\(\int \frac {(d+c x^2) (a x+\sqrt {-b+a^2 x^2})^{5/4}}{x (-b+a^2 x^2)^{5/2}} \, dx\) [3136]

Optimal result

Integrand size = 49, antiderivative size = 876 \[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\frac {\left (-97 b^2 c-a^2 b d+45 a^2 b c x^2-51 a^4 d x^2\right ) \sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}{96 a^2 b^{15/8} \left (-\sqrt {b}+a x\right ) \left (\sqrt {b}+a x\right )}+\frac {\sqrt {-b+a^2 x^2} \left (13 b^2 c x-83 a^2 b d x-45 a^2 b c x^3+51 a^4 d x^3\right ) \sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}{96 a b^{15/8} \left (-\sqrt {b}+a x\right )^2 \left (\sqrt {b}+a x\right )^2}+\frac {5 \left (-3 b c+29 a^2 d\right ) \arctan \left (\sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}\right )}{64 a^2 b^{15/8}}-\frac {\sqrt {2-\sqrt {2}} d \arctan \left (\frac {-\frac {1}{\sqrt {2-\sqrt {2}}}+\frac {\sqrt {\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}{\sqrt {2-\sqrt {2}}}}{\sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}\right )}{b^{15/8}}-\frac {\sqrt {2+\sqrt {2}} d \arctan \left (\frac {-\frac {1}{\sqrt {2+\sqrt {2}}}+\frac {\sqrt {\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}{\sqrt {2+\sqrt {2}}}}{\sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}\right )}{b^{15/8}}+\frac {5 \left (-3 b c+29 a^2 d\right ) \text {arctanh}\left (\sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}\right )}{64 a^2 b^{15/8}}-\frac {5 (-1)^{3/4} \left (-3 b c+29 a^2 d\right ) \text {arctanh}\left (\sqrt [4]{-1} \sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}\right )}{64 a^2 b^{15/8}}-\frac {5 \sqrt [4]{-1} \left (-3 b c+29 a^2 d\right ) \text {arctanh}\left ((-1)^{3/4} \sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}\right )}{64 a^2 b^{15/8}}-\frac {\sqrt {2-\sqrt {2}} d \text {arctanh}\left (\frac {\frac {1}{\sqrt {2-\sqrt {2}}}+\frac {\sqrt {\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}{\sqrt {2-\sqrt {2}}}}{\sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}\right )}{b^{15/8}}-\frac {\sqrt {2+\sqrt {2}} d \text {arctanh}\left (\frac {\frac {1}{\sqrt {2+\sqrt {2}}}+\frac {\sqrt {\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}{\sqrt {2+\sqrt {2}}}}{\sqrt [4]{\frac {a x+\sqrt {-b+a^2 x^2}}{\sqrt {b}}}}\right )}{b^{15/8}} \] Output:

1/96*(-51*a^4*d*x^2+45*a^2*b*c*x^2-a^2*b*d-97*b^2*c)*((a*x+(a^2*x^2-b)^(1/ 
2))/b^(1/2))^(1/4)/a^2/b^(15/8)/(-b^(1/2)+a*x)/(b^(1/2)+a*x)+1/96*(a^2*x^2 
-b)^(1/2)*(51*a^4*d*x^3-45*a^2*b*c*x^3-83*a^2*b*d*x+13*b^2*c*x)*((a*x+(a^2 
*x^2-b)^(1/2))/b^(1/2))^(1/4)/a/b^(15/8)/(-b^(1/2)+a*x)^2/(b^(1/2)+a*x)^2+ 
5/64*(29*a^2*d-3*b*c)*arctan(((a*x+(a^2*x^2-b)^(1/2))/b^(1/2))^(1/4))/a^2/ 
b^(15/8)-(2-2^(1/2))^(1/2)*d*arctan((-1/(2-2^(1/2))^(1/2)+((a*x+(a^2*x^2-b 
)^(1/2))/b^(1/2))^(1/2)/(2-2^(1/2))^(1/2))/((a*x+(a^2*x^2-b)^(1/2))/b^(1/2 
))^(1/4))/b^(15/8)-(2+2^(1/2))^(1/2)*d*arctan((-1/(2+2^(1/2))^(1/2)+((a*x+ 
(a^2*x^2-b)^(1/2))/b^(1/2))^(1/2)/(2+2^(1/2))^(1/2))/((a*x+(a^2*x^2-b)^(1/ 
2))/b^(1/2))^(1/4))/b^(15/8)+5/64*(29*a^2*d-3*b*c)*arctanh(((a*x+(a^2*x^2- 
b)^(1/2))/b^(1/2))^(1/4))/a^2/b^(15/8)-5/64*(-1)^(3/4)*(29*a^2*d-3*b*c)*ar 
ctanh((-1)^(1/4)*((a*x+(a^2*x^2-b)^(1/2))/b^(1/2))^(1/4))/a^2/b^(15/8)-5/6 
4*(-1)^(1/4)*(29*a^2*d-3*b*c)*arctanh((-1)^(3/4)*((a*x+(a^2*x^2-b)^(1/2))/ 
b^(1/2))^(1/4))/a^2/b^(15/8)-(2-2^(1/2))^(1/2)*d*arctanh((1/(2-2^(1/2))^(1 
/2)+((a*x+(a^2*x^2-b)^(1/2))/b^(1/2))^(1/2)/(2-2^(1/2))^(1/2))/((a*x+(a^2* 
x^2-b)^(1/2))/b^(1/2))^(1/4))/b^(15/8)-(2+2^(1/2))^(1/2)*d*arctanh((1/(2+2 
^(1/2))^(1/2)+((a*x+(a^2*x^2-b)^(1/2))/b^(1/2))^(1/2)/(2+2^(1/2))^(1/2))/( 
(a*x+(a^2*x^2-b)^(1/2))/b^(1/2))^(1/4))/b^(15/8)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(12230\) vs. \(2(876)=1752\).

Time = 52.47 (sec) , antiderivative size = 12230, normalized size of antiderivative = 13.96 \[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[((d + c*x^2)*(a*x + Sqrt[-b + a^2*x^2])^(5/4))/(x*(-b + a^2*x^2) 
^(5/2)),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 3.90 (sec) , antiderivative size = 1407, normalized size of antiderivative = 1.61, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {7293, 2009}

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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {8 c \left (a x+\sqrt {a^2 x^2-b}\right )^{17/4}}{3 a^2 \left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^3}-\frac {5 c \left (a x+\sqrt {a^2 x^2-b}\right )^{9/4}}{6 a^2 \left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^2}+\frac {8 d \left (a x+\sqrt {a^2 x^2-b}\right )^{9/4}}{3 \left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^3}+\frac {15 c \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{16 a^2 \left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )}+\frac {39 d \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{16 b \left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )}-\frac {7 d \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{2 \left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^2}-\frac {2 d \arctan \left (\frac {\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{-b}}\right )}{(-b)^{15/8}}-\frac {15 c \arctan \left (\frac {\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}\right )}{64 a^2 b^{7/8}}+\frac {145 d \arctan \left (\frac {\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}\right )}{64 b^{15/8}}+\frac {\sqrt {2} d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{-b}}\right )}{(-b)^{15/8}}-\frac {\sqrt {2} d \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{-b}}+1\right )}{(-b)^{15/8}}+\frac {15 c \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}\right )}{64 \sqrt {2} a^2 b^{7/8}}-\frac {145 d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}\right )}{64 \sqrt {2} b^{15/8}}-\frac {15 c \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}+1\right )}{64 \sqrt {2} a^2 b^{7/8}}+\frac {145 d \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}+1\right )}{64 \sqrt {2} b^{15/8}}-\frac {2 d \text {arctanh}\left (\frac {\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{-b}}\right )}{(-b)^{15/8}}-\frac {15 c \text {arctanh}\left (\frac {\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}\right )}{64 a^2 b^{7/8}}+\frac {145 d \text {arctanh}\left (\frac {\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{\sqrt [8]{b}}\right )}{64 b^{15/8}}+\frac {d \log \left (\sqrt {a x+\sqrt {a^2 x^2-b}}-\sqrt {2} \sqrt [8]{-b} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\sqrt [4]{-b}\right )}{\sqrt {2} (-b)^{15/8}}-\frac {d \log \left (\sqrt {a x+\sqrt {a^2 x^2-b}}+\sqrt {2} \sqrt [8]{-b} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\sqrt [4]{-b}\right )}{\sqrt {2} (-b)^{15/8}}+\frac {15 c \log \left (\sqrt {a x+\sqrt {a^2 x^2-b}}-\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\sqrt [4]{b}\right )}{128 \sqrt {2} a^2 b^{7/8}}-\frac {145 d \log \left (\sqrt {a x+\sqrt {a^2 x^2-b}}-\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\sqrt [4]{b}\right )}{128 \sqrt {2} b^{15/8}}-\frac {15 c \log \left (\sqrt {a x+\sqrt {a^2 x^2-b}}+\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\sqrt [4]{b}\right )}{128 \sqrt {2} a^2 b^{7/8}}+\frac {145 d \log \left (\sqrt {a x+\sqrt {a^2 x^2-b}}+\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\sqrt [4]{b}\right )}{128 \sqrt {2} b^{15/8}}\)

Input:

Int[((d + c*x^2)*(a*x + Sqrt[-b + a^2*x^2])^(5/4))/(x*(-b + a^2*x^2)^(5/2) 
),x]
 

Output:

(8*d*(a*x + Sqrt[-b + a^2*x^2])^(9/4))/(3*(b - (a*x + Sqrt[-b + a^2*x^2])^ 
2)^3) + (8*c*(a*x + Sqrt[-b + a^2*x^2])^(17/4))/(3*a^2*(b - (a*x + Sqrt[-b 
 + a^2*x^2])^2)^3) - (7*d*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(2*(b - (a*x + 
 Sqrt[-b + a^2*x^2])^2)^2) - (5*c*(a*x + Sqrt[-b + a^2*x^2])^(9/4))/(6*a^2 
*(b - (a*x + Sqrt[-b + a^2*x^2])^2)^2) + (15*c*(a*x + Sqrt[-b + a^2*x^2])^ 
(1/4))/(16*a^2*(b - (a*x + Sqrt[-b + a^2*x^2])^2)) + (39*d*(a*x + Sqrt[-b 
+ a^2*x^2])^(1/4))/(16*b*(b - (a*x + Sqrt[-b + a^2*x^2])^2)) - (2*d*ArcTan 
[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/(-b)^(1/8)])/(-b)^(15/8) - (15*c*ArcTan[ 
(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)])/(64*a^2*b^(7/8)) + (145*d*ArcTa 
n[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)])/(64*b^(15/8)) + (Sqrt[2]*d*Ar 
cTan[1 - (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)])/(-b)^(15/ 
8) - (Sqrt[2]*d*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b) 
^(1/8)])/(-b)^(15/8) + (15*c*ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2] 
)^(1/4))/b^(1/8)])/(64*Sqrt[2]*a^2*b^(7/8)) - (145*d*ArcTan[1 - (Sqrt[2]*( 
a*x + Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)])/(64*Sqrt[2]*b^(15/8)) - (15*c*A 
rcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)])/(64*Sqrt[2] 
*a^2*b^(7/8)) + (145*d*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4 
))/b^(1/8)])/(64*Sqrt[2]*b^(15/8)) - (2*d*ArcTanh[(a*x + Sqrt[-b + a^2*x^2 
])^(1/4)/(-b)^(1/8)])/(-b)^(15/8) - (15*c*ArcTanh[(a*x + Sqrt[-b + a^2*x^2 
])^(1/4)/b^(1/8)])/(64*a^2*b^(7/8)) + (145*d*ArcTanh[(a*x + Sqrt[-b + a...
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {\left (c \,x^{2}+d \right ) \left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {5}{4}}}{x \left (a^{2} x^{2}-b \right )^{\frac {5}{2}}}d x\]

Input:

int((c*x^2+d)*(a*x+(a^2*x^2-b)^(1/2))^(5/4)/x/(a^2*x^2-b)^(5/2),x)
 

Output:

int((c*x^2+d)*(a*x+(a^2*x^2-b)^(1/2))^(5/4)/x/(a^2*x^2-b)^(5/2),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.47 (sec) , antiderivative size = 3342, normalized size of antiderivative = 3.82 \[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((c*x^2+d)*(a*x+(a^2*x^2-b)^(1/2))^(5/4)/x/(a^2*x^2-b)^(5/2),x, a 
lgorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a x + \sqrt {a^{2} x^{2} - b}\right )^{\frac {5}{4}} \left (c x^{2} + d\right )}{x \left (a^{2} x^{2} - b\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((c*x**2+d)*(a*x+(a**2*x**2-b)**(1/2))**(5/4)/x/(a**2*x**2-b)**(5 
/2),x)
 

Output:

Integral((a*x + sqrt(a**2*x**2 - b))**(5/4)*(c*x**2 + d)/(x*(a**2*x**2 - b 
)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (c x^{2} + d\right )} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {5}{4}}}{{\left (a^{2} x^{2} - b\right )}^{\frac {5}{2}} x} \,d x } \] Input:

integrate((c*x^2+d)*(a*x+(a^2*x^2-b)^(1/2))^(5/4)/x/(a^2*x^2-b)^(5/2),x, a 
lgorithm="maxima")
 

Output:

integrate((c*x^2 + d)*(a*x + sqrt(a^2*x^2 - b))^(5/4)/((a^2*x^2 - b)^(5/2) 
*x), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x^2+d)*(a*x+(a^2*x^2-b)^(1/2))^(5/4)/x/(a^2*x^2-b)^(5/2),x, a 
lgorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{5/4}\,\left (c\,x^2+d\right )}{x\,{\left (a^2\,x^2-b\right )}^{5/2}} \,d x \] Input:

int(((a*x + (a^2*x^2 - b)^(1/2))^(5/4)*(d + c*x^2))/(x*(a^2*x^2 - b)^(5/2) 
),x)
 

Output:

int(((a*x + (a^2*x^2 - b)^(1/2))^(5/4)*(d + c*x^2))/(x*(a^2*x^2 - b)^(5/2) 
), x)
 

Reduce [F]

\[ \int \frac {\left (d+c x^2\right ) \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{x \left (-b+a^2 x^2\right )^{5/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+d)*(a*x+(a^2*x^2-b)^(1/2))^(5/4)/x/(a^2*x^2-b)^(5/2),x)
 

Output:

(32*sqrt(a**2*x**2 - b)*(sqrt(a**2*x**2 - b) + a*x)**(1/4)*a**2*d*x - 32*s 
qrt(a**2*x**2 - b)*(sqrt(a**2*x**2 - b) + a*x)**(1/4)*b*c*x + 44*int((sqrt 
(a**2*x**2 - b) + a*x)**(1/4)/(4*sqrt(a**2*x**2 - b)*a**8*x**8 - 13*sqrt(a 
**2*x**2 - b)*a**6*b*x**6 + 15*sqrt(a**2*x**2 - b)*a**4*b**2*x**4 - 7*sqrt 
(a**2*x**2 - b)*a**2*b**3*x**2 + sqrt(a**2*x**2 - b)*b**4 + 4*a**9*x**9 - 
15*a**7*b*x**7 + 21*a**5*b**2*x**5 - 13*a**3*b**3*x**3 + 3*a*b**4*x),x)*a* 
*6*b**3*d*x**4 - 32*int((sqrt(a**2*x**2 - b) + a*x)**(1/4)/(4*sqrt(a**2*x* 
*2 - b)*a**8*x**8 - 13*sqrt(a**2*x**2 - b)*a**6*b*x**6 + 15*sqrt(a**2*x**2 
 - b)*a**4*b**2*x**4 - 7*sqrt(a**2*x**2 - b)*a**2*b**3*x**2 + sqrt(a**2*x* 
*2 - b)*b**4 + 4*a**9*x**9 - 15*a**7*b*x**7 + 21*a**5*b**2*x**5 - 13*a**3* 
b**3*x**3 + 3*a*b**4*x),x)*a**4*b**4*c*x**4 - 88*int((sqrt(a**2*x**2 - b) 
+ a*x)**(1/4)/(4*sqrt(a**2*x**2 - b)*a**8*x**8 - 13*sqrt(a**2*x**2 - b)*a* 
*6*b*x**6 + 15*sqrt(a**2*x**2 - b)*a**4*b**2*x**4 - 7*sqrt(a**2*x**2 - b)* 
a**2*b**3*x**2 + sqrt(a**2*x**2 - b)*b**4 + 4*a**9*x**9 - 15*a**7*b*x**7 + 
 21*a**5*b**2*x**5 - 13*a**3*b**3*x**3 + 3*a*b**4*x),x)*a**4*b**4*d*x**2 + 
 64*int((sqrt(a**2*x**2 - b) + a*x)**(1/4)/(4*sqrt(a**2*x**2 - b)*a**8*x** 
8 - 13*sqrt(a**2*x**2 - b)*a**6*b*x**6 + 15*sqrt(a**2*x**2 - b)*a**4*b**2* 
x**4 - 7*sqrt(a**2*x**2 - b)*a**2*b**3*x**2 + sqrt(a**2*x**2 - b)*b**4 + 4 
*a**9*x**9 - 15*a**7*b*x**7 + 21*a**5*b**2*x**5 - 13*a**3*b**3*x**3 + 3*a* 
b**4*x),x)*a**2*b**5*c*x**2 + 44*int((sqrt(a**2*x**2 - b) + a*x)**(1/4)...