\(\int \frac {x^5}{(c+d x)^3 (a+b x^2)^2} \, dx\) [210]

Optimal result

Integrand size = 20, antiderivative size = 279 \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=\frac {c^5}{2 d^2 \left (b c^2+a d^2\right )^2 (c+d x)^2}-\frac {c^4 \left (b c^2+5 a d^2\right )}{d^2 \left (b c^2+a d^2\right )^3 (c+d x)}-\frac {a^2 \left (c \left (b c^2-3 a d^2\right )-d \left (3 b c^2-a d^2\right ) x\right )}{2 b \left (b c^2+a d^2\right )^3 \left (a+b x^2\right )}-\frac {a^{3/2} d \left (15 b^2 c^4-10 a b c^2 d^2-a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{3/2} \left (b c^2+a d^2\right )^4}+\frac {2 a c^3 \left (b c^2-5 a d^2\right ) \log (c+d x)}{\left (b c^2+a d^2\right )^4}-\frac {a c^3 \left (b c^2-5 a d^2\right ) \log \left (a+b x^2\right )}{\left (b c^2+a d^2\right )^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=\frac {\frac {c^5 \left (b c^2+a d^2\right )^2}{d^2 (c+d x)^2}-\frac {2 \left (b c^2+a d^2\right ) \left (b c^6+5 a c^4 d^2\right )}{d^2 (c+d x)}-\frac {a^2 \left (b c^2+a d^2\right ) \left (b c^2 (c-3 d x)+a d^2 (-3 c+d x)\right )}{b \left (a+b x^2\right )}+\frac {a^{3/2} d \left (-15 b^2 c^4+10 a b c^2 d^2+a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+4 a c^3 \left (b c^2-5 a d^2\right ) \log (c+d x)+2 a c^3 \left (-b c^2+5 a d^2\right ) \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^4} \] Input:

Integrate[x^5/((c + d*x)^3*(a + b*x^2)^2),x]
 

Output:

((c^5*(b*c^2 + a*d^2)^2)/(d^2*(c + d*x)^2) - (2*(b*c^2 + a*d^2)*(b*c^6 + 5 
*a*c^4*d^2))/(d^2*(c + d*x)) - (a^2*(b*c^2 + a*d^2)*(b*c^2*(c - 3*d*x) + a 
*d^2*(-3*c + d*x)))/(b*(a + b*x^2)) + (a^(3/2)*d*(-15*b^2*c^4 + 10*a*b*c^2 
*d^2 + a^2*d^4)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2) + 4*a*c^3*(b*c^2 - 5* 
a*d^2)*Log[c + d*x] + 2*a*c^3*(-(b*c^2) + 5*a*d^2)*Log[a + b*x^2])/(2*(b*c 
^2 + a*d^2)^4)
 

Rubi [A] (verified)

Time = 1.87 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {601, 2160, 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.

\(\displaystyle \int \frac {x^5}{\left (a+b x^2\right )^2 (c+d x)^3} \, dx\)

\(\Big \downarrow \) 601

\(\displaystyle -\frac {\int \frac {\frac {a^3 d \left (3 b c^2-a d^2\right ) c^3}{b \left (b c^2+a d^2\right )^3}+\frac {a^2 \left (2 b^2 c^4+3 a b d^2 c^2-3 a^2 d^4\right ) x c^2}{b \left (b c^2+a d^2\right )^3}-\frac {a^3 d^3 \left (7 b c^2+3 a d^2\right ) x^2 c}{b \left (b c^2+a d^2\right )^3}-\frac {a \left (2 b^3 c^6+6 a b^2 d^2 c^4+9 a^2 b d^4 c^2+a^3 d^6\right ) x^3}{b \left (b c^2+a d^2\right )^3}}{(c+d x)^3 \left (b x^2+a\right )}dx}{2 a}-\frac {a^2 \left (c \left (b c^2-3 a d^2\right )-d x \left (3 b c^2-a d^2\right )\right )}{2 b \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}\)

\(\Big \downarrow \) 2160

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^5/((c + d*x)^3*(a + b*x^2)^2),x]
 

Output:

-1/2*(a^2*(c*(b*c^2 - 3*a*d^2) - d*(3*b*c^2 - a*d^2)*x))/(b*(b*c^2 + a*d^2 
)^3*(a + b*x^2)) - (-((a*c^5)/(d^2*(b*c^2 + a*d^2)^2*(c + d*x)^2)) + (2*a* 
c^4*(b*c^2 + 5*a*d^2))/(d^2*(b*c^2 + a*d^2)^3*(c + d*x)) + (a^(5/2)*d*(15* 
b^2*c^4 - 10*a*b*c^2*d^2 - a^2*d^4)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b^(3/2)* 
(b*c^2 + a*d^2)^4) - (4*a^2*c^3*(b*c^2 - 5*a*d^2)*Log[c + d*x])/(b*c^2 + a 
*d^2)^4 + (2*a^2*c^3*(b*c^2 - 5*a*d^2)*Log[a + b*x^2])/(b*c^2 + a*d^2)^4)/ 
(2*a)
 

Defintions of rubi rules used

rule 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 2009

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

rule 2160

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, 
 d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {a \left (\frac {-\frac {a d \left (a^{2} d^{4}-2 b \,c^{2} d^{2} a -3 b^{2} c^{4}\right ) x}{2 b}+\frac {a c \left (3 a^{2} d^{4}+2 b \,c^{2} d^{2} a -b^{2} c^{4}\right )}{2 b}}{b \,x^{2}+a}+\frac {\frac {\left (20 a \,b^{2} c^{3} d^{2}-4 b^{3} c^{5}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (a^{3} d^{5}+10 a^{2} b \,c^{2} d^{3}-15 a \,b^{2} c^{4} d \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{2 b}\right )}{\left (a \,d^{2}+b \,c^{2}\right )^{4}}-\frac {c^{4} \left (5 a \,d^{2}+b \,c^{2}\right )}{d^{2} \left (a \,d^{2}+b \,c^{2}\right )^{3} \left (d x +c \right )}+\frac {c^{5}}{2 d^{2} \left (a \,d^{2}+b \,c^{2}\right )^{2} \left (d x +c \right )^{2}}-\frac {2 c^{3} a \left (5 a \,d^{2}-b \,c^{2}\right ) \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{4}}\)	\(288\)
risch	\(\text {Expression too large to display}\)	\(4207\)

Input:

int(x^5/(d*x+c)^3/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1124 vs. \(2 (265) = 530\).

Time = 2.33 (sec) , antiderivative size = 2272, normalized size of antiderivative = 8.14 \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x^5/(d*x+c)^3/(b*x^2+a)^2,x, algorithm="fricas")
 

Output:

[-1/4*(2*a*b^3*c^9 + 22*a^2*b^2*c^7*d^2 + 14*a^3*b*c^5*d^4 - 6*a^4*c^3*d^6 
 + 2*(2*b^4*c^8*d + 12*a*b^3*c^6*d^3 + 7*a^2*b^2*c^4*d^5 - 2*a^3*b*c^2*d^7 
 + a^4*d^9)*x^3 + 2*(b^4*c^9 + 10*a*b^3*c^7*d^2 + 4*a^2*b^2*c^5*d^4 - 6*a^ 
3*b*c^3*d^6 - a^4*c*d^8)*x^2 + (15*a^2*b^2*c^6*d^3 - 10*a^3*b*c^4*d^5 - a^ 
4*c^2*d^7 + (15*a*b^3*c^4*d^5 - 10*a^2*b^2*c^2*d^7 - a^3*b*d^9)*x^4 + 2*(1 
5*a*b^3*c^5*d^4 - 10*a^2*b^2*c^3*d^6 - a^3*b*c*d^8)*x^3 + (15*a*b^3*c^6*d^ 
3 + 5*a^2*b^2*c^4*d^5 - 11*a^3*b*c^2*d^7 - a^4*d^9)*x^2 + 2*(15*a^2*b^2*c^ 
5*d^4 - 10*a^3*b*c^3*d^6 - a^4*c*d^8)*x)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqr 
t(-a/b) - a)/(b*x^2 + a)) + 2*(2*a*b^3*c^8*d + 11*a^2*b^2*c^6*d^3 + 4*a^3* 
b*c^4*d^5 - 5*a^4*c^2*d^7)*x + 4*(a^2*b^2*c^7*d^2 - 5*a^3*b*c^5*d^4 + (a*b 
^3*c^5*d^4 - 5*a^2*b^2*c^3*d^6)*x^4 + 2*(a*b^3*c^6*d^3 - 5*a^2*b^2*c^4*d^5 
)*x^3 + (a*b^3*c^7*d^2 - 4*a^2*b^2*c^5*d^4 - 5*a^3*b*c^3*d^6)*x^2 + 2*(a^2 
*b^2*c^6*d^3 - 5*a^3*b*c^4*d^5)*x)*log(b*x^2 + a) - 8*(a^2*b^2*c^7*d^2 - 5 
*a^3*b*c^5*d^4 + (a*b^3*c^5*d^4 - 5*a^2*b^2*c^3*d^6)*x^4 + 2*(a*b^3*c^6*d^ 
3 - 5*a^2*b^2*c^4*d^5)*x^3 + (a*b^3*c^7*d^2 - 4*a^2*b^2*c^5*d^4 - 5*a^3*b* 
c^3*d^6)*x^2 + 2*(a^2*b^2*c^6*d^3 - 5*a^3*b*c^4*d^5)*x)*log(d*x + c))/(a*b 
^5*c^10*d^2 + 4*a^2*b^4*c^8*d^4 + 6*a^3*b^3*c^6*d^6 + 4*a^4*b^2*c^4*d^8 + 
a^5*b*c^2*d^10 + (b^6*c^8*d^4 + 4*a*b^5*c^6*d^6 + 6*a^2*b^4*c^4*d^8 + 4*a^ 
3*b^3*c^2*d^10 + a^4*b^2*d^12)*x^4 + 2*(b^6*c^9*d^3 + 4*a*b^5*c^7*d^5 + 6* 
a^2*b^4*c^5*d^7 + 4*a^3*b^3*c^3*d^9 + a^4*b^2*c*d^11)*x^3 + (b^6*c^10*d...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**5/(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. 704 vs. \(2 (265) = 530\).

Time = 0.13 (sec) , antiderivative size = 704, normalized size of antiderivative = 2.52 \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (a b c^{5} - 5 \, a^{2} c^{3} d^{2}\right )} \log \left (b x^{2} + a\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {2 \, {\left (a b c^{5} - 5 \, a^{2} c^{3} d^{2}\right )} \log \left (d x + c\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} - \frac {{\left (15 \, a^{2} b^{2} c^{4} d - 10 \, a^{3} b c^{2} d^{3} - a^{4} d^{5}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (b^{5} c^{8} + 4 \, a b^{4} c^{6} d^{2} + 6 \, a^{2} b^{3} c^{4} d^{4} + 4 \, a^{3} b^{2} c^{2} d^{6} + a^{4} b d^{8}\right )} \sqrt {a b}} - \frac {a b^{2} c^{7} + 10 \, a^{2} b c^{5} d^{2} - 3 \, a^{3} c^{3} d^{4} + {\left (2 \, b^{3} c^{6} d + 10 \, a b^{2} c^{4} d^{3} - 3 \, a^{2} b c^{2} d^{5} + a^{3} d^{7}\right )} x^{3} + {\left (b^{3} c^{7} + 9 \, a b^{2} c^{5} d^{2} - 5 \, a^{2} b c^{3} d^{4} - a^{3} c d^{6}\right )} x^{2} + {\left (2 \, a b^{2} c^{6} d + 9 \, a^{2} b c^{4} d^{3} - 5 \, a^{3} c^{2} d^{5}\right )} x}{2 \, {\left (a b^{4} c^{8} d^{2} + 3 \, a^{2} b^{3} c^{6} d^{4} + 3 \, a^{3} b^{2} c^{4} d^{6} + a^{4} b c^{2} d^{8} + {\left (b^{5} c^{6} d^{4} + 3 \, a b^{4} c^{4} d^{6} + 3 \, a^{2} b^{3} c^{2} d^{8} + a^{3} b^{2} d^{10}\right )} x^{4} + 2 \, {\left (b^{5} c^{7} d^{3} + 3 \, a b^{4} c^{5} d^{5} + 3 \, a^{2} b^{3} c^{3} d^{7} + a^{3} b^{2} c d^{9}\right )} x^{3} + {\left (b^{5} c^{8} d^{2} + 4 \, a b^{4} c^{6} d^{4} + 6 \, a^{2} b^{3} c^{4} d^{6} + 4 \, a^{3} b^{2} c^{2} d^{8} + a^{4} b d^{10}\right )} x^{2} + 2 \, {\left (a b^{4} c^{7} d^{3} + 3 \, a^{2} b^{3} c^{5} d^{5} + 3 \, a^{3} b^{2} c^{3} d^{7} + a^{4} b c d^{9}\right )} x\right )}} \] Input:

integrate(x^5/(d*x+c)^3/(b*x^2+a)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 526, normalized size of antiderivative = 1.89 \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (a b c^{5} - 5 \, a^{2} c^{3} d^{2}\right )} \log \left (b x^{2} + a\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {2 \, {\left (a b c^{5} d - 5 \, a^{2} c^{3} d^{3}\right )} \log \left ({\left | d x + c \right |}\right )}{b^{4} c^{8} d + 4 \, a b^{3} c^{6} d^{3} + 6 \, a^{2} b^{2} c^{4} d^{5} + 4 \, a^{3} b c^{2} d^{7} + a^{4} d^{9}} - \frac {{\left (15 \, a^{2} b^{2} c^{4} d - 10 \, a^{3} b c^{2} d^{3} - a^{4} d^{5}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (b^{5} c^{8} + 4 \, a b^{4} c^{6} d^{2} + 6 \, a^{2} b^{3} c^{4} d^{4} + 4 \, a^{3} b^{2} c^{2} d^{6} + a^{4} b d^{8}\right )} \sqrt {a b}} - \frac {a b^{3} c^{9} + 11 \, a^{2} b^{2} c^{7} d^{2} + 7 \, a^{3} b c^{5} d^{4} - 3 \, a^{4} c^{3} d^{6} + {\left (2 \, b^{4} c^{8} d + 12 \, a b^{3} c^{6} d^{3} + 7 \, a^{2} b^{2} c^{4} d^{5} - 2 \, a^{3} b c^{2} d^{7} + a^{4} d^{9}\right )} x^{3} + {\left (b^{4} c^{9} + 10 \, a b^{3} c^{7} d^{2} + 4 \, a^{2} b^{2} c^{5} d^{4} - 6 \, a^{3} b c^{3} d^{6} - a^{4} c d^{8}\right )} x^{2} + {\left (2 \, a b^{3} c^{8} d + 11 \, a^{2} b^{2} c^{6} d^{3} + 4 \, a^{3} b c^{4} d^{5} - 5 \, a^{4} c^{2} d^{7}\right )} x}{2 \, {\left (b c^{2} + a d^{2}\right )}^{4} {\left (b x^{2} + a\right )} {\left (d x + c\right )}^{2} b d^{2}} \] Input:

integrate(x^5/(d*x+c)^3/(b*x^2+a)^2,x, algorithm="giac")
 

Output:

-(a*b*c^5 - 5*a^2*c^3*d^2)*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*(a*b*c^5*d - 5*a^2*c^3*d^3 
)*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*(15*a^2*b^2*c^4*d - 10*a^3*b*c^2*d^3 - a^4*d 
^5)*arctan(b*x/sqrt(a*b))/((b^5*c^8 + 4*a*b^4*c^6*d^2 + 6*a^2*b^3*c^4*d^4 
+ 4*a^3*b^2*c^2*d^6 + a^4*b*d^8)*sqrt(a*b)) - 1/2*(a*b^3*c^9 + 11*a^2*b^2* 
c^7*d^2 + 7*a^3*b*c^5*d^4 - 3*a^4*c^3*d^6 + (2*b^4*c^8*d + 12*a*b^3*c^6*d^ 
3 + 7*a^2*b^2*c^4*d^5 - 2*a^3*b*c^2*d^7 + a^4*d^9)*x^3 + (b^4*c^9 + 10*a*b 
^3*c^7*d^2 + 4*a^2*b^2*c^5*d^4 - 6*a^3*b*c^3*d^6 - a^4*c*d^8)*x^2 + (2*a*b 
^3*c^8*d + 11*a^2*b^2*c^6*d^3 + 4*a^3*b*c^4*d^5 - 5*a^4*c^2*d^7)*x)/((b*c^ 
2 + a*d^2)^4*(b*x^2 + a)*(d*x + c)^2*b*d^2)
 

Mupad [B] (verification not implemented)

Time = 7.87 (sec) , antiderivative size = 1332, normalized size of antiderivative = 4.77 \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^5/((a + b*x^2)^2*(c + d*x)^3),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1560, normalized size of antiderivative = 5.59 \[ \int \frac {x^5}{(c+d x)^3 \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^5/(d*x+c)^3/(b*x^2+a)^2,x)
 

Output:

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