\(\int \frac {1}{x^2 (c+d x)^{5/2} (a+b x^2)} \, dx\) [639]

Optimal result

Integrand size = 22, antiderivative size = 639 \[ \int \frac {1}{x^2 (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=-\frac {2 d^3}{3 c^2 \left (b c^2+a d^2\right ) (c+d x)^{3/2}}-\frac {4 d^3 \left (2 b c^2+a d^2\right )}{c^3 \left (b c^2+a d^2\right )^2 \sqrt {c+d x}}-\frac {\sqrt {c+d x}}{a c^3 x}+\frac {b^{7/4} d \left (3 b c^2-a d^2-2 \sqrt {b} c \sqrt {b c^2+a d^2}\right ) \arctan \left (\frac {\sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}-\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}\right )}{\sqrt {2} a \left (b c^2+a d^2\right )^{5/2} \sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}-\frac {b^{7/4} d \left (3 b c^2-a d^2-2 \sqrt {b} c \sqrt {b c^2+a d^2}\right ) \arctan \left (\frac {\sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}+\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}\right )}{\sqrt {2} a \left (b c^2+a d^2\right )^{5/2} \sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}+\frac {5 d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a c^{7/2}}-\frac {b^{7/4} d \left (3 b c^2-a d^2+2 \sqrt {b} c \sqrt {b c^2+a d^2}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}} \sqrt {c+d x}}{\sqrt {b c^2+a d^2}+\sqrt {b} (c+d x)}\right )}{\sqrt {2} a \left (b c^2+a d^2\right )^{5/2} \sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.33 (sec) , antiderivative size = 357, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^2 (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\frac {-\frac {\sqrt {a} \left (3 b^2 c^4 (c+d x)^2+2 a b c^2 d^2 \left (3 c^2+19 c d x+15 d^2 x^2\right )+a^2 d^4 \left (3 c^2+20 c d x+15 d^2 x^2\right )\right )}{c^3 \left (b c^2+a d^2\right )^2 x (c+d x)^{3/2}}-\frac {3 i b^2 \arctan \left (\frac {\sqrt {-b c-i \sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+i \sqrt {a} d}\right )}{\left (\sqrt {b} c+i \sqrt {a} d\right )^2 \sqrt {-b c-i \sqrt {a} \sqrt {b} d}}+\frac {3 i b^2 \arctan \left (\frac {\sqrt {-b c+i \sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-i \sqrt {a} d}\right )}{\left (\sqrt {b} c-i \sqrt {a} d\right )^2 \sqrt {-b c+i \sqrt {a} \sqrt {b} d}}+\frac {15 \sqrt {a} d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{7/2}}}{3 a^{3/2}} \] Input:

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

Output:

(-((Sqrt[a]*(3*b^2*c^4*(c + d*x)^2 + 2*a*b*c^2*d^2*(3*c^2 + 19*c*d*x + 15* 
d^2*x^2) + a^2*d^4*(3*c^2 + 20*c*d*x + 15*d^2*x^2)))/(c^3*(b*c^2 + a*d^2)^ 
2*x*(c + d*x)^(3/2))) - ((3*I)*b^2*ArcTan[(Sqrt[-(b*c) - I*Sqrt[a]*Sqrt[b] 
*d]*Sqrt[c + d*x])/(Sqrt[b]*c + I*Sqrt[a]*d)])/((Sqrt[b]*c + I*Sqrt[a]*d)^ 
2*Sqrt[-(b*c) - I*Sqrt[a]*Sqrt[b]*d]) + ((3*I)*b^2*ArcTan[(Sqrt[-(b*c) + I 
*Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c - I*Sqrt[a]*d)])/((Sqrt[b]*c 
 - I*Sqrt[a]*d)^2*Sqrt[-(b*c) + I*Sqrt[a]*Sqrt[b]*d]) + (15*Sqrt[a]*d*ArcT 
anh[Sqrt[c + d*x]/Sqrt[c]])/c^(7/2))/(3*a^(3/2))
 

Rubi [A] (verified)

Time = 2.33 (sec) , antiderivative size = 825, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {561, 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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1610

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

2*d*(-1/3*d^2/(c^2*(b*c^2 + a*d^2)*(c + d*x)^(3/2)) - (2*d^2*(2*b*c^2 + a* 
d^2))/(c^3*(b*c^2 + a*d^2)^2*Sqrt[c + d*x]) - Sqrt[c + d*x]/(2*a*c^3*d*x) 
+ (5*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(2*a*c^(7/2)) - (b^(7/4)*(3*b*c^2 - a 
*d^2 - 2*Sqrt[b]*c*Sqrt[b*c^2 + a*d^2])*ArcTanh[(Sqrt[Sqrt[b]*c + Sqrt[b*c 
^2 + a*d^2]] - Sqrt[2]*b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[b*c^2 
+ a*d^2]]])/(2*Sqrt[2]*a*(b*c^2 + a*d^2)^(5/2)*Sqrt[Sqrt[b]*c - Sqrt[b*c^2 
 + a*d^2]]) + (b^(7/4)*(3*b*c^2 - a*d^2 - 2*Sqrt[b]*c*Sqrt[b*c^2 + a*d^2]) 
*ArcTanh[(Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]] + Sqrt[2]*b^(1/4)*Sqrt[c + 
 d*x])/Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]])/(2*Sqrt[2]*a*(b*c^2 + a*d^2 
)^(5/2)*Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]) + (b^(7/4)*(3*b*c^2 - a*d^2 
 + 2*Sqrt[b]*c*Sqrt[b*c^2 + a*d^2])*Log[Sqrt[b*c^2 + a*d^2] - Sqrt[2]*b^(1 
/4)*Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]]*Sqrt[c + d*x] + Sqrt[b]*(c + d*x 
)])/(4*Sqrt[2]*a*(b*c^2 + a*d^2)^(5/2)*Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2 
]]) - (b^(7/4)*(3*b*c^2 - a*d^2 + 2*Sqrt[b]*c*Sqrt[b*c^2 + a*d^2])*Log[Sqr 
t[b*c^2 + a*d^2] + Sqrt[2]*b^(1/4)*Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]]*S 
qrt[c + d*x] + Sqrt[b]*(c + d*x)])/(4*Sqrt[2]*a*(b*c^2 + a*d^2)^(5/2)*Sqrt 
[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]]))
 

Defintions of rubi rules used

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 = 1.19 (sec) , antiderivative size = 1043, normalized size of antiderivative = 1.63

method	result	size
pseudoelliptic	\(\text {Expression too large to display}\)	\(1043\)
derivativedivides	\(\text {Expression too large to display}\)	\(2616\)
default	\(\text {Expression too large to display}\)	\(2616\)
risch	\(\text {Expression too large to display}\)	\(7399\)

Input:

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

Output:

5/(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c)^(1/2)/(d 
*x+c)^(3/2)*(-1/20*(ln((-d*x-c)*b^(1/2)+(d*x+c)^(1/2)*(2*((a*d^2+b*c^2)*b) 
^(1/2)+2*b*c)^(1/2)-(a*d^2+b*c^2)^(1/2))-ln(b^(1/2)*(d*x+c)+(d*x+c)^(1/2)* 
(2*((a*d^2+b*c^2)*b)^(1/2)+2*b*c)^(1/2)+(a*d^2+b*c^2)^(1/2)))*x*(d*x+c)^(1 
/2)*((-2*b^(3/2)*(c^(9/2)*d*x+c^(11/2))*(a*d^2+b*c^2)^(1/2)+b*(a*d^3*x*c^( 
7/2)+a*c^(9/2)*d^2-3*c^(11/2)*b*d*x-3*c^(13/2)*b))*((a*d^2+b*c^2)*b)^(1/2) 
+2*b^(5/2)*(c^(13/2)+c^(11/2)*d*x)*(a*d^2+b*c^2)^(1/2)-(a*c^(11/2)*d^2+a*d 
^3*x*c^(9/2)-3*b*(c^(13/2)*d*x+c^(15/2)))*b^2)*(4*(a*d^2+b*c^2)^(1/2)*b^(1 
/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c)^(1/2)*(2*((a*d^2+b*c^2)*b)^(1/2)+2*b* 
c)^(1/2)+d*((d*x*(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)- 
2*b*c)^(1/2)*(d*x+c)^(3/2)*(a*d^2+b*c^2)^2*arctanh((d*x+c)^(1/2)/c^(1/2))- 
2/5*d*b^(5/2)*x*(arctan((-2*b^(1/2)*(d*x+c)^(1/2)+(2*((a*d^2+b*c^2)*b)^(1/ 
2)+2*b*c)^(1/2))/(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)- 
2*b*c)^(1/2))-arctan((2*b^(1/2)*(d*x+c)^(1/2)+(2*((a*d^2+b*c^2)*b)^(1/2)+2 
*b*c)^(1/2))/(4*(a*d^2+b*c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b* 
c)^(1/2)))*(c^(9/2)*d*x+c^(11/2))*(d*x+c)^(1/2)-(2/5*c^(11/2)*b^2*d*x+2/5* 
d^2*(1/2*b*x^2+a)*b*c^(9/2)+1/5*c^(13/2)*b^2+d^3*(38/15*c^(7/2)*b*x+d*((2* 
b*x^2+1/5*a)*c^(5/2)+4/3*c^(3/2)*a*d*x+c^(1/2)*a*d^2*x^2))*a)*(4*(a*d^2+b* 
c^2)^(1/2)*b^(1/2)-2*((a*d^2+b*c^2)*b)^(1/2)-2*b*c)^(1/2))*(a*d^2+b*c^2)^( 
1/2)-1/5*d*x*(d*x+c)^(1/2)*(arctan((-2*b^(1/2)*(d*x+c)^(1/2)+(2*((a*d^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5521 vs. \(2 (528) = 1056\).

Time = 15.36 (sec) , antiderivative size = 11051, normalized size of antiderivative = 17.29 \[ \int \frac {1}{x^2 (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/x^2/(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^2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1337 vs. \(2 (528) = 1056\).

Time = 0.27 (sec) , antiderivative size = 1337, normalized size of antiderivative = 2.09 \[ \int \frac {1}{x^2 (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-(2*(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*d*abs(b) + (3*sqrt(-a*b)*b^3*c^6*d + 5*sqrt(-a*b)*a*b^2*c^4*d^3 + 
sqrt(-a*b)*a^2*b*c^2*d^5 - sqrt(-a*b)*a^3*d^7)*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) - (a*b^6*c^11*d + 3 
*a^2*b^5*c^9*d^3 + 2*a^3*b^4*c^7*d^5 - 2*a^4*b^3*c^5*d^7 - 3*a^5*b^2*c^3*d 
^9 - a^6*b*c*d^11)*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)) + (2*(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*d*abs(b) - (3*sqrt(-a*b)*b^3 
*c^6*d + 5*sqrt(-a*b)*a*b^2*c^4*d^3 + sqrt(-a*b)*a^2*b*c^2*d^5 - sqrt(-a*b 
)*a^3*d^7)*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) - (a*b^6*c^11*d + 3*a^2*b^5*c^9*d^3 + 2*a^3*b^4*c^7*d^5 
 - 2*a^4*b^3*c^5*d^7 - 3*a^5*b^2*c^3*d^9 - a^6*b*c*d^11)*sqrt(-b^2*c + sqr 
t(-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)...
 

Mupad [B] (verification not implemented)

Time = 12.98 (sec) , antiderivative size = 39881, normalized size of antiderivative = 62.41 \[ \int \frac {1}{x^2 (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((a^7*b^7*(400*a^4*c^3*d^17 - 845*b^4*c^11*d^9 + 1712*a*b^3*c^9*d^11 
+ 2500*a^3*b*c^5*d^15 + 4001*a^2*b^2*c^7*d^13) + 5*a^3*b^15*c^19*d - 75*a^ 
12*b^6*c*d^19 - 25*a^4*b^14*c^17*d^3 + 631*a^5*b^13*c^15*d^5 - 1968*a^6*b^ 
12*c^13*d^7)*(a^4*b^15*c^22*(-(a^2*d^5*(-a^7*b^7)^(1/2) + a^3*b^6*c^5 + 5* 
a^5*b^4*c*d^4 - 10*a^4*b^5*c^3*d^2 + 5*b^2*c^4*d*(-a^7*b^7)^(1/2) - 10*a*b 
*c^2*d^3*(-a^7*b^7)^(1/2))/(a^11*d^10 + a^6*b^5*c^10 + 5*a^10*b*c^2*d^8 + 
5*a^7*b^4*c^8*d^2 + 10*a^8*b^3*c^6*d^4 + 10*a^9*b^2*c^4*d^6))^(1/2)*(c + d 
*x)^(1/2)*2i + a^7*b^14*c^27*(-(a^2*d^5*(-a^7*b^7)^(1/2) + a^3*b^6*c^5 + 5 
*a^5*b^4*c*d^4 - 10*a^4*b^5*c^3*d^2 + 5*b^2*c^4*d*(-a^7*b^7)^(1/2) - 10*a* 
b*c^2*d^3*(-a^7*b^7)^(1/2))/(a^11*d^10 + a^6*b^5*c^10 + 5*a^10*b*c^2*d^8 + 
 5*a^7*b^4*c^8*d^2 + 10*a^8*b^3*c^6*d^4 + 10*a^9*b^2*c^4*d^6))^(3/2)*(c + 
d*x)^(1/2)*5i + a^10*b^13*c^32*(-(a^2*d^5*(-a^7*b^7)^(1/2) + a^3*b^6*c^5 + 
 5*a^5*b^4*c*d^4 - 10*a^4*b^5*c^3*d^2 + 5*b^2*c^4*d*(-a^7*b^7)^(1/2) - 10* 
a*b*c^2*d^3*(-a^7*b^7)^(1/2))/(a^11*d^10 + a^6*b^5*c^10 + 5*a^10*b*c^2*d^8 
 + 5*a^7*b^4*c^8*d^2 + 10*a^8*b^3*c^6*d^4 + 10*a^9*b^2*c^4*d^6))^(5/2)*(c 
+ d*x)^(1/2)*3i - a^15*b^4*d^22*(-(a^2*d^5*(-a^7*b^7)^(1/2) + a^3*b^6*c^5 
+ 5*a^5*b^4*c*d^4 - 10*a^4*b^5*c^3*d^2 + 5*b^2*c^4*d*(-a^7*b^7)^(1/2) - 10 
*a*b*c^2*d^3*(-a^7*b^7)^(1/2))/(a^11*d^10 + a^6*b^5*c^10 + 5*a^10*b*c^2*d^ 
8 + 5*a^7*b^4*c^8*d^2 + 10*a^8*b^3*c^6*d^4 + 10*a^9*b^2*c^4*d^6))^(1/2)*(c 
 + d*x)^(1/2)*25i + a^23*c^6*d^26*(-(a^2*d^5*(-a^7*b^7)^(1/2) + a^3*b^6...
 

Reduce [F]

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

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

Output:

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