\(\int \frac {(a+b x^2)^{5/2}}{c+d x^2} \, dx\) [80]

Optimal result

Integrand size = 21, antiderivative size = 157 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{c+d x^2} \, dx=-\frac {b (4 b c-7 a d) x \sqrt {a+b x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right )^{3/2}}{4 d}+\frac {\sqrt {b} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 d^3}-\frac {(b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} d^3} \] Output:

-1/8*b*(-7*a*d+4*b*c)*x*(b*x^2+a)^(1/2)/d^2+1/4*b*x*(b*x^2+a)^(3/2)/d+1/8* 
b^(1/2)*(15*a^2*d^2-20*a*b*c*d+8*b^2*c^2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2 
))/d^3-(-a*d+b*c)^(5/2)*arctanh((-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a)^(1/2) 
)/c^(1/2)/d^3
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{c+d x^2} \, dx=\frac {b d x \sqrt {a+b x^2} \left (-4 b c+9 a d+2 b d x^2\right )-\frac {8 (-b c+a d)^{5/2} \arctan \left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{\sqrt {c}}-\sqrt {b} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 d^3} \] Input:

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

Output:

(b*d*x*Sqrt[a + b*x^2]*(-4*b*c + 9*a*d + 2*b*d*x^2) - (8*(-(b*c) + a*d)^(5 
/2)*ArcTan[(-(d*x*Sqrt[a + b*x^2]) + Sqrt[b]*(c + d*x^2))/(Sqrt[c]*Sqrt[-( 
b*c) + a*d])])/Sqrt[c] - Sqrt[b]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*Log 
[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*d^3)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {318, 25, 403, 25, 398, 224, 219, 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.

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

\(\Big \downarrow \) 318

\(\displaystyle \frac {\int -\frac {\sqrt {b x^2+a} \left (b (4 b c-7 a d) x^2+a (b c-4 a d)\right )}{d x^2+c}dx}{4 d}+\frac {b x \left (a+b x^2\right )^{3/2}}{4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\int \frac {\sqrt {b x^2+a} \left (b (4 b c-7 a d) x^2+a (b c-4 a d)\right )}{d x^2+c}dx}{4 d}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\frac {\int -\frac {b \left (8 b^2 c^2-20 a b d c+15 a^2 d^2\right ) x^2+a \left (4 b^2 c^2-9 a b d c+8 a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 d}+\frac {b x \sqrt {a+b x^2} (4 b c-7 a d)}{2 d}}{4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\frac {b x \sqrt {a+b x^2} (4 b c-7 a d)}{2 d}-\frac {\int \frac {b \left (8 b^2 c^2-20 a b d c+15 a^2 d^2\right ) x^2+a \left (4 b^2 c^2-9 a b d c+8 a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 d}}{4 d}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\frac {b x \sqrt {a+b x^2} (4 b c-7 a d)}{2 d}-\frac {\frac {b \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{d}-\frac {8 (b c-a d)^3 \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{d}}{2 d}}{4 d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\frac {b x \sqrt {a+b x^2} (4 b c-7 a d)}{2 d}-\frac {\frac {b \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{d}-\frac {8 (b c-a d)^3 \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{d}}{2 d}}{4 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\frac {b x \sqrt {a+b x^2} (4 b c-7 a d)}{2 d}-\frac {\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right )}{d}-\frac {8 (b c-a d)^3 \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{d}}{2 d}}{4 d}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\frac {b x \sqrt {a+b x^2} (4 b c-7 a d)}{2 d}-\frac {\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right )}{d}-\frac {8 (b c-a d)^3 \int \frac {1}{c-\frac {(b c-a d) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{d}}{2 d}}{4 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac {\frac {b x \sqrt {a+b x^2} (4 b c-7 a d)}{2 d}-\frac {\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right )}{d}-\frac {8 (b c-a d)^{5/2} \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} d}}{2 d}}{4 d}\)

Input:

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

Output:

(b*x*(a + b*x^2)^(3/2))/(4*d) - ((b*(4*b*c - 7*a*d)*x*Sqrt[a + b*x^2])/(2* 
d) - ((Sqrt[b]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*ArcTanh[(Sqrt[b]*x)/S 
qrt[a + b*x^2]])/d - (8*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqr 
t[c]*Sqrt[a + b*x^2])])/(Sqrt[c]*d))/(2*d))/(4*d)
 

Defintions of rubi rules used

rule 25

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

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 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 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 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 318

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

rule 398

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

rule 403

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

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(-\frac {\frac {2 \left (a d -b c \right )^{3} \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )}{\sqrt {\left (a d -b c \right ) c}}+\frac {b \left (-d \sqrt {b \,x^{2}+a}\, \left (2 b d \,x^{2}+9 a d -4 b c \right ) x -\frac {\left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{\sqrt {b}}\right )}{4}}{2 d^{3}}\)	\(136\)
risch	\(\frac {b x \left (2 b d \,x^{2}+9 a d -4 b c \right ) \sqrt {b \,x^{2}+a}}{8 d^{2}}+\frac {\frac {\sqrt {b}\, \left (15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d}+\frac {4 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (\frac {\frac {2 a d -2 b c}{d}-\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x +\frac {\sqrt {-c d}}{d}\right )^{2} b -\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x +\frac {\sqrt {-c d}}{d}}\right )}{\sqrt {-c d}\, d \sqrt {\frac {a d -b c}{d}}}-\frac {4 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (\frac {\frac {2 a d -2 b c}{d}+\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x -\frac {\sqrt {-c d}}{d}\right )^{2} b +\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x -\frac {\sqrt {-c d}}{d}}\right )}{\sqrt {-c d}\, d \sqrt {\frac {a d -b c}{d}}}}{8 d^{2}}\)	\(462\)
default	\(\text {Expression too large to display}\)	\(2048\)

Input:

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

Output:

-1/2/d^3*(2*(a*d-b*c)^3/((a*d-b*c)*c)^(1/2)*arctan(c*(b*x^2+a)^(1/2)/x/((a 
*d-b*c)*c)^(1/2))+1/4*b*(-d*(b*x^2+a)^(1/2)*(2*b*d*x^2+9*a*d-4*b*c)*x-(15* 
a^2*d^2-20*a*b*c*d+8*b^2*c^2)/b^(1/2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 935, normalized size of antiderivative = 5.96 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{c+d x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*((8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt 
(b*x^2 + a)*sqrt(b)*x - a) + 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt((b*c - 
 a*d)/c)*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*(a*c^2*x + (2*b*c^2 - a*c*d)*x^3)*sqrt(b*x^2 + a)* 
sqrt((b*c - a*d)/c))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 2*(2*b^2*d^2*x^3 - (4* 
b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x^2 + a))/d^3, -1/8*((8*b^2*c^2 - 20*a*b*c* 
d + 15*a^2*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 2*(b^2*c^2 - 
 2*a*b*c*d + a^2*d^2)*sqrt((b*c - a*d)/c)*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*(a*c^2*x + (2*b*c 
^2 - a*c*d)*x^3)*sqrt(b*x^2 + a)*sqrt((b*c - a*d)/c))/(d^2*x^4 + 2*c*d*x^2 
 + c^2)) - (2*b^2*d^2*x^3 - (4*b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x^2 + a))/d^ 
3, 1/16*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-(b*c - a*d)/c)*arctan(1/2 
*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)*sqrt(-(b*c - a*d)/c)/((b^2*c - 
a*b*d)*x^3 + (a*b*c - a^2*d)*x)) + (8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*s 
qrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^2*d^2*x^3 
- (4*b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x^2 + a))/d^3, -1/8*((8*b^2*c^2 - 20*a 
*b*c*d + 15*a^2*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 4*(b^2* 
c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-(b*c - a*d)/c)*arctan(1/2*((2*b*c - a*d)* 
x^2 + a*c)*sqrt(b*x^2 + a)*sqrt(-(b*c - a*d)/c)/((b^2*c - a*b*d)*x^3 + (a* 
b*c - a^2*d)*x)) - (2*b^2*d^2*x^3 - (4*b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{c+d x^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.12 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{c+d x^2} \, dx=\frac {-8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b \,x^{2}+a}-\sqrt {d}\, \sqrt {b}\, x}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2}+16 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b \,x^{2}+a}-\sqrt {d}\, \sqrt {b}\, x}{\sqrt {c}\, \sqrt {b}}\right ) a b c d -8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b \,x^{2}+a}-\sqrt {d}\, \sqrt {b}\, x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2}-8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b \,x^{2}+a}+\sqrt {d}\, \sqrt {b}\, x}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2}+16 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b \,x^{2}+a}+\sqrt {d}\, \sqrt {b}\, x}{\sqrt {c}\, \sqrt {b}}\right ) a b c d -8 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b \,x^{2}+a}+\sqrt {d}\, \sqrt {b}\, x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2}+9 \sqrt {b \,x^{2}+a}\, a b c \,d^{2} x -4 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} d x +2 \sqrt {b \,x^{2}+a}\, b^{2} c \,d^{2} x^{3}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c \,d^{2}-20 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} d +8 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} c^{3}}{8 c \,d^{3}} \] Input:

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

Output:

( - 8*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x 
**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**2*d**2 + 16*sqrt(c)*sqrt(a 
*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt( 
b)*x)/(sqrt(c)*sqrt(b)))*a*b*c*d - 8*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a* 
d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)) 
)*b**2*c**2 - 8*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sq 
rt(a + b*x**2) + sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**2*d**2 + 16*sqrt 
(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x**2) + sqr 
t(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a*b*c*d - 8*sqrt(c)*sqrt(a*d - b*c)*ata 
n((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x**2) + sqrt(d)*sqrt(b)*x)/(sqrt(c 
)*sqrt(b)))*b**2*c**2 + 9*sqrt(a + b*x**2)*a*b*c*d**2*x - 4*sqrt(a + b*x** 
2)*b**2*c**2*d*x + 2*sqrt(a + b*x**2)*b**2*c*d**2*x**3 + 15*sqrt(b)*log((s 
qrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*c*d**2 - 20*sqrt(b)*log((sqrt(a 
 + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b*c**2*d + 8*sqrt(b)*log((sqrt(a + b*x* 
*2) + sqrt(b)*x)/sqrt(a))*b**2*c**3)/(8*c*d**3)