/( 
d*x^2+c)^2/c^2/(a*d-b*c)^4/a^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1105 vs. \(2 (264) = 528\).

Time = 2.65 (sec) , antiderivative size = 2250, normalized size of antiderivative = 7.71 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

[1/96*(3*(48*a^4*b^2*c^4*d^2 - 16*a^5*b*c^3*d^3 + 3*a^6*c^2*d^4 + (48*a^2* 
b^4*c^2*d^4 - 16*a^3*b^3*c*d^5 + 3*a^4*b^2*d^6)*x^8 + 2*(48*a^2*b^4*c^3*d^ 
3 + 32*a^3*b^3*c^2*d^4 - 13*a^4*b^2*c*d^5 + 3*a^5*b*d^6)*x^6 + (48*a^2*b^4 
*c^4*d^2 + 176*a^3*b^3*c^3*d^3 - 13*a^4*b^2*c^2*d^4 - 4*a^5*b*c*d^5 + 3*a^ 
6*d^6)*x^4 + 2*(48*a^3*b^3*c^4*d^2 + 32*a^4*b^2*c^3*d^3 - 13*a^5*b*c^2*d^4 
 + 3*a^6*c*d^5)*x^2)*sqrt(b*c^2 - a*c*d)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2 
*d^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 + 4*((2*b*c - a*d)*x^3 
 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a))/(d^2*x^4 + 2*c*d*x^2 + c^2) 
) + 4*((16*b^6*c^5*d^2 - 104*a*b^5*c^4*d^3 + 46*a^2*b^4*c^3*d^4 + 51*a^3*b 
^3*c^2*d^5 - 9*a^4*b^2*c*d^6)*x^7 + (32*b^6*c^6*d - 184*a*b^5*c^5*d^2 + 8* 
a^2*b^4*c^4*d^3 + 75*a^3*b^3*c^3*d^4 + 87*a^4*b^2*c^2*d^5 - 18*a^5*b*c*d^6 
)*x^5 + (16*b^6*c^7 - 56*a*b^5*c^6*d - 152*a^2*b^4*c^5*d^2 + 96*a^3*b^3*c^ 
4*d^3 + 84*a^4*b^2*c^3*d^4 + 21*a^5*b*c^2*d^5 - 9*a^6*c*d^6)*x^3 + 3*(8*a* 
b^5*c^7 - 40*a^2*b^4*c^6*d + 32*a^3*b^3*c^5*d^2 - 16*a^4*b^2*c^4*d^3 + 21* 
a^5*b*c^3*d^4 - 5*a^6*c^2*d^5)*x)*sqrt(b*x^2 + a))/(a^4*b^5*c^10 - 5*a^5*b 
^4*c^9*d + 10*a^6*b^3*c^8*d^2 - 10*a^7*b^2*c^7*d^3 + 5*a^8*b*c^6*d^4 - a^9 
*c^5*d^5 + (a^2*b^7*c^8*d^2 - 5*a^3*b^6*c^7*d^3 + 10*a^4*b^5*c^6*d^4 - 10* 
a^5*b^4*c^5*d^5 + 5*a^6*b^3*c^4*d^6 - a^7*b^2*c^3*d^7)*x^8 + 2*(a^2*b^7*c^ 
9*d - 4*a^3*b^6*c^8*d^2 + 5*a^4*b^5*c^7*d^3 - 5*a^6*b^3*c^5*d^5 + 4*a^7*b^ 
2*c^4*d^6 - a^8*b*c^3*d^7)*x^6 + (a^2*b^7*c^10 - a^3*b^6*c^9*d - 9*a^4*...
 

Sympy [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{2}} \left (c + d x^{2}\right )^{3}}\, dx \] Input:

integrate(1/(b*x**2+a)**(5/2)/(d*x**2+c)**3,x)
 

Output:

Integral(1/((a + b*x**2)**(5/2)*(c + d*x**2)**3), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} {\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(5/2)*(d*x^2 + c)^3), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1010 vs. \(2 (264) = 528\).

Time = 0.43 (sec) , antiderivative size = 1010, normalized size of antiderivative = 3.46 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c)^3,x, algorithm="giac")
 

Output:

1/3*((2*b^10*c^5 - 19*a*b^9*c^4*d + 56*a^2*b^8*c^3*d^2 - 74*a^3*b^7*c^2*d^ 
3 + 46*a^4*b^6*c*d^4 - 11*a^5*b^5*d^5)*x^2/(a^2*b^9*c^8 - 8*a^3*b^8*c^7*d 
+ 28*a^4*b^7*c^6*d^2 - 56*a^5*b^6*c^5*d^3 + 70*a^6*b^5*c^4*d^4 - 56*a^7*b^ 
4*c^3*d^5 + 28*a^8*b^3*c^2*d^6 - 8*a^9*b^2*c*d^7 + a^10*b*d^8) + 3*(a*b^9* 
c^5 - 8*a^2*b^8*c^4*d + 22*a^3*b^7*c^3*d^2 - 28*a^4*b^6*c^2*d^3 + 17*a^5*b 
^5*c*d^4 - 4*a^6*b^4*d^5)/(a^2*b^9*c^8 - 8*a^3*b^8*c^7*d + 28*a^4*b^7*c^6* 
d^2 - 56*a^5*b^6*c^5*d^3 + 70*a^6*b^5*c^4*d^4 - 56*a^7*b^4*c^3*d^5 + 28*a^ 
8*b^3*c^2*d^6 - 8*a^9*b^2*c*d^7 + a^10*b*d^8))*x/(b*x^2 + a)^(3/2) - 1/8*( 
48*b^(5/2)*c^2*d^2 - 16*a*b^(3/2)*c*d^3 + 3*a^2*sqrt(b)*d^4)*arctan(1/2*(( 
sqrt(b)*x - sqrt(b*x^2 + a))^2*d + 2*b*c - a*d)/sqrt(-b^2*c^2 + a*b*c*d))/ 
((b^4*c^6 - 4*a*b^3*c^5*d + 6*a^2*b^2*c^4*d^2 - 4*a^3*b*c^3*d^3 + a^4*c^2* 
d^4)*sqrt(-b^2*c^2 + a*b*c*d)) - 1/4*(24*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b 
^(5/2)*c^2*d^3 - 16*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(3/2)*c*d^4 + 3*(s 
qrt(b)*x - sqrt(b*x^2 + a))^6*a^2*sqrt(b)*d^5 + 112*(sqrt(b)*x - sqrt(b*x^ 
2 + a))^4*b^(7/2)*c^3*d^2 - 136*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a*b^(5/2)* 
c^2*d^3 + 66*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(3/2)*c*d^4 - 9*(sqrt(b 
)*x - sqrt(b*x^2 + a))^4*a^3*sqrt(b)*d^5 + 88*(sqrt(b)*x - sqrt(b*x^2 + a) 
)^2*a^2*b^(5/2)*c^2*d^3 - 64*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*b^(3/2)*c 
*d^4 + 9*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^4*sqrt(b)*d^5 + 14*a^4*b^(3/2)* 
c*d^4 - 3*a^5*sqrt(b)*d^5)/((b^4*c^6 - 4*a*b^3*c^5*d + 6*a^2*b^2*c^4*d^...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{5/2}\,{\left (d\,x^2+c\right )}^3} \,d x \] Input:

int(1/((a + b*x^2)^(5/2)*(c + d*x^2)^3),x)
 

Output:

int(1/((a + b*x^2)^(5/2)*(c + d*x^2)^3), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^3} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (d \,x^{2}+c \right )^{3}}d x \] Input:

int(1/(b*x^2+a)^(5/2)/(d*x^2+c)^3,x)
 

Output:

int(1/(b*x^2+a)^(5/2)/(d*x^2+c)^3,x)