\(\int \frac {x^{17/2} (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [261]

Optimal result

Integrand size = 28, antiderivative size = 251 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}+\frac {15 b (13 b B-11 A c) \sqrt {b x^2+c x^4}}{77 c^4 \sqrt {x}}-\frac {9 (13 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^3}+\frac {(13 b B-11 A c) x^{7/2} \sqrt {b x^2+c x^4}}{11 b c^2}-\frac {15 b^{7/4} (13 b B-11 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{154 c^{17/4} \sqrt {b x^2+c x^4}} \] Output:

-(-A*c+B*b)*x^(15/2)/b/c/(c*x^4+b*x^2)^(1/2)+15/77*b*(-11*A*c+13*B*b)*(c*x 
^4+b*x^2)^(1/2)/c^4/x^(1/2)-9/77*(-11*A*c+13*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1 
/2)/c^3+1/11*(-11*A*c+13*B*b)*x^(7/2)*(c*x^4+b*x^2)^(1/2)/b/c^2-15/154*b^( 
7/4)*(-11*A*c+13*B*b)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x) 
^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c 
^(17/4)/(c*x^4+b*x^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.53 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2} \left (195 b^3 B+2 c^3 x^4 \left (11 A+7 B x^2\right )-2 b c^2 x^2 \left (33 A+13 B x^2\right )+b^2 \left (-165 A c+78 B c x^2\right )+15 b^2 (-13 b B+11 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{77 c^4 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[(x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(x^(3/2)*(195*b^3*B + 2*c^3*x^4*(11*A + 7*B*x^2) - 2*b*c^2*x^2*(33*A + 13* 
B*x^2) + b^2*(-165*A*c + 78*B*c*x^2) + 15*b^2*(-13*b*B + 11*A*c)*Sqrt[1 + 
(c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(77*c^4*Sqrt[x 
^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1943, 1429, 1429, 1429, 1431, 266, 761}

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.

\(\displaystyle \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1943

\(\displaystyle \frac {(13 b B-11 A c) \int \frac {x^{13/2}}{\sqrt {c x^4+b x^2}}dx}{2 b c}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {(13 b B-11 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \int \frac {x^{9/2}}{\sqrt {c x^4+b x^2}}dx}{11 c}\right )}{2 b c}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {(13 b B-11 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\right )}{11 c}\right )}{2 b c}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {(13 b B-11 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\right )}{11 c}\right )}{2 b c}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {(13 b B-11 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )}{2 b c}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(13 b B-11 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )}{2 b c}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(13 b B-11 A c) \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )}{2 b c}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\)

Input:

Int[(x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

-(((b*B - A*c)*x^(15/2))/(b*c*Sqrt[b*x^2 + c*x^4])) + ((13*b*B - 11*A*c)*( 
(2*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(11*c) - (9*b*((2*x^(3/2)*Sqrt[b*x^2 + c*x 
^4])/(7*c) - (5*b*((2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (b^(3/4)*x*(Sqr 
t[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*Ar 
cTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4])))/( 
7*c)))/(11*c)))/(2*b*c)
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1429

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b 
*d^2*((m + 2*p - 1)/(c*(m + 4*p + 1)))   Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ 
p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[m + 2*p - 1, 
 0] && NeQ[m + 4*p + 1, 0]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

rule 1943

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j 
+ 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( 
m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1)))   Int[(e*x)^(m 
 - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, 
n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 
] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (28 B \,c^{4} x^{7}+165 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c +44 A \,c^{4} x^{5}-195 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{3}-52 B b \,c^{3} x^{5}-132 A b \,c^{3} x^{3}+156 B \,b^{2} c^{2} x^{3}-330 A \,b^{2} c^{2} x +390 B \,b^{3} c x \right )}{154 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{5}}\)	\(281\)
risch	\(-\frac {2 \left (-7 B \,c^{2} x^{4}-11 A \,c^{2} x^{2}+20 x^{2} B b c +44 A b c -59 B \,b^{2}\right ) x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{77 c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b^{2} \left (\frac {121 A \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {c \,x^{3}+b x}}-\frac {136 B b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}-77 b \left (A c -B b \right ) \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{77 c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(462\)

Input:

int(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/154/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(28*B*c^4*x^7+165*A*(-b*c)^(1/ 
2)*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(- 
b*c)^(1/2))^(1/2)*(-c/(-b*c)^(1/2)*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/ 
(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c+44*A*c^4*x^5-195*B*(-b*c)^(1/2)*((c 
*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^( 
1/2))^(1/2)*(-c/(-b*c)^(1/2)*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c) 
^(1/2))^(1/2),1/2*2^(1/2))*b^3-52*B*b*c^3*x^5-132*A*b*c^3*x^3+156*B*b^2*c^ 
2*x^3-330*A*b^2*c^2*x+390*B*b^3*c*x)/c^5
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.62 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {15 \, {\left ({\left (13 \, B b^{3} c - 11 \, A b^{2} c^{2}\right )} x^{3} + {\left (13 \, B b^{4} - 11 \, A b^{3} c\right )} x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (14 \, B c^{4} x^{6} + 195 \, B b^{3} c - 165 \, A b^{2} c^{2} - 2 \, {\left (13 \, B b c^{3} - 11 \, A c^{4}\right )} x^{4} + 6 \, {\left (13 \, B b^{2} c^{2} - 11 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{77 \, {\left (c^{6} x^{3} + b c^{5} x\right )}} \] Input:

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

-1/77*(15*((13*B*b^3*c - 11*A*b^2*c^2)*x^3 + (13*B*b^4 - 11*A*b^3*c)*x)*sq 
rt(c)*weierstrassPInverse(-4*b/c, 0, x) - (14*B*c^4*x^6 + 195*B*b^3*c - 16 
5*A*b^2*c^2 - 2*(13*B*b*c^3 - 11*A*c^4)*x^4 + 6*(13*B*b^2*c^2 - 11*A*b*c^3 
)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^6*x^3 + b*c^5*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(x**(17/2)*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Giac [F]

\[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{17/2}\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int((x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-330 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,b^{2} c -66 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a b \,c^{2} x^{2}+22 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,c^{3} x^{4}+390 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{4}+78 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3} c \,x^{2}-26 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{4}+14 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}+165 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) a \,b^{4} c +165 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) a \,b^{3} c^{2} x^{2}-195 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b^{6}-195 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b^{5} c \,x^{2}}{77 c^{4} \left (c \,x^{2}+b \right )} \] Input:

int(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

( - 330*sqrt(x)*sqrt(b + c*x**2)*a*b**2*c - 66*sqrt(x)*sqrt(b + c*x**2)*a* 
b*c**2*x**2 + 22*sqrt(x)*sqrt(b + c*x**2)*a*c**3*x**4 + 390*sqrt(x)*sqrt(b 
 + c*x**2)*b**4 + 78*sqrt(x)*sqrt(b + c*x**2)*b**3*c*x**2 - 26*sqrt(x)*sqr 
t(b + c*x**2)*b**2*c**2*x**4 + 14*sqrt(x)*sqrt(b + c*x**2)*b*c**3*x**6 + 1 
65*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*a*b 
**4*c + 165*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x + 2*b*c*x**3 + c**2*x** 
5),x)*a*b**3*c**2*x**2 - 195*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x + 2*b* 
c*x**3 + c**2*x**5),x)*b**6 - 195*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x + 
 2*b*c*x**3 + c**2*x**5),x)*b**5*c*x**2)/(77*c**4*(b + c*x**2))