\(\int \frac {1}{x^2 (c+d x)^3 (a+b x^2)^{3/2}} \, dx\) [1255]

Optimal result

Integrand size = 22, antiderivative size = 301 \[ \int \frac {1}{x^2 (c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {b^2 \left (a d \left (3 b c^2-a d^2\right )+b c \left (b c^2-3 a d^2\right ) x\right )}{a^2 \left (b c^2+a d^2\right )^3 \sqrt {a+b x^2}}-\frac {\sqrt {a+b x^2}}{a^2 c^3 x}-\frac {d^5 \sqrt {a+b x^2}}{2 c^2 \left (b c^2+a d^2\right )^2 (c+d x)^2}-\frac {d^5 \left (11 b c^2+4 a d^2\right ) \sqrt {a+b x^2}}{2 c^3 \left (b c^2+a d^2\right )^3 (c+d x)}-\frac {3 d^4 \left (10 b^2 c^4+7 a b c^2 d^2+2 a^2 d^4\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{2 c^4 \left (b c^2+a d^2\right )^{7/2}}+\frac {3 d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2} c^4} \] Output:

-b^2*(a*d*(-a*d^2+3*b*c^2)+b*c*(-3*a*d^2+b*c^2)*x)/a^2/(a*d^2+b*c^2)^3/(b* 
x^2+a)^(1/2)-(b*x^2+a)^(1/2)/a^2/c^3/x-1/2*d^5*(b*x^2+a)^(1/2)/c^2/(a*d^2+ 
b*c^2)^2/(d*x+c)^2-1/2*d^5*(4*a*d^2+11*b*c^2)*(b*x^2+a)^(1/2)/c^3/(a*d^2+b 
*c^2)^3/(d*x+c)-3/2*d^4*(2*a^2*d^4+7*a*b*c^2*d^2+10*b^2*c^4)*arctanh((-b*c 
*x+a*d)/(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/c^4/(a*d^2+b*c^2)^(7/2)+3*d*a 
rctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(3/2)/c^4
 

Mathematica [A] (verified)

Time = 2.71 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^2 (c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {\frac {c \left (4 b^4 c^6 x^2 (c+d x)^2+2 a b^3 c^5 (c+d x)^2 (c+3 d x)+a^4 d^6 \left (2 c^2+9 c d x+6 d^2 x^2\right )+a^3 b d^4 \left (6 c^4+24 c^3 d x+19 c^2 d^2 x^2+9 c d^3 x^3+6 d^4 x^4\right )+a^2 b^2 c^2 d^2 \left (6 c^4+10 c^3 d x+8 c^2 d^2 x^2+22 c d^3 x^3+17 d^4 x^4\right )\right )}{a^2 \left (b c^2+a d^2\right )^3 x (c+d x)^2 \sqrt {a+b x^2}}-\frac {6 d^4 \left (10 b^2 c^4+7 a b c^2 d^2+2 a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )}{\left (-b c^2-a d^2\right )^{7/2}}+\frac {12 d \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}}{2 c^4} \] Input:

Integrate[1/(x^2*(c + d*x)^3*(a + b*x^2)^(3/2)),x]
 

Output:

-1/2*((c*(4*b^4*c^6*x^2*(c + d*x)^2 + 2*a*b^3*c^5*(c + d*x)^2*(c + 3*d*x) 
+ a^4*d^6*(2*c^2 + 9*c*d*x + 6*d^2*x^2) + a^3*b*d^4*(6*c^4 + 24*c^3*d*x + 
19*c^2*d^2*x^2 + 9*c*d^3*x^3 + 6*d^4*x^4) + a^2*b^2*c^2*d^2*(6*c^4 + 10*c^ 
3*d*x + 8*c^2*d^2*x^2 + 22*c*d^3*x^3 + 17*d^4*x^4)))/(a^2*(b*c^2 + a*d^2)^ 
3*x*(c + d*x)^2*Sqrt[a + b*x^2]) - (6*d^4*(10*b^2*c^4 + 7*a*b*c^2*d^2 + 2* 
a^2*d^4)*ArcTan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2) - a* 
d^2]])/(-(b*c^2) - a*d^2)^(7/2) + (12*d*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^ 
2])/Sqrt[a]])/a^(3/2))/c^4
 

Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 601, normalized size of antiderivative = 2.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {617, 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.

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

\(\Big \downarrow \) 617

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2} c^4}-\frac {2 b x}{a^2 c^3 \sqrt {a+b x^2}}-\frac {6 b d^4 \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{c^2 \left (a d^2+b c^2\right )^{5/2}}-\frac {3 b d^4 \left (4 b c^2-a d^2\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{2 c^2 \left (a d^2+b c^2\right )^{7/2}}-\frac {3 d^4 \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{c^4 \left (a d^2+b c^2\right )^{3/2}}-\frac {3 d}{a c^4 \sqrt {a+b x^2}}-\frac {1}{a c^3 x \sqrt {a+b x^2}}+\frac {d^2 (a d+b c x)}{a c^2 \sqrt {a+b x^2} (c+d x)^2 \left (a d^2+b c^2\right )}+\frac {b d^3 \sqrt {a+b x^2} \left (2 b c^2-13 a d^2\right )}{2 a c (c+d x) \left (a d^2+b c^2\right )^3}+\frac {d^3 \sqrt {a+b x^2} \left (2 b c^2-3 a d^2\right )}{2 a c^2 (c+d x)^2 \left (a d^2+b c^2\right )^2}+\frac {3 d^2 (a d+b c x)}{a c^4 \sqrt {a+b x^2} \left (a d^2+b c^2\right )}+\frac {2 d^2 (a d+b c x)}{a c^3 \sqrt {a+b x^2} (c+d x) \left (a d^2+b c^2\right )}+\frac {2 d^3 \sqrt {a+b x^2} \left (b c^2-2 a d^2\right )}{a c^3 (c+d x) \left (a d^2+b c^2\right )^2}\)

Input:

Int[1/(x^2*(c + d*x)^3*(a + b*x^2)^(3/2)),x]
 

Output:

(-3*d)/(a*c^4*Sqrt[a + b*x^2]) - 1/(a*c^3*x*Sqrt[a + b*x^2]) - (2*b*x)/(a^ 
2*c^3*Sqrt[a + b*x^2]) + (3*d^2*(a*d + b*c*x))/(a*c^4*(b*c^2 + a*d^2)*Sqrt 
[a + b*x^2]) + (d^2*(a*d + b*c*x))/(a*c^2*(b*c^2 + a*d^2)*(c + d*x)^2*Sqrt 
[a + b*x^2]) + (2*d^2*(a*d + b*c*x))/(a*c^3*(b*c^2 + a*d^2)*(c + d*x)*Sqrt 
[a + b*x^2]) + (d^3*(2*b*c^2 - 3*a*d^2)*Sqrt[a + b*x^2])/(2*a*c^2*(b*c^2 + 
 a*d^2)^2*(c + d*x)^2) + (b*d^3*(2*b*c^2 - 13*a*d^2)*Sqrt[a + b*x^2])/(2*a 
*c*(b*c^2 + a*d^2)^3*(c + d*x)) + (2*d^3*(b*c^2 - 2*a*d^2)*Sqrt[a + b*x^2] 
)/(a*c^3*(b*c^2 + a*d^2)^2*(c + d*x)) - (3*b*d^4*(4*b*c^2 - a*d^2)*ArcTanh 
[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(2*c^2*(b*c^2 + a*d 
^2)^(7/2)) - (6*b*d^4*ArcTanh[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + 
b*x^2])])/(c^2*(b*c^2 + a*d^2)^(5/2)) - (3*d^4*ArcTanh[(a*d - b*c*x)/(Sqrt 
[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(c^4*(b*c^2 + a*d^2)^(3/2)) + (3*d*ArcT 
anh[Sqrt[a + b*x^2]/Sqrt[a]])/(a^(3/2)*c^4)
 

Defintions of rubi rules used

rule 617

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[(a + b*x^2)^p, x^m*(c + d*x)^n, x], x] /; FreeQ[{ 
a, b, c, d, p}, x] && ILtQ[n, 0] && IntegerQ[m] && IntegerQ[2*p]
 

rule 2009

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1171\) vs. \(2(275)=550\).

Time = 0.52 (sec) , antiderivative size = 1172, normalized size of antiderivative = 3.89

method	result	size
risch	\(\text {Expression too large to display}\)	\(1172\)
default	\(\text {Expression too large to display}\)	\(1854\)

Input:

int(1/x^2/(d*x+c)^3/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-(b*x^2+a)^(1/2)/a^2/c^3/x-1/a/c^3*(b*d*c*a/((-a*b)^(1/2)*d+b*c)/((-a*b)^( 
1/2)*d-b*c)*(-1/2/(a*d^2+b*c^2)*d^2/(x+c/d)^2*(b*(x+c/d)^2-2*b*c/d*(x+c/d) 
+(a*d^2+b*c^2)/d^2)^(1/2)+3/2*b*c*d/(a*d^2+b*c^2)*(-1/(a*d^2+b*c^2)*d^2/(x 
+c/d)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-b*c*d/(a*d^2+b 
*c^2)/((a*d^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2* 
((a*d^2+b*c^2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^ 
(1/2))/(x+c/d)))+1/2*b/(a*d^2+b*c^2)*d^2/((a*d^2+b*c^2)/d^2)^(1/2)*ln((2*( 
a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2)^(1/2)*(b*(x+c/d)^2- 
2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d)))-b^2*a*d^3*((-a*b)^(3/2 
)*c*d^3+(-a*b)^(1/2)*a*b*c*d^3+3*a^2*b*d^4+9*a*b^2*c^2*d^2+10*b^3*c^4)/((- 
a*b)^(1/2)*d+b*c)^3/((-a*b)^(1/2)*d-b*c)^3/c/((a*d^2+b*c^2)/d^2)^(1/2)*ln( 
(2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2)^(1/2)*(b*(x+c/d 
)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d))-3*d/c/a^(1/2)*ln((2 
*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+1/2*b^3*c^3/((-a*b)^(1/2)*d+b*c)^3/a/(x-( 
-a*b)^(1/2)/b)*((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b))^ 
(1/2)-1/2*b^3*c^3/((-a*b)^(1/2)*d-b*c)^3/a/(x+(-a*b)^(1/2)/b)*((x+(-a*b)^( 
1/2)/b)^2*b-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b))^(1/2)-2*b^2*d^2*a*(a*d^2+2* 
b*c^2)/((-a*b)^(1/2)*d+b*c)^2/((-a*b)^(1/2)*d-b*c)^2*(-1/(a*d^2+b*c^2)*d^2 
/(x+c/d)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-b*c*d/(a*d^ 
2+b*c^2)/((a*d^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1358 vs. \(2 (276) = 552\).

Time = 5.81 (sec) , antiderivative size = 5501, normalized size of antiderivative = 18.28 \[ \int \frac {1}{x^2 (c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x^2/(d*x+c)^3/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(1/x**2/(d*x+c)**3/(b*x**2+a)**(3/2),x)
                                                                                    
                                                                                    
 

Output:

Integral(1/(x**2*(a + b*x**2)**(3/2)*(c + d*x)**3), x)
 

Maxima [F]

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

integrate(1/x^2/(d*x+c)^3/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(3/2)*(d*x + c)^3*x^2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 828 vs. \(2 (276) = 552\).

Time = 0.22 (sec) , antiderivative size = 828, normalized size of antiderivative = 2.75 \[ \int \frac {1}{x^2 (c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/x^2/(d*x+c)^3/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

-((a*b^7*c^9 - 6*a^3*b^5*c^5*d^4 - 8*a^4*b^4*c^3*d^6 - 3*a^5*b^3*c*d^8)*x/ 
(a^3*b^6*c^12 + 6*a^4*b^5*c^10*d^2 + 15*a^5*b^4*c^8*d^4 + 20*a^6*b^3*c^6*d 
^6 + 15*a^7*b^2*c^4*d^8 + 6*a^8*b*c^2*d^10 + a^9*d^12) + (3*a^2*b^6*c^8*d 
+ 8*a^3*b^5*c^6*d^3 + 6*a^4*b^4*c^4*d^5 - a^6*b^2*d^9)/(a^3*b^6*c^12 + 6*a 
^4*b^5*c^10*d^2 + 15*a^5*b^4*c^8*d^4 + 20*a^6*b^3*c^6*d^6 + 15*a^7*b^2*c^4 
*d^8 + 6*a^8*b*c^2*d^10 + a^9*d^12))/sqrt(b*x^2 + a) - 3*(10*b^2*c^4*d^4 + 
 7*a*b*c^2*d^6 + 2*a^2*d^8)*arctan(((sqrt(b)*x - sqrt(b*x^2 + a))*d + sqrt 
(b)*c)/sqrt(-b*c^2 - a*d^2))/((b^3*c^10 + 3*a*b^2*c^8*d^2 + 3*a^2*b*c^6*d^ 
4 + a^3*c^4*d^6)*sqrt(-b*c^2 - a*d^2)) - (10*(sqrt(b)*x - sqrt(b*x^2 + a)) 
^3*b^2*c^3*d^5 + 3*(sqrt(b)*x - sqrt(b*x^2 + a))^3*a*b*c*d^7 + 22*(sqrt(b) 
*x - sqrt(b*x^2 + a))^2*b^(5/2)*c^4*d^4 - 3*(sqrt(b)*x - sqrt(b*x^2 + a))^ 
2*a*b^(3/2)*c^2*d^6 - 4*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*sqrt(b)*d^8 - 
34*(sqrt(b)*x - sqrt(b*x^2 + a))*a*b^2*c^3*d^5 - 13*(sqrt(b)*x - sqrt(b*x^ 
2 + a))*a^2*b*c*d^7 + 11*a^2*b^(3/2)*c^2*d^6 + 4*a^3*sqrt(b)*d^8)/((b^3*c^ 
9 + 3*a*b^2*c^7*d^2 + 3*a^2*b*c^5*d^4 + a^3*c^3*d^6)*((sqrt(b)*x - sqrt(b* 
x^2 + a))^2*d + 2*(sqrt(b)*x - sqrt(b*x^2 + a))*sqrt(b)*c - a*d)^2) - 6*d* 
arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a*c^4) + 2*sqrt( 
b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a*c^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 (c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=\int \frac {1}{x^2\,{\left (b\,x^2+a\right )}^{3/2}\,{\left (c+d\,x\right )}^3} \,d x \] Input:

int(1/(x^2*(a + b*x^2)^(3/2)*(c + d*x)^3),x)
 

Output:

int(1/(x^2*(a + b*x^2)^(3/2)*(c + d*x)^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 3931, normalized size of antiderivative = 13.06 \[ \int \frac {1}{x^2 (c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(1/x^2/(d*x+c)^3/(b*x^2+a)^(3/2),x)
 

Output:

(6*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d 
+ b*c*x)*a**5*c**2*d**8*x + 12*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)* 
sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**5*c*d**9*x**2 + 6*sqrt(a*d**2 + b* 
c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**5*d**10 
*x**3 + 21*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2 
) - a*d + b*c*x)*a**4*b*c**4*d**6*x + 42*sqrt(a*d**2 + b*c**2)*log(sqrt(a 
+ b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**4*b*c**3*d**7*x**2 + 27* 
sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b 
*c*x)*a**4*b*c**2*d**8*x**3 + 12*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2 
)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**4*b*c*d**9*x**4 + 6*sqrt(a*d**2 
+ b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**4*b 
*d**10*x**5 + 30*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + 
b*c**2) - a*d + b*c*x)*a**3*b**2*c**6*d**4*x + 60*sqrt(a*d**2 + b*c**2)*lo 
g(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b**2*c**5*d** 
5*x**2 + 51*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c** 
2) - a*d + b*c*x)*a**3*b**2*c**4*d**6*x**3 + 42*sqrt(a*d**2 + b*c**2)*log( 
sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b**2*c**3*d**7* 
x**4 + 21*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) 
 - a*d + b*c*x)*a**3*b**2*c**2*d**8*x**5 + 30*sqrt(a*d**2 + b*c**2)*log(sq 
rt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**2*b**3*c**6*d**4...