\(\int \frac {1}{(c+d x)^3 (a+b x^2)^2} \, dx\) [215]

Optimal result

Integrand size = 17, antiderivative size = 263 \[ \int \frac {1}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=-\frac {d^3}{2 \left (b c^2+a d^2\right )^2 (c+d x)^2}-\frac {4 b c d^3}{\left (b c^2+a d^2\right )^3 (c+d x)}+\frac {b \left (a d \left (3 b c^2-a d^2\right )+b c \left (b c^2-3 a d^2\right ) x\right )}{2 a \left (b c^2+a d^2\right )^3 \left (a+b x^2\right )}+\frac {b^{3/2} c \left (b^2 c^4+10 a b c^2 d^2-15 a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (b c^2+a d^2\right )^4}+\frac {2 b d^3 \left (5 b c^2-a d^2\right ) \log (c+d x)}{\left (b c^2+a d^2\right )^4}-\frac {b d^3 \left (5 b c^2-a d^2\right ) \log \left (a+b x^2\right )}{\left (b c^2+a d^2\right )^4} \] Output:

-1/2*d^3/(a*d^2+b*c^2)^2/(d*x+c)^2-4*b*c*d^3/(a*d^2+b*c^2)^3/(d*x+c)+1/2*b 
*(a*d*(-a*d^2+3*b*c^2)+b*c*(-3*a*d^2+b*c^2)*x)/a/(a*d^2+b*c^2)^3/(b*x^2+a) 
+1/2*b^(3/2)*c*(-15*a^2*d^4+10*a*b*c^2*d^2+b^2*c^4)*arctan(b^(1/2)*x/a^(1/ 
2))/a^(3/2)/(a*d^2+b*c^2)^4+2*b*d^3*(-a*d^2+5*b*c^2)*ln(d*x+c)/(a*d^2+b*c^ 
2)^4-b*d^3*(-a*d^2+5*b*c^2)*ln(b*x^2+a)/(a*d^2+b*c^2)^4
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=\frac {-\frac {d^3 \left (b c^2+a d^2\right )^2}{(c+d x)^2}-\frac {8 b c d^3 \left (b c^2+a d^2\right )}{c+d x}+\frac {b \left (b c^2+a d^2\right ) \left (-a^2 d^3+b^2 c^3 x+3 a b c d (c-d x)\right )}{a \left (a+b x^2\right )}+\frac {b^{3/2} c \left (b^2 c^4+10 a b c^2 d^2-15 a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}+4 b d^3 \left (5 b c^2-a d^2\right ) \log (c+d x)-2 b \left (5 b c^2 d^3-a d^5\right ) \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^4} \] Input:

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

Output:

(-((d^3*(b*c^2 + a*d^2)^2)/(c + d*x)^2) - (8*b*c*d^3*(b*c^2 + a*d^2))/(c + 
 d*x) + (b*(b*c^2 + a*d^2)*(-(a^2*d^3) + b^2*c^3*x + 3*a*b*c*d*(c - d*x))) 
/(a*(a + b*x^2)) + (b^(3/2)*c*(b^2*c^4 + 10*a*b*c^2*d^2 - 15*a^2*d^4)*ArcT 
an[(Sqrt[b]*x)/Sqrt[a]])/a^(3/2) + 4*b*d^3*(5*b*c^2 - a*d^2)*Log[c + d*x] 
- 2*b*(5*b*c^2*d^3 - a*d^5)*Log[a + b*x^2])/(2*(b*c^2 + a*d^2)^4)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {496, 25, 657, 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.

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

\(\Big \downarrow \) 496

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 657

\(\displaystyle \frac {\int \left (-\frac {4 a b \left (a d^2-5 b c^2\right ) d^4}{\left (b c^2+a d^2\right )^3 (c+d x)}-\frac {b c \left (b c^2-11 a d^2\right ) d^2}{\left (b c^2+a d^2\right )^2 (c+d x)^2}+\frac {b^2 \left (c \left (b^2 c^4+10 a b d^2 c^2-15 a^2 d^4\right )-4 a d^3 \left (5 b c^2-a d^2\right ) x\right )}{\left (b c^2+a d^2\right )^3 \left (b x^2+a\right )}-\frac {2 \left (b c^2 d^2-2 a d^4\right )}{\left (b c^2+a d^2\right ) (c+d x)^3}\right )dx}{2 a \left (a d^2+b c^2\right )}+\frac {a d+b c x}{2 a \left (a+b x^2\right ) (c+d x)^2 \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b^{3/2} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-15 a^2 d^4+10 a b c^2 d^2+b^2 c^4\right )}{\sqrt {a} \left (a d^2+b c^2\right )^3}+\frac {b c d \left (b c^2-11 a d^2\right )}{(c+d x) \left (a d^2+b c^2\right )^2}+\frac {d \left (b c^2-2 a d^2\right )}{(c+d x)^2 \left (a d^2+b c^2\right )}-\frac {2 a b d^3 \left (5 b c^2-a d^2\right ) \log \left (a+b x^2\right )}{\left (a d^2+b c^2\right )^3}+\frac {4 a b d^3 \left (5 b c^2-a d^2\right ) \log (c+d x)}{\left (a d^2+b c^2\right )^3}}{2 a \left (a d^2+b c^2\right )}+\frac {a d+b c x}{2 a \left (a+b x^2\right ) (c+d x)^2 \left (a d^2+b c^2\right )}\)

Input:

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

Output:

(a*d + b*c*x)/(2*a*(b*c^2 + a*d^2)*(c + d*x)^2*(a + b*x^2)) + ((d*(b*c^2 - 
 2*a*d^2))/((b*c^2 + a*d^2)*(c + d*x)^2) + (b*c*d*(b*c^2 - 11*a*d^2))/((b* 
c^2 + a*d^2)^2*(c + d*x)) + (b^(3/2)*c*(b^2*c^4 + 10*a*b*c^2*d^2 - 15*a^2* 
d^4)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*(b*c^2 + a*d^2)^3) + (4*a*b*d^3 
*(5*b*c^2 - a*d^2)*Log[c + d*x])/(b*c^2 + a*d^2)^3 - (2*a*b*d^3*(5*b*c^2 - 
 a*d^2)*Log[a + b*x^2])/(b*c^2 + a*d^2)^3)/(2*a*(b*c^2 + a*d^2))
 

Defintions of rubi rules used

rule 25

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

rule 496

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 
 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a 
 + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 
*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad 
raticQ[a, 0, b, c, d, n, p, x]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {b^{2} \left (\frac {\frac {c \left (3 a^{2} d^{4}+2 b \,c^{2} d^{2} a -b^{2} c^{4}\right ) x}{2 a}+\frac {d \left (a^{2} d^{4}-2 b \,c^{2} d^{2} a -3 b^{2} c^{4}\right )}{2 b}}{b \,x^{2}+a}+\frac {\frac {\left (-4 a^{2} d^{5}+20 d^{3} a \,c^{2} b \right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (15 a^{2} c \,d^{4}-10 a \,c^{3} d^{2} b -c^{5} b^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{2 a}\right )}{\left (a \,d^{2}+b \,c^{2}\right )^{4}}-\frac {d^{3}}{2 \left (a \,d^{2}+b \,c^{2}\right )^{2} \left (d x +c \right )^{2}}-\frac {2 d^{3} b \left (a \,d^{2}-5 b \,c^{2}\right ) \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{4}}-\frac {4 b c \,d^{3}}{\left (a \,d^{2}+b \,c^{2}\right )^{3} \left (d x +c \right )}\)	\(268\)
risch	\(\text {Expression too large to display}\)	\(4123\)

Input:

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

Output:

-b^2/(a*d^2+b*c^2)^4*((1/2*c*(3*a^2*d^4+2*a*b*c^2*d^2-b^2*c^4)/a*x+1/2*d*( 
a^2*d^4-2*a*b*c^2*d^2-3*b^2*c^4)/b)/(b*x^2+a)+1/2/a*(1/2*(-4*a^2*d^5+20*a* 
b*c^2*d^3)/b*ln(b*x^2+a)+(15*a^2*c*d^4-10*a*b*c^3*d^2-b^2*c^5)/(a*b)^(1/2) 
*arctan(b*x/(a*b)^(1/2))))-1/2*d^3/(a*d^2+b*c^2)^2/(d*x+c)^2-2*d^3*b*(a*d^ 
2-5*b*c^2)/(a*d^2+b*c^2)^4*ln(d*x+c)-4*b*c*d^3/(a*d^2+b*c^2)^3/(d*x+c)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1056 vs. \(2 (249) = 498\).

Time = 3.05 (sec) , antiderivative size = 2139, normalized size of antiderivative = 8.13 \[ \int \frac {1}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(6*a*b^3*c^6*d - 14*a^2*b^2*c^4*d^3 - 22*a^3*b*c^2*d^5 - 2*a^4*d^7 + 
2*(b^4*c^5*d^2 - 10*a*b^3*c^3*d^4 - 11*a^2*b^2*c*d^6)*x^3 + 4*(b^4*c^6*d - 
 5*a*b^3*c^4*d^3 - 7*a^2*b^2*c^2*d^5 - a^3*b*d^7)*x^2 - (a*b^3*c^7 + 10*a^ 
2*b^2*c^5*d^2 - 15*a^3*b*c^3*d^4 + (b^4*c^5*d^2 + 10*a*b^3*c^3*d^4 - 15*a^ 
2*b^2*c*d^6)*x^4 + 2*(b^4*c^6*d + 10*a*b^3*c^4*d^3 - 15*a^2*b^2*c^2*d^5)*x 
^3 + (b^4*c^7 + 11*a*b^3*c^5*d^2 - 5*a^2*b^2*c^3*d^4 - 15*a^3*b*c*d^6)*x^2 
 + 2*(a*b^3*c^6*d + 10*a^2*b^2*c^4*d^3 - 15*a^3*b*c^2*d^5)*x)*sqrt(-b/a)*l 
og((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 2*(b^4*c^7 + 4*a*b^3*c^5* 
d^2 - 7*a^2*b^2*c^3*d^4 - 10*a^3*b*c*d^6)*x - 4*(5*a^2*b^2*c^4*d^3 - a^3*b 
*c^2*d^5 + (5*a*b^3*c^2*d^5 - a^2*b^2*d^7)*x^4 + 2*(5*a*b^3*c^3*d^4 - a^2* 
b^2*c*d^6)*x^3 + (5*a*b^3*c^4*d^3 + 4*a^2*b^2*c^2*d^5 - a^3*b*d^7)*x^2 + 2 
*(5*a^2*b^2*c^3*d^4 - a^3*b*c*d^6)*x)*log(b*x^2 + a) + 8*(5*a^2*b^2*c^4*d^ 
3 - a^3*b*c^2*d^5 + (5*a*b^3*c^2*d^5 - a^2*b^2*d^7)*x^4 + 2*(5*a*b^3*c^3*d 
^4 - a^2*b^2*c*d^6)*x^3 + (5*a*b^3*c^4*d^3 + 4*a^2*b^2*c^2*d^5 - a^3*b*d^7 
)*x^2 + 2*(5*a^2*b^2*c^3*d^4 - a^3*b*c*d^6)*x)*log(d*x + c))/(a^2*b^4*c^10 
 + 4*a^3*b^3*c^8*d^2 + 6*a^4*b^2*c^6*d^4 + 4*a^5*b*c^4*d^6 + a^6*c^2*d^8 + 
 (a*b^5*c^8*d^2 + 4*a^2*b^4*c^6*d^4 + 6*a^3*b^3*c^4*d^6 + 4*a^4*b^2*c^2*d^ 
8 + a^5*b*d^10)*x^4 + 2*(a*b^5*c^9*d + 4*a^2*b^4*c^7*d^3 + 6*a^3*b^3*c^5*d 
^5 + 4*a^4*b^2*c^3*d^7 + a^5*b*c*d^9)*x^3 + (a*b^5*c^10 + 5*a^2*b^4*c^8*d^ 
2 + 10*a^3*b^3*c^6*d^4 + 10*a^4*b^2*c^4*d^6 + 5*a^5*b*c^2*d^8 + a^6*d^1...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 660 vs. \(2 (249) = 498\).

Time = 0.13 (sec) , antiderivative size = 660, normalized size of antiderivative = 2.51 \[ \int \frac {1}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (5 \, b^{2} c^{2} d^{3} - a b d^{5}\right )} \log \left (b x^{2} + a\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {2 \, {\left (5 \, b^{2} c^{2} d^{3} - a b d^{5}\right )} \log \left (d x + c\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {{\left (b^{4} c^{5} + 10 \, a b^{3} c^{3} d^{2} - 15 \, a^{2} b^{2} c d^{4}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{4} c^{8} + 4 \, a^{2} b^{3} c^{6} d^{2} + 6 \, a^{3} b^{2} c^{4} d^{4} + 4 \, a^{4} b c^{2} d^{6} + a^{5} d^{8}\right )} \sqrt {a b}} + \frac {3 \, a b^{2} c^{4} d - 10 \, a^{2} b c^{2} d^{3} - a^{3} d^{5} + {\left (b^{3} c^{3} d^{2} - 11 \, a b^{2} c d^{4}\right )} x^{3} + 2 \, {\left (b^{3} c^{4} d - 6 \, a b^{2} c^{2} d^{3} - a^{2} b d^{5}\right )} x^{2} + {\left (b^{3} c^{5} + 3 \, a b^{2} c^{3} d^{2} - 10 \, a^{2} b c d^{4}\right )} x}{2 \, {\left (a^{2} b^{3} c^{8} + 3 \, a^{3} b^{2} c^{6} d^{2} + 3 \, a^{4} b c^{4} d^{4} + a^{5} c^{2} d^{6} + {\left (a b^{4} c^{6} d^{2} + 3 \, a^{2} b^{3} c^{4} d^{4} + 3 \, a^{3} b^{2} c^{2} d^{6} + a^{4} b d^{8}\right )} x^{4} + 2 \, {\left (a b^{4} c^{7} d + 3 \, a^{2} b^{3} c^{5} d^{3} + 3 \, a^{3} b^{2} c^{3} d^{5} + a^{4} b c d^{7}\right )} x^{3} + {\left (a b^{4} c^{8} + 4 \, a^{2} b^{3} c^{6} d^{2} + 6 \, a^{3} b^{2} c^{4} d^{4} + 4 \, a^{4} b c^{2} d^{6} + a^{5} d^{8}\right )} x^{2} + 2 \, {\left (a^{2} b^{3} c^{7} d + 3 \, a^{3} b^{2} c^{5} d^{3} + 3 \, a^{4} b c^{3} d^{5} + a^{5} c d^{7}\right )} x\right )}} \] Input:

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

Output:

-(5*b^2*c^2*d^3 - a*b*d^5)*log(b*x^2 + a)/(b^4*c^8 + 4*a*b^3*c^6*d^2 + 6*a 
^2*b^2*c^4*d^4 + 4*a^3*b*c^2*d^6 + a^4*d^8) + 2*(5*b^2*c^2*d^3 - a*b*d^5)* 
log(d*x + c)/(b^4*c^8 + 4*a*b^3*c^6*d^2 + 6*a^2*b^2*c^4*d^4 + 4*a^3*b*c^2* 
d^6 + a^4*d^8) + 1/2*(b^4*c^5 + 10*a*b^3*c^3*d^2 - 15*a^2*b^2*c*d^4)*arcta 
n(b*x/sqrt(a*b))/((a*b^4*c^8 + 4*a^2*b^3*c^6*d^2 + 6*a^3*b^2*c^4*d^4 + 4*a 
^4*b*c^2*d^6 + a^5*d^8)*sqrt(a*b)) + 1/2*(3*a*b^2*c^4*d - 10*a^2*b*c^2*d^3 
 - a^3*d^5 + (b^3*c^3*d^2 - 11*a*b^2*c*d^4)*x^3 + 2*(b^3*c^4*d - 6*a*b^2*c 
^2*d^3 - a^2*b*d^5)*x^2 + (b^3*c^5 + 3*a*b^2*c^3*d^2 - 10*a^2*b*c*d^4)*x)/ 
(a^2*b^3*c^8 + 3*a^3*b^2*c^6*d^2 + 3*a^4*b*c^4*d^4 + a^5*c^2*d^6 + (a*b^4* 
c^6*d^2 + 3*a^2*b^3*c^4*d^4 + 3*a^3*b^2*c^2*d^6 + a^4*b*d^8)*x^4 + 2*(a*b^ 
4*c^7*d + 3*a^2*b^3*c^5*d^3 + 3*a^3*b^2*c^3*d^5 + a^4*b*c*d^7)*x^3 + (a*b^ 
4*c^8 + 4*a^2*b^3*c^6*d^2 + 6*a^3*b^2*c^4*d^4 + 4*a^4*b*c^2*d^6 + a^5*d^8) 
*x^2 + 2*(a^2*b^3*c^7*d + 3*a^3*b^2*c^5*d^3 + 3*a^4*b*c^3*d^5 + a^5*c*d^7) 
*x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.86 \[ \int \frac {1}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (5 \, b^{2} c^{2} d^{3} - a b d^{5}\right )} \log \left (b x^{2} + a\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {2 \, {\left (5 \, b^{2} c^{2} d^{4} - a b d^{6}\right )} \log \left ({\left | d x + c \right |}\right )}{b^{4} c^{8} d + 4 \, a b^{3} c^{6} d^{3} + 6 \, a^{2} b^{2} c^{4} d^{5} + 4 \, a^{3} b c^{2} d^{7} + a^{4} d^{9}} + \frac {{\left (b^{4} c^{5} + 10 \, a b^{3} c^{3} d^{2} - 15 \, a^{2} b^{2} c d^{4}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{4} c^{8} + 4 \, a^{2} b^{3} c^{6} d^{2} + 6 \, a^{3} b^{2} c^{4} d^{4} + 4 \, a^{4} b c^{2} d^{6} + a^{5} d^{8}\right )} \sqrt {a b}} + \frac {3 \, a b^{3} c^{6} d - 7 \, a^{2} b^{2} c^{4} d^{3} - 11 \, a^{3} b c^{2} d^{5} - a^{4} d^{7} + {\left (b^{4} c^{5} d^{2} - 10 \, a b^{3} c^{3} d^{4} - 11 \, a^{2} b^{2} c d^{6}\right )} x^{3} + 2 \, {\left (b^{4} c^{6} d - 5 \, a b^{3} c^{4} d^{3} - 7 \, a^{2} b^{2} c^{2} d^{5} - a^{3} b d^{7}\right )} x^{2} + {\left (b^{4} c^{7} + 4 \, a b^{3} c^{5} d^{2} - 7 \, a^{2} b^{2} c^{3} d^{4} - 10 \, a^{3} b c d^{6}\right )} x}{2 \, {\left (b c^{2} + a d^{2}\right )}^{4} {\left (b x^{2} + a\right )} {\left (d x + c\right )}^{2} a} \] Input:

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

Output:

-(5*b^2*c^2*d^3 - a*b*d^5)*log(b*x^2 + a)/(b^4*c^8 + 4*a*b^3*c^6*d^2 + 6*a 
^2*b^2*c^4*d^4 + 4*a^3*b*c^2*d^6 + a^4*d^8) + 2*(5*b^2*c^2*d^4 - a*b*d^6)* 
log(abs(d*x + c))/(b^4*c^8*d + 4*a*b^3*c^6*d^3 + 6*a^2*b^2*c^4*d^5 + 4*a^3 
*b*c^2*d^7 + a^4*d^9) + 1/2*(b^4*c^5 + 10*a*b^3*c^3*d^2 - 15*a^2*b^2*c*d^4 
)*arctan(b*x/sqrt(a*b))/((a*b^4*c^8 + 4*a^2*b^3*c^6*d^2 + 6*a^3*b^2*c^4*d^ 
4 + 4*a^4*b*c^2*d^6 + a^5*d^8)*sqrt(a*b)) + 1/2*(3*a*b^3*c^6*d - 7*a^2*b^2 
*c^4*d^3 - 11*a^3*b*c^2*d^5 - a^4*d^7 + (b^4*c^5*d^2 - 10*a*b^3*c^3*d^4 - 
11*a^2*b^2*c*d^6)*x^3 + 2*(b^4*c^6*d - 5*a*b^3*c^4*d^3 - 7*a^2*b^2*c^2*d^5 
 - a^3*b*d^7)*x^2 + (b^4*c^7 + 4*a*b^3*c^5*d^2 - 7*a^2*b^2*c^3*d^4 - 10*a^ 
3*b*c*d^6)*x)/((b*c^2 + a*d^2)^4*(b*x^2 + a)*(d*x + c)^2*a)
 

Mupad [B] (verification not implemented)

Time = 7.89 (sec) , antiderivative size = 1275, normalized size of antiderivative = 4.85 \[ \int \frac {1}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(log(144*a^9*d^14*(-a^3*b^3)^(1/2) + a^3*b^9*c^14*x + 144*a^10*b^2*d^14*x 
+ a^2*b^7*c^14*(-a^3*b^3)^(1/2) - 5455*a^3*c^6*d^8*(-a^3*b^3)^(3/2) - 22*b 
^3*c^12*d^2*(-a^3*b^3)^(3/2) - 111*a*b^2*c^10*d^4*(-a^3*b^3)^(3/2) - 3460* 
a^2*b*c^8*d^6*(-a^3*b^3)^(3/2) - 927*a^8*b*c^2*d^12*(-a^3*b^3)^(1/2) + 101 
4*a^7*b^2*c^4*d^10*(-a^3*b^3)^(1/2) + 22*a^4*b^8*c^12*d^2*x + 111*a^5*b^7* 
c^10*d^4*x + 3460*a^6*b^6*c^8*d^6*x + 5455*a^7*b^5*c^6*d^8*x + 1014*a^8*b^ 
4*c^4*d^10*x - 927*a^9*b^3*c^2*d^12*x)*(b^2*((c^5*(-a^3*b^3)^(1/2))/4 - 5* 
a^3*c^2*d^3) + b*(a^4*d^5 + (5*a*c^3*d^2*(-a^3*b^3)^(1/2))/2) - (15*a^2*c* 
d^4*(-a^3*b^3)^(1/2))/4))/(a^7*d^8 + a^3*b^4*c^8 + 4*a^6*b*c^2*d^6 + 4*a^4 
*b^3*c^6*d^2 + 6*a^5*b^2*c^4*d^4) - ((a^2*d^5 - 3*b^2*c^4*d + 10*a*b*c^2*d 
^3)/(2*(a*d^2 + b*c^2)*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) - (b*x^3*(b^2* 
c^3*d^2 - 11*a*b*c*d^4))/(2*a*(a^3*d^6 + b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2 
*b*c^2*d^4)) + (x^2*(a^2*b*d^5 - b^3*c^4*d + 6*a*b^2*c^2*d^3))/(a*(a*d^2 + 
 b*c^2)*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) - (c*x*(b^3*c^4 - 10*a^2*b*d^ 
4 + 3*a*b^2*c^2*d^2))/(2*a*(a*d^2 + b*c^2)*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2* 
d^2)))/(a*c^2 + x^2*(a*d^2 + b*c^2) + b*d^2*x^4 + 2*b*c*d*x^3 + 2*a*c*d*x) 
 + (log(a^3*b^9*c^14*x - 144*a^9*d^14*(-a^3*b^3)^(1/2) + 144*a^10*b^2*d^14 
*x - a^2*b^7*c^14*(-a^3*b^3)^(1/2) + 5455*a^3*c^6*d^8*(-a^3*b^3)^(3/2) + 2 
2*b^3*c^12*d^2*(-a^3*b^3)^(3/2) + 111*a*b^2*c^10*d^4*(-a^3*b^3)^(3/2) + 34 
60*a^2*b*c^8*d^6*(-a^3*b^3)^(3/2) + 927*a^8*b*c^2*d^12*(-a^3*b^3)^(1/2)...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1514, normalized size of antiderivative = 5.76 \[ \int \frac {1}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 30*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c**3*d**4 - 60 
*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c**2*d**5*x - 30*sqr 
t(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c*d**6*x**2 + 20*sqrt(b) 
*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**5*d**2 + 40*sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**4*d**3*x - 10*sqrt(b)*sqr 
t(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**3*d**4*x**2 - 60*sqrt(b)*s 
qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**2*d**5*x**3 - 30*sqrt(b) 
*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c*d**6*x**4 + 2*sqrt(b)*s 
qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**3*c**7 + 4*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*a*b**3*c**6*d*x + 22*sqrt(b)*sqrt(a)*atan((b*x)/( 
sqrt(b)*sqrt(a)))*a*b**3*c**5*d**2*x**2 + 40*sqrt(b)*sqrt(a)*atan((b*x)/(s 
qrt(b)*sqrt(a)))*a*b**3*c**4*d**3*x**3 + 20*sqrt(b)*sqrt(a)*atan((b*x)/(sq 
rt(b)*sqrt(a)))*a*b**3*c**3*d**4*x**4 + 2*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt 
(b)*sqrt(a)))*b**4*c**7*x**2 + 4*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt( 
a)))*b**4*c**6*d*x**3 + 2*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b* 
*4*c**5*d**2*x**4 + 4*log(a + b*x**2)*a**4*b*c**2*d**5 + 8*log(a + b*x**2) 
*a**4*b*c*d**6*x + 4*log(a + b*x**2)*a**4*b*d**7*x**2 - 20*log(a + b*x**2) 
*a**3*b**2*c**4*d**3 - 40*log(a + b*x**2)*a**3*b**2*c**3*d**4*x - 16*log(a 
 + b*x**2)*a**3*b**2*c**2*d**5*x**2 + 8*log(a + b*x**2)*a**3*b**2*c*d**6*x 
**3 + 4*log(a + b*x**2)*a**3*b**2*d**7*x**4 - 20*log(a + b*x**2)*a**2*b...