\(\int \frac {1}{x (c+d x)^{5/2} (a-b x^2)} \, dx\) [602]

Optimal result

Integrand size = 23, antiderivative size = 232 \[ \int \frac {1}{x (c+d x)^{5/2} \left (a-b x^2\right )} \, dx=-\frac {2 d^2}{3 c \left (b c^2-a d^2\right ) (c+d x)^{3/2}}-\frac {2 d^2 \left (3 b c^2-a d^2\right )}{c^2 \left (b c^2-a d^2\right )^2 \sqrt {c+d x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a c^{5/2}}+\frac {b^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{a \left (\sqrt {b} c-\sqrt {a} d\right )^{5/2}}+\frac {b^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{a \left (\sqrt {b} c+\sqrt {a} d\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x (c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\frac {2 a d^4 (4 c+3 d x)-2 b c^2 d^2 (10 c+9 d x)}{3 \left (b c^3-a c d^2\right )^2 (c+d x)^{3/2}}-\frac {b \sqrt {-b c-\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{a \left (\sqrt {b} c+\sqrt {a} d\right )^3}+\frac {b \sqrt {-b c+\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{a \left (-\sqrt {b} c+\sqrt {a} d\right )^3}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a c^{5/2}} \] Input:

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

Output:

(2*a*d^4*(4*c + 3*d*x) - 2*b*c^2*d^2*(10*c + 9*d*x))/(3*(b*c^3 - a*c*d^2)^ 
2*(c + d*x)^(3/2)) - (b*Sqrt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b* 
c) - Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c + Sqrt[a]*d)])/(a*(Sqrt[ 
b]*c + Sqrt[a]*d)^3) + (b*Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-( 
b*c) + Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c - Sqrt[a]*d)])/(a*(-(S 
qrt[b]*c) + Sqrt[a]*d)^3) - (2*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(a*c^(5/2))
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {561, 25, 27, 1610, 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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1610

\(\displaystyle -2 \int \left (\frac {\left (\left (b c^2+a d^2\right ) (c+d x)-c \left (b c^2+3 a d^2\right )\right ) b^2}{a \left (b c^2-a d^2\right )^2 \left (b c^2-2 b (c+d x) c-a d^2+b (c+d x)^2\right )}-\frac {1}{a c^2 d x}+\frac {a d^4-3 b c^2 d^2}{c^2 \left (b c^2-a d^2\right )^2 (c+d x)}-\frac {d^2}{c \left (b c^2-a d^2\right ) (c+d x)^2}\right )d\sqrt {c+d x}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-2*(d^2/(3*c*(b*c^2 - a*d^2)*(c + d*x)^(3/2)) + (d^2*(3*b*c^2 - a*d^2))/(c 
^2*(b*c^2 - a*d^2)^2*Sqrt[c + d*x]) + ArcTanh[Sqrt[c + d*x]/Sqrt[c]]/(a*c^ 
(5/2)) - (b^(5/4)*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[a] 
*d]])/(2*a*(Sqrt[b]*c - Sqrt[a]*d)^(5/2)) - (b^(5/4)*ArcTanh[(b^(1/4)*Sqrt 
[c + d*x])/Sqrt[Sqrt[b]*c + Sqrt[a]*d]])/(2*a*(Sqrt[b]*c + Sqrt[a]*d)^(5/2 
)))
 

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 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1610

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a 
+ b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 
*a*c, 0] && IntegerQ[q] && IntegerQ[m]
 

rule 2009

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

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(-2 d^{2} \left (\frac {\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a \,d^{2} c^{\frac {5}{2}}}-\frac {1}{3 c \left (a \,d^{2}-b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}-\frac {a \,d^{2}-3 b \,c^{2}}{c^{2} \left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}-\frac {b^{3} \left (\frac {\left (-2 a b c \,d^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}-\sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {\left (2 a b c \,d^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}-\sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )^{2}}\right )\)	\(305\)
default	\(2 d^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a \,d^{2} c^{\frac {5}{2}}}-\frac {-a \,d^{2}+3 b \,c^{2}}{c^{2} \left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}+\frac {1}{3 c \left (a \,d^{2}-b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}+\frac {b^{3} \left (\frac {\left (-2 a b c \,d^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}-\sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}-\frac {\left (2 a b c \,d^{2}-\sqrt {a b \,d^{2}}\, a \,d^{2}-\sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )^{2}}\right )\)	\(306\)
pseudoelliptic	\(\frac {-\left (\left (a \,c^{\frac {11}{2}} d^{2}+a \,d^{3} x \,c^{\frac {9}{2}}+c^{\frac {13}{2}} \left (d x +c \right ) b \right ) \sqrt {a b \,d^{2}}+2 d^{2} \left (c^{\frac {13}{2}}+c^{\frac {11}{2}} d x \right ) b a \right ) \sqrt {d x +c}\, b^{2} \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )+\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (\sqrt {d x +c}\, b^{2} \left (\left (a \,c^{\frac {11}{2}} d^{2}+a \,d^{3} x \,c^{\frac {9}{2}}+c^{\frac {13}{2}} \left (d x +c \right ) b \right ) \sqrt {a b \,d^{2}}-2 d^{2} \left (c^{\frac {13}{2}}+c^{\frac {11}{2}} d x \right ) b a \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )+\frac {8 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (-\frac {3 \left (a \,d^{2}-b \,c^{2}\right )^{2} \left (d x +c \right )^{\frac {3}{2}} c^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{4}+\left (-\frac {5 c^{\frac {11}{2}} b}{2}+d \left (-\frac {9 c^{\frac {9}{2}} b x}{4}+d \left (c^{\frac {7}{2}}+\frac {3 d x \,c^{\frac {5}{2}}}{4}\right ) a \right )\right ) d^{2} a \right )}{3}\right )}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, c^{\frac {9}{2}} \sqrt {a b \,d^{2}}\, \left (d x +c \right )^{\frac {3}{2}} \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (a \,d^{2}-b \,c^{2}\right )^{2} a}\)	\(393\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5410 vs. \(2 (184) = 368\).

Time = 6.90 (sec) , antiderivative size = 10829, normalized size of antiderivative = 46.68 \[ \int \frac {1}{x (c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{x (c+d x)^{5/2} \left (a-b x^2\right )} \, dx=- \int \frac {1}{- a c^{2} x \sqrt {c + d x} - 2 a c d x^{2} \sqrt {c + d x} - a d^{2} x^{3} \sqrt {c + d x} + b c^{2} x^{3} \sqrt {c + d x} + 2 b c d x^{4} \sqrt {c + d x} + b d^{2} x^{5} \sqrt {c + d x}}\, dx \] Input:

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

Output:

-Integral(1/(-a*c**2*x*sqrt(c + d*x) - 2*a*c*d*x**2*sqrt(c + d*x) - a*d**2 
*x**3*sqrt(c + d*x) + b*c**2*x**3*sqrt(c + d*x) + 2*b*c*d*x**4*sqrt(c + d* 
x) + b*d**2*x**5*sqrt(c + d*x)), x)
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1291 vs. \(2 (184) = 368\).

Time = 0.24 (sec) , antiderivative size = 1291, normalized size of antiderivative = 5.56 \[ \int \frac {1}{x (c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-((a*b^2*c^4*d - 2*a^2*b*c^2*d^3 + a^3*d^5)^2*sqrt(-b^2*c - sqrt(a*b)*b*d) 
*(b*c^2 + a*d^2)*abs(b) - (sqrt(a*b)*b^3*c^7 + sqrt(a*b)*a*b^2*c^5*d^2 - 5 
*sqrt(a*b)*a^2*b*c^3*d^4 + 3*sqrt(a*b)*a^3*c*d^6)*sqrt(-b^2*c - sqrt(a*b)* 
b*d)*abs(a*b^2*c^4*d - 2*a^2*b*c^2*d^3 + a^3*d^5)*abs(b) + 2*(a^2*b^5*c^10 
*d^2 - 4*a^3*b^4*c^8*d^4 + 6*a^4*b^3*c^6*d^6 - 4*a^5*b^2*c^4*d^8 + a^6*b*c 
^2*d^10)*sqrt(-b^2*c - sqrt(a*b)*b*d)*abs(b))*arctan(sqrt(d*x + c)/sqrt(-( 
a*b^3*c^5 - 2*a^2*b^2*c^3*d^2 + a^3*b*c*d^4 + sqrt((a*b^3*c^5 - 2*a^2*b^2* 
c^3*d^2 + a^3*b*c*d^4)^2 - (a*b^3*c^6 - 3*a^2*b^2*c^4*d^2 + 3*a^3*b*c^2*d^ 
4 - a^4*d^6)*(a*b^3*c^4 - 2*a^2*b^2*c^2*d^2 + a^3*b*d^4)))/(a*b^3*c^4 - 2* 
a^2*b^2*c^2*d^2 + a^3*b*d^4)))/((sqrt(a*b)*a*b^5*c^10 - 5*sqrt(a*b)*a^2*b^ 
4*c^8*d^2 + 10*sqrt(a*b)*a^3*b^3*c^6*d^4 - 10*sqrt(a*b)*a^4*b^2*c^4*d^6 + 
5*sqrt(a*b)*a^5*b*c^2*d^8 - sqrt(a*b)*a^6*d^10)*abs(a*b^2*c^4*d - 2*a^2*b* 
c^2*d^3 + a^3*d^5)) + ((a*b^2*c^4*d - 2*a^2*b*c^2*d^3 + a^3*d^5)^2*sqrt(-b 
^2*c + sqrt(a*b)*b*d)*(b*c^2 + a*d^2)*abs(b) + (sqrt(a*b)*b^3*c^7 + sqrt(a 
*b)*a*b^2*c^5*d^2 - 5*sqrt(a*b)*a^2*b*c^3*d^4 + 3*sqrt(a*b)*a^3*c*d^6)*sqr 
t(-b^2*c + sqrt(a*b)*b*d)*abs(a*b^2*c^4*d - 2*a^2*b*c^2*d^3 + a^3*d^5)*abs 
(b) + 2*(a^2*b^5*c^10*d^2 - 4*a^3*b^4*c^8*d^4 + 6*a^4*b^3*c^6*d^6 - 4*a^5* 
b^2*c^4*d^8 + a^6*b*c^2*d^10)*sqrt(-b^2*c + sqrt(a*b)*b*d)*abs(b))*arctan( 
sqrt(d*x + c)/sqrt(-(a*b^3*c^5 - 2*a^2*b^2*c^3*d^2 + a^3*b*c*d^4 - sqrt((a 
*b^3*c^5 - 2*a^2*b^2*c^3*d^2 + a^3*b*c*d^4)^2 - (a*b^3*c^6 - 3*a^2*b^2*...
 

Mupad [B] (verification not implemented)

Time = 13.08 (sec) , antiderivative size = 24512, normalized size of antiderivative = 105.66 \[ \int \frac {1}{x (c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((b^22*c^44*d^10*(c + d*x)^(1/2)*18304i)/((c^5)^(1/2)*(18304*b^22*c^4 
2*d^10 - 203456*a*b^21*c^40*d^12 + 1259264*a^2*b^20*c^38*d^14 - 5151296*a^ 
3*b^19*c^36*d^16 + 15203456*a^4*b^18*c^34*d^18 - 34125504*a^5*b^17*c^32*d^ 
20 + 60310016*a^6*b^16*c^30*d^22 - 85934784*a^7*b^15*c^28*d^24 + 100287616 
*a^8*b^14*c^26*d^26 - 96709184*a^9*b^13*c^24*d^28 + 77232896*a^10*b^12*c^2 
2*d^30 - 50857664*a^11*b^11*c^20*d^32 + 27310976*a^12*b^10*c^18*d^34 - 117 
45344*a^13*b^9*c^16*d^36 + 3938304*a^14*b^8*c^14*d^38 - 989696*a^15*b^7*c^ 
12*d^40 + 175104*a^16*b^6*c^10*d^42 - 19456*a^17*b^5*c^8*d^44 + 1024*a^18* 
b^4*c^6*d^46 - (576*b^23*c^44*d^8)/a)) + (b^23*c^46*d^8*(c + d*x)^(1/2)*57 
6i)/((c^5)^(1/2)*(576*b^23*c^44*d^8 - 18304*a*b^22*c^42*d^10 + 203456*a^2* 
b^21*c^40*d^12 - 1259264*a^3*b^20*c^38*d^14 + 5151296*a^4*b^19*c^36*d^16 - 
 15203456*a^5*b^18*c^34*d^18 + 34125504*a^6*b^17*c^32*d^20 - 60310016*a^7* 
b^16*c^30*d^22 + 85934784*a^8*b^15*c^28*d^24 - 100287616*a^9*b^14*c^26*d^2 
6 + 96709184*a^10*b^13*c^24*d^28 - 77232896*a^11*b^12*c^22*d^30 + 50857664 
*a^12*b^11*c^20*d^32 - 27310976*a^13*b^10*c^18*d^34 + 11745344*a^14*b^9*c^ 
16*d^36 - 3938304*a^15*b^8*c^14*d^38 + 989696*a^16*b^7*c^12*d^40 - 175104* 
a^17*b^6*c^10*d^42 + 19456*a^18*b^5*c^8*d^44 - 1024*a^19*b^4*c^6*d^46)) - 
(a*b^21*c^42*d^12*(c + d*x)^(1/2)*203456i)/((c^5)^(1/2)*(18304*b^22*c^42*d 
^10 - 203456*a*b^21*c^40*d^12 + 1259264*a^2*b^20*c^38*d^14 - 5151296*a^3*b 
^19*c^36*d^16 + 15203456*a^4*b^18*c^34*d^18 - 34125504*a^5*b^17*c^32*d^...
 

Reduce [F]

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

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

Output:

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